by David Levy
updated August 10, 1998
The search for literature about historical games leads to some fascinating side trips. The theory of mathematical probability arose almost entirely as a result of early interest in games of chance. So the early works of Pascal, Fermat, De Moivre and others have much interesting material on early games.
One early work on probability has a small section on trictrac, consisting of three problems. The first and third problem are trivial; the second caused me to consider some fascinating issues about preserving a grand jan.
Pierre Remond de Montmort lived from 1678 to 1719. His work on probability was originally published in 1708 and a second greatly expanded edition appeared five years later:
[Montmort] Essay D'Analyse sur Les Jeux de Hazard. Seconde Edition. Revue & augmentée de plusieurs Lettres. Paris: Jacque Quillau. 1713. xlii+414pp.
The book poses problems in probability and combinations about the games of the day and attempts to develop mathematical solutions. In solving the problems, Montmort extended the knowledge of probability beyond what had been published earlier by Pascal and Fermat in the mid-17th century. For a full account of the historical importance of the work, see Todhunter's 1865 History of the Theory of Probability (or the 1949 reprint). While I will focus on the trictrac problems, you will get a flavor of the style of the early works on probabilities.
In Section 148 (page 189), Montmort notes that to play trictrac well, it would be useful to determine quickly the chances of hitting a checker or filling a jan. He gives an example where Black is trying to fill the petit jan. Recall that filling the petit jan is worth four points each way for non-doublets and six points each way for doublets
Montmort comments that there are few players who can determine at a glance that there is one roll which wins twelve points, ten which win eight, three which win six, sixteen which win four and six which fail to fill the petit jan. But these details which are overlooked by ordinary players are nonetheless important for good play.
The table below supplies the details (which, thanks to Witold Szczeponik, have been reviewed and corrected).
Number
of ThrowsNumber
of PointsThrows 1 12 2-2 10 8 5-4 5-2 4-2 4-1 3-2 3 6 5-5 4-4 1-1 16 4 6-5 6-4 6-2 5-3 5-1 4-3 3-1 2-1 6 0 6-6 6-3 6-1 3-3
What are the chances of making the coin de repos in the first two throws of the game?
Recall that you can only make the coin de repos with two checkers together and that you can make the coin par puissance by having two checkers land on the opponent's empty coin.
Montmort notes that there are three sequences which make the coin.
Thus the chances are 2/36 * 2/36 + 1/36 * 1/36 + 1/36 * 2/36 = 7/1296.
Montmort gives the position at left and asks what are the chances that White can play two throws without breaking the grand jan?Assume that White cannot play to the other side of the board (See the Lois de Retour). So if White rolls 6-6, the only play is W6 to W12. The other six cannot be played.
Before reviewing the calculations, let us look at some of the subtleties of checker play. How would you play each of the following rolls as white in this position?
Before you move, you have only one way to play a six -- from W6 to W12. By using 4-1 to move the checker from W6, if your next roll contains any six you can move only the lower die and will preserve your board.
The alternative W7 to W12 is better only if you roll 4-2 or 3-3 (three chances in 36) on the next roll, but is worse if you roll 6-5, 6-4, 6-3, 6-2 or 6-1 (10 chances in 36). Killing sixes is a common technique to preserve a board in backgammon and applies equally here. Kill the six!
The alternative W6 to W10 and W7 to W8 maintains the jan with 4-2 (two chances), but breaks on 6-5 and 5-1 (four chances). Keeping the spare on the highest possible point will help maintain the board. Preserve a five!
Moving both checkers to W9 loses on 6-5 and 5-1 (four chances), gaining only on 3-3 and 3-2 (three chances). Preserve the five!
W7 to W11 gives up 6-5, 6-4, 6-3 and 6-2 and gains only on 5-2 and 4-2. W6 to W9 and W7 to W8 gives up 6-5 and 5-2 and 5-1 and gains on 4-3 and 3-3.
The alternative W6 to W11 and W7 to W9 gives up 6-4 and 2-2 and does not gain on any rolls. Preserve the four!
The alternative, two checkers to W10 gains on 2-2, but loses on 6-3 and 3-1. Preserve the three!
Preserve the two! The alternative, two checkers to W11 would force you to break with 6-2.
So, now that we know how to play the first roll, we can look at Montmort's calculations. I have added some clarifying detail.
