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\bigskip
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:
\smallskip
\disleft 25pt:
(1):a maximizing node, with successor chosen by the maximizing player;
\disleft 25pt:
(2):a minimizing node, with successor chosen by the minimizing player;
\disleft 25pt:
(3):a randomizing node, with each successor chosen with a specified
probability;
\disleft 25pt:
(4):a terminal node, with payoff determined by the rules of the game.
\smallskip
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 ({\sl 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 $\pm 1$ (``{\sl plain games\/}''),
$\pm 2$ ({\sl gammons\/}), or $\pm 3$ ({\sl backgammons\/}).
We shall also
consider a simplification in which all payoffs are $\pm 1$ times the stake;
we shall call it {\sl bearoff\/}, as it models the phase of backgammon in
which both players are bearing checkers off the board.
For our purposes, a {\sl backgammon-like\/} game is one with alternating
plays and
a doubling rule, with payoffs of $\pm 1$, $\pm 2$, or
$\pm 3$ times the stake. A~{\sl bearoff-like\/} game
is one with alternating plays and a doubling rule, with
payoffs of $\pm 1$ times the stake.
A position is specified by three components:
\smallskip
\disleft 25pt:
(1):A checker position $C$
\disleft 25pt:
(2):a doubling status $D$, always a power of 2
\disleft 25pt:
(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 to double first; and $\pm$ if either may double first. If
$D=1$, $R =\pm$. If $D>1$, $R = +$ or~$-$.
\smallskip
At times we will consider versions of backgammon-like games where doubling
is not allowed; we shall say in this case that $R = \emptyset$.
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~$C$,
$$\vcenter{\halign{$\rt{#}\;$&$\ctr{#}\;$&$\ctr{#}\;$&$\ctr{#}\;$&$\lft{#}$\cr
E(C,-)&≤&E(C,\pm)&≤&E(C,+)\quad\hbox{and}\cr
E(C,-)&≤&E(C,\emptyset)&≤&E(C,+)\,.\cr}}$$
We shall also use $E↓+(C)$ for $E(C,+)$, etc.
\vfill\eject
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:
\figbox 200pt:
\noindent
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.
\bigskip
\noindent
{\bf Bearoff-like games.}
We shall show some relations between $E↓-(C)$, $E↓{\pm}(C)$, $E↓+(C)$ and
$E↓{\emptyset}(C)$ 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
$P(C)=0.5\bigl(E↓{\emptyset}(C)+1\bigr)$ as the probability
that the maximizing player wins when doubling is disallowed.
\vfill\eject
\noindent{\bf Theorem:} $E↓+≥2P-1$, $E↓+≤1$, $E↓+≤8P/3-1$.
\figbox 180pt:
\noindent{\bf 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 $E↓+ ≤ 8P/3-1$.
The three extreme points are achieved by these three games:
\bigskip
$$\vcenter{\halign{\ctr{#}\qquad\qquad
&\rt{#}\quad&\ctr{#}\quad&\lft{#}\qquad\qquad
&\ctr{#}\cr
max&&max&&max\cr
\noalign{\bigskip\bigskip}
$-1$&&ran&&$+1$\cr
\noalign{\medskip}
&0.25&&0.75\cr
\noalign{\smallskip}
&$-1$&&$+1$\cr}}$$
\bigskip
\noindent
All other points in the shaded region may be achieved by setting the
probabilities $\alpha$, $\beta$, $\gamma$ appropriately in this game:
\bigskip
$$\vcenter{\halign{\rt{#}\q&\rt{#}\q&\ctr{#}&\ctr{#}\q&\lft{#}\q&\lft{#}\cr
&&ran\cr
\noalign{\medskip}
$\alpha$&&$\beta$&&&$\gamma$\cr
\noalign{\medskip}
$-1$&&max&&&$+1$\cr
\noalign{\bigskip\bigskip}
&&ran\cr
\noalign{\medskip}
&025&&&0.75\cr
\noalign{\smallskip}
&$-1$&&&$+1$\cr}}$$
\vfill\eject
\noindent
{\bf Theorem:} $E↓+≥E↓{\emptyset}$, $E↓+≤4E↓{\emptyset}/3+1/3$.
