perm filename BACK[1,RWF]7 blob
sn#545528 filedate 1980-10-29 generic text, type C, neo UTF8
COMMENT ⊗ VALID 00094 PAGES
C REC PAGE DESCRIPTION
C00001 00001
C00010 00002 DON'T FORGET TO TYPE TTY NO TABS
C00013 00003 EXPECTATIONS -- ADVANCED.
C00016 00004 (3) If E[-] <= 0 , a double is never correct.
C00019 00005 (10) It is always correct to double if P >= .8125 .
C00021 00006 (17) If you will never have another chance to double, and the cube is in
C00023 00007 (20) If every outcome of your and your opponent's next roll leaves you with
C00024 00008 In an earlier chapter, you have seen expectations and strategies for doubling
C00025 00009 ↑
C00027 00010 Accepting Doubles: 1-4 Roll Situations.
C00029 00011 ↑
C00030 00012 Accepting Doubles: 4-6 Roll Situations.
C00032 00013 Similar analyses give E[-] for other values of C :
C00033 00014 ↑
C00034 00015 Summing up, for C >= 8 , threshold values of B{*} for accepting a double
C00035 00016 Doubling: 1-4 Roll Situations.
C00037 00017 D{0*}(A,B)
C00040 00018 In a 2-roll situation, the expected gain from doubling is
C00042 00019 ↑
C00043 00020 ↑
C00044 00021 (C = 10)
C00046 00022 Situations where either player can miss are more complex. If each can only
C00049 00023 V 4 6 8
C00050 00024 An advanced problem.
C00052 00025 ↑
C00053 00026 Another Advanced Problem
C00054 00027 ↑
C00056 00028 RUNNING FROM THE GAMMON.
C00061 00029 Now we see why slotting the 3-point on the previous roll was important.
C00066 00030 My highest priorities in gammon-avoiding situations are these:
C00068 00031 ONE CHECKER, CLOSED OUT
C00070 00032 When O has no more than four checkers off, X should play
C00073 00033 When O has come in, and moved to X's outer board, with X to play, as
C00074 00034 NEWSPAPER ARTICLE: AVOIDING A GAMMON.
C00077 00035 NEWPAPER COLUMN -- RUNNING TO AVOID BEING GAMMONED
C00080 00036 Exercise. How should O play a 2 to minimize the chance of being
C00081 00037 AVOIDING GAMMON
C00084 00038 NEWSPAPER COLUMN -- AVOIDING THE GAMMON
C00087 00039 A NEWSPAPER COLUMN -- AVOIDING BLOTS ON 6-6 .
C00090 00040 ↑
C00092 00041 PLAY AFTER CLOSING OUT ONE CHECKER -- THE RACE.
C00095 00042 Now X needs, on the average, a shade under 6 rolls. O needs, on
C00097 00043 GETTING AN EXTRA CHECKER OFF BEFORE BEING CLOSED OUT.
C00099 00044 Example:
C00100 00045 BREAKING POINTS, AGAINST OPPOSITION.
C00102 00046 Since you and I are under no such pressure, let's look at the two positions
C00105 00047 In Diagram (A), O is struggling to avoid being gammoned. On his
C00108 00048 The net gain of playing 2/1 is 24 + 40 - 27 - 16 = 21 every
C00109 00049 AVOIDING THE GAMMON -- (NEWPAPER COLUMN)
C00111 00050 WAIT OR RUN?
C00113 00051 ↑ (C)
C00116 00052 ↑ (E)
C00117 00053 ACE-POINT RACES.
C00120 00054 To fill out more of this table, let us now look at A = 4 , B = 4 .
C00124 00055 Positions closely related to ace-point race positions can be analyzed by
C00127 00056 O can miss with rolls of 1-2 , 1-3 , and 2-3 . This reduces the
C00130 00057 DOUBLING THEORY -- DOUBLING ON A KEY ROLL.
C00134 00058 If, however, the probabilities are these:
C00137 00059 DOUBLE OR WAIT? COUNT THE KILLERS.
C00141 00060 The second situation favors an immediate double. You become virtually
C00145 00061 Let's try the Killer Rule on a few actual positions.
C00147 00062 (C)↑
C00149 00063 What about the Killer Rule when you already own the cube? We can go
C00152 00064 Your killer rolls are the rolls which speed you up by a full turn, or slow
C00154 00065 APPENDIX
C00158 00066 Where do these numbers come from? They can be calculated readily, starting
C00159 00067 ↑
C00161 00068 Answer: 20/12/8/4 guards against rolling a 6-4 . Keep your spare men
C00162 00069 A NEWSPAPER COLUMN -- 3 MAN BEAROFF.
C00164 00070 Exercise: How should O play a 1 ?
C00165 00071 A 5-ROLL, PIP COUNT, DOUBLING SITUATION.
C00167 00072 A 3-ROLL DOUBLE.
C00169 00073 AN END GAME DOUBLE.
C00172 00074 NEWSPAPER FILLER -- A 2-ROLL DOUBLE.
C00173 00075 ↑
C00175 00076 ↑
C00177 00077 O
C00179 00078 ACE POINT RACES.
C00181 00079 the probability that I win in Position A is 1/6 of the probability that
C00184 00080 P(2R,1R) = 1: P*(0R,1R) = 1
C00187 00081 If you decline the double, you will be behind 5-6 in the match. Later in
C00191 00082 Generally, the conclusions to be drawn from Table ? are:
C00194 00083 Notice that at five rolls, you are favored to roll at least one doubleton
C00196 00084 WHEN TO DOUBLE AGAINST A BACK GAME.
C00200 00085 These positions are too good to double:
C00201 00086 TAKE RISKS FOR POSITION.
C00206 00087 ADJUSTING YOUR PIP COUNT FOR CHECKERS STACKED ON LOW POINTS..
C00209 00088 Should you ever double when you are behind in the current game? It is
C00213 00089
C00217 00090 Today's column is for the novice backgammon player.
C00220 00091
C00224 00092 QUANTITATIVE STUDY.
C00230 00093 (Intermediates)
C00233 00094 Backgammon column--send to Backgammon Guide-for intermediates.
C00237 ENDMK
C⊗;
DON'T FORGET TO TYPE TTY NO TABS
In a two-roll situation, my position is identical to my opponent's, and
it is his roll. We each have chance x of going off in 1 roll (large
doubletons) and chance y of missing and going off in 3 rolls (taking
into account the possibility of a doubleton on the second roll). Then, if
z is the chance of going off in 2 rolls, x+y+z = 1 .
The probability that I will win is P = 1/2 - 1/2(x{2}+y{2}+z{2}) ;
the threshold for my taking a double is about P = .25 , or
x{2}+y{2}+z{2} = 1/2 .
The locus of x{2}+y{2}+z{2} = 1/2 is a sphere. That of x+y+z = 1
is a plane. The intersection is a (tilted) circle. The projection on the
x-y plane is an ellipse.
Here is a graph of the ellipse (E = - 1/2) .
@
In the interior of the ellipse, P > .25 , and I should take. Obviously,
if I have three or four men, x <= 1/6 . If y <= 6/36 , it is a drop..
If y > 6/36 , the pip count is greater than 4 , and x <= 5/36 .
Typically, x is around 4/36 . Here, if 9/36 <= y <= 23/36 ,
I should take. Here are examples of such distributions:
@
Additionally, I can redouble if y < 1/2 and my opponent misses on my
first roll. If y >
?
EXPECTATIONS -- ADVANCED.
[An advanced section, suitable for an appendix, with results which must be
used in certain derivations.]
The values of E and P for a position can be given a mathematical
definition, although in most cases the evaluation of the resulting formulae
is feasible neither for men nor computers. If A is a position,
@
Move [-](A,i,j) in that position B , such that the play of i,j can change
A to B , for which E{*}[-](B) is greatest.
@
@
@
with similar equations for E{*} , etc.
@
These formulae need not be memorized; it is not even essential that they be
understood. They do, however, determine the value of each position in
principle; a perfect player, playing against a perfect player, would always
move to the position with the largest expectation.
By mathematical induction on the defining equations, assuming that gammons
are out of the question, we may relate E , E[-] , E{+} , E[0] , etc.
(1) @
Proof: Obviously true of all terminal positions; assume true on the right
side of the defining equations, and it can be seen true on the left.
(2) If E[-] >= .5 , E[+] = E[0] = E[D] = 1 , and a double is always correct.
(Note: If opponent is not playing rationally, e.g. steaming, it may be
correct to wait until E[-] is even bigger, if you think he'll take.)
(3) If E[-] <= 0 , a double is never correct.
Proof: E[D] = 2E[-] , while E[ow][+] >= E[-] ; E[D]-E[ow][+] <= E[-] ,
so at least {-}E[-] is lost by doubling.
(4) E = 2P-1 .
Proof: True for terminal conditions; preserved by linearity of equations
defining E and P .
(5) E[+] <= -1 + 8/3 . P .
Proof: If P >= .75 , immediate. If P < .75 , E[-] <= E = 2P-1 < 1 ,
and E[D] = 2.E[-] <= 4P-2 <= -1 + 8/3 . P . Equality is achieved only by
doubling with E[-] = 1 and P{-} = .75 , so that E = 1 and the cube has
no further value for X .
(6) Corollary. E[0] <= -1 + 8/3 P .
(7) E[-] >= - 5/3 + 8/3 P .
(Similar to (5), looked at from X's point of view.)
(8) Corollary. E[0] >= - 5/3 + 8/3 P .
@ diagram
By Equation (5), E[+] is bounded above by AB , the upper heavy line,
and E[-] below by CD . Thus all the E's lie between these two lines.
E is the line AD . E[0] and E[+] are bounded below by FG , so a lower
bound for E[0] is CFG , and a lower bound for E[+] is AHG . Similarly,
an upper bound for E{*}[0] and E{*}[-] is JI , so JIC is an upper
bound for E{*}[0] , and JKD for E{*}[-] .
(9) It is always correct to decline or double if P <= 3/16 = .1875 .
Proof: 2E[+] <= 2(-1 + 8/3 . P) <= -1 .
(10) It is always correct to double if P >= .8125 .
Proof: {-}P <= .1875 , and my opponent should drop.
(11) It is always correct to accept a double if P >= 1/4 = .25 .
Proof: 2E[+] >= wE = 2(2P-1) >= -1 .
(12) It is always correct to decline a double if P <= 1/4 = .25 , and
you can't redouble.
Proof: If you take, your expectation is 2E = 2(2P-1) <= -1 .
(13) If no further doubling by either side is possible, it is correct to
double if P >= .50 .
Proof: If you double, your expectation will be 2E ; if not, it will be E .
Your gain by doubling is E = 2P-1 >= 0 .
(14) It is always correct not to double if P <= .5 .
Proof: By (3), since E[-] <= 0 .
(15) It is always correct to beaver if P >= .5 .
Proof: Doubling gains E[+] >= 0 .
(16) It is always correct not to beaver if P <= .375 = 3/8 .
Proof: Doubling gains E[+] <= (-1 + 8/3 . P) <= 0 .
Example: Beaver with
↑
not with
↑
(To beaver is correct if E[+] >= 0 .)
(17) If you will never have another chance to double, and the cube is in
the center, it is always correct to double if E[-] >= 0 . Therefore, it
is always correct to double in this situation if P >= 5/8 = .625 , since
E[-] >= - 5/3 + 8/3 P .
Examples are the same as for (16); the first is not a double for X , the
second is.
(18) If you will never have another chance to double, and you own the cube,
it is always correct to double if 2E[-] - E >= 0 . Since
E[-] >= - 5/3 + 8/3 P , the condition becomes 2(- 5/3 + 8/3 P) - (2P-1)
>= 0 , or - 7/3 + 10/3 P >= 0 , or P >= .7 .
Then it is not correct to redouble with:
↑
, but it is with:
↑
(19) The value of the cube is never more than 2/3 .
Proof: The value of the cube is, presumably, E[+] - E[-] , which is never
more than 2/3 = .67 .
[unchecked]
An example where it is worth nearly this amount is:
↑
where O will miss on the next two rolls about half the time. If he misses,
X will double him, and he has a bare take. If he does not miss, O doubles
X , who will have a bare take or a bare drop, depending on whether O still
has a checker in the 3-point.
(20) If every outcome of your and your opponent's next roll leaves you with
E[-] <= 1/2 , it is correct not to double.
Proof: You could instead defer doubling one turn. If you then want to,
you can still double and your opponent will still take, so you have lost
nothing. You have the option, however, to wait longer, which may have value,
and you have deprived your opponent of the doubling option for one turn.
In an earlier chapter, you have seen expectations and strategies for doubling
and taking doubles when all checkers are on their own 1-points. We now look
at situations similar to such ace-point races, but complicated by moving one
or more checkers back to other points, so that not all doubletons take off
four checkers, and some numbers may fail to take off even two checkers.
Examples:
↑
O doubles -- should X accept?
↑
Should O double? Redouble?
↑
O doubles -- should X accept?
We will find general rules applying to these positions and many others like
them.
Assume that O and X each have C checkers, where C is, typically,
even. As usual, it is O's roll. O has G good rolls, bearing off
four checkers; M medium rolls, bearing off two (or three) checkers;
and B bad rolls, bearing off no more than one checker. If O has
this position,
↑
G = 2 , M = 24 , B = 10 .
We shall in most cases assume that on all subsequent rolls after the first
roll, G = 6 , M = 30 , B = 0 , so that the 1-point race expectations can
be used.
If not all of X's checkers are on the 1-point, we shall use G{*} , M{*} ,
and B{*} analogously as the numbers of good, medium, and bad rolls for X .
↑
G{*} = 3 , M{*} = 27 , B{*} = 6 .
Accepting Doubles: 1-4 Roll Situations.
Here, if all O's checkers are on the 1-point, X will not accept a double.