Case
Chances in 36
first roll First Roll
Location of Spares
after first RollChances in 36
second rollSecond Roll
Product
1
2
6-5
W12 W12
1
6-6
2
2
3
6-4 5-5
W11 W12
3
6-6 6-1
9
3
4
6-3 5-4
W10 W12
6
6-6 6-2 6-1
1-124
4
4
6-2 5-3
W9 W12
10
6-6 6-3 6-2
6-1 2-1 1-140
9
1
4-4
W10 W11
8
6-6 6-2 6-1
2-1 1-18
5
2
4-3
W9 W11
12
6-6 6-3 6-2
6-1 3-1 2-1
1-124
6
4
6-1 5-2
W8 W12
15
6-6 6-4 6-3
6-2 6-1 3-1
2-2 2-1 1-160
7
6
6-6 5-1 4-2
3-3W7 W12
12
6-6 6-5 6-4
6-3 6-2 6-1
4-1 3-2 3-1
2-2 2-1 1-1126
8
4
4-1 3-2
W7 W11
23
6-6 6-5 6-4
6-3 6-2 6-1
5-1 4-1 3-2
3-1 2-2 2-1
1-192
10
3
3-1 2-2
W7 W10
27
6-6 6-5 6-4
6-3 6-2 6-1
5-2 5-1 4-2
4-1 3-2 3-1
2-2 2-1 1-181
11
2
2-1
W7 W9
32
breaks on
5-5 5-4 4-464
12
1
1-1
W7 W8
35
breaks on
5-535
total
36
565
So with careful play, White can preserve the grand jan in 565 of 1296 cases.
The backgammon writer Danny Kleinman has noted on many occasions, "It's not how many, but how good or bad." With that thought, let's take another look at the position:
Assume the score stands at 11 trous to 11 and that Black has ten points and White has two toward the final trou. Also assume that Black cannot score on his own throw of the dice.You roll 4-1 and score another four points, making six. Do you still play W6-W11, killing sixes and preserving fives?
Check the scoring rules carefully!
If you do, you will win the partie next term if you roll 6-6, 2-2 or 1-1, scoring six for preserving the board by doublets. However, with any other six, you will score only four points but Black will win two points and the partie for the six you are unable to play (nombre non jouer)!
The play we rejected earlier, W7 to W12 has four match winning doublets -- the three above plus 3-3. It does not immediately lose the match on 6-5, 6-4, 6-3, 6-2 or 6-1 since those rolls can all be played. Thus considering only the immediate match-ending rolls, preserving sixes is better after all.
Montmort discusses the impossibility of examining a situation in Trictrac and determine which player has the best position. The only solvable problems are those in which both of the players have already borne off many of their checkers. He gives the following example:
Pierre, White, is on roll, playing against Paul, Black. What is the value of the Pierre?
I have to say that Montmort has a convoluted way to solve this, which may explain why he ended up with the wrong answer!
Let the stake be A and S be the value of the game to Pierre. Pierre wins A if he does not roll 2-1 or 1-1. Let x be the value to Pierre after rolling 2-1 and y be the value to Pierre after rolling 1-1. Then:
After rolling 2-1, let u be the value to Pierre if Paul does not roll 2-1 or 1-1. Let h be the value if Paul rolls 2-1 and t be the value if Paul rolls 1-1. Then:
1 S = (33A + 2x + y)/36
To find the value of u, note Pierre wins if he does not roll 1-1, so
2 x = (33u + 2h + t)/36
To find the value of h, note that Pierre wins if he does not roll 1-1, or if he rolls 1-1 and so does Paul.
3 u = 35/36*A
To find the value of t, note that Pierre wins if does not roll 1-1, or if he rolls 1-1 and Paul rolls 2-2, 2-1 or 1-1. So:
4 h = (35*36 + (1*1)/(36*36) ) * A
Substituting equations 3, 4 and 5 into 2, both Montmort and I get
5 t = (35*36 + (1*4)/(36*36) ) * A
Recall that y is the value after he rolls 1-1. Again there are three cases, depending on whether Paul rolls 2-1, 1-1 or another throw. Let q be the value after another throw, p after 2-1 and n after 1-1. Then:
6 x = 45466 / 46656 * A
since Pierre wins unless he rolls 2-2, 2-1 or 11.
7 y = (33q + 2p + n)/36
After 1-1, 2-1, Pierre wins if he does not roll 2-2 2-1 or 1-1 or, if he does, if Paul rolls 1-1. Thus
8 q = 32/36 A
After 1-1, 1-1, Pierre wins if he does not roll 2-2, 2-1 or 1-1, or, if he does, if Paul rolls, 2-2, 2-1 or 1-1. Thus:
9 p = (32/36 + (4*1)/(36*36) )* A
Substituting 8,9 and 10 into 7, both Montmort and I get
10 p = (32/36 + (4*4)/(36*36) ) * A
Substituting 6 and 11 into 1, Montmort (or his printer!) gets
11 y = 41496 / 46656 * A
which fails the reasonableness test. Clearly Paul wins more than 15 times in 46656. My math yields
12 S = 46641 / 46656 * A
which I also get by taking a more rational (meaning verifiable) approach to the problem! I also note that carrying the stake throughout the problem is confusing and misleading. The chances of Pierre winning are 46441 / 46656. The value of the position to him is (46441 - 215 ) / 46656 times the stake.
13 S + 46441 / 46656 * A
