Here's an analysis that tries to take into account redoublings and
cube ownership. It is based on completely ignoring the actual
gameplay of backgammon and substituting the (quite counterfactual)
assumption that the underlying game is a simple Brownian motion: The
game starts by placing a counter a point 0.5 on a scale from 0 to
1. The counter then performs a one-dimensional continuous-time random
walk. When it reaches either 1 or 0, the game ends and player A or
B, respectively, is declared the winner. While the game is ongoing,
the two players have the options of doubling and redoubling at
arbitrary times, but otherwise according to the backgammon doubling
rules.
In this game, the only choice a player has to make is when to offer
the cube. His strategy can be summarized by two numbers $k$ and
$\lambda$. When a player owns the doubling cube, he will offer a
redouble as soon as the position is $k$ or more; he will offer the
first doubling when the position reaches $\lambda$ for the first
time. The situation before the first doubling is appreciably different
from when the cube has an owner, so $\lambda$ can differ from $k$, but
since a random walk is symmetric under time shifts, there is no reason
to consider strategies where $k$ changes as the game ages.
Let's find the optimal $k$ first. Consider two functions $f$ and $g$
such that $f(p)$ is the expected value of the game (for the player
who wins at $p=1$, and assuming optimal play) at position $p$, given
that the player owns the cube, and $g(p)$ is the expected value when
the player doesn't own the cube. These expected values are always
between $0$ and $1$; we imagine that we have already paid $\frac 12$
into the pot that the winner will take home.
Because of symmetry we must have (if both players play optimally,
which in particular means that their $k$s are the same):
$$g(p)=1-f(1-p)$$
Look at the value of the game at position $k$ when we're just about to
offer a redouble; let's call this value $v$. Then
$$v=f(k) = \min(1, 2 g(k) - \frac12)$$
The $\min$ is because the opponent will only accept the doubling if
doing so will be more advantageous to him than refusing. Subtracting
$\frac 12$ accounts for our share of doubling the pot (or in other
words, for the risk that we may eventually lose 2 units rather than
1).
Now, it is clear that we should redouble at least as soon as the
point where a rational opponent would refuse it -- from that point,
waiting any longer is not going to yield us anything. So we can do
away with the $\min(1,\ldots)$ and just remember that $k$ must be
chosen such that $v\le 1$. We then have
$$\tag{1} v = 2g(k)-\frac12 = 2(1 - f(1-k))-\frac 12 = \frac32 - 2f(1-k)$$
When $p$ is between $0$ and $k$, the value of the game depends on the
probability that the position will reach $k$ before it reaches $0$. By
a wonderful property of Brownian motion, this probability is simply
$p/k$, so we have
$$\tag{2} f(p) = \frac pk f(k) = \frac vk p$$
Clearly, for optimal play we must choose $k$ such that the
proportionality constant $\frac vk$ is as large as possible.
To find the relation between $v$ and $k$, specialize (2) to $p=1-k$:
$$ f(1-k) = \frac{1-k}{k} v$$
and telescope that into (1):
$$ v = \frac 32 - 2\frac{1-k}{k} v \quad\Longrightarrow\quad v = \frac{3k}{4-2k}$$
Thus $\frac vk$, which we're trying to maximize, is
$\frac{3}{4-2k}$. This increases monotonically with $k$, so we want to
have $k$ as large as possible. But, as argued previously, we cannot
have $v>1$, so we find the optimal $k$ by solving
$1=v=\frac{3k}{4-2k}$ for $k$. This yields
$$k=0.8$$
for optimal play once the doubling cube is owned.
We're now ready to find $\lambda$. At the beginning of the game the
situation is symmetric, so if both players follow the same (optimal)
strategy, each player will make the first doubling offer with probability
$\frac 12$. The objective is then to maximize the value of the game after
that first doubling happens:
$$ g(\lambda) = 1-f(1-\lambda) = 1 - \frac{1-\lambda}k = 1.25\lambda -
0.25$$
which is maximized by choosing $\lambda$ as large as possible. But as
before, choosing a $\lambda$ so large that the opponent refuses our
doubling is just a waste. So in fact $\lambda$ should be chosen just
at the threshold where the opponent would start refusing the
doubling. But that happens to be the same criterion as was used to
find $k$, so in fact $\lambda=k$ is optimal.
Conclusion:
For optimal play with doublings and redoublings, assuming that
backgammon can be modeled as a Brownian motion:
- Offer to double or redouble as soon as your position is 80% or more.
- Accept a doubling or redoubling offer if your position is better than 20%.
Exercise: prove that with this strategy, no matter which strategy
your opponent follows, your net expected outcome of a game is never
negative.
There is no simplified description of the Nash equilibrium of this game.
You can compute the best strategy starting from positions where both players are about to win and going backwards from there.
Let $p(Y,O,P)$ the probability that you win if you are at the situation $(Y,O,P)$ and if you make the best choices. The difficulty is that to compute the strategy and probability to win at some situation $(Y,O,P)$, you make your choice depending on the probability $p(O,Y,0)$. So you have a (piecewise affine and contracting) decreasing function $F_{(Y,O,P)}$ such that $p(Y,O,P) = F_{(Y,O,P)}(p(O,Y,0))$, and in particular, you need to find the fixpoint of the composition $F_{(Y,O,0)} \circ F_{(O,Y,0)}$ in order to find the real $p(O,Y,0)$, and deduce everything from there.
After computing this for a 100 points game and some inspecting, there is no function $g(Y,O)$ such that the strategy simplifies to "stop if you accumulated $g(Y,O)$ points or more". For example, at $Y=61,O=62$,you should stop when you have exactly $20$ or $21$ points, and continue otherwise.
If you let $g(Y,O)$ be the smallest number of points $P$ such that you should stop at $(Y,O,P)$, then $g$ does not look very nice at all. It is not monotonous and does strange things, except in the region where you should just keep playing until you lose or win in $1$ move.
Best Answer
It's easy to compute the optimal strategy for $N$ reasonably small (say, up to $N \le 10000$). You just iterate backwards from the end of the game. There are three possible doubling states:
Note that the number currently showing on the doubling die -- to use the backgammon term -- is irrelevant to your strategy (although not irrelevant to your expected gain/loss).
So for each doubling state $s \in \{1,2,3\}$ you want to compute the expected gain $E(s,x,y)$, expressed as a multiple of the number on the doubling die, where $s$ is the doubling state, $x$ is the number of points you need to win, and $y$ is the number of points your opponent needs to win.
$x$ and $y$ can't both be zero, so we can start the ball rolling with $E(s,0,y) = 1$, $E(s,x,0) = 0$. Then you have to set up a number of simple but tedious equations expressing $E(s,x,y)$ in terms of $E(s', x', y)$ for $x' < x$ and all values of $s$ and $s'$. Then you can work backwards to calculate $E(s,x,y)$ for all $s,x,y$.
The details are, as I said, tedious, and I'm not going to spell them out unless you pay me. Just be aware that you have to store the intermediate values in a table of size $3N^2$. Don't be seduced by the obvious recursive solution -- its asymptotic complexity is exponentially worse.