If O has bad rolls, however, it may be correct for X to take, in the hope
that O will miss and be redoubled out.
If C = 2 , the decision is well known: X accepts if B > 9 , and
(optionally) if B = 9 .
If C > 2 , we want to calculate 2E{-} , the expectation when O doubles
and X accepts. X should accept if E{-} < .50 . The formula for E{-} ,
assuming X is on the 1-point, is
G: E{*-}(C-4,C)
E = M: E{*-}(C-2,C)
B: E{*-}(C,C)
If C = 4 , E{-} = 1/36(G . 1 + M . .67 + B . -1.00) , this is less than
.50 if
G + .67(36 - G - B) - B < 18 ;
.33G - 1.67B < -6 ;
G - 5B < .18 ;
5B > 18 + G .
Since G is ordinarily at least 2 , and never more than 6 , X should
take if B >= 5 .
↑
G = 3 , M = 27 , B = 6 .
If O doubles, X should take.
↑
B = 4 B = 2 B = 4
If O doubles, X should drop.
If C = 6 , X should accept a double, analogously, if
1/36(G . 1.00 + M . .49 + B . -1) < .5 ,
or, simplifying,
if B > -.24 + .34G .
This condtion reduces to B >= 2 .
(If C = 8 , we know that X should accept a double even if B = 0 .)
Summing up, the threshold B for accepting a double, for each C , is
C 2 4 6 8
B 9 5 2 0
Accepting Doubles: 4-6 Roll Situations.
With 4 or more rolls to go, X will accept a double even if B = 0 ,
unless X may himself miss (B{*} > 0 , or G{*} < 6) . Suppose O has
all checkers on the 1-point. For such situations, look ahead two plays,
and form a weighted average of the expectation E{-} at that time.
That average is
E{-} = G{*}: 5: E{-}(C-2,C-4)
1: E{-}(C-4,C-4)
M{*}: 5: E{-}(C-2,C-2)
1: E{-}(C-4,C-2)
B{*}: 5: E{-}(C-2,C)
1: E{-}(C-4,C) .
X should accept if E{-} < 0.50 .
If C = 8 , the formula for E{-} reduces to
1/36(G{*} . -.48 + M{*} . .65 + B{*} . .97) ,
which we will also write as <G{*},M{*},B{*}>.<-.48,.65,.97> . If G{*} = 6
and B{*} = 0 m E{-} = .46 . As the coefficient of G{*} is
-(.48 + .65)/36 = -.03 , and that of B{*} is .32/36 = .01 .
The worst position at C = 8 in which X should accept a double is this
one:
↑
G{*} = 5 , B{*} = 2' .
Similar analyses give E[-] for other values of C :
If C = 10 , E{-} = <G{*},M{*},B{*}>.<-.53,.54,.92> .
If C = 12 , E{-} = <G{*},M{*},B{*}>.<-.51,4.6,.89> .
(Notice that E{-} is much more sensitive to G{*} than to B{*} .)
In practice, the threshold takes are:
C = 10 , B{*} = 6
C = 12 , B{*} = 10
exemplified by these positions
↑
C = 10 , G{*} = 4 , M{*} = 26 , B{*} = 6 , E{-} = .487 .
↑
C = 12 , G{*} = 3 , M{*} = 23 , B{*} = 10 , E[-] = .495 .
X takes a double -- barely.
↑
C = 12 , G{*} = 3 , M{*} = 21 , B{*} = 12 , E{-} = .52
X drops.
↑
C = 12 , G{*} = 5 , M{*} = 21 , B{*} = 10 , E{-} = .491 .
X takes.
↑
C = 12 , G{*} = 5 , M{*} = -20 , B{*} == 11 , E{-} = .504 .
X drops, in part because he may miss on his second roll as well.
Summing up, for C >= 8 , threshold values of B{*} for accepting a double
are:
C 8 10 12
B{*} 2 6 10
Doubling: 1-4 Roll Situations.
As everyone knows, in a 1-roll situation (C = 2) , it is correct to double,
or redouble, if B < 18 . For situations not so close to the end of the game,
it is correct to double more conservatively, because the doubling cube will
continue to be of value on later plays. The expectation from a position if
doubles (and X takes) is
G: E{*-}(C-4,C)
2 M: E{*-}(C-2,C)
B: E{*-}(C,C)
while, if O waits, the expectation is
G: E{*0}(C-4,C)
M: E{*0}(C-2,C)
B: E{*0}(C,C)
If we use the notation
D{*0}(U,V) =2E{*-}(U,V)-E{*0}(U,V)
as the gain from having already doubled, after O rolls and endsup with
U checkers vs. V checkers, the gain of doubling is
G: D{*0}(C-4,C)
M: D{*0}(C-2,C)
B: D{*0}(C,C)
The following tables give values of D{*0} , the gain (with hindsight) of
having doubled, and the analogous values of D{*+} , the gain of having
redoubled;
D{*+}(u,v) = 2E{*-}(U,V)-E{*+}(U,V)
D{0*}(A,B)
B 2 4 6 8 10 12
A
2 -1.00 .67 1.00 1.00 1.00 1.00
4 -1.00 -1.00 .26 .90 1.00 1.00
6 -1.00 -1.00 -1.00 .00 .80 .99
8 -1.00 -1.00 -1.00 - .92 - .13 .69
10 -1.00 -1.00 -1.00 -1.00 - .78 - .13
12 -1.00 -1.00 -1.00 -1.00 -1.00 - .71
P{*}(A,B)
B 2 4 6 8 10 12
A
2 .00 .83 1.00 1.00 1.00 1.00
4 .00 .14 .74 .98 1.00 1.00
6 .00 .00 .22 .69 .95 1.00
8 .00 .00 .02 .26 .67 .92
10 .00 .00 .00 .04 .29 .65
12 .00 .00 .00 .00 .07 .30
D{+*}(A,B) = 2{-}E{*}-E{+*}
2 4 6 8 10 12
2 -1.00 .67 1.00 1.00 1.00 1.00
4 -1.00 -1.28 + .26 .90 1.00 1.00
6 -1.00 -1.00 -1.43 - .03 .80 1.00
8 -1.00 -1.00 -1.04 -1.37 - .19 .68
10 -1.00 -1.00 -1.00 -1.09 -1.12 - .21
12 -1.00 -1.00 -1.00 -1.01 -1.14 -1.01
P(A,B)
B 2 4 6 8 10 12
A
2 1.00 1.00 1.00 1.00 1.00 1.00
4 .17 .86 1.00 1.00 1.00 1.00
6 0 .26 .78 .98 1.00 1.00
8 0 .02 .31 .74 .96 1.00
10 0 0 .05 .33 .71 .93
12 0 0 0 .08 .35 .70
@
@
In a 2-roll situation, the expected gain from doubling is
<G,M,B>.<D{*0}(0,4),D{*0}(2,4),D{*0}(4,4)>
= <G,M,B>.<1.00,.67,-1.00> ;
the corresponding expected gain from redoubling is
<G,M,B>.<1.00,.67,-1.28> .
↑
G = 2 , M = 19 , B = 15
In this position, the gain from doubling is <2,19,15>.<1,99m,56,-1.00> = .00 ,
an optional double; <2,19,15>.<1.00,.67,-1.28> = .12 , definitely not a
redouble. (In fact, since rolls of 3-1 and 1-1 do not guarantee that O
will bear off in one more roll, it is not correct to double.)
↑
<2,21,13>. <1.00,.67,-1.00> = .09
<1.00,.67,-1.28> = -.03 .
This position is a double, but not a redouble.
↑
<2,24,10>.<1.00,.67,-1.28> = .15 ; a definite redouble.
The analogous gains for doubling in three-, four-, and five-roll situations
are:
Double Redouble
C = 6 <1.00,.26,-1.00> <1.00,.26,-1.43>
C = 8 <.90,.00,-.92> <.90,-.03,-1.37>
C = 10 <.98,.67,.29> < >
Examples follow:
(C = 6)
↑
<4,26,6>.<1.00,.26,-1.43> = .06 , a bare redouble.
↑
<3,21,12>.<1.00,.26,-1.00> = -.19 , not a double.
↑
<4.5,21.5,10>. <1.00,.26,-1.00> = .002 ,
<1.00,.26,-1.43> = -.12 ,
a marginal double, not a redouble.
(P = .62)
(C = 8)
↑
<5,29,2>.<.90,-.03,-1.37> = .02 ,
a bare redouble.
(P = <5,29,2>.<.98,.69,.26> = .71)
↑
<4,26,6>.<.90,.00,-.92> = -.003 , not quite a double.
(P = .64)
(C = 10)
↑
<5,29,2>.<.80,-.13,-.78> = -.04 ,
not quite a double; it is a double if a checker is moved
to the 2-point.
(P = .69)
At C = 10 , not even a perfect <6,30,0> is a redouble.
These positions accord well with the rule of thumb that, in an R-roll
position, P must be at least .8 - .3/(sqrt R) to double.
Summing up, we give threshold values of B to double and redouble for
C between 2 and 8 :
C 2 4 6 8
B (double) 18 15 10 4
B (redouble) 18 13 8 2
This graph shows threshold values of B for taking a double (D) ,
doubling (X) , and redoubling ($) for 1-4 roll situations
↑
Roughly, the pattern is:
Doubles: B = 18 on last roll; 4 fewer for each additional roll.
Redoubles: B = 18 on last roll; 5 fewer for each additional roll.
Accepts: B = 9 on last roll; 3 fewer for each additional roll.
Situations where either player can miss are more complex. If each can only
miss on his next roll, the gain from doubling is obtained by looking ahead
two ply, taking weighted averages of the gain resulting from having doubled,
as a function of the number of checkers remaining (presumably on the 1-point)
for each player. This gain is
D{0}(U,V) = 2E{-}(U,V)-E{+}(U,V) , or,
for redoubles,
D{+}(U,V) = 2E{-}(U,V)-E{+}(U,V) .
D{0} = 2E{-}-E{0} , the gain resulting from having doubled on the previous
turn.
V 2 4 6 8 10 12
U
2 1.00 1.00 1.00 1.00 1.00 1.00
4 - .67 .44 1.00 1.00 1.00 1.00
6 -1.00 - .72 .15 .92 1.00 1.00
8 -1.00 -1.00 - .79 .00 .83 .99
10 -1.00 -1.00 -1.00 - .76 .00 .73
12 -1.00 -1.00 -1.00 - .99 - .67 .00
↑
G{*} = 4 , M{*} = 26 , B{*} = 6 ,
G = 4 , M = 26 , B = 6 .
Should O double?
We see that if O misses, X will not immediately redouble. Therefore,
if after O plays and X plays, O has U checkers remaining and X
has V checkers remaining, the gain from having doubled will be
D{0}(U,V) . By taing the appropriate weighted average over the several
values of U and V , we calculate the expected gain from doubling. In
a convenient tabular notation,
V 4 6 8
U
4 .44 1.00 1.00 G = 3
6 - .72 .15 .92 M = 23
8 -1.00 - .79 .00 B = 10
Weighted - .70 - .04 .67
average
<4,26,6>.<-.70,-.04,.67> = .00 , an optional double, not a redouble.
If the players each have the same position,
↑
the gain from redoubling is:
4 6 8
4 .44 1.00 1.00 G = 3
6 - .95 .15 .92 M = 23
8 -1.05 -1.12 0 B = 10
Average - .86 - .13 .67
<3,23,10>.<-.86,-.13,.67> = .03 , correct to redouble.
An advanced problem.
↑
Should O redouble?
Each player has 5 good rolls (G) , 21 medium (M) , and 10 bad (B) .
Each player is off in:
1 roll (G)
2 rolls (MM, MG, or BG)
or 3 rolls (BB, BM, MB) .
The gains of redoubling are tabulated below:
X G M B
O
G = 5 1.00 1.00 1.00
M = 21 -1.00 0 (alpha) .84 (beta)
B = 10 -1.00 -1.01 (gamma) .52 (delta)
Average - .72 - .14 .77
<5,21,10>.<-.72,-.14,.77> = .03 , a redouble.
(alpha)
↑
O will still redouble, X will still take. No change in expectation.
↑
(beta)
P{*} = 10/36 . 5/36 = 50/1296
E{-} = 1-2P{*} = .92
D = 2E{-}-E[+] = .84
↑
(gamma)
P{*} = 31/36 . 21/36 = 806/1296 = .62 ;
E{+} = E = 1-2P{*} = -.24
E{-} = -.62 (see Chapter ?)
D = 2E{-}-E{+} = -1.01
↑
(delta)
Another Advanced Problem
Mel Drapkin ↑
Bill Robertie
G{*} = 5 , M{*} = 31 , B{*} = 0
G = 2 , M == 25 , B == 9
(Counting 12 , 13 , and 23 as 1/3 bad rolls.)
X G = 5 M = 31
O
G = 2 1.00 1.00
M = 25 - .72 (alpha) .44 (beta)
B = 9 -1.00 -1.00
Average - .69 .11 <-.69,11><5,31> = -.00
an incorrect double (marginal)
↑
(alpha)
D = ?
P = 5/36 ; E = - 26/36 = -.72
↑
(beta)
E{0}= 1.00 , P{*} = 31/36 . 6/36 = .14
E{-} = 1-2P{*} = .72
D = .44
(The above analysis does not take into account that on his second roll,
O probably does not bear off four men with 11 or 22 ; thus there is
another reason not to double.)
Detailed Study of Previous Problem.
↑
How many misses on next 2 rolls? How many doubletons that bear 4 off?
↑
for medium moves (31 of them):
sigma misses = 86 == 2{+} bad rolls
sigma good doubletons = 120 == average of 4 .
In effect, B = 9 , G = 0 ; R = 3 . <0,27,9>.<1.00,.26,-1.00> = -2/36 = -.05 ,
not a double. Even if we set G = 2 , we get
<2,25,9>.<1.00,.26,-1.00> = -.5/36 = .01 , not a double.
