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C00015 00004 PAULM
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BGMESS
About your BG message of 6 October 82, at match point it's a very close decision
whether to put a second checker behind your 7-2 prime; I would do it because
the opponent may not be able to link on my 1-point. If my prime were 10-5,
I would abstain; what I fear is that my opponent will link on the 3 or 4 point,
and run with 6-6 or 5-5 when I open up. I estimate that one checker, closed
out, can win less than 5% of the time. One expert says 3%. A strong player
I know tried to get me to hit a second checker when I had an 8-3 prime. It's close.
What you call "games that are all defence" (I didn't know you were British),
they are called prime-vs-prime games. They are notoriously difficult: when to
double, when to take, whether to split the back men, whether to flee or stay, etc.
As a rule of thumb, build your own prime very strong (e.g. 7-3, opponent with two
checkers on the ace) before making any escape play where you might be closed out.
Timing is everything. Often I slot my bar point in situations like the one you
describe, if I don't think I can make it naturally; if not hit, I can double, or
cover and then double.
Directory for BKG
DOUBLG
Doubling Policy in the Racing Game (c) 1981
Bob Floyd
Every basic backgammon book has a different formula for doubling and
taking doubles when the game has become a race to bear off. They can't
all be right; maybe all of them are wrong. I decided to find the most
ccnvincing evidence I could for correct doubling, redoubling, and drop
points, during the bearoff.
First, I simplified the game to one where no pips are wasted, as if
each player had only one checker left on an enormous backgammon board.
Let's call this the pip-count race, or PCR. I programmed a computer to
calculate, for every pair of pip counts in the PCR, the expectation
(equity, fair settlement value) with X on roll for four situations: (1) X
has the cube (2) centered cube (3) O has the cube (4) cube not in use.
From the equity with the cube out of play, it is also easy to calculate
the raw probability of X winning the game, by the formula P=(E+1)/2.
I found the doubling points summarized in Table 1. For example, when
the player on roll has 80 pips to go, he should double if his lead
(difference in pip counts) is more than 6.90, at which point he has a
71.5% chance of winning if the game is rolled out to the end. The
fractional pip count lead of 6.90 is, of course, a fiction; it represents
an interpolation between X's loss of equity on doubling with a 6-pip lead,
and his gain of equity on doubling with a 7-pip lead. The last column of
the table shows, then, that X, with 80 pips to go, doubles with a 7-pip
lead, redoubles with a 9-pip lead, and doubles O out with a 12-pip lead.
Table 1: Doubling in the PCR
Pip count of
player on roll 20 30 40 50 60 70 80
Lead in pips to
justify double -1.43 0.60 2.21 3.56 4.71 5.75 6.90
Raw probability at
doubling point 0.631 0.661 0.681 0.693 0.701 0.707 0.715
Lead in pips to
justify redouble -.83 1.43 3.29 4.75 5.95 7.08 8.06
Raw probability at
redoubling point 0.655 0.687 0.708 0.719 0.726 0.731 0.734
Lead in pips to
justify dropping 2.58 4.63 6.42 7.81 9.03 10.12 11.21
Raw probability at
drop point 0.766 0.771 0.777 0.779 0.780 0.781 0.782
In a real bearoff, however, the underdog's chances are not as good as
in a PCR. In a PCR, when X has 20 pips and O has 22, X has a 75.2% chance
to win. In a real bearoff, taking as X's position a composite of
positions (A) and (B),
(A) (B)
and taking as O's position a composite of (C) and (D),
(C) (D)
X has a 76.1% chance of winning. The difference is that the underdog
needs large doubletons to catch up, but they can't be used efficiently.
If O rolls a 4-4 in position C, or a 5-5 in position D, he still needs two
more rolls to bear off.
To correct the PCR table for use in real backgammon races, I
investigated the difference in likelihood for the underdog to catch up in
the PCR and in real races at a given pip count, using the same dice rolls
for both. (For connoiseurs of statistical methodology, I used stratified
sampling to avoid having any one roll that plays inefficiently occur too
often or too seldom.) If we say the advantage to X is his lead in pips,
plus four pips because he is on roll, then X needs only about 90% as much
advantage in real backgammon as in the PCR to have a given probability of
winning. So, the table on doubling in the PCR should have all the
advantages scaled down by about 10%. This gives Table 2, which I think is
a good guide to doubling in real backgammon positions where both sides are
well placed for the bearoff, like position (E).
(E)
Table 2: Doubling in the Bearoff
Pip count of
player on roll 20 30 40 50 60 70 80
Lead in pips to
justify double -1.69 0.14 1.59 2.80 3.84 4.87 5.81
Lead in pips to
to justify redouble -1.15 0.89 2.56 3.88 4.96 5.97 6.85
Lead in pips to
justify dropping 1.92 3.77 5.38 6.63 7.73 8.71 9.69
Yes, I understand. You don't want to memorize this table. You have
several things you'd rather do, such as reporting your true winnings for
the past year to the IRS, or telling your wife about your true losses, or
standing in line to buy an AMTRAK ticket, or rereading your high school
trigonometry book. All right. If the pip count of the player on roll is
between 20 and 80 pips, the following formulas are accurate to within one
pip.
Drop if the opponent's lead is more than P/8, where P is
his pip count. Double if you are within four pips of the
opponent's drop point. Redouble if you are within three
pips of the opponent's drop point.
For example, in position (E), P is 70; P/8 is a bit under 9, and with a
10-pip lead X should double and O should drop. Move O's checker from the
9-point to the 4-point, giving X a 5-pip lead, and X should double (5 is
more than P/8-4) but not redouble (5 is less than P/8-3).
(RWF: rewrite next paragraph)
Be careful, though; if the underdog has a position where large
doubletons don't play efficiently, like position (F), the leader should be
much more ready to double, and the trailer, to drop.
(F)
In fact, when you are far behind, your checker play must assume that
you will eventually roll a large doubleton. In position (E), for example,
X has a 14-pip advantage: a ten pip lead and the roll. Since the average
roll is 8 pips, O is still 6 pips behind if he rolls 4-4. While O's
chance in position (E) is better than 20%, he has only a 5% chance to win
the game without rolling a 6-6 or 5-5. He must therefore strip the
4-point and try to reach a position like (G)
(RWF: correct this example)
(G) (H)
rather than (H), although if the race were close, (H) would be better.
PAULM
How Paul Magriel almost gave up backgammon,
as the result of Wayne McClintock's misplayed ace.
by Bob Floyd
copyright 1982
It happened in Round 5 of the Black & White Chicago Classic, in October
1979. Magriel had beaten Gillman, Anderson, Yen, and Boyd. Wayne
McClintock was hoping to avenge his friend Bill Boyd, who had been tied
12-12 in a seventeen point match with Magriel until swept away by a gammon
with the cube at four. The first game had become a straight race; not a
checker had been hit. Both players were tired and making mistakes. Twice
McClintock had failed to double in positions where Magriel should have
dropped. Magriel had made a minor error or two in his usually impeccable
checker play. Then, in Position A, McClintock rolled 1-1, playing 23/off,
19/21.
McClintock
A
Magriel
This play is a technical error; in case he rolls 1-2 next time, he should
have played 23/off, 20/22, as we shall see later. As the game proceeded,
in Position B, Magriel overlooked an obvious double (he is 67% to win),
and rolled a 6-3. McClintock was then punished for misplaying the aces by
also rolling a 6-3. Magriel won the point and later the match.
McClintock
B
Magriel
I like to imagine, though, a scenario in which McClintock is rewarded for
the misplay of the double aces, and Magriel gives the correct double, but
is punished for it. In my fantasy, the match continues from Position B.
Magriel correctly doubles and rolls 1-2, grimaces, and plays 2/off, 4/3.