\figbox 170pt:
\noindent{\bf Proof:}
Use the relation $E↓{\emptyset}=2P-1$ in the established relation between
$E↓+$ and~$P$.
\bigskip\noindent
{\bf Theorem:} $E↓-≤2P-1$, $E↓-≥-1$, $E↓-≥8P/3-5/3$.
\figbox 170pt:
\noindent{\bf Proof:}
Analogous to the proof for the relation between $E↓+$ and~$P$,
interchanging max and min, and reversing the signs of the payoffs.
\vfill\eject
\noindent
{\bf Theorem:} $E↓-≤E↓{\emptyset}$, $E↓-≥4E↓{\emptyset}/3-1/3$.
\figbox 170pt:
\noindent{\bf Proof:}
Use the relation $E↓{\emptyset}=2P-1$.
\vfill\eject
\noindent
{\bf Theorem:} $E↓+≤E↓-+0.5$.
\figbox 170pt:
\noindent{\bf 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 argument is that even in the positions
where the maximizing player will double, he gains no more than half a point
by having the ability to do so.
The games below correspond to the extreme points in the graph of $E↓+$
vs.~$E↓-$:
\bigskip
$$\vcenter{\halign{\ctr{#}\qq&\rt{#}\q&\ctr{#}\q&\lft{#}\qq\qq
&\rt{#}\q&\ctr{#}\q&\lft{#}\qq&\ctr{#}\cr
$-1$&&min&&&max&&$+1$\cr
\noalign{\bigskip\bigskip}
&&ran&&&ran\cr
\noalign{\medskip}
&0.75&&0.25&0.25&&0.75\cr
\noalign{\smallskip}
&$-1$&&$+1$&$-1$&&$+1$\cr}}$$
\vfill\eject
All other points in the graph may be reached by one of the two games
below, with appropriate values of $\alpha$, $\beta$, $\gamma$,~$\delta$:
\bigskip
\noindent
(Case where $E↓{\emptyset}≤ 0$):
$$\vcenter{\halign{$\ctr{#}$\q&$\rt{#}$\q&\ctr{#}\q&$\lft{#}$\q&\ctr{#}\q
&$\rt{#}$\q&\ctr{#}\q&$\lft{#}$\q&$\ctr{#}$\cr
&&&&max\cr
\noalign{\bigskip\bigskip}
&&&&ran\cr
\noalign{\medskip}
&\alpha&&\beta&&\gamma&&\delta\cr
\noalign{\medskip}
-1&&min&&&&min&&+1\cr
\noalign{\bigskip\bigskip}
&&ran&&&&max\cr
\noalign{\medskip}
&0.75&&0.25\cr
\noalign{\medskip}
&-1&&+1&&&ran\cr
\noalign{\medskip}
&&&&&0.25&&0.75\cr
\noalign{\medskip}
&&&&&-1&&+1\cr}}$$
\bigskip
\noindent
(Case where $E↓{\emptyset}> 0$):
$$\vcenter{\halign{$\ctr{#}$\q&$\rt{#}$\q&\ctr{#}\q&$\lft{#}$\q&\ctr{#}\q
&$\rt{#}$\q&\ctr{#}\q&$\lft{#}$\q&$\ctr{#}$\cr
&&&&min\cr
\noalign{\bigskip\bigskip}
&&&&ran\cr
\noalign{\medskip}
&\alpha&&\beta&&\gamma&&\delta\cr
\noalign{\medskip}
-1&&max&&&&max&&+1\cr
\noalign{\bigskip\bigskip}
&&min&&&&ran\cr
\noalign{\medskip}
&&&&&0.75&&0.25\cr
\noalign{\medskip}
&&ran&&&-1&&+1\cr
\noalign{\medskip}
&0.75&&0.25\cr
\noalign{\medskip}
&-1&&+1\cr}}$$
\vfill\eject
\noindent
{\bf Theorem:} $8P/3-5/3≤E↓{\pm}≤8P/3-1$.