RUNNING FROM THE GAMMON.
In position A , I was one of an inharmonious team of three players in
a chouette, playing the O pieces. The captain rolled 4-1 , and I urged
caution. He was listening to no Cassandras, however, and played 12/17 ,
reaching position B , without waiting to hear why it was dangerous. He
muttered something about having a sure gammon.
The attentive reader will have noticed that the captain's play leaves blots
on rolls of 6-6 , 6-5 , 6-4 , 6-3 , 5-5 , and 5-4 . Had he played 12/16 ,
23/24 , 6-3 and 5-4 would be safe.
I believed that our gammon chances were too small to justify risking a
loss.
Wanting to capture the captain's attention, I offered to bet him ten
dollars at even money that we would not win a gammon. He hesitated until the
third partner, another evangelist for the gospel according to St. Gammon,
offered to take half the bet.
Why did I make such an apparently foolhardy bet? Watch a typical play.
Say O rolls 5-1 , playing 17/22 , 24/off . Then he rolls 6-4 ,
playing 6/off , 6/2 . X rolls 3/3 , staying out. O rolls 6-1 , playing
5/off , 5/4 . In position C , X rolls 6/3 . He has been lucky:
X could have come in one roll earlier. In the absence of doubletons,
O will go off in six more rolls (12 dice). I will win the bet if the
X pieces go off in 12 dice, including the current two. What should
X's strategy be?
X can expect to get two of each number on the twelve dice. The back man,
starting from the bar, can reach home using four dice (a distance of 19 ).
In fact, since he starts with a six, the other three dice used must add up to
13 . By using his fours, fives, and sixes (he expects six of these, and only
needs three) to move back men, and smaller numbers to bear into his home
board, X will waste no dice, and can expect to come off in twelve dice:
four for the back man, seven to bring the rest in, and one to bear a man off.
With this plan, X becomes a favorite. Doubletons, large or small, can be
played to good advantage, speeding up the bearoff by a whole turn.
Under this plan, X plays bar/19 , 9/6 .
The game proceeds:
O: 4-5 4/off , 4/off (Position D)
X: 3-1 8/5 , 7/6
(Notice the loss of a pip, for good cause.)
O: 3-3 4/1 , 3/off , 3/off , 3/off (Position E)
(Notice that this does not gain a roll. Only large
doubletons help O .)
X: 4-2 19/15 , 8/6
O: 6-5 2/off , 2/off (Position F)
X: 6-4 15/9 , 7/3 (Important!)
O: 2-3 2/off , 2/off (Position G)
X: 3-1 9/6 , 7/6
Now we see why slotting the 3-point on the previous roll was important.
X wants to be able to handle the rolls of 1-1 , 1-2 , 1-3 , and 2-3 on his
next turn. To have no checkers on the 1 , 2 , or 3 points on the last roll
exposes him to missing on all the above rolls. With a checker on the 3-point,
X succeeds unless he rolls 1-2 .
O: 4-2 1/off , 1/off
X: 4-2 7/5 , 4/off , avoiding the gammon.
(An experiment showed that X lost the gammon, with correct play, only
twelve times out of thirty trials, so I got even money on a 3-2 proposition.)
In avoiding a gammon, one must play to take into account the need for pips,
the need for crossovers, and, often, the need for sixes. In position H ,
O needs 8 crossovers and at least 23 pips on his next four rolls
(including the current one) to avoid being gammoned. To play 8/6 , 7/4
wastes two pips to no purpose, losing needlessly if the next three rolls total
18 or 19 (12% of the time). To play 15/12/10 wastes a crossover,
requiring doubletons to catch up. To play 15/12 , 14/12 has a less
obvious defect: in the absence of doubletons, it loses unless O gets
two or more sixes on the next three rolls, which only happens 27% of the time.
(Chance of no sixes: (5/6){6} = .33 . Chance of one six:
({6}[1]).(5/6){5}.1/6 = (5/6){5} = .40 . Chance of 2 or more sixes:
1-.33-.40 = .27 .) A better play is 15/12 , 8/6 , but this still makes
O dependent on rolling a six or a doubleton.
The best play is 14/11 (!), 8/6 , (Position I) which requires no
subsequent sixes, wastes no pips, achieves two crossovers, and handles all
subsequent numbers well. Subsequent numbers are played as follows:
1 7/6
2 7/5 , wasting a pip for the sake of an essential crossover.
3 15/12 or 7/4 , depending on comparative risks of shortage
of sixes and shortage of pips.
4 15/11
5 11/6
6 15/9 : Better than 11/5 . After 11/5 , one needs the
following large numbers on the next rolls to avoid missing
a crossover; 6-3 , 6-4 , 6-5 , 6-6 , 5-4 , or 5-5 .
After 15/9 , all of these still work, and so does 5-3 .
In general, the rules for bearing in to avoid gammons are these:
(1) Don't waste pips in your home board without solid reason,
such as an essential crossover. Usually it is wrong to waste
any until your last two rolls, but if you have many men on your
seven point, it may be correct much earlier.
(2) Don't pile up men on your 12 and 18 points unless only double
sixes can save you. You will waste crossovers on these men
eventually, unless you roll many sixes.
(3) Try to spread your men out in your outer board; avoid duplication.
(4) Cross over into a new board when you can.
(5) Occupy your 11 point, to make good use of fives and sixes later on.
(6) Occupy your 7 point, so that ones will always play well.
My highest priorities in gammon-avoiding situations are these:
(1) Bear in to the 6-point.
(2) Play a 6 from 13 to 7 so that ones will bear in, if the 7
point is vacant.
(3) Bring a man down to a vacant point in your outer board, with the
exception of the 12-point.
(4) Bring a man down to an occupied point.
(5) Bring a man into your opponent's outer board, especially to points
13-15 .
My usual preferences for playing the individual numbers are:
To play a 1: 7/6 , 13/12 , 19/18 , 12/11 , diversify.
2: 8/6 , 13/11 , 19/17 , 14/12 , 20/18 .
3: 9/6 , 14/11 , 13/10 , 19/16 , 20/17 .
4: 10/6 , 15/11 , 14/10 , 19/15 , 20/14 , 16/12 .
5: 11/6 , 16/11 , 17/12 , 15/10 .
6: 12/6 , 18/12 , 13/7 .
with exceptions to meet special conditions.
ONE CHECKER, CLOSED OUT
A back game often ends with one player (O , say) being hit and closed
out after taking several men off. If all his other men are on the 1-point,
no trap play to get another man back is possible; the best X can do is to
try to bear off quickly, while keeping O on the bar as long as possible.
↑
(A)
A typical position with one checker closed out.
In situations like position A , O has the advantage if he has taken 8
or more men off. With 10 men off, he can double from the bar; X should
drop. With 5 men off, X can double O out. The table below gives an
experimental estimate of the number of times X will win, depending on the
number of men O has borne off, if the game is played to completion.
Wins for X in Correct style of
Number of men off 36 games play for X
0 36
1 36
2 36 Conservative
3 34
4 32
5 31
6 25 Moderate
7 21
8 16
9 12
10 8
11 5 Radical
12 2
13 1
14 1
15 0
When O has no more than four checkers off, X should play
conservatively, avoiding gaps and odd extra checkers on the high point.
He should, however, try to bear off a few checkers before opening up if
possible; O has a very fast board. When O has 5 to 7 checkers off,
experiments show that X should take moderate chances. He should avoid
opening points on which O can come in, even at the expense of leaving
gaps and odd extra checkers on high points.
When O has 8 or more checkers off, X must take desperate chances until
he is within about 5 checkers of having as many off.
↑ FILL IN EXAMPLES OF PLAY
As X bears off, if O stays on the bar, the positions are about even
when X has four more men left than O , with X to play.
(B) ↑
In position (B), if the game is played to completion 36 times, X will
win about 20 of them. However, if O has the cube, O will be a slight
favorite (his expectation is around 0 to 5% of the cube). Each extra
checker makes about a 20% difference in O's chance to win. If O has
one more checker on the 1-point, X should double, and O barely has a
take. In general, with a smooth position X should double when he has
five more checkers than O who is still on the bar.
When O has come in, and moved to X's outer board, with X to play, as
in position (C),
↑
(C)
X to play
the position is even if X has 4 more men than O . If O had 7 men
on the 1-point, X should double him out. If it were O's roll in
position D , he could double X out. (The expected number of rolls is
5.0 for X , 4.5 for O .)
NEWSPAPER ARTICLE: AVOIDING A GAMMON.
↑ (A)
Here you are trying to avoid being gammoned. The 6 must be played
12/6 . Many players would play the 2 by 7/5 , without thinking. Should you
ever make such a move, look at the resulting position before you pick up your
dice, and work out what happens to you if your next roll consists of the
numbers of empty points -- in this case, 3's and 4's . Doubletons 3's
and 4's can't hurt, but a 3-4 on the next roll fails to take a man off.
You should look for a safer play.
The play 12/6/4 , leaving gaps on the 3 , 5 , and 6 , is perfectly safe;
every roll consisting of 3's , 5's , and 6's takes a man off.
When you are racing to avoid being gammoned, it is tempting to play fast,
and to blame a later miss on bad luck. By anticipating your bad rolls, you will
have bad luck less often.
Exercise.
↑ (B)
O to play 2-1 .
Answer: If you played 8/6 , 7/6 , you left yourself open to a 4-4 , 4-5,
or 5-5 on the next roll. Play 7/5 , 6/5 to guarantee bearing a man off
on the next turn. To play 8/6/5 is not correct; you need the man on the
8 point in case of a 4-4 . To play 7/5/4 loses a gammon if you roll 5-5 .
NEWPAPER COLUMN -- RUNNING TO AVOID BEING GAMMONED
Here O is running to avoid being gammoned. How should he play a 1 ?
↑
He must bear off in two more rolls. If he plays the back man 18/17 , he
risks getting a 1 on the second roll, and not being able to bear a man off.
If he plays it 2/1 , he risks not reaching his home board on the first
three dice. A detailed analysis counts the rolls that make one or the
other move better:
If the first roll totals: 5 6 7 8 9 10
Probability: 4/36 4/36 6/36 5/36 4/36 2/36
Second rolls good only with 6-1 5-1 4-1 3-1 2-1
2/1 6-1 5-1 4-1 3-1
6-1 5-1 4-1
6-1 5-1
6-1
Second rolls good only with 6-2 5-2 4-2 3-2
18/17 6-3 5-3 4-3
6-4 5-4
6-5
The better move is 2/1 ; it bears a man off in two rolls an extra 2% of
the time. The decision is a close one: if the back man were on the 19
point, 19/18 would be the correct move. In the situation of the diagram,
O is likely to need an odd number of dice to bring the back man in, and the
extra die may well be a 1 .
In situations of this type, it is correct to slot on the 1 point if the
back man is left on the 18 , 17 , 16 , or 15 points; also on the 11 , 10 ,
and 9 points.
Exercise. How should O play a 2 to minimize the chance of being
gammoned?
↑
Answer: If O plays 10/8 , any roll containing a 1 , except 1-1 will
lose the gammon. If he plays the correct 3/1 , only 1-1 , 1-2 , 1-3 ,
and 2-3 lose the gammon.
This position is closely akin to the choice between having two men on
the 2 point, and having a man on the 4 and a man on the 1 point,
in a bearing-off situation. It is usually incorrect to waste two pips inside
the home board when playing to avoid a gammon; this position is exceptional.
AVOIDING GAMMON
↑
O will probably get three more turns. Assume he plays 2/1 . In three
turns, he must use at least 19 pips to bring men home, and then bear off
a man with his smallest die of the final roll. The average value of the
smallest die is
(11 . 1 + 9 . 2 + 7 . 3 + 5 . 4 + 3 . 5 + 1 . 6)/36 = 2.53 ,
so the total number of pips needed is 21.53 on the average. If he plays
16/15 , 10/9 , or 8/7 , he will still need at least 18 pips to bring
men home, and then must bear off a man with the smallest die other than 1 .
In the 10 cases where one of the dice is 1 , the average value of the
other die is 4 , raising the average number used to bear off by
11(4-1)/36 = .92 , to 3.45 , so the total number of pips needed is 21.45 ,
on the average.
This suggests that, contrary to much expert opinion, it is not worth while
to use 1's to slot the 1-point in general. Naturally, there are special
situations which call for slotting the 1-point. If there is a single outside
man on the 9 or 10 point, or between the 14- and 18- points, slotting
the 1 is helpful.
In this particular position, unless he gets doubletons, O must cross
into a new board with each subsequent move. To play 14/13 allows use of
even a 1 to enter a new board.
An experiment confirmed the correctness of 14/13 . The position was
played out 36 times with both initial moves 2/1 and 14/13 , using the
same rolls of the dice. O took a man off in 3 rolls 17 times after 2/1 ,
19 times after 14/13 . The main reason for the difference was the fact
that 1's could be used to cross into a new board after 14/13 .
NEWSPAPER COLUMN -- AVOIDING THE GAMMON
Ordinarily, when trying to avoid being gammoned, you should avoid wasting
pips within your inner board. An important exception arises when your
pip count is low compared to the number of men you have to bring in. In
this position, O has a 4 to play.
↑
He is likely to get only four more turns. He can expect an average of
32 pips in four turns -- more than enough (22 could do it), but, in the
absence of doubletons, he must cross into a new board with each move
(including the current one) to get off in four. Playing 17/13 is a
mistake: he can get off in four only with the aid of doubletons. He
must play 8/4 (or 7/3 ), saving the man on the 17 point for 5's
and 6's . If O gets doubletons, he is farly safe. If not, he will
get two 6's , or a 5 and a 6 , among the next four rolls nearly 75%
of the time.