McClintock
C
Magriel
McClintock sees that he is a clear favorite, and that only one roll
permits Magriel to redouble. At the match score, the redoubling decision
is like that in a money game, so McClintock correctly redoubles, rolls
1-2, and without hesitation plays 21/24.
McClintock
D DD
Magriel
Had McClintock played his aces correctly in Position A, the position would
now be DD, and Magriel could not have redoubled to eight. In Position D,
however, Magriel, tilting his head sideways, chewing his tongue and
looking at the stars, estimates that cube handling should still be much as
in a money game, and calculates that his average gain of points will be
1.30 if he holds the cube, 1.43 if he turns it. Although fearing the
redouble, he turns the cube to eight. McClintock, envisioning a grim
struggle if Magriel should bear off for eight points, knows he must
nevertheless take. After rechecking his arithmetic, he pulls the cube to
his side of the board. Magriel rolls 1-2, taking a checker off.
McClintock
E EE
Magriel
Had McClintock played his aces correctly in Position A, the position would
have been EE, and McClintock, not having the cube, would simply roll. In
Position E, however, McClintock sees that he is a strong favorite, and
that Magriel gets negligible value from possession of the cube at sixteen.
He places the cube, at sixteen, on the board.
Magriel estimates, using tables memorized years before, that a drop would
leave him no better than a 17% chance for the match against an equal
opponent. A take gives him a 34% chance. He respects McClintock's play
and trusts his own arithmetic, so, without rechecking, he slides the cube
toward himself. McClintock rolls and plays a 1-2.
McClintock
F FF
Magriel
If McClintock had not misplayed his aces, the position would have been FF;
Magriel would have redoubled to eight and McClintock would have grumbled
and dropped. In Position F, Magriel sees that the seventeenth point to
win the match is probably his on a seventeen-to-one shot. He correctly
turns the cube to thirty two.
McClintock tries to imagine fighting Magriel from a 0-16 score. He does
not know exactly what his chances for the match would be, but his estimate
is not far from the 7/10 of 1% Magriel is estimating. He takes the cube.
Perhaps he envisions a pair of dice coming to rest with small numbers
showing. Magriel shakes his cup several times, otherwise showing no
concern, and rolls a 1-2.
McClintock goes on to beat Barbara Glazer in the semifinal and Roger Low
in the final. A well known publisher asks him to coauthor a backgammon
book. A collection of his favorite games, coauthored by George Plimpton,
wins acclaim everywhere. He subsequently lends his name to popular books
on bridge, poker, blackjack, and cribbage.
Magriel, after offering McClintock a firm handshake, murmurs a few words
to an admiring kibitzer, apparently explaining why McClintock took the
cube at thirty-two. He makes no mention of the double aces. He writes a
short note, perhaps a list of some sort, which he gives to a young woman
who has followed the game with more than average interest, and walks
toward the travel counter. For a long time, players ask if Magriel ever
plays backgammon any more. As the years pass, the answer, as often as
not, becomes ``Who?''
Author's note: The game, as actually played, follows. McClintock should
have doubled before his twelfth and fifteenth rolls, and Magriel probably
should have dropped. 18/15, 12/6 is a shade better on Magriel's eighth
roll. Magriel's play of the 3-3 on his seventh roll fascinates and
mystifies me.
The fantasy, beginning with Position B, represents correct play by both
sides for two players of equal strength in the first game of a 17-point
match; if the players roll 1-2 five consecutive times, the game decides
the match.
The odds were at least 1889567 to 1 that Magriel would not give up backgammon.
Magriel McClintock
1. 6-1 13/7, 8/7 5-4 1/10
2. 6-1 13,7, 6/5 6-5 10/16, 12/17
3. 6-3 8/5, 24/18 5-4 16/21, 17/21
4. 5-4 83, 7/3 4-3 1/8
5. 6-2 24/18, 6/4 6-3 8/17
6. 6-4 13/3 5-5 12/17 (4)
7. 3-3 18/12, 13/10, 7/4 (??) 5-3 17/22, 17/20
8. 6-3 18/9 (?) 2-1 17/20
9. 4-3 9/5, 13/10 3-1 17/20, 19/20
10. 3-3 7/4, 10/7, 12/6 5-2 17/19, 17/22
11. 5-4 10/5, 7/3 6-6 17/23 (2), 19/off (2)
12. 4-1 4/off, 3/2 3-2 22/off, 23/off (??)
l3. 6-4 6/off, 4/off 2-2 23/off, 21/off, 19/21
14. 5-4 5/off, 4/off 6-3 19/off, 22/off
15. 6-1 6/off, 5/4 1-1 21/off (??)
16. 6-1 6/off, 5/4 4-3 21/off, 20/23
17. 6-2 6/off, 2/off 2-1 23/off, 20/21
18. 3-1 3/off, 3/2 4-3 21/off, 20/23
19. 5-3 5/off, 3/off 1-1 23/off, 19-21 (?)
20. 6-3 4/off, 4/1 (???) 6-3
Wins
3chkrs
Doubles, Redoubles, Takes, and Beavers:
Three Checkers vs. Checkers on Ace Point
Suppose you have three checkers to bear off, and your opponent has three
or four on the ace point. If your chance to be off in two rolls is more
than 0.90, he should drop your double. A pipcount of ten is marginal. (In
the diagrams, p2 is X's chance to bear off in two rolls; p is his chance
to win the game if played to completion).
A drop: p2=.92 A take: p2=.81,
(1-3 and 1-4 miss completely.)
You should double or redouble with a good twelve-pip position:
A redouble: p2=.76 Not a double: p2=.69
(P=.64) (P= )
The opponent should beaver all but one of the thirteen-pip doubles:
Not a beaver: p2=.67 Beaver: p2=.64 (P=.54)
Now suppose you have three checkers to bear off and your opponent has five
or six on the ace point. You should always double; if your pip count is
less than sixteen, your opponent should drop.
A drop: p2=.49 A bare take: p2=.39,
p3=.96 p3=.93
(P=.82) (P=.76)
What if the shoe is on the other foot? If your opponent is on roll with
five or six checkers on the ace point, to take a double you need a .36
chance of being off in two rolls; a sixteen pip position is good enough:
A drop A take
The opponent should redouble at fifteen pips:
A redouble Not a double
You should beaver with a good 12 pip position:
A beaver Not quite a beaver
Summing up: RD = Redouble, dropped
RT = Redouble, taken
DT = Double, but no redouble
W = Wait
B = Beavered
? = Depends on distribution
Your-three Your action Opponent's Your action
checker vs 3-4 action with vs 5-6
pip count on ace 5-6 or ace on ace
3-9 RD B RD
10 R(T?) B RD
11 RT B RD
12 R(?)T B(?) RD
13 W(B?) W RD
14 B W RD
15 B RT RD
16 B RT RT
17 B RD RT
18 B RD RT
dbgbkg.tex
\input basic
\input smacros
\ctrline{Doubling a Back Game}
\par \vskip .5in
\par
Division of risk works in your favor when your opponent is playing a well-timed
back game, so long as he does not own the cube. Suppose that, if the game is played
out to the end, you would lose two thirds of the time, and win a gammon the
other third. Without the cube, you would just break even. If you have access
to the cube, however, you can play to a point in the bearoff at which one third
of your games will have been lost; at that point, you are equally likely to
lose the game and to win the gammon, so you have a near-perfect double; your
opponent has nothing to gain by taking, and if he drops, you win two thirds
of your games.
Let's estimate the correct point at which to double while bearing in
against a well-timed back game. Suppose that your raw (no-cube) expectation
in a certain position is 0.2; that is, you will win a gammon 40\%\ of the time,
and lose the game 60\%\ of the time (assume the few backgammons and single games
cancel out).
Typically, you have a 30\%\ chance of being
lethally hit before your next roll.
If you play ten games to completion in the position you will lose three immediately,
three more later, and will gammon in four. If you survive the next roll, you can
double your opponent out.