\figbox 174pt:
\noindent{\bf Proof:}
$$\vcenter{\halign{$\rt{#}\;$&$\ctr{#}\;$&$\ctr{#}\;$&$\ctr{#}\;$&$\lft{#}$\cr
E↓{\pm}&≤&E↓+&≤&8P/3-1\,;\cr
E↓{\pm}&≥&E↓-&≥&8P/3-5/3\,.\cr}}$$
\bigskip\noindent
{\bf Theorem:} $4E↓{\emptyset}/3+1/3≤E↓{\pm}≤4E↓{\emptyset}-1/3$.
\figbox 174pt:
\noindent{\bf Proof:} Obvious.
\vfill\eject
\noindent
{\bf Theorem:} $E↓{\pm}≤E↓+≤3E↓{\pm}/4+1/4$.
\figbox 174pt:
\noindent{\bf Proof:}
$E↓+≥ E↓{\pm}$. To show that $E↓+≤ 3E↓{\pm}/4+1/4$, 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.
\bigskip\noindent
{\bf Theorem:} $3E↓{\pm}/4-1/4≤E↓-≤E↓{\pm}$.
\figbox 174pt:
\noindent{\bf Proof:} Analogous.
\vfill\eject
\noindent
{\bf Theorem:} It is always correct to decline a double when
$E↓+ ≤ -0.5$; when $P ≤ 3/16$; or when $E↓{\emptyset}≤ -5/8$.
\bigskip\noindent
{\bf Proof:}
If $E↓+ ≤ -0.5$, by accepting the double, one has expectation
$2\times E↓+ ≤\,-1$; by declining, one has payoff $-1$,
which is no worse. By previous
results, if $P ≤ 3/16$ or $E↓{\emptyset}≤ -5/8$,
$E↓+ ≤ -0.5$. The extreme situation is realized in this game:
$$\vcenter{\halign{\rt{#}\q&\ctr{#}\q&\ctr{#}\q&\lft{#}\cr
&min\cr
\noalign{\bigskip\bigskip}
&ran\cr
\noalign{\medskip}
0.75&&0.25\cr
\noalign{\smallskip}
$-1$&&max\cr
\noalign{\bigskip\bigskip}
&&ran\cr
\noalign{\medskip}
&0.25&&0.75\cr
\noalign{\smallskip}
&$-1$&&$+1$\cr}}$$
%\vfill\eject
\noindent
corresponding to this backgammon position:
%\figbox 140pt:
\vfill\eject
\noindent{\bf Theorem:}
It is always correct to accept a double when $E↓+≥-0.5$; when $P≥1/4$;
or when $E↓{\emptyset}≥-0.5$.
\bigskip\noindent
{\bf Proof:} Analogous.
\bigskip\noindent
{\bf Theorem:}
It is always correct to double when $E↓- ≥0.5$; when $P≥13/16$;
when $E≥5/8$.
\bigskip\noindent
{\bf 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:
\bigskip
$$\vcenter{\halign{\rt{#}\qq\qq&\ctr{#}\q&\rt{#}&\ctr{#}\q&\lft{#}\q&\ctr{#}\qq
&\rt{#}\cr
&max\cr
\noalign{\bigskip\bigskip}
&ran\cr
\noalign{\medskip}
$\alpha$&&&&$1-\alpha$\cr
\noalign{\smallskip}
$-1$&&&&&min\cr
\noalign{\bigskip\bigskip}
&&&&&max\cr
\noalign{\bigskip\bigskip}
&&&&&ran\cr
\noalign{\medskip}
&&&0.25&&&0.75\cr
\noalign{\smallskip}
&&&min&&&$+1$\cr
\noalign{\bigskip\bigskip}
&&&ran\cr
\noalign{\medskip}
&&0.75&&0.25\cr
\noalign{\smallskip}
&&$-1$&&$+1$\cr}}$$
\bigskip
\noindent
{\bf Theorem:} It is always correct to double with $R=\pm$, if there is probability
$≥0.5$ of reaching a winning position (a terminal position with payoff $=+1$)
before the next turn.