Problem: Should O play 8/4 or 7/3 in the above position?
Answer: 7/3 , the better move, hurts O only if he rolls 1's on each
of his first three moves, a chance of less than .03 ; playing 8-4 is
likely to hurt him if his fourth roll is 3-1 , or if it is 6-3 or 5-3
and he needs the larger die to bring the back man in.
A NEWSPAPER COLUMN -- AVOIDING BLOTS ON 6-6 .
↑
In this position, O will almost surely win unless he has to leave a
blot. On the current move, only 6-6 will force him to leave a blot. In
bearing-in situations, with an enemy man on the bar and a closed board,
6-6 is often a dangerous roll. The position shown is identified as a
bad one by counting an odd number of men between the 6 and 11 points,
the region from which a 6 can be played once. The rule in such situations
is to keep an even number of men on points 6 to 11 .
Double 5's and 4's are also sometimes hard to play. These tend to be
less dangerous because after all the men are on the home board, remaining
moves can be played in more than one way. It is still advisable to check
that they are safe. Double 5's are always safe if you have an even
number of men between the 5 and 9 points, and are sometimes unsafe if
this number is odd. Double 4's are always safe with an even number of men
between the 4 and 7 points, and otherwise may be unsafe. The position
above is safe, by the even-number rule, against rolls of 5-5 and 4-4 .
Exercise: Play a 1 .
↑
4/3 makes 6-6 catastrophic. 12/11 makes 5-5 dangerous.
6/5 is correct.
↑
In a 9-point match, X has 6 , O has 3 .
Should O redouble?
Yes, even though he only has a 10/36 = 28% chance of winning the game.
If he loses the game without redoubling he would have to win the Crawford game
and three doubled games to win the match, so his chances are about 6% ;
redoubling only loses this small equity. If he wins the game, redoubling
takes him from being still an underdog with about a 40% chance to being in
the lead with about a 60% chance. His net gain in chance of winning the
match is 28% times .20 - 72% times .06 , or 1.3% . If X is the
stronger player, the redouble is even more imperative.
PLAY AFTER CLOSING OUT ONE CHECKER -- THE RACE.
↑
O has seven checkers off, and is closed out; X has a near-perfect
position for bearing off safely. Experience shows that this is a nearly
even game; X probably has a slight advantage, but if O enters before
X has safely removed about three checkers, O is a favorite. It is
therefore important for X to avoid opening points until he has three
checkers off, even if he thereby increases the risk of being forced to
leave blots. In fact, at times he may voluntarily leave a blot rather
than open another point.
In the initial position these are my plays with each roll of the dice:
1-1 3/0 , 5/4 2-2 6/0 , 3/1 3-3 6/0 , 3/0 , 5/2
1-2 3/0 2-3 3/0 , 6/4 3-4 3/0 , 6/2
1-3 3/0 , 5/4 or 5/0 3-5 5/0 , 3/0
1-4 5/0 2-4 6/0 3-6 6/0 , 3/0
1-5 6/0 2-5 5/0 , 6/4 4-4 4/0(2) , 6/2 5/1
1-6 6/0 , 3/2 2-6 6/0 , 3/1 4-5 5/0 , 6/2
4-6 6/0 , 5/1
5-5 5/0(3) , 6/1
5-6 5/0 , 6/0
6-6 6/0(3) , 5/0
A typical game might proceed:
X O
4-1: 5/0 ----
4-5: 5/0 , 5/1 6-6: (0)
1-1: 1/0 , 3/0 3-2: (0)
3-2: 6/1 X avoids opening up another
point by 3/0 , 3/1 ; his move
delays O and does not slow
X up much.
5-2: 0/5/7
Now X needs, on the average, a shade under 6 rolls. O needs, on
the average, 2 rolls to get his back man to (say) the 21-point, then
3 & 1/2 more to bear off, for a total of 5 & 1/2 . Since X is on roll,
the position is quite equal.
6-5: 6/0 , 6/1 3-2: 7/12
5-4: 4/0(2) 6-1: 12/19
Doubles Drops
X has rolled very well, and O poorly.
If O's back man were on the 22-point he could take.
An extensive stratified experiment (108 games) shows that if O owns the
cube, his expectation in Position (A) is about -.22 of the cube.
Other experiments show that if O has no more than 5 checkers off,
X should play for maximum safety. On the other hand, if O has 9 or
more checkers off, X must take large risks if necessary. For example,
↑
X to play 6-2 .
Here X must open the 6-point. If O rolls a 6 , X is lost. X must
assume O will not roll a 6 , and play 6/0 , 2/0 . When he has another
man or two off, X will pick up his blots, as he can hope to win even if
O enters.
If O has 10 or more checkers off, he can safely double from the bar.
If he has no more than 5 off, X should double and O should decline.
GETTING AN EXTRA CHECKER OFF BEFORE BEING CLOSED OUT.
In Position (A) Dan Harrington, a strong Boston player, played 4/1 ,
4/off to minimize the number of rolls that hit.
↑ (A)
O to play 4-3 .
By this play, he will be hit in 11 games out of every 36 . Of those,
having seven checkers off, he will lose about 60% , or 6.6 games, and win
4.4 , for a net score of
25 gammons 50
4.4 wins 4.4
6.6 losses - 6.6
47.8 points.
By playing 4/0 , 3/0 , he will be hit in 12 games out of 36 , but,
with eight checkers off, will lose only about 44% of those, achieving a
net score of
24 gammons 48
6.7 wins 6.7
5.3 losses - 5.3
49.4 points.
The expected gain from bearing one more man off is (49.4-47.8)/36 = .04
in this position. If X could not be gammoned the saving would be
2.6/36 = .08 . It is usually correct to take a slight additional risk
(say, a 17 to 1 shot) in order to bear off an additional checker in
situations like this one, if you will then have 6 to 10 checkers off.
Outside this range, much depends on the gammon equity and the possession
of the doubling cube.
Example:
↑ (B)
O to play 4-1 .
O should play 4/0 , 4/3 to maximize his gammon chances. If hit, he will
double X out, so he need not bear off an extra checker.
BREAKING POINTS, AGAINST OPPOSITION.
A Little Learning ...
Position A arose in a recent low-stakes chouette.
(A) ↑
O to play 4-5 .
One of my partners remembered a similar position she had recently seen in
Magriel's book, probably like position B ,
(B) ↑
O to play 4-5 .
where clearing the 8-point leaves O in danger if he then rolls 6-4 ,
6-5 , or 5-5 . She urged playing 7/2 , 7/3 .
I pointed out, or tried to, that the resulting gap on the 7-point was
very dangerous because of X's men on the 5-point, in direct range. My
partners, MOLTO CRESCENDO, became very agitated about the fact that 6-4
leaves a double direct shot; one said that a double shot is much worse
than a single shot. I asked, "How much worse? How many times worse is
a double shot than a single one?" She said, "I don't know how many times,
I just know I don't want to leave a double shot." Eventually the captain
yielded to majority sentiment, playing 7/2 , 7/3 ; the pressure to make
the move with which most of the players had identified their egos had
precluded taking time to count all the rolls that leave shots.
Since you and I are under no such pressure, let's look at the two positions
O can choose.
↑ (C) (D)
After 8/3 , 8/4 After 7/2 , 7/3
In position C , the rolls to examine are those which do not clear the
7-point, i.e., those containing a 2 or 6 . Those which can not be played
safely elsewhere are 6-3 , 6-4 , 6-5 , and 2-5 . Each leaves a double shot
which is hit 20/36 of the time; the total chance of being hit is
8 * 20 = 160 out of every 36 * 36 = 1296 .
In position D , the dangerous rolls which fail to clear the 8 point
contain a 1 or a 3 . They are 6-1 , 6-3 , 5-1 , 5-3 , 4-1 , and 3-1 .
Rolls which leave a single blot on the 8-point are 6-1 , 5-1 , 4-1 , and
3-1 . These are hit 15/36 of the time, by a combination of 11 direct
and 4 indirect shots. Worse are the rolls of 6-3 and 5-3 , which
leave blots on both the 6- and 8-points. These are hit by all direct
1's , 3's , and 5's , or 27/36 of the time. The total risk of being
hit is 8 * 15 + 4 * 27 = 120 + 108 = 228 out of every 1296 rolls.
My partners were so concerned about the double shot in position C , that
they did not look for the triple shot in position D . After his next
roll, O will be hit 68/1296 = 5% more often in position D than
position C , and, if not hit, the gap may continue to plague him.
This position entails a conflict between two desirable goals; one of not
leaving gaps and one of safely breaking points at the holding distance of
six from an opponent's point. Argument can not resolve such conflicts
about minimizing risk; a careful player will count the dangerous numbers.
... is a dangerous thing. Drink deeply at the ? spring or not at all ...
In Diagram (A), O is struggling to avoid being gammoned. On his
next turn, he must bring in the man from the 10-point and bear off one man.
If he plays 10/9 now, he will need one die of 3 or more to bring the last
man in, and one of 2 or more to bear off. These are exactly the rolls
needed to bear off two men from the 3 and 2 points. Any 1 misses,
for a total of eleven bad rolls.
If he plays 2/1 now, he will need one roll of 4 or more to come in,
after which any roll bears off. As in bearing off from the 4 and 1 points,
only seven rolls fail.
To minimize his risk, he must play 2/1 . If the outside man were on the
9-point, 2/1 is again the correct play. The expected gains and losses in 36
trials from playing 2/1 , depending on the location of the outside man, are:
location 8 9 10 11 12 13 14 15 16
gain 0 8 4 0 -4 -9 0 0 -1
↑ (A)
O to play a 1 .
Going back one roll, to Diagram (B), where the outside man is on the
17-point, O with a 1 to play should move 2/1 if his next roll will
add up to 7 or 8 , but should play 17/16 if the next roll will total
4 or 5 . To decide the correct play, multiply the gain for each roll
by the probability of that roll.
roll 4 5 6 7 8
probability 3/36 4/36 6/36 5/36
gain -9/36 -4/36 0 4/36 8/36
product -27 -16 24 40
The net gain of playing 2/1 is 24 + 40 - 27 - 16 = 21 every
36{2} = 1296 trials. The same kind of analysis shows that the correct
play is found to be 2/1 if the outside man is anywhere from the 14-
to the 18-point. On the 13- and 19-points, the conventional wisdom
of crossing over into a new board is upheld.
↑ (B)
O to play a 1 .
Analogous situations arise when running with two men. In Diagram (C),
O should play 2/1 if the outside man is on the 8- to the 12-points.
↑ (C)
AVOIDING THE GAMMON -- (NEWPAPER COLUMN)
↑ (A)
O to play 6-3 .
In Position (A), O's only concern is to avoid being gammoned; since
X will bear off in two more turns, O must try to go off in one more. His
greatest danger lies in the gaps on the 4- and 5-points, so he should play
to bear a man off even if he rolls 4-4 , 4-5 , or 5-5 . If he plays the
plausible 8/5 , 8/2 , a roll of 4-4 is disastrous, because he still has
four men on points higher than the 4 . The correct play is 7-4 , 8-2 ,
guaranteeing a man off on the next play. In such situations, playing to
the lower gap tends to be right.
↑ (B)
O to play a 1 .
In Position (B), it is tempting to play 7/6 , but you should first
look at the gaps on the 3- , 4- , and 5-points. Rolls of 3-4 , 3-5 , and
4-5 would be disastrous. If you play 6/5 , no subsequent roll can prevent
your bearing a man off. Naturally, if you roll a 6-4 , you must play
7/3 , 6/off , and similarly for 6-3 , but you know even without checking
that the gap on the 6-point can't hurt you when there is only one man on
a higher point.
WAIT OR RUN?
↑ (A)
O to play 5-1 .
In this all-too-familiar situation, O can play 24/19/18 , conceding
the gammon for an expectation of -2 . He should, however, hang back in the
hope of a shot. The expectation is:
5 (2-2 or better): -3
1 (1-1): 27 (6 or more): -2
9 (5 or less): -3
= -2.25
20 (2-3, etc.; 2 men off): 23 (6 or more, no hit): -2
2 (2-3): -3
11 (hit): -1
(-.88 if O owns the cube)
= - 63/36 = -1.75
10 (1-2, etc.; 1 man off): 20 (O hits and doubles X out(?)): +1
16 -2
= - 16/36 = -.44
= - 56.65/36 = -1.57 .
↑ (B)
Removing a man from the 23-point, O's chances to win vanish. The chance
of avoiding gammon by hitting must be balanced against the chance of a
backgammon. The chance of avoiding the gammon is, at best,
10/36 (X rolls a 1) times 11/36 (O rolls a 1) = .08 .
The chance of being backgammoned is at least 26/36 = .81 . O must run
if his roll totals 6 or more.
↑ (C)
When X has 4 men on the 23-point, the chance of being backgammoned
drops to about 5/36 (X rolls 2-2 or better) + (26/36){2} (X goes off
in 2) times 9/36 (O rolls 5 or less) = .26 . The chance of avoiding
the gammon by a hit remains .08 , since O must certainly run on his next
turn. Again, O must run.
↑ (D)
Now if O runs, he can expect two more turns. If all his other men are in
his home board, and he gets a typical 8-point roll, played 24-16 , he has a
good chance to avoid the gammon by running. He needs 10 more pips to come
in, and an average of 2.5 pips to bear off. His chance to get 12.5
pips in two rolls is .85 . The probability of going off in one roll is
3/36 = .08 . The chance that X goes off in two rolls is about 10/36 ,
or .27 . By the usual convolutions, O's chance to avoid gammon is
.08 + .77 * .73 = .64 , for an expectation of -1.36 .
