On Optimal Doubling in Backgammon
|Originally published in Management Science, vol. 23, no. 8, April 1977.|
A rolls first and starts out xA spaces from home, while B begins as distance xB from home. (xA is called player A’s count). Each player’s count is reduced by the total of his roll. Thus if A’s count is 45, and he rolls 5 and 3, his count becomes 37. Player A wins if and only if his count becomes ≤ 0 before B’s count becomes ≤ 0.
Since gammons and backgammons (these are bonuses) rarely occur in running games, they will be disregarded.
Players are allowed to double the stakes if they follow betting rules which will be explained in the next section. We give the optimal doubling and redoubling strategies for race-type positions. The strategies we present differ from those given by Keeler and Spencer (Keeler et al. 1975) because they assumed that the total of a player’s dice could be 5 with probability 11/24, 10 with probability 11/24, and 20 with probability 2/24. In reality, for race purposes, the dice can show any one of 13 numbers.Keeler et al. 1975). Observe that A wins if p(t) becomes 1 before it becomes 0.
Lemma 1. In a continuous game, let the probability of A winning be x, and let a, b > 0 be such that 0 ≤ x − a < x < x + b ≤ 1. Let E be the event that p(t) becomes x + b before it becomes x − a. The the probability of E is a/(a + b).
Proof. We have, by conditioning on whether p(t) first reaches x + b or x − a, x = p(E) · (x + b) + (1 − p(E)) · (x − a). This yields
x − (x − a)
(x + b) − (x − a)
a + b
Example 1. Suppose that x = 0.75. Then the probability that A wins (p = 1 before p = 0) is 0.75/1 = 0.75.
Example 2. Suppose that x = 0.6. Then the probability that x = 0.8 before x = 0 is 0.6/(0.6 + 0.2) = 3/4.
Lemma 1 may be extended to any linear function of p(t), such as A’s expectation. In other words, the probability that A’s expectation rises by b before it falls by a is also a/(a + b).
It is shown in Keeler and Spencer (Keeler et al. 1975) that in a continuous game, A should double or redouble when p = 0.8. When A doubles it does not matter whether B accepts or declines, his expectation in both cases is −(the current value of the game). To see this, let K denote the value of the game before A doubles. If B accepts, he will own the cube at 2K. He will win when p = 0.2 and lose when p = 1. Using Lemma 1, his probability of winning is 0.2/(0.6 + 0.2) = 1/4. Therefore his expectation is 1/4·(2K) − 3/4·(2K) = −K, which is what he would lose if he did not accept the double.
In a noncontinuous game, the cube loses some of its value because one might have to reach an 85% position part of the time to win by doubling instead of an 80% position. (Having to reach 70% to double may be shown to be roughly equivalent to having to reach some point > 0.80 to double.) We may measure this loss of value by approximating our discontinuous game by a continuous one with doubling points DA for A and DB for B. Both are functions of the state of the game, i.e., the position of the checkers, whose roll it is, etc. DA lies between 0.80 and 1. DB lies between 0.20 and 0. A wins when p(t) = DA, B wins when p(t) = DB.
The loss of the cube value to A is measured by the magnitude of DA − 0.80. When DA = 0.80, as is the case in a continuous game, the cube has a maximum value since A must travel the minimum distance to win. As DA moves upward, the cube loses value. When DA = 1.0 the cube has no value to A.
In general, as the race nears the end, DA will increase since p(t) will become “less continuous.” In an even race with 40 spaces to go and the cube in the middle, our results indicate that DA ≈ 0.88 and DB ≈ 0.12. With 20 pips to go, DA ≈ 0.90 and DB ≈ 0.10.