\bigskip\noindent
{\bf 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
$E↓{\pm}(C)$. The minimum value of $2E↓-(C)-E↓{\pm}(C)$is~$-1$,
achieved when $P=0$. Thus
the gain of doubling is at least $q-(1-q)=2q-1≥1$.
\vfill\eject
\noindent
{\bf 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.
\bigskip\noindent
{\bf 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 $q≥0.6$.
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:
\figbox 140pt:
\ctrline{$X$ should redouble $-$ barely.}
\bigskip
\noindent
{\bf Theorem:} If you will never have another chance to double, and
$R=\pm$, it is always correct to double if $P≥5/8$.
\bigskip\noindent
{\bf Proof:} By doubling, your expectation becomes $2E↓-$. By waiting, your
expectation is, in effect,~$E↓-$. If $P≥5/8$, $E↓-≥0$, so $2E↓->E↓-$.
\vfill\eject
\noindent
{\bf Theorem:} If you will never have another chance to redouble, and
$R=+$, it is always correct to redouble if $P≥0.7$.
\bigskip\noindent
{\bf Proof:} By doubling, your expectation becomes $2E↓-$. By waiting, your
expectation is, in effect,~$E↓{\emptyset}$.
Since $E↓-≥4E↓{\emptyset}/3-1/3$,
$2E↓--E↓{\emptyset} ≥5/3E↓{\emptyset}-2/3$, which is positive if
$E↓{\emptyset}≥0.4$, i.e., if $P≥0.7$.
The following game, with $\alpha <0.6$, shows that if
$P<0.7$ it may be incorrect to redouble even on your last turn:
\bigskip
$$\vcenter{\halign{\rt{#}\q&\rt{#}\q&\lft{#}\q&\ctr{#}\qq\qq&&\lft{#}\cr
&&&max\cr
\noalign{\bigskip\bigskip}
&&&ran\cr
\noalign{\medskip}
&$1-\alpha$&&&$\alpha$\cr
\noalign{\medskip}
&min&&&$+1$\cr
\noalign{\bigskip\bigskip}
&ran\cr
\noalign{\medskip}
0.75&&0.25\cr
\noalign{\smallskip}
$-1$&&$+1$\cr}}$$
\vfill\eject
\figbox 140pt:
\ctrline{X on roll. The player with the right to double has the advantage.}
%\vfill\eject
\bigskip
$$\vcenter{\halign{\lft{#}\q&\lft{#}\q&\lft{#}%
&\lft{#}\q&\lft{#}%
&\ctr{#}\q&\lft{#}&\ctr{#}\q&\rt{#}%
&\lft{#}\q&\lft{#}&\ctr{#}%
&\rt{#}&\lft{#}\cr
&&\phantom{ran}&&&&&&&\multispan2 $E↓-=-0.020$\hfill\cr
%&&&&\phantom{ran}&&&&&\multispan2 $E↓-=-0.020$\hfill\cr
&&&&&&&&&\multispan2 $E↓+=0.149$\hfill\cr
&&&&&&&&max\cr
\noalign{\bigskip\bigskip}
&&&&&&&&ran\cr
\noalign{\bigskip}
&&&&\multispan2 10/36\hfill&&&&21/36&&5/36\cr
&&\multispan2(e.g., 2-1)\hfill&&&&&&\multispan2 (e.g., 3-2)\hfill&&(e.g., 2-2)\cr
\noalign{\bigskip}
&min&&&&&&&min&&&&$+1$\cr
&&\multispan2 $E↓-=-1.000$\hfill&&&&&&\multispan2 $E↓-=0.204$\hfill
&&&$E↓-=1.0$\cr