Assume instead that O waits a turn. His expectation is
10 (X rolls 1-2, etc.): 11/36 (O hits): -1
25/36: about -1.85
= -1.44
5 (X rolls 2-2, etc.): 11/36 (O hits): -1
25/36: -2
= -1.7
21 (X rolls 2-3, etc.): O's chance to avoid a gammon by running is
about 21/36 (X goes off in 2) times
.2 (O goes off in 2) = .17 , for an
expectation of -1.83.
= 61.33/36 = -1.7 .
Again, O must run. If he has several men still outside, however, he
should probably bring them home and hope for a shot even if he must waste
pips.
↑ (E)
Here, assuming all O's other men are home, his expectation on running is
almost exactly -1 . If he waits one turn, his expectation becomes
.08 (he hits on his next turn): at best, +1
.92 (missing, he runs): -1.36
= -1.33
Again, running is preferable.
RULE: If X has 8 or more men, O can afford to wait without serious risk
of a gammon. Otherwise, X should wait only if O has 3 men on the 2-point.
ACE-POINT RACES.
Occasionally, toward the end of a running game, both players will have
checkers remaining only on their 1-points. More often, they come down
to positions which play as if all the checkers were on the 1-points. Take
this position, for example,
↑ (A)
The only combinations of rolls which fail to bear X off as fast as
they would if all the checkers were on the 1-point are 1-1 preceded or
followed by 1-2 , 1-3 , 1-4 , 1-5 , or 1-6 , a total of 20 rolls
out of 1296 . For O , only rolls of 2-1 preceded or followed by
1-1 are worse than they would be if he had five or six checkers on the
1-point, a total of 4 rolls out of 1296 . As a practical consequence,
correct doubling strategy for this position is identical to that where
all the checkers are on the 1-point.
We now look at the calculation of expectations for all such ace-point
races. The reslts are summed up in the tables on page ---. We use
E(A,B) to mean the expectation when the player has A checkers and
the opponent has B checkers left.
If A = 2 : The player must win and E = 1.00 .
If A = 4 , B = 2 : Neither player can usefully double, so
E[+] = E[0] = E[-] = E .
E = 1/6 . +1 + 5/6 . -1 = -.67 .
If A > 4 , B = 2 : E = -1 .
A partial table of expectations can now be filled in:
B 2 4 6 8
A
2 1.00 1.00 1.00 1.00
4 - .67
6 -1.00
8 -1.00
To fill out more of this table, let us now look at A = 4 , B = 4 .
First consider E[0W] ,the expectation if the player waits rather than
doubling. In 5/6 of the plays, the player takes off two checkers,
resulting in E{*}[0](2,4) = -E[0](4,2) = .67 ; in the other 1/6 of
the plays, the player takes off four checkers, resulting in E(0,4) = 1.00 ;
the weighted average is
5: .67
1: 1.00
= .72
This is also the value of E(4,4) and E[-](4,4) . On the other hand, to
double gives the player the expectation E[0] = min(1, 2.E[-](4,4)) =
min(1,1.44) = 1 . It is therefore correct for the player to double and
the opponent to drop. Thus E[0](4,4) = E[+](4,4)= 1.00 . Obviously,
if A = 4 and B > 4 , E = E[-] = E[0] = E[+] = 1.00 .
When we get to E[0](8,8) , it is no longer correct for the opponent to
drop a double. If he accepts, the player's expectation is
2 * E[-](8,8) = 2 * 5: E{*}[-](6,8)
1: E{*}[-](4,8)
= 2 * =
Since this is less than 1 , the opponent should accept the double.
Continuing in this fashion, all the numbers in the table may be filled in.
The equations governing the table are these:
E(O,B) = E[-](O,B) = E[0](O,B) = E[+](O,B) = 1.00
E(A,O) = E[-](A,O) = E[0](A,O) = E[+](A,O) = -1.00
E[-](A,B) = 5: E{*][-](A-2,B)
1: E{*}[-](A-4,B)
E{*}[C](A,B) = -E[-C](B,A)
E[+][0](A,B) = max(E[+][0W](A,B) , E[D](A,B))
E[+][0W](A,B) = 5: E{*}[+][0](A-2,B)
1: E{*][+][0](A-4,B)
E[D](A,B) = min(1, 2 * E[-](A,B)) .
A computer was programmed to do the arithmetic implied by these equations,
although the tables could be obtained by hand in a few hours of patient
calculation. The tables indicate that if A < B , the player should always
double or redouble; if A = B , he should double; and if A = B <= 8 , he
should redouble. The opponent should take if A = B >= 8 .
Positions closely related to ace-point race positions can be analyzed by
reference to expectations in ace-point races. Take this position:
↑ (B)
If O doubles, should X take? It would be a drop if O had all six
checkers on the 1-point since 2 * E[-] = 1.16 . Here, however, O can
miss with a 1-2 , and be redoubled out of the game. Thus in two cases
out of 36 , O's expectation is changed from
2 * E{*}[-](4,6) = 2 * -E[+](6,4) = 2 * .49 = .98 , which it would be in
an ace-point game, to 2 * -1 . The net change in E[D] resulting is
2/36 * 2.98 = .17 ; 2 * E[-] = 1.16-.17 = .99 , and it is correct for X
to take. Naturally, it is correct for O to double; he will seldom find
a position so close to the ideal doubling point E[-] = .5 . (We have failed
to consider a few other rolls, like an initial 2-2 for 0 , which do not
play as well as they would if all checkers were on the 1-point; the roll of
2-1 is the only one which makes a major difference in the expectations.)
If the odd checker in the previous position is moved to the 4-point:
↑ (C)
O can miss with rolls of 1-2 , 1-3 , and 2-3 . This reduces the
expectation after doubling, E[D] , to 1.16 - (6/36 * 2.98) = .66 .
If instead O waits, his expectation is
6: -1
4: E{*}[0](2,6) = +1 {double 1's , 4's , 5's , 6's}
26: E{*}[0](4,6) = .72
= .46
so even this position is a double.
Over the board, one could try to make a correct decision on doubling by
estimating P . In a 3-roll ace point race, P = .79 . In Position (C),
however, six bad rolls give the opponent P = .79 , for a shift of about
.79-.21 = .58 ; thus P is close to .79-(6/36 * .58) = .69 . The
rule of thumb for doubling at T turns, P >= .8 - .3/(sqrt T) , with
T = 3 , gives P >= .63 as the doubling threshold, so a double is
plausible.
If the back checker is on the 5-point,
↑ (D)
O misses with twelve rolls: 1-2 , 1-3 , 1-4 , 2-3 , 2-4 , and 3-4 .
The expectation after doubling is 1.16-(12/36 * 2.98) = .17 . If he waits,
his expectation is at least half this, or .08 . In addition, if he waits,
rolls a number which takes two checkers off (21/36) and X fails to roll
doubletons (30/36) , his expectation is better by
E[0](4,4) - E[-](4,4) = 1 - .72 = .28 , for a net gain of
21/36 * 30/36 * .28 = .14 ; thus E[D] = .17 , but
E[W] >= .08 + .14 = .22 , and waiting is correct.
DOUBLING THEORY -- DOUBLING ON A KEY ROLL.
Suppose the next roll will certainly decide the outcome of the game.
Naturally, if your chance P of a favorable roll is greater than .5 ,
you should double. If you don't double, E is 2 * (P-.5) ; if you do,
E is min(4 * (P-.5),1) . In the range .5 <= P <= .75 , you gain
2 * (P-.5) by doubling, whether or not you own the cube.
In most late game situations, however, victory is not so certain. Perhaps
the roll that favors you will allow your opponent to play a well-timed ace
point back game, so your probability will be about .8 . Perhaps the roll
that favors your opponent will still allow you one more direct shot, giving
you a probability of .3 , or perhaps you will find yourself in an unfavorable
race, with a normalized lead of -.8 or -1 , and a probability of .2 or
.15 . We will analyze several simplified versions of this kind of situation,
and determine the correct doubling decision. In this situation:
P = .7 (my good roll) .8 (game goes well): 1
.2 (game goes badly): 0
.8
.3 (my bad roll) .2 (game goes well): 1
.8 (game goes badly): 0
.2
= .56 + .06 = .62
E = 2 * P-1 = .24
so I have a definite lead. If I double, my expectation is
2 * E[-] = 2 * (.7 * .6 + .3 * -1) = .24 . If I leave the cube in the
center, my expectation is E[0W] = .7 * 1 + .3 * -1 = .4 . If I own
the cube, my expectation is E{+W} = .7 * 1 + .3 * -.6 = .52 .
I lose .16 units of equity by doubling, .28 units by redoubling. As
a rough rule of thumb, it is unwise to double on P < .65 unless the cube
will be of no further use to either player.
If my good roll has probability .8 , with everything else the same,
2 * E[-] = .56 , E[0W] = .60 , and E[0+] = .68 , so I should still wait,
even though P = .68 .
If, however, the probabilities are these:
P = .7 (my good roll) .9 (I win)
.1 (I lose)
= .9
.3 (my bad roll) .1 (I win)
.9 (I lose)
= .1
.66
we get
2E[-] = 2(.7 * .8 + .3 * -1) = .52 ,
E[0W] = .7 * .1 + .3 * -.1 = .4
E[+W] = .7 * 1 + .3 * -.8 = .46
and a double, or redouble, has a positive payoff, even though P is only
.7 * .9 + .3 * .1 = .66 , less than in the previous situation.
The reason for this seemingly paradoxical result is that the advantage
of an immediate double is zero if, after your favorable roll, your opponent
would still take a double. If P[1] > .75 is your probability of winning
after a favorable roll, your gain in expectation from having doubled is
2 * (P[1]-.75) . In the first two examples, P[1] was .8 , and the gain
from an early double, assuming a favorable roll, was only .2 . If P[2]
is the probability of the opponent winning after an unfavorable roll, your
loss in expectation from having doubled is 1 if P[2] > .75 , and
4 * (P[2]-.5) otherwise, when the opponent redoubles. The moral of this
analysis is, that, if your opponent will still take a double after your
favorable move, you should always wait. If P[1] = .8 , you should have
at least a 5/6 = .83 chance of a favorable roll to double, and more to
redouble. If P[1] = .9 and P[2] is large, you might double with as
little as a .625 chance of a favorable roll.
As applied to common late game board situations, in the absence of
gammon chances, never double if the opponent has a good holding point,
such as his bar point against your midpoint. Don't double on the
prospect that the opponent will play an ace-point back game if he has as
much as one chance in six of an immediate reversal.
DOUBLE OR WAIT? COUNT THE KILLERS.
Suppose that you and your opponent are bearing off in a five-roll
situation. Most of the men are on low numbered points, so you are counting
men, not pips. Your positions are very similar, but you have the advantage
of the roll. You are thinking about doubling.
Your chance of winning is enough to justify a double. Your opponent can
win only by rolling a doubleton (ignoring misses for the moment), and in a five
roll situation, his chance of a doubleton on his first four turns is
1 - (5/6){4} = 1 - 625/1296 = 671/1296 = .52 .
If he gets a doubleton, you may also get one. Even if he gets one on his
first roll, you get to roll twice before being doubled out. Your chance for
failing to get a doubleton, assuming that on the average you will get three
rolls before being doubled out, is
(5/6){3} = 125/216 = .58 .
To win, he must get a doubleton and you must not; his chance to win is
.52 * .58 = .30 . This is small enough that you intend to double him, but
large enough that he must take.
To decide between doubling and waitingone more roll, you have to look
at three possibilities which arise before your next chance to double, and
which affect the desirability of the double:
(1) You and your opponent may each bear two men off on your next rolls.
(2) You may get a doubleton, or your opponent may miss an essential
bearoff, leaving you with such a strong position that your opponent
will no longer accept a double.
(3) You may miss an essential bearoff, or your opponent may get a
doubleton. When your turn comes again, you will no longer be
thinking of doubling. Rather, you will be hoping to roll a
doubleton before you yourself are redoubled out.
Let's look at the profit and loss figures on each of these situations,
depending on whether you double or wait.
In the first situation, which happens most of the time, the outcome is
exactly the same; whether you double this turn or next has no effect on the
dice, the positions, and your opponent's acceptance of the double.
Situation (1) therefore doesn't have any effect on your decision whether to
double.
The second situation favors an immediate double. You become virtually
certain to win, and if you wait you only win one point. Had you doubled
immediately, you would expect two points (actually about 2 * .96* = 1.92
[footnote: All starred numbers come from the table in the Appendix.] points,
taking into account the possibility of a small miracle), for a net profit of
.92 points.
On the other hand, the third situation would make you regret an immediate
double. You become an underdog, with your expectation being about
2 * -.75* = -1.5 points. Had you waited, your expectation would be
-.71* points. An early double loses you .79 points of equity in the game.
Putting this all together, the profit you can expect to make from doubling
early, if you play this position 36 times, is .92 times the number of
rolls which kill your opponent, minus .79 times the number of rolls which
kill you.
This, then, is the way to decide.
(A) Count the rolls which kill your opponent, either by speeding you up
one roll, or by slowing your opponent down one roll. These are
your killer rolls.
(B) Count the rolls which kill you, either by slowing you down one roll,
or by speeding your opponent up one roll. These are your opponent's
killer rolls.
(C) Unless your opponet's killer rolls, multiplied by .79/.92 = .86 ,
about 6/7 , outweigh yours, you should double immediately.
This is the Killer Rule. It applies in non-contact bearoff positions, from
five- to seven-, with the cube in the center, when you have the advantage
of being on roll in otherwise nearly equal positions. The rule can be
adapted to three- and four-roll situations as well.
We have assumed a 5-roll position. Let's see how much difference being in
a 7-roll position makes.
In a 7-roll position, the profit from doubling before a move which kills
your opponent is 2 * .86* - 1 = .72 ; the loss by doubling before a move which
kills you is -(2 * .58 - .51) = -.66 ; in this situation your opponent's
killers are worth .66/.72 = .92 times as much as yours. Ordinarily you
double if the killers are equal, but not if your opponent has more assassins
looking for you than you have for him.