Suppose first that you retain the cube on your side. In ten
games, your expectation is:
\ip{} Minus one point in three games where you are hit
\ip{} Plus one point in seven games where you double your opponent out\par\noindent
for a total of four points.
If instead you double immediately, your expectation is:
\ip{} Minus two points in six games
\ip{} Plus four points in four games\par\noindent
for a total of four points, also.
So the situation where you will win
gammons 40\%\ of the time, and lose the game 60\%\ of the time, is the break-even
point for doubling. This conclusion, you should notice, does not depend on
whether the Jacoby Rule is in effect; no gammons will be won in undoubled
games in any event.
If you reach a situation where you will win gammons in more than two thirds
of your games, you are too good to double, and should play on, unless the
Jacoby Rule requires that you turn the cube to score the gammon.
If you own the cube, the arithmetic is a little different. Suppose you will
still have a 20\%\ chance to win the game even if hit, and your chance to win
the gammon without being hit is 42\%\ . By retaining the cube, your expectation
in one hundred games is:
\ip{} Minus one point in 24 games where you are hit and lose;
\ip{} Plus one point in 6 games where you are hit and win;
\ip{} Plus one point in 70 games where you double your opponent out next turn
\par\noindent
for a total of 52 points.
By doubling immediately, your expectation is:
\ip{} Minus two points in 58 games where you are hit and doubled out
\ip{} Plus four points in 42 games where you win a gammon\par\noindent
for a total of 52 points.
If you own the cube, a 42\%\ gammon chance is the breakeven point for redoubling.
\par\vfill\eject
Experiment suggests that the correct point to double, when bearing in
smoothly against a back game, is in positions like these:
\par \vskip 1in
\display{\tabskip 10pt plus 100pt minus 100pt
{\halign to 6.5in{\lft{#} \lft{#}\cr
E F\cr
\space{2in}
G H\cr
}}}
\par\vskip .5in
\par \noindent
You should have no more than three points to clear against the 1-3 and 1-2
backgames, and only two points against the 2-3 backgame.
This analysis {\it only} holds in money play; in match play your opponent can
take away the value of your gammons by re-redoubling if you are fewer than eight
points from match.
\par\vfill\end
MARKET
The Market
Bob Floyd
Copyright 1982
``If we don't double now, we'll lose our market''
You've heard it many times. You've probably said it many times. It's the
national anthem of backgammon; sing it and all differences of opinion are
forgotten, everyone stands up and salutes the flag. Still, it ain't
necessarily so. Next time you hear it, reply ``Did you say that if we
don't double now, we'll lose some equity?''
Equity: your average winnings when the position is played many times. In
the long run, the player who maximizes equity wins more money. Sometimes
this means foregoing the use of the cube when you're a definite favorite.
Often it means taking the cube when you have only a 25% chance to win.
Sometimes it even means making a play that doesn't have the best chance to
win the game. Let's look at the equity in a late bearoff position where
you can lose your market.
Are you a favorite here? Yes. Will your opponent take a double next time?
No. Should you double? No.
If this position is played out to the end, your chance of winning is 62%.
Out of 100 games, you will win 62 and lose 38, for a net gain of 62-38=24
times the stake, if your opponent doesn't redouble. But your opponent will
double you out if you roll any of your ten bad numbers. This cuts your
chance to 56%; if you double, your net gain in 100 games is still
only 56-44=12 times the doubled stake, or 24 times the initial stake.
If you wait, your winning chance is more than 56%, because you will now
double your opponent out in positions where if he owned the cube he would
come back to beat you. After average rolls by both players, you may get
this position:
MARKET
where you can redouble him out. The fraction of games you save this way is:
17/36 (your rolls of 2-2, 3-4, etc.)
X 30/36 (his non-doubletons)
X 30/36 (your non-doubletons)
X 6/36 (his doubletons)
= 0.055
You may also get this position:
You would have two ways to lose this if you couldn't turn the cube. You
could roll 1-2 and be redoubled out, with a chance of
8/36 (Your initial rolls of 1-3, 1-4, 1-5, or 1-6)
X 30/36 (His non-doubletons)
X 2/36 (Your 1-2)
= 0.010
or he could come off with a final doubleton, with a chance
8/36 (Your initial rolls of 1-3, etc.)
X 30/36 (His non-doubletons)
X 30/36 (Your average rolls: 1-3, 2-2, 5-6, etc.)
X 5/36 (His doubletons)
= 0.026
So, by keeping access to the cube, you will win 0.56+0.05+0.01+0.03=0.65
of your games. Now in 100 games, your net gain is 65-35=30 times the
initial stake, better than the 24 points you stand to win if you double
hastily. At a ten dollar stake, you are giving away sixty cents every
time you give away a centered cube in this position. If you owned the
cube initially, your redouble costs you $1.80, because you are giving up
your chance to redeem yourself by rolling a doubleton after your initial
roll misses.
All right, you admit you shouldn't always double in a market-loser
situation, but you can't always stop to calculate the exact odds in a hot
chouette. We've thought of that. The rest of this article gives rules
for doubling, taking, and beavering in positions where only the player on
roll can miss.
MARKET
Two Rolls
Everyone knows the answer in a one-roll position. In a two-roll position,
double with fourteen bad numbers; redouble with twelve bad numbers
X should not double
(15 bad numbers)
X should redouble
(12 bad numbers)
The opponent should drop unless there are four bad numbers:
O should barely take
(4 bad numbers)
With fifteen bad rolls, a double should be beavered.
O should beaver
(thirteen bad numbers
and many repeaters)
(I lost twenty-four points in a chouette by giving a correct beaver in
this position. One of the captain's dice came to rest on a three. The
other spun, spun some more, and then paired the first one.)
MARKET
Three Rolls
In a three-roll position, double with eight bad numbers; redouble with six
bad numbers.
X should not double,
(seven bad numbers plus too many repeaters)
X should redouble
(six bad numbers)
The opponent should take if there is at least one bad number:
O has a bare take
MARKET
Four Rolls
In a four-roll position, the opponent has a take even if no numbers miss,
and he will almost have a take next roll if each player takes two checkers
off, so doubling, and especially redoubling, should be conservative.
Double, but don't redouble, with two bad numbers:
X should not double
(four bad numbers)
X should double but not redouble
(two bad numbers)
Eleven bad rolls calls for a beaver:
O should beaver
(twelve bad numbers)
MARKET
Five Rolls or More
With more than five rolls to go, don't double if any roll misses; don't
redouble at all.
X should double but not redouble
MARKET
Summing Up
Numbers of rolls: 1 2 3 4 5
Bad rolls, to take
the cube: 9 4 2 0 0
Bad rolls,
to redouble 17 12 6 0 -
Bad rolls, to double: 17 14 8 2 0
Bad rolls, to beaver 19 15 13 12 11
A rule of thumb: On the last roll, double or redouble with any advantage.
For each additional roll remaining, you must have five fewer bad numbers
to double, and six fewer to redouble.
BGGUID
A Killer Problem
Bob Floyd
Copyright 1982
Should X double?
He leads 9-7 in
an 11-point match.
This artificial-looking position arose for San Jose's Killer Joe Glazier
in a recent Louisville tournament. In a money game, it is a double, and
only a bare take because of the redouble equity; O's winning chance if the
game is played to completion is 24.96%. At the match score, though, the
redouble is a wicked weapon. Joe doubled, missed, and dropped the
redouble. Surprisingly, this is the right strategy even though it gives
him only a 53% chance to win the game, while waiting would give him a 75%
chance for the game.
First, a little background. If X wins one point, his chance to win the
match is 82%; O must win three consecutive games, unless he wins a gammon
on the first or second. If X wins two points, he wins the match. If O wins
one point, X has a 59% chance for the match (there is argument about this
figure; some place it as low as 55%). If O wins two points, the chances
are equal. If O wins four points, X loses the match.