&&\multispan2 $E↓+=-0.768$\hfill&&&&&&\multispan2 $E↓+=0.574$\hfill
&&&$E↓+=1.0$\cr
\noalign{\bigskip\bigskip}
&ran&&&&&&&ran\cr
\noalign{\medskip}
6/36&&&30/36&&&&6/36&&&30/36\cr
\noalign{\bigskip}
$-1$&&&&max&&&$-1$&&&max\cr
&&&&&\multispan2 $E↓-=-0.722$\hfill&&&&&\multispan3 $E↓-=0.444$\hfill\cr
&&&&&\multispan2 $E↓+=-0.722$\hfill&&&&&\multispan3 $E↓+=0.889$\hfill\cr
\noalign{\bigskip\bigskip}
&&&&ran&&&&&&ran\cr
\noalign{\medskip}
&&&31/36&&&5/36&&&10/36&&26/36\cr
\noalign{\medskip}
&&&$-1$&&&$+1$&&&\q$-1$&&$+1$\cr}}$$
\vfill\eject
\noindent{\bf Backgammon-like Games} (European form).
Now we consider games in which some terminal positions specify payoffs
of $\pm 2$, but not $\pm 3$. This corresponds to the European rules.
There is no longer in general a well-defined probability of winning,
as shown by this game:
$$\vcenter{\halign{$\lft{#}$\qq&\ctr{#}\qq&$\rt{#}$\qq&\ctr{#}\qq
&$\lft{#}$\qq&\ctr{#}\qq&$\lft{#}$\cr
&&&max\cr
\noalign{\bigskip}
&ran&&&&ran\cr
\noalign{\medskip}
1/3&&2/3&&1/2&&1/2\cr
\noalign{\medskip}
+2&&-1&&+1&&-1\cr}}$$
\bigskip\noindent
where the maximizing player has two equally good alternatives, both
with payoff~0.0; one has a winning probability of~1/3, the other
of~1/2. If the max and min nodes have no branching, however, the
probability~$P$ of winning is well-defined.
\figbox 250pt:
\noindent
{\bf Proof:}
$$\eqalign{E↓{\emptyset}&≤2\cdot P-(-1)(1-P)\cr
E↓{\emptyset}&≥1\cdot P+(-2)(1-P)\cr}$$
\vfill\eject
The game below achieves all points in the shaded region by appropriate
settings of the probabilities $\alpha$, $\beta$, $\gamma$,~$\delta$:
$$\vcenter{\halign{\rt{#}\q&\rt{#}\q&\ctr{#}\q&\lft{#}\q&\lft{#}\cr
&&ran\cr
\noalign{\medskip}
$\alpha$&$\beta$&&$\gamma$&$\delta$\cr
\noalign{\medskip}
$-2$&$-1$&&$+2$&$+1$\cr}}$$
\figbox 250pt:
\noindent
{\bf Proof:} Obviously $E↓+≥E↓{\emptyset}$. With $R=+$, the minimizing
player can hold the maximizing player's expectation below the upper
boundary of the region by accepting doubles if $E↓{\emptyset}≤1/2$,
declining if $E↓{\emptyset}>1/2$, and never redoubling.
The game below, with appropriate weights, reaches the entire shaded region.
$$\vcenter{\halign{\rt{#}\q&\rt{#}\q&\ctr{#}&\lft{#}\q&\lft{#}\cr
&&ran\cr
$\alpha$&&$\beta$&&$\gamma$\cr
\noalign{\medskip}
$-2$&&max&&$+2$\cr
\noalign{\medskip}
&3/4&&1/4\cr
\noalign{\medskip}
&$+1$&&$-1$\cr}}$$
\vfill\eject
\figbox 250pt:
\noindent
{\bf Proof:} Analogous to the previous.
