Mathematics of Backgammon 
Bob Floyd
When you ride the tiger,
getting off is the hard part.
The problem in Position A was composed by U.S. Amateur Champion Bill Kennedy. Kennedy's opinion was that the position is a perpetual redouble and take; that is, that each player should redouble until one of them comes in from the bar. The truth is much more complicated than that.
Correct cube handling in Position A depends on how wealthy you and your opponent are, and whether he pays up when he loses a bundle (I assume that you do, of course).

Position A. Should the player on roll redouble? 
The chance that the first player comes in on his first turn is 11/36; that of the first player failing and the second player coming in on his first turn is (25/36) × (11/36). The 36 to 25 ratio between these chances is repeated for all subsequent turns until one player has come in, so the odds are 36 to 25 that the player on roll will come in first.
If the first player keeps the cube, we would expect that if the position were played sixtyone times, the first player would win twice the cube in thirtysix of them, the second player would win twice the cube in twentyfive of them, and the expectation for the player on roll would be
2 ×

If the first player redoubles and the second player keeps the cube, the stakes are twice as high with everything else the same, and the expectation for the first player would be 72% of the original cube, so it is clearly better to redouble if your opponent will not redouble, but it is also a clear take.
If the first player redoubles and the second player reredoubles, and they keep this up until one or the other comes in, the first player's expectation is some fraction E of the cube. The equation for E is:
E = 2 × (

where the first 2 results from the cube turn, the second is from the gammon, and the minus sign is because if the first player fails, the second player's positive expectation is the first player's negative expectation.
The solution of this equation is E = 51% of the original cube. (We shall see later on that there is a logical fallacy in applying this equation to this problem.) So, apparently, the player on roll does best to double, and his opponent holds the first player's expectation down to 51% of the cube by taking and redoubling.
However, this position has a great likelihood of reaching very high stakes; the chance of a 1024cube, with each player dancing four times, is (25/36)^{8} = 5.4%. Eventually, one will decide not to redouble because he can't cover his potential loss, or because he can't afford to take the reredouble, or because he doesn't trust his opponent to pay.
Let's let E(i) be the first player's expectation as a multiple of the original cube if the cube is turned one time then held. We already know that E(0) = .36, E(1) = .72. The equation that gives all the E's is
E(i + 1)
= 2 × (

This table gives some of the solutions:
i  Doubler  Cube  E(i)  




If the first player redoubles on his first three turns, then chickens out when faced with turning the cube from 128 to 256, his expectation is minus 58% of the cube, far worse than the 36% he expects if he doesn't double at all. If the first player is the one who eventually keeps the cube, he should not double the first time; the longer he doubles, the more he can expect to lose, unless he can afford to keep it up longer than his opponent. When you ride the tiger, getting off is the hard part.
If the cube is in the middle, though, the first player must double, under the Jacoby rule, to activate the gammon. By doubling, accepting the redouble, and holding the cube, his expectation is E(2) = .22; if he declines to double, accepts a double from the second player, and holds the cube, his expectation is

Because of the potential for very high stakes in this position, I would ask any opponent to agree to put cash on the table to back each redouble and each take.
Now, let's go back and look again at the equation for the expectation when the game is played as an infinite redouble (bearing in mind that it can't really happen because no backgammon player is infinitely rich). The true expectation to the first player (let's say Black is on roll) is an infinite sum:
4 × 11/36  (Black enters on his first roll)  
−  8 × 25/36 × 11/36  (White enters on his first roll) 
+  16 × (25/36)^{2} × 11/36  (Black enters on his second roll) 
−  32 × (25/36)^{3} × 11/36  (White enters on his second roll) 
+  32 × (25/36)^{4} × 11/36  (Black enters on his third roll) 
etc. 
which equals 1.22 − 1.70 + 2.36 − 3.27 + 4.55 − 6.32 + . . . , a series which doesn't add up to .51 times the cube. It doesn't "add up" to anything. It alternates between being larger and smaller than .51, getting further and further away. Technically, this is a "divergent series," and the usual formula for the sum of an infinite series doesn't apply to it. The equation for expectation when both players redouble gave a number that would be the sum if the series had one, but it doesn't have one. The average amount you win in this position, assuming you do win, is infinite; so is the average amount you lose.
The expectation for the first player if the game is played as a perpetual redouble is not .51 times the cube, or any other number; it is infinity minus infinity. If you know how much that is, please tell me, and I'll notify the mathematicians. The practical reality of the position lies in the analysis that takes the players' limited resources into account; the player with the larger bankroll, if both are honest, or the player who is prepared to welsh on a large bet, has the advantage.
In all such positions, a player who can't otherwise afford to give a mathematically correct redouble should sell a share of his equity in the game. However, in Position A, no finite amount of financial backing is certain to be enough. If your backing is better than your opponent's, you should redouble, but be careful; he may have rich friends you don't know about.

Position B. Should Black redouble to 32? 


In Positions C and D, Black should only take a redouble if he can afford to reredouble.

Position E. Should Black take White's redouble? 
In Position E, Black has about a 24 percent chance of winning if the game is played to the end. If he can afford to reredouble, his cube equity gives him a take; if not, he should drop.


In Positions F and G, Black should only beaver if he can afford to reredouble later.