If X decides not to double, and then misses, O must not double; to do so
gives up an 18% match equity when he loses the game, and gains no more
than 9% match equity when he wins. O should not give even a last-roll
double unless his winning chance is more than 2 to 1. O could barely
double on the last roll with checkers on his 2- and 3-points.
So if X does not double, the game will be played out for one point, X will
win it almost exactly 3/4 of the time, and his match chance will be
3/4 X .82 + 1/4 X .59 = 76.25%.
If X doubles and misses, he must drop the redouble; it is better to even
the match score at 9-9 than to play on for match as a slight underdog. So
if X doubles, his match chance will be 19/36 + 17/36 X .50 = 76.39%. Joe's
double gained him an extra 1/7 of one percent chance of winning the match.
With that kind of precision in his doubling, is it any wonder that they
call him Killer?
When you are two points away from match and your opponent is four or more
away, a last-roll double is a powerful weapon. You should double with even
a 36% chance to win the game, as with checkers on the 2- and 6-points, or
the 3- and 5-points. Your opponent should drop if your chance is more than
64%; checkers on the 1- and 5-points, or the 2- and 4-points, are marginal
drops.
On the other hand, before your last roll, because your opponent can give
the cube back and put the match on the line, your opponent will take even
if you have an 82% chance; you should not even consider doubling and
taking the redouble unless your chance is 77%. If each player has eight
checkers on the 1-point, a bare take in a money game, you should not
double at a two-away-four-away match score; in fact, mathematically you
should wait until each player has just four checkers on the ace point. In
practice, even a 5% chance that your opponent will drop justifies a double
with six checkers on each ace point.
When both players have most of their checkers on the low-numbered points
of the home board, and the player on roll has a few numbers that waste a
whole roll, while his opponent can't miss, what is correct doubling/
redoubling/taking/beavering strategy?
Two Rolls
Everyone knows the answer in a one-roll position. In a two-roll position,
don't double with fourteen bad numbers:
X should not double
Double, but don't redouble, with twelve bad numbers:
X should not redouble
The opponent should drop unless there are six bad numbers:
O should take
With fifteen bad rolls, a double should be beavered.
O should beaver
(fourteen bad numbers
and repeaters)
(I lost forty-eight points in a chouette by giving a correct beaver in
this position. One of the opponent's dice came to rest on a three. The
other spun, spun some more, and then paired the first one.)
Three Rolls
In a three-roll position, don't double with ten bad numbers:
X should not double
Double, but don't redouble, with eight bad numbers:
X should not redouble
(six bad numbers plus repeaters)
X should redouble
(six bad numbers)
The opponent should drop unless there is at least one bad number:
O has a bare take
Four Rolls
In a four-roll position, the opponent has a take even if no numbers miss,
and he will almost have a take next roll if each player takes two checkers
off, so doubling, and especially redoubling, should be conservative.
Double, but don't redouble, with four bad numbers:
X should not redouble
Redouble with two bad numbers:
X should redouble
Eleven bad rolls calls for a beaver:
O should beaver
Five Rolls or More
With more than five rolls to go, don't double if any roll misses; don't
redouble at all.
X should double but not redouble
Summing Up
Number of rolls: 1 2 3 4 5
Bad rolls,
to redouble: 17 11 6 2 -
Bad rolls, to double: 17 13 8 4 0
Bad rolls, to take
the cube: 9 6 2 0 0
Bad rolls, to beaver: 19 15 13 11
A rule of thumb: On the last roll, double or redouble with any advantage.
For each additional roll remaining, you must have four fewer bad numbers
to double, and five fewer to redouble. On the last roll, take if your
opponent has nine bad numbers. For each additional roll remaining, three
fewer bad numbers are needed to justify a take.
ROLLS
When both players have most of their checkers on the low-numbered points
of the home board, and the player on roll has a few numbers that waste a
whole roll, while his opponent can't miss, what is correct doubling/
redoubling/taking/beavering strategy?
Two Rolls
Everyone knows the answer in a one-roll position. In a two-roll position,
don't double with fourteen bad numbers:
X should not double
Double, but don't redouble, with twelve bad numbers:
X should not redouble
The opponent should drop unless there are six bad numbers:
O should take
With fifteen bad rolls, a double should be beavered.
O should beaver
(fourteen bad numbers
and repeaters)
(I lost forty-eight points in a chouette by giving a correct beaver in
this position. One of the opponent's dice came to rest on a three. The
other spun, spun some more, and then paired the first one.)
Three Rolls
In a three-roll position, don't double with ten bad numbers:
X should not double
Double, but don't redouble, with eight bad numbers:
X should not redouble
(six bad numbers plus repeaters)
X should redouble
(six bad numbers)
The opponent should drop unless there is at least one bad number:
O has a bare take
Four Rolls
In a four-roll position, the opponent has a take even if no numbers miss,
and he will almost have a take next roll if each player takes two checkers
off, so doubling, and especially redoubling, should be conservative.
Double, but don't redouble, with four bad numbers:
X should not redouble
Redouble with two bad numbers:
X should redouble
Eleven bad rolls calls for a beaver:
O should beaver
Five Rolls or More
With more than five rolls to go, don't double if any roll misses; don't
redouble at all.
X should double but not redouble
Summing Up
Number of rolls: 1 2 3 4 5
Bad rolls,
to redouble: 17 11 6 2 -
Bad rolls, to double: 17 13 8 4 0
Bad rolls, to take
the cube: 9 6 2 0 0
Bad rolls, to beaver: 19 15 13 11
A rule of thumb: On the last roll, double or redouble with any advantage.
For each additional roll remaining, you must have four fewer bad numbers
to double, and five fewer to redouble. On the last roll, take if your
opponent has nine bad numbers. For each additional roll remaining, three
fewer bad numbers are needed to justify a take.
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\centerline{The Principle of Divided Risk}
\vskip .125in
\centerline{Bob Floyd}
\vskip 12pt
\centerline{$\copyright 1982$}
\vskip .125in
When you play backgammon using the doubling cube, successful money management
is not just a matter of playing to win more games than your opponent. It is
also a matter of winning games with the cube high, and losing them with the
cube low.
Suppose you have your choice of two plays in a certain position. By making play A,
you put the game on the line; you win (and double your opponent out on your next
turn) 47\%.\ of the time; you lose, and get doubled out , 53\%\ of the time.
In a hundred plays of this position, you will lose six points.
By making play B, you divide your risk. You run a 40\%\ risk of being doubled out
after your first play, and, if you survive, a 30\%\ chance of being doubled out
after the second; if you get through both risks, you can claim the game.
Your chance to win the game is then only 60\%\ $\times 70$\%\ = 42\%; doesn't sound
very promising, does it? Look again. In a hundred games, your earnings will be:
\outlineone : Minus 1 point in 40 games, where you drop after one roll.
\outlineone : Minus 2 points in 18 games, where you double after the first roll and
drop after the second.
\outlineone : Plus 2 points in 42 games, where you double after the first roll
and go on to win.
\noindent
Your net is $-1\times 40-2\times 18+2\times 42= + 8$ points.
The play to win the game is play A; the play to win the money is play B.
This illustrates a general principle:
{\narrower\smallskip\noindent
{\bf If you have a certain total risk to run, and have choices
about how to divide it between successive moves, try to divide it so as to
give yourself a useful double at some stage.\smallskip}}
\noindent
I call this the principle of divided risk. It can be used in many areas of
backgammon: contact and non-contact bearoffs, holding games, backgames,
openings, escaping-checker games, probably any other kind of game you like.
\vskip .75in
(A)
\vskip .75in\par
\centerline{X to play 5-1}
\vskip 12pt
Position A is a dramatic illustration.
X had had a deuce-point game, and hit O's last loose checker. He took a double
after closing O out, if he hadn't already taken it; such a closeout is a
definite take for X, although probably a good double for O. When X opened
up, O did not come in immediately, and X is planning to redouble when he
gets down to ten checkers, the theoretical last take point in such a bearoff.