Let's try the Killer Rule on a few actual positions.
(A) ↑
Here you (X) don't even have to think. The positions are identical, you
have the same number of killers as your opponent, and you should double
immediately.
(B) ↑
Here you (X) have only two killers, 6-6 and 5-5 ; nothing else takes
off four of your men, and even if your opponent rolls 1-2 , he is still very
much in the game, because missing once does not kill him.
Your opponent, however, has a number of killers which take three or four
men off and leave him in a sound three-roll position: 6-6 , 5-5 , 4-4 , 3-3 ,
and 2-2 . Even 1-1 is dangerous. Comparing your two killers to .86 of
his five, you are ougunned and should keep a low profile, at least until
the next turn. If it were O's turn in the same position, he would double
with great speed, and X should drop.
(C)↑
(D)↑
In position (C), X , who is five pips ahead, can be killed by his own
rolls of 3-2 and 3-1 , and by O's rolls of 6-6 , 5-5 , and 4-4 , for a
total of seven rolls; O can be killed only if X rolls 6-6 , 5-5 , or
4-4 , totalling three rolls. X must wait to double. If we move some of
X's men back five pips, we get position (D), in which each player has the
same killers and X should double right away.
You might think X's postion in (C) is worse than that in (D), because of
the gap on the four point. An experiment, bearing off from both positions
using the same rolls 72 times, shows that the position in (C) is actually
better, bearing off faster by a total of 4 rolls in 72 tries. The only
likely dangers to X , in (C), occur on his first turn, and he can afford to
wait until he is past them to double. In (D) he must double right away; if
he waits until he is past all danger he will probably score only one point.
As this example illustrates, the Killer Rule is not based on your eventual
likelihood of winning; it is based only on what could happen before your
next chance to double.
What about the Killer Rule when you already own the cube? We can go
through the same argument, but this time the numbers are different when your
opponent gets a killer. Had you waited to redouble, your possession of the
cube would give you an expectation of -.36* of the cube's original value.
Had you redoubled early, your expectation would be 2 * -.75* = -1.50 .
The cost of an early redouble, when your opponent rolls a killer, is therefore
1.14 times the value of the cube. His killers are worth 1.14/.92 = 1.24 ,
or about 5/4 , as much as yours are. This gives us the Killer Rule for
redoubling:
Don't redouble unless your killer rolls outnumber your opponent's
by 5 to 4 .
In positions (A) and (B), X should not redouble. In position (B), if it
is O's turn, he should redouble quite happily.
The Killer Rule does not apply directly to four-roll situations. The
reason is that ordinarily, the four-roll situation is the last one in which your
opponent will take your double. If neither player rolls a killer, you will
double your opponent out, but you would prefer that he owned the cube at two;
your equity would then be 2 * .58* = 1.16 , rather than 1 . Because of
this, you should double a shade more aggressively in four-roll bearing off
positions. For example, X should double in position (E):
(E) ↑
with only three killers to his opponent's six. If each rolls a 5-2 , say,
O will no longer be interested in accepting X's double. X could even
redouble, in position (E); he no longer has much reason to fear O's
possession of the cube.
Summing up:
The Killer Rule applies in 5- to 7-roll checker counting bearoff situations,
where you are considering doubling.
Your killer rolls are the rolls which speed you up by a full turn, or slow
your opponent down by a full turn. Your opponent's killer rolls are those
which speed him up by a full turn, or slow you down by a full turn. If you
have more killer rolls than your opponent, you should usually double or
redouble. If you have fewer killer rolls than your opponent, you should
usually refrain from doubling. If you are even in killers, you should
usually double but not redouble. The exceptions:
(1) If you have more than 6 killer rolls, double even if your opponent
has one more than you do.
(2) If your opponent has more than 4 killer rolls, don't redouble unless
you have two more than he has.
APPENDIX
This table shows your expected winnings, using the current value of the cube
as the unit, depending on the number of men you and your opponent have to
bear off from low points, and on the location of the cube. It is assumed to
be your turn. In each position, the three numbers represent, from top to
bottom, your expectation with the cube on your opponent's side, in the center,
and on your side. For example, if you have 9 or 10 men, and your opponent
has 7 or 8 , your expectation is -.70 , -.65 , or -.26 times the cube
depending on whether the cube is your opponent's, in the center, or yours.
Opponent's Men
Your 1-2 3-4 5-6 7-8 9-10 11-12 13-14 15
men
1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
1-2 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
- .67 .72 1.00 1.00 1.00 1.00 1.00 1.00
3-4 - .67 1.00 1.00 1.00 1.00 1.00 1.00 1.00
- .67 1.00 1.00 1.00 1.00 1.00 1.00 1.00
-1.00 - .72 .58 .96 1.00 1.00 1.00 1.00
5-6 -1.00 - .72 1.00 1.00 1.00 1.00 1.00 1.00
-1.00 - .49 1.00 1.00 1.00 1.00 1.00 1.00
-1.00 -1.00 - .75 .46 .92 .99 1.00 1.00
7-8 -1.00 -1.00 - .71 .92 1.00 1.00 1.00 1.00
-1.00 - .95 - .36 .92 1.00 1.00 1.00 1.00
-1.00 -1.00 -1.00 - .70 .36 .86 .98 1.00
9-10 -1.00 -1.00 -1.00 - .65 .73 1.00 1.00 1.00
-1.00 -1.00 - .90 - .26 .75 1.00 1.00 1.00
-1.00 -1.00 -1.00 - .99 - .58 .29 .81 .97
11-12 -1.00 -1.00 -1.00 - .99 - .50 .59 1.00 1.00
-1.00 -1.00 - .99 - .83 - .19 .65 1.00 1.00
-1.00 -1.00 -1.00 -1.00 - .96 - .51 .25 .77
13-14 -1.00 -1.00 -1.00 -1.00 - .95 - .41 .51 1.00
-1.00 -1.00 -1.00 - .98 - .78 - .15 .59 1.00
-1.00 -1.00 -1.00 -1.00 -1.00 - .94 - .46 .22
15 -1.00 -1.00 -1.00 -1.00 -1.00 - .93 - .35 .45
-1.00 -1.00 -1.00 -1.00 - .96 - .73 - .13 .54
Where do these numbers come from? They can be calculated readily, starting
from the upper left. For example, with 10 men each and the cube on your
side, the table gives .75 . This reflects the fact that 5/6 of the time
you take off 2 men, giving your opponent an expectation -.70 , and 1/6
of the time you take off 4 men, giving your opponent an expectation of
-1.00 . His average expectation after your move is
5/6 * -.70 + 1/6 * -1.00 = -.75 , and his expected loss is your expected gain.
↑
O to play 6-1 .
It is tempting to play 16/10/9 in the diagrammed position, for several
reasons. It is quicker to move one man than two. The move to the 9-point
hastens bearing off. The alternative 16/10 , 2/1 wastes a pip on low
points.
These reasons are all irrelevant to winning. O will probably win unless
he leaves a blot and is hit, in which case he will probably lose. The only
important difference between the two moves is that 16/10/9 leaves O with
no safe way to play 6-4 or 5-5 . By plaing 16/10 , 2/1 , O can be
forced to leave a blot only by 6-6 and 6-5 , which are bad in any case.
When you are stripped on the higher numbered points of your home board, you
are in danger, even if you still have a man in your outer board. Take a
little time to plan what to do if the next roll is large. Backgammon is a
race, but it is not just a race.
Problem:
↑
O to play 4-4 .
Answer: 20/12/8/4 guards against rolling a 6-4 . Keep your spare men
in the home board on high points for flexibility.
(Rearrange above problem so both men must be moved -- e.g. outside men on
13 and 8 , to play a 6-4 .
A NEWSPAPER COLUMN -- 3 MAN BEAROFF.
With 3 checkers to bear off, you can afford to miss once. If you expect
to have two turns, you should play to avoid missing twice.
(A) ↑ (B)↑
In position (A), every roll takes at least one man off. The worst roll,
1-2 , leaves O on the 5 and 2 poins, with a 19/36 chance of coming
off on the next roll. In position (B), a 2-3 roll is catastrophic.
While there are rolls (like 1-2 followed by 4-2 ) which play better in
position (B), (A) is 1% more likely to bear off in two turns and just as
likely as (B) to bear off in one turn. Therefore, if you have a 6-5 to
play in this position,
(C) ↑
the best play is 10/5/0 , leaving you with position (A) on your next turn.
The rule in such situations is to look at the gaps in your board (in this
case 2 and 3 ) and make sure you can bear a man off if both your dice fall
in the gaps. This is a good rule whenever you have an odd number of men,
with a small pip count, and are ahead in the race.
Exercise: How should O play a 1 ?
↑
Answer: 2/1 makes 2-3 catastrophic. 4/3 makes 1-4 catastrophic.
6/5 is correct.
A 5-ROLL, PIP COUNT, DOUBLING SITUATION.
↑
Should O double?
Definitely. In fact, X only barely has a take. An experiment using the
same dice rolls for both sides 36 times found that O came off in 3
rolls 3 times; 4 rolls 11 times; 5 rolls 15 times; 6 rolls
7 times. The corresponding numbers for X were 3 , 7 , 16 , and 9 ;
once, X came off in 7 rolls. The chance that X wins, in 36 * 36
tries, is 3 * (11 + 15 + 7) + 7 * (15 + 7) + 16 * 7 = 365 ; the
probability for X is .28 ; that for O is .72 . A good rule of thumb
for doubling in running games with T turns to go is to double if your
chance to win is .8 - .3/(sqrt T) . Here T is 4.7 , and the doubling
threshold is .8 - .3/2.2 = .66 , so O should certainly double.
Over the board, I would base a double on this pip count formula: is my
lead, in pips, greater than (my pip count/8)-3 ? Here my pip count is
32 , my lead is 2 pips, and the answer is yes. Were X's position
identical to O's , a double would probably be premature.
A 3-ROLL DOUBLE.
↑
X doubles. Should O take?
It is not likely that anyone could work out the exact odds for O to win,
over the board. It isn't necessary. Barring a miracle 6-6 for O , or
repeated 1-2's for X , O can win only if he goes off in two turns and
X goes off in three. If P is the probability that O goes off in three
turns, the probability that X does not go off in two is 1-P , and O's
probability of winning is P(1-P) , which is at its largest value, .25 ,
when P = .5 .
Accepting the double can not be profitable. In a situation where each
player will be off in either two rolls or three, it is never advisable to take
a double unless your board is better than your opponent's. If the man on O's
6-point were on the 4-point instead, O would have a definite take; he
would be a slight favorite to be off in 2 , and X would not.
AN END GAME DOUBLE.
↑
X doubles. Should O take?
If all the checkers were on the 1-points, O would have a definite take.
Here, however, doubletons, the underdog's weapons, are not so helpful. Only
3-3 or better will help O on his first turn; 1-1 will never help him.
On the average, only about 4 & 1/2 of the 36 rolls help O on each turn.
While X has the same problem, X expects to win without doubletons.
Let's look at the ways O might win the game, using X2 as a shorthand
for X bearing two men off. The only ways O can win, with their
probabilities, are:
X2 ; O4 ; X2 ; O doubles; X drops.
.88 * .12 * .88 = .093
X2 ; O4 ; X4 ; O4
.88 * .12 * .12 * .12 = .002
X2 ; O2 ; X2 ; O4 ; X2
.88 * .88 * .88 * .12 = .072
X2 ; O2 ; X2 ; O2 ; X2 ; O4
.88 * .88 * .88 * .88 * .88 * .12 = .063
X4 O4 X2 O4 .002
.232
With less than a 25% chance, O must drop. Move an O checker from the
3-point to the 1- or 2-point, and O has a take.
The moral: in a 4-roll situation, with equal positions, and no likelihood
of your opponent missing, don't take a double unless at least five of your
doubletons are winners.
NEWSPAPER FILLER -- A 2-ROLL DOUBLE.
↑
X doubles. Should O take?
Emphatically yes. If X rolls 1-3 , 1-4 , 2-3 , 2-4 , or 3-4 , O will
redouble X out, so O has at least a 10/36 chance. Additionally,
possession of the cube gives O substantial equity if X should roll 1-1 ,
1-5 or 1-6 . On the other hand, X was correct to double, for, if X
rolls a 5-2 or better (19 rolls), his expectation is doubled, from
28/36 to 56/36 .
↑
Should O double? Should X take?
Look at the second answer first. If O rolls a total of 5 or less
(9 rolls) the best thing to do with it is to play to the 4-5 points.
Then X doubles O out. So it is a clear take.
If O doubles, rolls of 5 or less lose. Rolls of 6 play, at best,
to the 5-3 , where X doubles, and O has only a bare take. So O
has 13 losing rolls. A roll of 7 , played to the 5-2 , leaves O
a slight underdog. A roll of 8 , played to the 5-1 or 4-2 , leaves
O a slight favorite. Thus there are 11 neutral rolls. The twelve
rolls of 9 or more are likely, but not certain, winners. In short,
if O doubles, he has a negative expectation. If O waits, then rolls
1-6 , 3-4 , or any 8 , 9 , or 10 , his access to the cube will increase
his expectation by 4/36 , 18/36 , 10/36 , and 4/36 respectively
(X will have doubled him in the first case). This is a much better use
to make of the cube.
↑
Now O has only 9 losing rolls (5 or less), and 17 winners. X still
has a definite take, considering that he may roll doubletons. To estimate
O's chances, treat 44 , 55 , and 66 as quick winners, rolls of 6 or
less (9 of them) as more likely than not to lose, and the other 24 rolls
as likely winners. His total chance to win is roughly
4/36 + 6/36 * 24/36 = 24/36 = .67 , a shade over the .65 we recommend for
a two-roll double.