\figbox 250pt:
\noindent
{\bf Proof:} Obviously, $E↓+≥E↓-$.
We establish the upper boundary by induction on the depth of the tree.
\vfill\eject
At a randomizing node, the value of $\langle E↓-,E↓+\rangle$ is a weighted
average of vectors which, by induction, are in the shaded region. At a
maximizing node, three cases arise:
\disleft 25pt:(1):
It is incorrect to double. Then the maximizing player must choose between
(say) two nodes as shown below
$$\vcenter{\halign{$\lft{#}$\q&\ctr{#}\qq\qq\qq\qq&$\lft{#}$\cr
&max\cr
\noalign{\bigskip}
E↓-=a&&E↓-=c\cr
E↓+=b&&E↓+=d\cr}}$$
\bigskip
\disleft 25pt:{}:
where, by induction, $\langle a,b\rangle$ and $\langle c,d\rangle$ are in the
shaded region. The value of $\langle E↓1,E↓+\rangle$ is
$\langle\max(a,c),\max(b,d)\rangle$, which is the upper right corner
of the rectangle whose two opposite corners are $\langle a,b\rangle$
and $\langle c,d\rangle$. This point is in the shaded region.
\smallskip
\disleft 25pt:(2):
It is corrct for the maximizer to double, and the minimizer to decline.
Then $E↓-≥1/2$, and $E↓+=1$, so the node lies below the upper boundary
of the shaded region.
\smallskip
\disleft 25pt:(3):
It is correct for the maximizer to double, and the minimizer to accept.
Then $E↓+=2E↓-≤1$, and again the node lies below the upper boundary.
\smallskip\noindent
At a minimizing node, the same three cases must be enumerated, with the
same result.
The game below achieves all points in the shaded region.
$$\vcenter{\halign{$\rt{#}$\q&$\lft{#}$\q&\rt{#}\q&$\lft{#}$\q
&\ctr{#}\q&$\lft{#}$\q&\lft{#}\q&$\lft{#}$\q&$\lft{#}$\cr
&&&&ran\cr
\noalign{\medskip}
\alpha&&$\beta$&&&&$\gamma$&&\beta\cr
\noalign{\medskip}
-2&&min&&&&max&&+2\cr
\noalign{\medskip}
&3/4&&1/4&&1/4&&3/4\cr
\noalign{\medskip}
&-1&&+1&&-1&&+1\cr}}$$
\vfill\eject
\figbox 250pt:
\noindent
{\bf Proof:} The upper boundary follows because $E↓{\pm}≤E↓+$, which has
already been shown to lie below the upper boundary. The lower boundary
similarly follows from $E↓{\pm}≥E↓-$.
\figbox 250pt:
\noindent
{\bf Proof:} Very much like all the others.
\vfill\eject
\bigskip\noindent
{\bf Theorem.} It is always correct to double or redouble when $E↓-=0.5$.
\medskip\noindent
{\bf Proof:} If $E↓-=0.5$, we know already that $E↓+≤1$. By doubling,
we achieve an expected value of 1.0 whether the minimizing player accepts
or declines; this is the best one can do.
\bigskip\noindent
{\bf Theorem.} It s always correct to redouble against a team of
opponents if at least ??? of them will decline, and at least ??? of
them will accept.
\medskip\noindent
{\bf Proof:} Assume a fraction $F$ will accept.
\smallskip
{\bf Case 1.} They should accept. Then $E↓-≤0.5$, and $E↓+≤1$.
Presumably it is incorrect to double, so $E↓2≥2E↓-$
[RWF: fill in details.]
\smallskip
{\bf Case 2.} They should decline
[RWF: fill in details.]
\vfill\eject\end