Here he has rolled 5-1. He could play 5/off, 5/4, and hope that O stays off
for two more rolls, or comes in and rolls small numbers. The correct play,
however, is 5/off, l/off. This play gives X equity of about +.25\%\ of the
cube; the cautious play gives him slightly negative equity.
\vfil\eject
\noindent
After the correct play, three possibilities occur:
\outlineone :(1) O comes in with a six (9 rolls). Because X took off his
odd checker, he had substantial chances to catch up in the race.
\outlineone :(2) O comes in with a five (11 rolls) hitting. The position is
nearly hopeless for X, but he wouldn't have had more than a 25\%\ chance with the
cautious play.
\outlineone :(3) O stays out (16 rolls). Because he took an extra checker off,
X has a powerful double, which is nearly as good as winning the game.
Had X made the more cautious play, he would not be good quite enough to double,
and by next roll, he would very likely be either so far beyond the doubling point
that the cube gains him little, or else ahead by too little to double.
The bold play of bearing two checkers off reaches a good doubling position in two
ways. If O stays out (42\%\ chance), X has an immediate double which may not even
be a take. If O comes in with a six, X rolls average numbers, and O fails to get
past the thirteen-point on his next roll (4\%\ chance), X again has a very good
double.
In this similar example from an actual game (Position B),
X correctly played a 1-4 by taking two
checkers off. O fanned, X correcty doubled, O correctly took, and X won
eight points.
\vskip 1in
(B)
\vskip 1in
\centerline{X to play 4-1}
\vskip 12pt
In Position C, X has an ace to play in a non-contact bearoff.
\vskip 1in
(C)
\vskip 1in
\centerline{X to play a 1}
\vfil\eject
\noindent
The play more likely to win, by 100/46656=0.002, is 2/1. Nevertheless,
X should play 3/2. If X plays 2/1, he will either not be good enough to redouble
on his last roll (if he rolls a 1 or 2 next time), or he will have a virtual
claimer. If he plays 3/2, he has divided his risk almost perfectly when he rolls
2-4, 2-5, 2-6, 3-4, 3-5, 3-6, 4-5, 4-6,or 5-6; he has a barely takable redouble.
As a result, 3/2 gives X higher average winnings, by 5\%\ of the value of the cube.
\vskip 1in
(D)
\vskip 1in
\centerline{X to play a 1}
\vskip 12pt
In Position D, X will be off in two rolls 45\%\ of the time after playing 2/1,
and only 43\%\ of the time after playing 6/5. Even after allowing for the
possibility of O rolling a doubleton and X rolling 5-5, 2/1 is the play to
win the game. Nonetheless, the strong cube action after X plays 6/5 and then rolls
2-2, 2-5, 2-6, 4-5, 4-6, or 5-6 gives the edge to 6/5.
\vskip 1in
(E)
\vskip 1in
\centerline{X to play 5-4}
\vskip 12pt
In Position E, X has a 5-4 to play. He can play it cautiously within his
board, hoping for a doubleton to clear the midpoint. However, X has a nineteen pip
lead (after the roll)
which would allow him a powerful double after successfully breaking the
midpoint. If X waits, he may have to pile up checkers wastefully on the lower
points in his board, losing his racing advantage. By breaking the midpoint now,
13/9, 13/8, X allows O a 30\%\ chance to win, but in effect wins almost all of the
other 70\%\ of games by doubling on the next roll. (Without X's access to the
cube, this play would give O nearly equal chances in the game, combining
hitting and racing chances.)
\vfil\eject
\noindent
[RWF: replace this by more appropriate material.]
In Position F (from Kent Goulding's Backgammon with the Champions series,
Vol. 1, Number 1, pg. 15; Magriel {\it vs} Eisenberg) Magriel, playing O,
has a five remaining to play.
If Eisenberg owned the cube, or would take the imminent double, Magriel might
well hit on the ace point to drive that checker forward. In the actual game,
Magriel may have felt that Eisenberg would not take a double unless he hit,
so Magriel played 13/8, leaving only a 5-4 to hit.
This is a typical conservative risk-division play, where either play allows a
double-out if it goes well, so the smaller immediate risk is chosen.
\vskip 1in
(F)
\vskip 1in
\centerline{O to play a 5}
\vskip 12pt
\noindent
This is probably the most familiar appliation of the strategy. When you will have
an untakable double, or a barely takable double, provided you are not hit, you
should play to leave the fewest shots, even if another play gives you a sounder
position.
In Position G, with a 3-3 to play, X has the choice between 24/18, leaving
all eights to hit, and 24/12, leaving all twos to hit. If the game is played
without the cube, or if O owns the cube, leaving the direct shot is better than
allowing likely repeated indirect shots. However, if X has access to the cube,
he can double O if not hit, and should lower his immediate risk by playing
24/18, 6/3 (2).
The play 24/12 runs a 33\%\ risk. The play 24/18, 6/3 (2) runs a total risk
of about 35\%\ , of which only 17\%\ is taken before he turns the cube. (I have
ignored O's tiny racing equity.)
\vskip 1in
(G)
\vskip 1in
\centerline{X to play 3-3}
\vfil\end
ZEROSM
Backgammon is a two-person zero-sum game of perfect information, in which
dice are used as a random element. There is no limit to the length of a
backgammon game. Such a game can be represented by a directed graph, where
each node is one of:
(1) a maximizing node, with successor chosen by the maximizing player;
(2) a minimizing node, with successor chosen by the minimizing player;
(3) a randomizing node, with each successor chosen with a specified
probability;
(4) a terminal node, with payoff determined by the rules of the game.
If the graph is finite and cycle-free, the game has an expected value; for
each non-terminal position, the expected value is the maximum, minimum, or
average of the expected value of its successors, according to whether it
is a maximizing, minimizing, or randomizing position.
Actual backgammon is not free of cycles, and allows positions which do not
have a well-defined expected value, as we shall see. In practice, most
positions have an expected value. We shall treat the game as if it were
cycle-free.
The characteristic element of backgammon is the doubling of the stake. The
game is played initially for one point. Each player, at his turn, has the
option of announcing a doubling of the stake (doubling); his opponent must
forfeit the previous stake, or agree to play for the doubled stake. After
an initial double, the right to double alternates between the players; a
player may not double twice unless his opponent has doubled in the
interim. The terminal positions specify payoffs of the stake multiplied by
+ 1(``plain games''), + 2(gammons), or + 3(backgammons). We shall also
consider a simplification in which all payoffs are + 1 times the stake; we
shall call it bearoff, as it models the phase of backgammon in which both
players are bearing checkers off the board.
For our purposes, a backgammon-like game is one with alternating plays and
a doubling rule, with payoffs of + 1, + 2, or + 3 times the stake. A
bearoff-like game is one with alternating plays and a doubling rule, with
payoffs of + 1 times the stake.
A position is specified by three components:
(1) A checker position C
(2) a doubling status D, always a power of 2
(3) a right to double, R, which is + if only the maximizing player has
the right to double first;- if only the minimizing player has the right
right to double first; and + if either may double first. If D=1, R = +.
If D>1, R = + or -.
At times we will consider versions of backgammon-like games where doubling
is not allowed; we shall say in this case that R = O.
A position has an expected value, E(C,D,R). For fixed C and R, E(C,D,R) is
proportional to D, so we will mainly use E(C,R)=E(C,D,R)/D. It is obvious
that for fixed C,
E(C,-) ≤ E(C,+) ≤ E(C,+) and
E(C,-) ≤ E(C,O) ≤ E(C,+).
We shall also use E (C) for E(C,+), etc.