↑
This is a definite double for O , of course. The hard question to answer
over the board is whether it is a take for X . Let us divide the number
line between the two players in proportion to their winning chances.
|---------------------------------------|
0 1
First we give 4/36 = .11 to O , for the possibility of immediately rolling
33 or better.
O
|--------------------------------|------|
.89
We also give 5/36 = .14 to X , who can effectively double O out if O
rolls a total of 3 or 4 .
X O
|------|-------------------------|------|
.14 .89
Of the remainder, we give 1/6 to X for doubletons.
X X
|------|------|------------------|------|
.14 .26 .89
Already X has more than .25 , so it is a clear take in a money game. For
the rest, of O's 27 initial rolls not already discussed, half the 5's
(i.e., 2 rolls) and a quarter of the 6's (one roll) lose; we give X
3/27 of the remaining segment of the line, the rest to O .
X X X O O
|------|------|----|-------------|------|
.26 .33
With a little practice, this method of dividing the line, remembering as
you go the endpoints of the no-man's-land in the middle, can be used over
the board to decide whether to take a double when several rolls remain to
be played.
ACE POINT RACES.
Suppose all my checkers, and all of my opponent's, are on the ace points.
All non-doubletons take off two checkers, and all doubletons take off four.
Suppose, say, that I have seven checkers, my opponent has six, and it is my
turn. I want to calculate my chance of winning.
Let P(4R,3R) mean the probability that I win if I need four (non-doubleton)
rolls, my opponent needs three, and it is my turn. Let P*(2R,3R) mean the
probability that I win if I need two rolls, my opponent needs three, and it
is his turn. Since
Position A:
(O to move)
leads one sixth of the time to position B :
(X to move)
and five-sixths of the time to position C :
(X to move)
the probability that I win in Position A is 1/6 of the probability that
I win in Position B , plus 5/6 of the probability that I win in Position C.
As a formula,
P(4R,3R) = 1/6 P*(2R,3R) + 5/6 P*(3R,3R) .
As an abbreviated notation for such weighted averages, we use
P(4R,3R) = 1: P*(2R,3R)
5: P*(3R,3R)
meaning that for every one time that my probability becomes P*(2R,3R) ,
there are five where it becomes P*(3R,3R) . The notation
A: X
B: Y
C: Z
is shorthand for the formula (AX + BY +CZ)/(A + B + C) .
If we try to actually compute P(4R,3R) , we need P*(3R,3R) . To compute
that, we need first to find P(3R,2R) . To compute that, we need to find
P*(1R,2R) . To work backward this way is awkward. When a problem of this
sort arises, it is usually most convenient to work forward, finding first
P(1R,1R) and P*(1R,1R) , then P(2R,1R) and P*(2R,1R) , etc. We will
calculate a table of these, up to P(8R,8R) . Because some situations are
sure wins, we can immediately fill in parts of the table:
Table of P(AR,BR)
A B 1 2 3 4 5 6 7 8
1 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
2 1.00 1.00 1.00 1.00 1.00 1.00
3 0 1.00 1.00 1.00 1.00
4 0 1.00 1.00
5 0 0
6 0 0
7 0 0 0
8 0 0 0
Table of P*(AR,BR)
A B 1 2 3 4 5 6 7 8
1 0 1 1 1 1 1 1
2 0 1 1 1 1
3 0 0 1 1
4 0 0
5 0 0 0
6 0 0 0
7 0 0 0 0
8 0 0 0 0
The rest of the table is filled in, starting at
P(2R,1R) = 1: P*(0R,1R) = 1
5: P*(1R,1R) = 0
= 1/6 = .167
P*(1R,2R) = 1: P(1R,0R) = 0
5: P(1R,1R) = 1
= 5/6 = .833
It is no accident that these two numbers add up exactly to one. The chance
that I win, when I need A rolls, and my opponent needs B , is P(AR,BR) ,
the chance that my opponent wins is P*(BR,AR) , and the two numbers add up
to 1 .
We need not actually compute the table of P*(AR,BR) ; we instead can take
1-P(BR,AR) , using the table of P . For example,
P(4R,3R) = 1: P*(2R,3R)
5: P*(3R,3R)
= 1: 1-P(3R,2R)
5: 1-P(3R,3R)
= 1- 1: P(3R,2R)
5: P(3R,3R)
The complete table of P(AR,BR) is now easy to fill in:
P(AR,BR)
A B 1 2 3 4 5 6 7 8
1 1 1 1 1 1 1 1 1
2 .167 .861 1 1 1 1 1 1
3 0 .255 .788 .981 1 1 1 1
4 0 .023 .301 .745 .957 .997 1 1
5 0 0 .051 .329 .718 .934 .992 1.000
6 0 0 .003 .078 .347 .698 .914 .985
7 0 0 0 .010 .102 .361 .682 .895
8 0 0 0 .000 .018 .124 .372 .669
Example calculation: P(4R,3R) = 1-P*(3R,4R)
= 1 - (5 times P(3R,3R) + 1 times P(3R,2R))/6
= 1 - (5 times .788 + 1 times .255)/6 = .301
Observation: P(nR,nR) is well approximated by 1/2 + 1/( 2 sqrt n) ,
and superbly by 1/2 + .478/sqrt n .
As an example of how you might use these numbers, suppose the score in a
seven point match is 5-5 , and your opponent doubles you in a 5-roll
situation. If you accept, the match hinges on this game, and your opponent's
chance is P(5R,5R) = .714.
If you decline the double, you will be behind 5-6 in the match. Later in
the book we will estimate that an equal opponent will have a chance of about
.70 to win the match, so you should decline the double, unless the opponent
is somewhat stronger in checker play than you. In a seven-roll situation,
you might prefer to take the double; your opponent will only have a .68
chance to win the race.
In a money game, you need to know the average amount you win, or EXPECTATION,
rather than your chance of winning. The expectation depends on the position
of the cube. We use E@0(4R,3R) [@ denotes subscript] to mean your
expectation with the cube in the center when it is your turn, you need four
rolls, and your opponent needs three. Similarly, E@+(4R,3R) is your
expectation when you own the cube (and can't be doubled out); E@-(4R,3R)
is your expectation when your opponent owns the cube. The expectation
multiplies the value of the cube; if your expectation is .46 , and the
cube is at four, your average winning is 4 times .46 = 1.84 . An
asterisk over the E shows that it is your opponent's roll.
In some situations, the player on roll has the option to double. We use
E@D to mean the expectation if he chooses to double, and E@W to mean
the expectation if he waits; presumably, he will choose the one which is
larger.
The equations connecting these formulas are:
E@-(AR,BR) = 5: E↑*@-((A-1)R,BR)
1: E↑*@-((A-2)R,BR)
E@0(AR,BR) = maximum of E@D(AR,BR)
E@W0(AR,BR)
E@D(AR,BR) = minimum of 1
2 times E@-(AR,BR)
E@W0(AR,BR) = 5: E↑*@0((A-1)R,BR)
1: E↑*@0((A-2)R,BR)
E↑*@-(AR,BR) = 1 - E@+(BR,AR)
(etc.)
These equations were evaluated by computer, and the results are shown in
Table ?. If you double your opponent in a four-roll position, your
expectation should he take, is the value of the cube times E@-(4R,4R) ,
or 2 times .46 = .92 . If he should drop, your expectation is 1 , so
he should take. On the other hand, if you double in a three-roll position,
or any position where E@- is more than .50 , your opponent should drop.
The fact that E@D(5R,5R) is less than E@+(5R,5R) shows that it is correct
to wait if you own the cube in a five roll situation. Since E@D(5R,5R) =
E@0(5R,5R) , you should double if the cube is in the center.
Generally, the conclusions to be drawn from Table ? are:
(1) Give an initial double whenever you are on roll in an equal ace-point
race.
(2) In such a race, don't redouble until you reach a four-roll position.
(3) Accept a double when your opponent is on roll in an equal ace-point
race if there are at least four rolls per player to go.
In Chapter ? we will use the numbers in Table ? to help decide whether
to double, or to take a double, in positions like:
where certain numbers miss.
In analyzing positions like:
it is important to know how likely the player on the ace point is to bear off
all his checkers in a particular number of rolls. We use P(4R,3T) to mean
the probability that a player in a 4-roll position will bear off in three
turns (or fewer). The relevant equation is:
P(AR,BT) = 5: P((A-1)R,(B-1)T)
1: P((A-2)R,(B-1)T)
Table ? shows the probability P(AR,BT) computed from this equation:
A B 1 2 3 4 5 6 7
1 1 1 1 1 1 1 1
2 .167 1 1 1 1 1 1
3 0 .306 1 1 1 1 1
4 0 .028 .422 1 1 1 1
5 0 0 .074 .518 1 1 1
6 0 0 .004 .132 .598 1 1
7 0 0 0 .016 .188 .665 1
A good approximation to P((A+I)R,AT) is A((17-A)/100) .
Notice that at five rolls, you are favored to roll at least one doubleton
among the first four rolls.
Should you need to know the probability of going off in exactly B turns,
use P(AR,BT)-P(AR,(B-1)T) . For example, the chance that bearing off takes
exactly three rolls in a four-roll position is .422 - .028 = .394 .
As an example of the use of this table,
As a rough estimate, X is unlikely to go off in two turns, and is very
likely to go off in three. We can estimate O's chance as the chance that
he will go off in three turns, or .422 . (The actual exact probability that
O wins can be calculated to be .48 .)
WHEN TO DOUBLE AGAINST A BACK GAME.
If your opponent has a well-timed back game, he will win more often than not.
Because you will usually gammon if not hit, your expectation, in the absence
of the doubler, is usually positive. If you wait too long to double as you
bear off, your expectation will be greater than 1 ; you might as well not
have access to the cube. On the other hand, there is no point to doubling if
your opponent will still take on the next turn, or if E@- is not substatially
positive.
The ideal point at which to double is that at which E@- = .5 , so that it is
indifferent whether your opponent drops or takes. Naturally, anywhere between
E@- = .5 and E@0 = 1.0 , you double. If E@- is between 0 and .5 , it
is hard to decide whether to double or to wait.
Let's suppose you have a position without major weaknesses, in which you are
trying to clear one point at each turn (Diagram). As rough rule of thumb,
you run a 30% chance to be hit on each turn, if you are stripped on your
high point, as you must eventually be in order to clear it. Say that
currently E is .2 , corresponding to P@s(-1) = .6 , P@s(+2) = .4 .
If E@next is your average expectation next time if not hit
.2 = E = .3: -1
.7: E@next ,
and E@next = .7 .
In this situation, if you double immediately you have E@- about
2 times E = .4 ; if you wait, and double your opponent out next time, you have
E@W = .3: -1
.7: +1
= .4
so this is the point at which it is indifferent whether you double or wait.
In brief, you want to double when your chance to bear off safely and win the
gammon is about .4 . For example, with three points to clear and a 30%
chance to be hit as you clear each one, your chance is about .7↑3 = .35 ,
perhaps too early to double. Allowing for the possibility that you may be
hit with nine or ten men off and still have enough winning chances to take
a double, this might be the doubling threshold.
So typical positions in which O should double are these:
(Here the 3 and 4 points are relatively easy to clear.)
These positions are too good to double:
Naturally, the expectations depend not only on the chance of being hit, but
on X's precision in
(1) Redoubling after a hit.
(2) Playing to capture a second checker after a hit.
(3) Deciding what risks to take of being backgammoned.
As a result, this analysis is only approximate. It is probably most useful
as a pattern for the reader to plan his own experiments.
TAKE RISKS FOR POSITION.
It is often tempting to minimize your immediate risks, hoping that something
will come up to solve the resulting positional problems. It is usually worth
a small additional risk of being dealt an immediate lethal blow, in order to
have a sound strucuture for the rest of the game. In team play, I find this
approach often agitates my partners, but there is good evidence for its
correctness.
A case in point comes from the final match of the recent World Championship,
where X in Position A has a 3-1 to play. In the match, X played 7/3 .
Bill Robertie, backgammon columnist for the Boston Globe and one of New
England's best players, analyzed the play in his column. Robertie observed
that in the resulting Position B , X is subject to four numbers that leave
two blots (6-5 and 5-4) . He recommends instead 6/5 , 6/3 , resulting
in Position C , where no roll leaves a blot on the next turn, and X has
numerous spare men to play freely.
I felt a few doubts about the position because of the gap on the six point,
so I played out the rest of the game 36 times each way. I used the same
dice rolls, and I made sure that on X's next roll, every number came up
the expected number of times (this entailed using tables of computer-generated
``dice rolls''). In Position B , the ``dangerous'' position, X was
evetually hit in 8 of the 36 games and lost 6 of them; in Position C ,
the ``safe'' position, X was hit in 14 of the games, losing 10 of them.
Perhaps this was a statistical fluke (I encourage the reader to make his own
experiments in the position), but I found why Position C is far more
dangerous than it looks. Almost without exception, it leads to a position
like Position D , where only 17 rolls clear the 7-point safely (1-1 and
all rolls consisting of 2's, 3's, 4's, and 6's), and eight force an
immediate shot (1-5, 1-6, 3-5, and 5-6). On the average, it takes about
1-1/2 turns to clear the point, so on the average twelve numbers leave
shots before the point is cleared. The resulting risk of being hit is
larger than the immediate risks in position B . A point above a gap is
twice as hard to clear safely as a point above a solid board, and there
are at least twice as many numbers that leave shots. I feel certain that
the unlucky player who rolled 3-1 and played 7/3 made a perfectly
reasonable play. He may not agree, however; he rolled 6-5 on the next
turn, leaving two blots. O was Luigi Villa, who then hit both blots
with 4-3 , won a gammon, and went on to become World Champion.
ADJUSTING YOUR PIP COUNT FOR CHECKERS STACKED ON LOW POINTS..