Lemma 2. In a continuous game let DA(t) and DB(t) represent the effective doubling points for A and B at time t. Let p(t) represent A’s chances of winning at t assuming a cube is not used. Let p(t, A) denote A’s chances when holding the cube, and let p(t, B) denote A’s chances when B holds the cube. Let p(t, M) denote A’s chances when the cube is in the middle. Then we have
|p(t, A) = p(t)/DA(t),||(1)|
|p(t, B) = (p(t) − DB(t))/(1 − DB(t)),||(2)|
|p(t, M) = (p(t) − DB(t))/(DA(t) − DB(t)).||(3)|
Proof. The proof is immediate from Lemma 1. For example, to prove (3), we take x = p(t), a = p(t) − DB(t), b = DA(t) − p(t). Then
a/(a + b) = (p(t) − DB(t))/(DA(t) − DB(t)).
Theorem 1. In a continuous game with effective doubling points DA(t), DB(t) for A and B respectively, let M(t) denote the greatest value of p(t) for which B can accept A’s double. M(t) is called the minimum take point and varies from 75% to 80%. Let r(t) and d(t) denote the minimum values of p(t) for which A can double and redouble respectively. Then we have
|M(t) = 1 − 1/4·(1 − DB(t)),||(4)|
Proof. We prove (4) first. After A doubles, B’s chances of winning given that he owns the cube are 1 − p(t, B), which by (2) of Lemma 2 equals
B’s minimum take point occurs when (7) = 25%, i.e., we have
Solving this for M(t) yields (4).
Proof of (5). r(t) is determined by the condition that A’s expectation after redoubling be equal to A’s expectation before redoubling. A’s expectation before redoubling equals
p(t, A)K + (1 − p(t, A)) (−K) = K(2p(t, A) − 1).
By (1) of Lemma 2, this equals
|K(2p(t)/DA(t) − 1).||(8)|
A’s expectation after doubling is
|2K(2p(t, B) − 1),||(9)|
which by (2) of Lemma 2 equals
When p(t) = r(t), we must have (8) = (10), or
Solving (11) for r(t) yields (5).
The proof of (6) is similar to the proof of (5) and involves equating 2p(t, M) − 1 to 2(2p(t, B) − 1) and solving for p(t).
Player A wins if, when he takes n rolls to get off, B takes at least n rolls. Let P(x, n) denote the probability that a player with count x gets off in exactly n rolls. Then the probability that A wins is simply
This quantity may be calculated exactly using convolutions. Let f(y) denote the probability that a player’s total in one roll is y. In backgammon, the minimum roll is 3 and the maximum roll is 24. Thus we have
|P(x, 1) =||
P(x, k + 1) may be computed using the recursion
|P(x, k + 1) =||
|f(y) · P(x − y, k).|
Table 1 was computed using the above method. It gives the probability (in percent) that a player will win when it is his roll and his opponent’s count is at various distances from his own. For example, when A’s count is 70 and he is 10 pips ahead, his probability of winning is 77.9%.
Table 1. Probability of winning the running game if it is your turn to roll.
Keeler and Spencer (Keeler et al. 1975) computed a similar table by using a method which involved rolling the dice 1,000 times to approximate P(x, n) for all values of x from 1 to 125. It is interesting to note that their results differ from Table 1 by no more than 1%.
Table 2. Results of the simulation.
|Player A||Player B||Not doubling||Doubling|
All entries in Table 2 assume that A rolls first. The first entry of each pair gives A’s expectation assuming he holds onto the cube for at least one roll. The second entry gives A’s expectation assuming he doubles immediately. The term “double” in the leftmost column means that the cube is initially in the middle. Thus the first set of entries in the table indicate that when A’s count is 20 and B’s count is 20, A’s expectation by holding onto the cube is 0.611 whereas his expectation by doubling before rolling is 0.685.
Table 3. Optimal doubling strategy in running games.
|Minimum take for opponent||+2||+5½||+10||+14½|
Table 3 is based on Table 2 and gives the optimal strategy for accepting, doubling, and redoubling. The first column indicates that when your count is 20, you should redouble if your opponent’s count is 19½ or more, and double if his count is 18½ or more. Essentially, this means that you should not redouble when his count is 19 unless there is a weakness in his position (a gap or men on the 1 and 2 points, etc.) Your opponent should drop when his count is 23 or higher and take otherwise.
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