ZEROSM
A simple game which approximates the backgammon bearoff is the random walk
of length n, where C is an integer in (0,n), and for 1≤C≤n-1 the successor
position is, with probability 1/2, C+1 or C-1. If C=0, the payoff is -1;
if C=n, the payoff is +1. If the random walk game is played without
doubling, the probability that the maximizing player will win is C/n, and
the expectation is 2(C/n)-1.
When n is a multiple of 5, the random walk game has a simple strategy. The
maximizing player should double if C≥0.8n, and should decline a double if
C<0.2n; the minimizing player should double if C≤0.2n, and should decline
a double if C>0.8n. This gives the expected values shown in the graph:
One can show the correctness of this graph by showing that if the
maximizing player adopts the strategy above, he guarantees an expected
payoff at least as great as shown, and similarly the minimizing player may
guarantee an expected payoff no greater than shown.
Bearoff-like games
We shall show some relations between E-(C), E+(C), E+(C) and E (C) for
bearoff-like games.
Typically, we prove linear inequalities among them by mathematical
induction, and define particular bearoff-like games to show that all
relations satisfying these inequalities may actually occur. We shall use
P(C)=0.5(E (C)+1) as the probability that the maximizing player wins when
doubling is disallowed.
ZEROSM
ZEROSM
Proof:
Since E+ ≥ E , E+ ≥ 2P-1, clearly E+ ≤ 1, since the minimizing player has
has the option of declining all doubles. The minimizing player may adopt
the strategy of declining doubles if P > 0.75, accepting if P ≥ 0.75, and
and never redoubling; an easy inductive proof then shows that E+ ≤ 4P/3-1.
The three extreme points are achieved by these three games:
All other points in the shaded region may be achieved by setting the
probabilities α,β, appropriately in this game:
ZEROSM
Proof: Use the relation E =2P-1 in the established relation between E+ and
P.
Proof: Analogous to the proof for the relation between E+ and P,
interchanging max and min, and reversing the signs of the payoffs.
ZEROSM
Proof: Use the relation E =2P-1.
ZEROSM
Proof: As already seen, E+ ≥ E-, and E+ ≤ 1. To show that E+ ≤ E +0.5, let
the minimizing player adopt the strategy of declining doubles when
E- ≥ 0.5, accepting when E- < 0.5, and never doubling; an inductive proof
gives the result. In essence, the arugment is that even in the positions
where the maximizing player will double, he gains no more than half a point
point by having the ability to do so.
ZEROSM
The games below correspond to the extreme points in the graph of E+ vs. E-:
All other points in the graph may be reached by one of the two games
below, with appropriate values of α, β, , :
(Case where E ≤ 0)
(Case where E > 0)
ZEROSUM
ZEROSM
Proof: -1 ≤ E+ ≤ +1;
E+ ≤ E+ ≤ 4P/3-1;
E+ ≥ E- ≥ 4P/3-1/3.
ZEROSM
Proof: Obvious
ZEROSM
ZEROSM
Proof: E+ ≥ E+. To show that E+ ≤ 4E+/3-0.5, let the maximizing player
adopt the policy of accepting doubles if E+ ≥ -0.5, declining doubles if
E+ < -0.5, and doubling if 2E- ≥ E+. An inductive proof shows the
inequality.
ZEROSM
Proof: Analogous.
ZEROSM
Theorem: It is always correct to decline a double when E+ ≤ -0.5; when
P ≤ 3/16; or when E ≤ -5/8.
Proof: If E+ ≤ -0.5, by accepting the double, one has expectation 2 x E+ ≤
-1; by declining, one has payoff -1, which is no worse. By previous
results, if P ≤ 3/16 or E ≤ -5/8, E+ ≤ -0.5. The extreme situation is
realized in this game:
corresponding to this backgammon position:
Theorem: It is always correct to accept a double when E+≥-0.5; when P≥1/4;
or when E ≥-0.5.
Proof: Analogous
Theorem: It is always correct to double when E-≥0.5; when P≥13/16; when E
≥5/8.
ZEROSM
Proof: When E-≥0.5, it is correct for the opponent to decline, so the
expected value after doubling must be +1; no larger expected value is
possible, because the opponent could adopt a policy of declining all
doubles. When P<13/16, the game below shows that it may be incorrect to
double:
Theorem: It is always correct to double with R=+, if there is probability
≥0.5 of reaching a winning position (a terminal position with payoff =+1)
before the next turn.
Proof: Let q≥0.5 be the probability of reaching a winning position. When a
winning position is reached, the double has gained one point. Otherwise,
doubling results in a position C with expected value 2E-(C) rather than
E+(C). The minimum value of 2E-(C)-E+(C) is -1, achieved when P=0. Thus
the gain of doubling isat least q-(1-q)=2q-1≥1.
Theorem: It is always correct to double with R=+ if there is a probability
≥0.6 of reaching a winning position before the next turn.
Proof: Let q≥0.6 be the probability of reaching a winning position. When a
winning position is reached, the redouble has gained one point. Otherwise,
the redouble results in a position C with expected value 2E-(C) rather
than E+(C). The minimum value of 2E-(C)-E+(C) is -1.5, achieved when
E+(C)=-0.5, E-(C)=-1.0. The gain of having redoubled is at least
q-1.5(1-q)=2.5q-1.5, which is positive if q≥0.6.
ZEROSM
As applied to actual backgammon, in positions where gammons are not likely
it is always correct to redouble if at least 22 of the 36 possible rolls
result in a certain win; the position below is an example:
X should redouble - barely.
Theorem: If you will never have another chance to double, and R=+, it is
always correct to double if P≥5/8.
Proof: By doubling, your expectation becomes 2E-. By waiting, your
expectation is, in effect, E-. If P≥5/8,d E-≥0,d so 2E->E-.
Theorem: If you will never have another chance to redouble, and R=+, it is
always correct to double if P≥0.7.
Proof: By doubling, your expectation becomes 2E-. By waiting, your
expectation is, in effect, E .
Since E-≥4E/3-1/3, 2E--E ≥5/3E -2/3, which is positive if E ≥0.4, i.e. if
P≥0.7.
The following game, with α<0.6, shows that if P<0.7 it may be incorrect to
redouble even on your last turn:
UNHAPP
Bob Floyd
Copyright 1983
If that unhappy fraternity whose only hope for the current game is to
avoid being gammoned, if, I say, it agrees on anything, that would be the
merit of crossovers into a new quadrant. Like the angels contemplating
the torments of the damned from the palisades of paradise, those of us who
are seldom gammoned may wonder, in a spirit of lazy curiosity, whether
their belief is a sustaining and nourishing one, or whether it is merely
an opiate to their pain, sustaining an illusion that they still have a
measure of control over their fate. How valuable, indeed, are crossovers,
and why, when at least one of them plays no special role in the definition
of the game, do they seem so important?
In extreme situations, taking a crossover which leaves an odd number of
outside crossovers to go is the only alternative to reliance on
doubletons. Naturally, in Position A,
(A)
X to play a 1
we move 13/12, so that most sixes bear a checker off next time; playing
the foolish 2/1 offers no hope but 3-3, 4-4, 5-5, or 6-6.
In Position B,
(B)
X to play a 2
UNHAPP
X plays 14/12; his chances of bearing off a checker are still not good,
but rolling 6-3 or better next time makes him a favorite; in fact, overall
he has a 42% chance to bear a checker off in two rolls, and no play that
requires a doubleton ever has more than a 31% chance to bear off in two
rolls. But what about positions where there is one or more crossover to
spare, like Position C,
(C)
X to play a 1
where X can play 13/12, and still bear off in two turns with such mediocre
rolls rolls as 3-2, 5-1? Even here, if we look at every one of the 1296
ways four dice can fall, the crossover is superior by 1047 to 1015, an
advantage of more than 2%.