In the position below, X doubles.
If you go by pip count, O has a great take; indeed, X only barely has a
double. If you play it out, however, you will find that O has only about a 6%
chance to win. If you want to compare pip counts, you must add something to O's
pip count for the pips wasted in bearing off the extra three men on the ace point.
An experiment, using the same dice rolls for both players, 36 times, showed that
on the average O needs 1.1 more rolls to bear off than X , the equivalent of
nine more pips. Thus each of the three extra checkers (they are "extra" because
in the six rolls it takes to get 45 pips, on the average, you can expect only
two 1's) on the ace point should be counted as three pips. In the same way, if
you have extra checkers on the 2-point, they should also be counted as three
pips. In the position above, O should consider his effective pip count to be
51, not 45; obviously, that is a drop.
There is a further disadvantage for O in this position, not reflected
in the pip count. If O gets larger-than-average numbers, he will soon clear the
6- and 5-points and will waste pips thereafter. Since O is behind, he must have
many large numbers (or doubletons) to catch up; however, he does not get full
value from the large numbers, so he is not as likely to catch up as if his
position were something like:
March 5, 1980
Copyright 1980, Robert W. Floyd
Should you ever double when you are behind in the current game? It is
often correct to do so in tournament play when you are also behind in the match
score. Suppose you need four points, and your opponent two, to win the match. If
he doubles you, you should not only take if your chance in the current game is more
than 18%, but turn the cube right back to 4 and let the entire match ride on this
game; once you have taken the cube at 2, you have nothing more to lose.
Surprisingly, though, you should sometimes double even when the match score
is tied and you are behind in the current game.
In Position A, you are O ; you and your opponent are each two points short of
match. Your chance to win the current game is only 40%, yet you must double.
The reason is that, if you fail to roll a doubleton, the cube will go to
2 anyway. Your opponent will double you, and you will accept. It makes no
difference which player has the cube at 2 because 2 is the match for both players.
If you double, it only affects the outcome if you roll a doubleton. If you knew
you were going to roll a doubleton, wouldn't you double?
Even Position B is a correct double; you have a 38% chance for match if
you double, a 37% chance (against an equal opponent) if you wait.
The same kind of situation arises whenever the cube has little or no value for your
opponent. If you are playing a satisfying backgame, you have the cube at 2, and you
are behind 3-4 in a 9-point match, you should double. If you lose without doubling,
you will probably lose a gammon; your chance of catching up at 3-8 is less than 10%,
so your opponent will not give the cube back unless he is an overwhelming favorite.
If you win, the extra two points put you well ahead.
March 5, 1980
Copyright 1980, Robert W. Floyd
Down 2-7 in the final match to 11 of a local tournament, I had doubled early in
hopes of a gammon to restore some balance to the score. My opponent held my ace
point, from which he hit me in the bearoff; later, I hit him from the bar as he
bore off. After a frustrating exchange of hits between our runners, he redoubled
me to 4 in the position above.
In a money game, this is a simple drop. O needs a large doubleton, or
a very lucky shot, to win. My chances were no better than one in five.
Nevertheless, I have a clear take, and the double itself is suspect.
Because my opponent has doubled himself into match, I must immediately
redouble, with a 20% chance to reach 10-7 , from which score I have about an 82%
chance to win the match. By taking and redoubling I give myself a 16% chance for
match.
If I drop, the score goes to 2-9; my chance of winning against an equal
player is 10%; the take and redouble offers me nearly twice as great a chance.
Had my opponent kept the cube until the end, my chance for the match would
be:
.20 x .30 (I win the game, then win the match from 4-7)
+ .80 x .10 (I lose the game, then win the match from 2-9)
= .14
Unless he assumed that I might mistakenly drop, he made a mistake to give me back
the cube. He should have waited a roll or two; at this score, the ideal doubling
point (and my drop point) is when O has a mere 12% chance. In a position like
this one,
X would have a good double, and I would probably drop.
The moral: when you are far ahead in a match (three or more points in
matches up to 11 points), it is almost always a mistake to redouble if your opponent
has much more than a 10% chance to win.
Perhaps you are curious how my match turned out. I took, of course, and my
opponent then rolled 6-6. Backgammon is that kind of game.
March 6, 1980
Copyright 1980, Robert W. Floyd
Today's column is for the novice backgammon player.
In the early stages of the game, your primary aim is to block your
opponent's back men from escaping. For example, if you roll 6-1, you should make
the seven point. However, not all rolls let you make a good blocking point.
Suppose you roll a 3-2. You could play it "safely" 13/8, taking no risks, but not
improving your position in any way.
It is much better to take some risks to improve your chances of building a
blockade next turn. Look at your position if you play 13/10, 13/11.
Your opponent may hit you with rolls of 3-3, 3-6, 4-5, and 4-6. These
are less than 20% of his possible rolls and it's not the end of the world if he does
hit you. If he doesn't hit you, you can make good blocking points with many rolls.
In addition to rolls like 1-6 that make a point even if you don't take the risk,
these are the rolls that make a good blocking point for you next time:
1-5, 3-5, 3-6, and 5-6 make the five point;
6-2 and 6-4 make the four point;
1-4 and 3-4 make the seven point. You have improved your chances of making a
good blocking point by 44%.
Backgammon is a game of risks and probabilities. The best players are
those who are willing to take a risk when the probabilities favor them.
Sometimes this takes a little courage. More often than not, thoughtfully
courageous play is rewarded.
March 6, 1980
Copyright 1980, Robert W. Floyd
Even though O has only a 45% chance of winning this game, doubling is
correct here unless X is a much weaker player. He must double because he
has nothing to lose, and something to gain, by doubling. There are two
possibilities:
(1) O rolls doubletons. In this case, had O doubled, he
would win the match.
(2) O does not roll doubletons. Now X will double, if O
has not already done so, and O will accept (unless he
thinks himself very much stranger than X ). X's next roll
decides the match whether or not O had doubled.
If O wins 65% or more of his games against X , he has a better chance in
this situation by not doubling, and declining X's double. Not many
matches are that lopsided, however, and the score suggests that X is no
pushover.
Because there are no useful redoubles when both players are two points
away from the match score, you should double on almost any excuse.
(1) Double if you are even slightly ahead in a race. When
it is your roll, I believe it is correct to double even
when you are two behind in the pip count, in typical racing
positions. The reason: if you fall slightly behind, your
opponent will turn the cube to 2 anyway and you will
accept. If you go further ahead, your opponent may no
longer accept a double.
(2) Double if your opponent has substantial gammon chances, even
if your chance of winning is less than 50%.
Your opponent's gammon will still be worth only two points
to him, but if you win, your victory will be worth twice
as much. Especially, double if you are playing a backgame
where losing will cost you a gammon.
(3) Double if, as in today's problem, you will still accept a double no
matter how badly you roll.
In other tied match situations where redoubles are meaningless, nearly the
same rules apply. For example, if you own the cube at 2, and the score is 3-3 or
4-4 in a 7-point match, you should be very quick to double.
March 6, 1980
Copyright 1980, Robert W. Floyd
QUANTITATIVE STUDY.
The chance for doubletons in n turns is:
n 1 2 3 4 5 6 7
P .17 .31 .42 .52 .60 .67 .70
P/n .17 .16 .14 .13 .12 .11 .10
So a good estimate is .17n - .01 n↑2 .
If, as in this position,
double 1s don't help, the chance for
22 or better at least once in n turns is:
n 1 2 3 4 5 6 7
P .14 .26 .36 .45 .53 .59 .65
(A good estimate is .14n - .01n↑2)
Here, your chance is roughly the chance that X does not roll a doubleton,
times the chance that O rolls a 22 or better, in three turns; this is
(l-.42) * .36 = .21 . The redouble equity if O rolls a doubleton the first
time raises the chances a little, but probably not enough to take a double, as
you would with five, or even four, of your checkers on the ace point.
When all your checkers are on the 1- and 2-points, a roll of 2-3
is as good as a 6-5 . In fact, there are only three kinds of rolls:
3-pip rolls like 1-2 and 1-6 (10 of them) ,
4-pip rolls like 1-1, 2-3 and 5-6 (21 of them) ,
and 8-pip rolls like 2-2 and 6-6 (5 of them) .
In one roll, then, your chance of bearing off when you need four pips is
(21+5)/36 = 26/36 = .72 . By calculating convolutions of this distribution,
we can find the exact probability of bearing off any set of checkers on the 1-
and 2-points in any given number of rolls. The distribution, in units of
1/36 , is:
d↓1(3) = 10, d↓1(4) = 21, d↓1(8) = 5 .
Let d↓2 = d↓1 * 2 . In units of 1/36↑2 , d↓2(6) = 100 , d↓2(7) = 420 ,
d↓2(8) = 441 , d↓2(11) = 100 , d↓2(12) = 210 , d↓2(16) = 25 .
The corresponding cumulative distribution is
C↓2(6) = 100, C↓2(7) = 520, C↓2(8) = 961, C↓2(11) = 1061, C↓2(12) = 1271,
C↓2(16) = 1296 .
So the chances of failing to get off in two rolls in these positions are:
C↓2(6) = C↓2(7) = C↓2(8) = C↓2(11) =
100/1296 = .08 520/1296 = .40 961/1296 = .74 1061/1296 = .82
For a three-roll situation, we calculate d↓3 = d↓1 * 3 . In units
of 1/36↑3 , d↓3 is:
n 9 90 91 12 14 15 16 19 20 24
d↓n 1000 6300 13230 9261 1500 6300 6615 750 1575 125
C↓n 1000 7300 20530 29791 31291 37591 44206 44956 46531 46656
C↓n/36↑3 .02 .16 .44 .64 .67 .81 .95 .96 .997 1.00
So the chances of failing to get off in three rolls in these positions
are:
.00 .02 .16 .44 .64 .81
As an example of the application of these numbers, what is your chance here?
P↓2(X) = .31 ; P↓3(X) = .69
P↓2(O) = .18 ; P↓3(O) = .38 ; P↓4(O) = .44
P = .18 * 1 + .38 * .69 = .44 ,
so don't double; you aren't even a favorite.
P↓2(X) = .31 , P↓3(X) = .69
P↓2(O) =.26 , P↓3(O) = .58 , P↓4(O) = .16
P = .26 * 1 + .58 * .69 = .66 ,
a reasonable double for O but a very definite take for X .
P↓2(X) = .31 P↓3(X) = .69
P↓2(O) = .26 P↓3(O) = .72 P↓4(O) = .02
P = .26 * 1 + .72 * .69 = .76 ,
a double and a (bare) drop.
May 21, 1980
Copyright 1980, Robert. W. Floyd
(Intermediates)
On his last roll, X ran 1/7/10 with a roll of 6-3. You can not hit
the escapee. Rather than play passively, however, you should hit X on the
1-point. This has numerous benefits:λ
(1) If X now rolls 4-4, 4-5, 4-6, 5-5, 5-6, or 6-6 (9 rolls) you have a
second chance to hit the checker on the 10-point, or to close your bar point
and
restrain the remaining checker on the 1-point.
(2) If X now rolls 1-1, 1-3, 1-6, or 3-6, he will not be able to make his 5-
point.
(3) If X now rolls 1-4, he will not be able to make his bar point.
Modern backgammon doctrine holds that a player who gets one of his back
men to safety has a significant advantage, other things being equal. In contrast,
having three or four men back in your opponent's board is not necessarily
disadvantageous; it offers improved chances to make an advanced anchor on the
18-, 20-, or 21-point. So if your opponent appears to be escaping with one
checker, be ready to stir up confusion in order to send another enemy checker back.
Remember, only the final stage of a backgammon game is a race. In the early
stages, it is a game of position and control. If you have flexibility and a solid
position, the race will later take care of itself.
October 29, 1980
Copyright 1980, Robert W. Floyd
Backgammon column--send to Backgammon Guide-for intermediates.
Should O double?
Most decidedly. In fact, X does not quite have a take. The best way to
recognize this in actual play is to memorize, for comparison, this position:
where X has almost exactly a 50% chance to bear off in two rolls. (This is
the best position for four checkers with a pip count of 14).
Obviously, X is much worse off in the first diagram, and O will win
by rolling doubletons a third of the time even if X goes off in two rolls.
Another quick way to see how poor X's chances are is to classify his
rolls into 4 good (3-3 or better), 20 bad (all 1's and 2's), and 12 medium (3-4,
etc.). The sequences that are sure to lose are Bad-Medium, Bad-Bad, and
Medium-Bad, a total of
2 * 20 * 12 + 20 * 20 = 880 ;
880/36↑2 = 68% , and of the remaining 32% , doubletons by O win about
1/4 of the time. The exact chance that O wins is a shade under 25%, and X
should drop, in a money game.
In a match against an equal opponent, if O is two away from match and
X is four away, not only should X take and immediately return the cube, O
should not double. By doubling, O gives X a 25% chance for match. If O
ignores the cube, X's chance is:
25% (X wins the game) x 40%↑* (X then wins the match) = .10
+ 75% (X loses the game) x 18%↑* (X then wins the match) = .135 for a total
of .235
so O should definitely not double right away. If X rolls a 1 or 2 on his
first turn, O can double him out, cutting X's chances in the game by
20/36 (X rolls a 1 or 2) x 4/36 (X then rolls 3-3 or better) x 3/4 (O
rolls no doubletons) = 5%, and thereby cutting X's chances in the match to 22.5%.
If you are an intermediate playing an expert opponent, though, give him the cube
even if you are two away from match and he is four away. If he takes your double,
he has a 25% chance for match. If he drops it, his chance for match is about 28%↑*.
If he knows what he is doing, he will drop.
* Starred numbera are from Norman Zadeh, "On Doubling in Tournament Backgammon",
Management Science, May 1977, p. 986.
October 29, 1980
Copyright 1980, Robert W. Floyd