The strongest clue to the value of the crossover
comes from finding the exceptional situations where the crossover is
wrong. In Position D,
(D)
X to play two 1's
the obvious 14/12 is worse than either 12/10 or 12/11, 14/13. If we look
ahead one roll, to avoid depending on a last roll doubleton X must next
turn bring one checker home, and the other as far forward as possible. If
he can play one checker exactly to his six point, he saves the maximum
possible number of pips to advance the other checker, so it is correct to
maximize, by diversification, the likelihood of playing exactly to the 6
point next turn. That chance is only 17/36 after 14/12, but 19/36 after
12/10, and also 19/36 after 12/11, 14/13. (The play of 14/12 has other
drawbacks; a subsequent roll of 5-4 forces burying a checker deep in the
home board)
UNHAPP
Similarly, in Position E,
(E)
X to play a 2
playing 11/9 is better than 13/11; again, the chance of bearing in exactly
to the six point is better with 11/9 (18/36 to 15/36).
We see the crossover, then, as desirable when it gives the effect of
diversification. By increasing the likelihood of bearing in exactly to
the sixpoint next turn, we reduce the likelihood of being forced to waste
pips later. Crossing to a new quadrant always increases the chance that
the moved checker will play to the six point next turn. From the
opponent's home board, the chance of playing to the six point is at most
1/36. From his outer board, it varies from 2/36 to 6/36. From one's own
outer board, the chance is at least 11/36.
This analysis suggests that a higher rule for running off the gammon than
taking crossovers is the rule of maximizing numbers that play to the six
point. Avoid positions like F,
(F)
with outside checkers six pips apart; they duplicate both good numbers
(such as 6-2 and 2-2 in Position F) and bad ones (any roll containing a 1
in Position F).
UNHAPP
When the positiions of other checkers are considered, it may be better to
move a checker within a board than take a crossover. In Position (G),
(G)
X to play a 1
to play 13/12 takes away 5-2 as a roll which plays to the 6-point; to play
12/11 adds 5-3. In Position H,
(H)
X to play a 3
playing 20/17 adds no number which plays to the six point; playing 16/13
adds 4-3.
(RWF: check out this position)
PEOPLE.TEX
\input basic
\input smacros
\ctrline{Advanced Gammon Avoidance}
\ctrline{Bob Floyd}
\ctrline{$\copyright$ 1982}
\par\vskip .5in\par
People who have given up on the game, and are racing home to save the gammon,
usually play very fast, as though there were no element of skill in bringing
the checkers home. Actually, there are many subtle issues in escaping the
gammon when you have three or more outside checkers, and many experts misplay
these races. This article explores some of the subtleties.
I assume you are familiar with the basics, as in Magriel's Chapter 23. When
you can, you bear in exactly to your six point; you take crossovers and diversify
when you can; when you cross over into a new board, you try to avoid going
deep into that board; you ordinarily don't waste pips inside your home board
without a strong reason.
When analyzing a gammon-avoidance position, you should first try to estimate how
likely the gammon is. Your correct play depends very much on whether you can be
optimistic or not. Estimate how many rolls it will take your opponent to bear
off. To avoid being gammoned, you must take a checker off in one roll fewer.
If you expect your opponent to be off in four rolls, you will have three. You
can reasonably expect to do so if you need no more than five outside crossovers,
and about twenty as an outside pip count. With more than five crossovers, you
must depend on rolling a doubleton; with exactly five, you must depend on not
missing a subsequent crossover.
If either the pip-count or the crossover count dictates pessimism, you must plan
to use large doubletons effectively. However, except for one-roll situations,
this is seldom best done by stacking your checkers on one point; you should be
sure that average rolls play well while you wait for the saving 5-5.
If the counts allow optimism, you should take precautions against rolling small
numbers; if you expect to have three more turns, for example, and you will need
four outside crossovers, try to play so that you will not miss
every time you roll an ace; otherwise even rolling 1-6 twice will destroy you.
In intermediate situations, you must balance many conflicting needs. You should
try to make crossovers where possible, but you should also try to protect yourself
against rolling specific bad numbers (normally 1s) or never rolling a specific
needed number (usually a 6), and you should try not to waste pips inside your
home board, because wasted pips return to haunt you as later missed crossovers.
As a general rule, you should consider wasting pips to get down to an odd number
of outside crossovers, but not to an even number; there are exceptions, though.
Most of the positions that follow are intermediate-to-optimistic; they call for
subtle play to guard against certain unobvious dangers.
In Position A, X has a 2 to play.
\par\vskip 1in\par
(A)
\par\vskip 1in
\ctrline{X to play a 2}
\par\vskip 12pt
\par \noindent
He has, in all likelihood, another two rolls to bear off. Nothing he does with
a 2 can affect his chance to be off in one roll, so he need only look at his
two-roll chances. When this position arose in a chouette, the captain
automatically began to play 7/5 for the crossover. I stopped him, noticing
that his play left the number of crossovers even, so that missing a crossover
now would not matter if we didn't miss again. After 9/7, X will not miss on
his first roll, and seems unlikely to on his second. I started looking for
bad sequences of rolls for either play.
When crossovers are important, two aspects of a position (other than just not
getting enough pip count) are likely to make you miss:
\ip{(1)} You may not have enough checkers on the 1, 7, 13, and 19 points to
cross over with all the 1's you will roll.
\ip{(2)} You may have too many checkers on the 12, 18, and 24 points, and not be
able to roll enough sixes to bring them into a new board each time they move.
\par \noindent
Here only the first danger is present. The needed pip count is only eight or
nine, easy to roll in two turns.
I looked at rolls containing ones. The play of 9/7 looked safe even if both
numbers contained a one. The first roll could be played 7/6, 9/?; the second,
7/6, ?/off. Only 1-2 seemed to offer difficulties. The play of 7/5, if
followed by any ace, would leave X with a checker in the 8 or 9 point and a
gap on the ace point, so that another ace would miss again. We played 9/7,
and sighed in relief when our next roll was 1-6. We had only two bad rolls,
(1-2 and 2-1), to worry about, rather than eleven.
Later, I looked at every two-roll sequence in this position; there is no
sequence that would have made us regret playing 9/7, while any of 1-1, 1-3, 1-4,
1-5, and 1-6 followed by another number of this same set misses if we play 7/5.
The moral: when you are a favorite to escape the gammon, look at your worst
rolls, especially those containing aces. If you don't have several checkers
which can cross over with aces, you are at risk of missing a crossover.
If you also have no checkers on the 2, 8, 14, and 20 points, as in position A,
you run the risk of missing twice in a row on rolls containing aces.
At the other end of the spectrum, some positions will miss unless you roll one
or more sixes, even though the total pip count is modest. In Position B, X
has a 3-2 to play.
\par\vskip 1in\par
(B)
\par\vskip 1in
\ctrline{X to play 3-2}
\par\vskip 12pt
\par\noindent
The obvious play is 15/12, 8/6. This play wastes no pips, bears in perfectly
to the 6-point, achieves two crossovers, and gets another checker ready to
bear in. It is also wrong. If X makes this play, he will fail to be off in
three rolls unless he rolls a six or a doubleton. He should rather bear in
both checkers from the 8-point. He will still be off the gammon in three
rolls if he rolls a six and any other number which is three or greater. He
will also escape the gammon if he rolls two fives, or a four and a five, even
with no sixes. The chance of missing with 15/12 is about $({20/36})↑3=17$\%,
since the chance that a roll will be a singleton with no sixes is twenty
out of thirtysix. The chance of missing after bearing the checkers in from
the eight and nine-points, while harder to estimate, is actually about 6\%.
Even if the checker were on the 16 point, with a 4 to play, it is slightly
better to avoid being dependent on rolling a doubleton or a six in the
next three rolls.
\par\vfill\eject
Position C, from Magriel's {\it Backgammon}, pg. 79,
is a close analogue of Position B.
\par\vskip 1in\par
(C)
\par\vskip 1in\par
\ctrline{X to play 4-3}
\par\vskip 12pt
\par \noindent
Magriel's play of 16/12, 9/6 is not as good as 9/6, 9/5, which would have
escaped the gammon in the game from which the position is taken.
Again, the reason is the
dependency on sixes, in a position where there are no crossovers to spare.
(The difference in the probability of bearing a checker off in three more
rolls is only a few percent.)
PEOPLE
Position D is also from Magriel, pg. 80, and is also misplayed there.
\par \vskip 1in\par
(D)
\par\vskip 1in\par
\ctrline{X to play 5-2}
\par\vskip 12pt
\par \noindent
Here O is at least 50-50 to be off in three more rolls, so X must plan to be
off in two. Barring doubletons, he can only do this by bringing two checkers
in. Magriel's play of 12/7, 8/6, wasting no pips, only has about a 28\% chance
to escape the gammon in two more rolls, while 9/4, 8/6 has about a 53\% chance.
When you have an odd number of crossovers remaining it is usually {\it essential}
to take two crossovers per roll if you can, even at the cost of wasting pips.
\par\vfill\eject
When you come down
to one last outside checker and you have not slotted
the one-point, you may have one of these positions:
\par \vskip 1in
\display{\tabskip 10pt plus 100pt minus 100pt
{\halign to 6.5in{\lft{#} \lft{#}\cr
E8 E9\cr
\space{2in}
E11 E12\cr
}}}
If the outside checker is on the 8 or 9 point, you will wish you had played
2/1. If it is on the 11 or 12 point, you will be glad you didn't. On the
7 or 10 point, it makes no difference. Until you can predict where the last
checker is likely to be, hold off on slotting inside points. Position F
(from Deyong's Playboy Book of Backgammon, Diagram 5-15), exemplifies this.
\par \vskip 1in\par
(F)
\par\vskip 1in\par
\ctrline{X to play a 1}
\par\vskip 12pt
\par \noindent
Deyong's play, 2/1, escaped the gammon in four rolls once {\it less} often
in 36 trials, using the same dice rolls for all positions, than 16/15 or 9/8.
\par\vfill\eject
PEOPLE.TEX
When there is a single outside checker, it is occasionally right to slot the
ace point before the next-to-last roll. In Position G:
\par \vskip 1in
(G)
\par\vskip 1in
\par \ctrline{X to play a 1}
\par \noindent
If X mistakenly plays 17-16, he is a shade more likely to be on the 8- or 9-point
next time (rolls of 3-5, 2-5, 3-4) than on the 11- or 12-point (rolls of
2-4, 2-3).
For similar reasons, he should slot the ace-point if he is on the
14 or 15 points (but not on the 16, because of the value of being off in one more
roll with 3-3).
In position (H), X, who had almost given up, rolls a 3-3. How many checkers
should he bring in? Calculation shows that his chance to be off in two more
rolls after playing 8/5 two, three, or four times is, respectively 55\%, 66\%
and 62\%. How do we explain these figures? How could we find the correct
play over the board?
\par \vskip 1in\par
(H)
\par\vfill\eject
To bring in none or one of the checkers on the 8-point leaves too many
crossovers, unless X rolls another doubleton. To bring two in and play
18/12 creates the double dangers of rolling a 1 on the next roll and of not
rolling a six in the next two rolls. So the candidate plays are to bring in
3 or 4,
\par \vskip 2.5in
\display{\tabskip 10pt plus 100pt minus 100pt
{\halign to 6.5in{\lft{#} \rt{#}\cr
(H3) (H4)\cr
}}}
\par\vskip 12pt
\par \noindent
resulting in positions H3 and H4. If the numbers on the next roll
total nine or more, X is virtually sure to be off on the following roll, in either
position. If the numbers total less than six, they require a subsequent
doubleton, except for 3-2. We need only look in detail at the remaining
numbers:
1-5 and 1-6 substantially favor H4.
2-3, 2-4, 2-5, and 2-6 substantially favor H3.
3-4 and 3-5 are indifferent, or nearly so.
\par\noindent
The preponderance favors H3, although they are close. The most important
decision, though, is to reject bringing in just two checkers.
\par\noindent
In a longer race to get off the gammon, it becomes important to know whether
or not you are favored. Uusually, in these situations, you can count pairs
of checkers as rolls for your opponent, and pairs of crossovers or eights of
pips (whichever is larger) as rolls for you. If you are favored, it becomes
most important to follow traditional principles: diversify in your outfield,
don't waste pips, take crossovers, bear in to your six-point. If you are
significantly behind, plan to take advantage of large numbers and especially
the large doubtletons; play your outside checkers to the 18, 16, 12, and 11
points. If the race is fairly even, correct play depends on whether you are
pip-count-bound or crossover-bound.
If pip-count-bound, assume that to have a chance you must roll 4's, 5's, and
6's anyway, so place checkers on the high points of your outfield. If
crossover-bound, avoid creating positions where too many aces or not enough
sixes will hurt you. Either way, in most races where you have a medium
chance, desperation plays of stacking several checkers on a single point
or bearing deep into your home board are very suspect.
Remember, when you are tempted to waste pips to save a crossover, that a bad
pip count can be expected to turn into missed crossovers later.
\par\vfill\eject
{\topinsert{\vskip 2.5in}}
\par \ctrline{X to play a 3}
\par \noindent
Your chances are only moderate, because of the pip count, but you have a crossover
to spare.
The crossover 15-12 creates insurance against a later miss.
Even rolling a subsequent ace and
no six is not necessarily catastrophic. To play 8-5 would aggravate your
pip-count problem.
\par \vskip 2.5in
\par \ctrline{X to play a 3}
The crossover is essential, but 15/12 makes you dependent on a six. You are
favored, but you need every crossover. Play 8/5.
\par\vskip 2in
\par \ctrline{X to play a 1}
\par\vfill\eject
\par \noindent
Diversification is a mistake here; 4-1, 5-1, and 6-1 next roll are three times
more likely than a 3-2. You are favored, but you need every crossover.
Guard against aces when your chances are good; play 8/7.
PEOPLE.TEX
\par\vskip 2.5in
\ctrline{X to play a 1}
Here you have no crossovers to spare, and a pip count which may lead to a
later miss.
You will need a good number of pips to be off in three rolls however you play,
so assume that you will be rolling a six and no aces; play 11-10 and pray.
If you roll a 4 next time, this play will save you two pips.
\par\vskip 2.5in
\par
\ctrline{X to play a 1}
\par\vfill\eject
Even though you are getting desperate here, don't play 10-9 to stack up for
double 3's; if you do get them next roll, you will still be an underdog to escape
the gammon. A play of 12-11 leaves you favored to escape after 4-4, 5-5, or 6-6.
Another play to consider is 8/7, in case you get an ace on the next roll
followed by doubletons, but it doesn't help unless you also get a six.
The gains from each play are these combinations:
\display{\tabskip 10pt plus 100pt minus 100pt
{\halign to 6.5in{\lft{#} \lft{#} \lft{#} \lft{#}\cr
Play First roll Second roll Total\cr
12-11 5-2, 5-3 4-4, 5-5\cr
5-4 3-3, 4-4, 5-5\cr
4-4 5-1, 5-2, 5-3, 5-4 43 (in 1296)\cr
5-5 twenty-one rolls\cr
\space{20 pt}
10-9 6-2, 6-3 3-3\cr
3-3 6-1, 6-2, 6-3, 6-4, 6-5 14\cr
\space{20 pt}
8-7 1-6 4-4, 5-5, 6-6\cr
1-1, 1-3, 1-4, 1-5 6-6 13\cr
}}}
Notice that X gains more (14 chances in 1296) on 12/11 by not missing on an
initial 5-2, 5-3, or 5-4 than he does on 8/7 by rolling a 3-3 (only 10 in 1296).
The moral: even when your position is desperate, if there is more than one
roll remaining you should prepare for normal rolls rather than for a
specific doubleton.
\par \vfill \end