Two players play a game of three points ie they play an even

Two players play a game of three points [i.e. they play an even game repeatedly, the first to win three games gets the stake] and each player has staked 32 pistoles. How should the sum be divided if they break off at any stage? [discuss all the possible situations. For extra credit you may go into the more general situation of more than two players, etc., as Pascal and Fermat did in subsequent letters. E.g. discuss, like Pascal did, three players A,B,C, where A needs 1 more point, and both B and C need two more points].

Verify Pascal\'s above calculation. Also calculate the probability of throwing \"Sonnez\" in 24, and in 25 throws

Solution

Tutor\'s note: I am giving the answer from web, http://www.maths.tcd.ie/pub/HistMath/People/Pascal/RouseBall/RB_Pascal.html, and not sure about the correctness.

Suppose that the first player has gained two points, and the second player one point; they have now to play for a point on this condition, that, if the first player gain, he takes all the money which is at stake, namely, 64 pistoles; while, if the second player gain, each player has two points, so that there are on terms of equality, and, if they leave off playing, each ought to take 32 pistoles. Thus if the first player gain, then 64 pistoles belong to him, and if he lose, then 32 pistoles belong to him. If therefore the players do not wish to play this game but to separate without playing it, the first player would say to the second, ``I am certain of 32 pistoles even if I lose this game, and as for the other 32 pistoles perhaps I will have them and perhaps you will have them; the chances are equal. Let us then divide these 32 pistoles equally, and give me also the 32 pistoles of which I am certain.\'\' Thus the first player will have 48 pistoles and the second 16 pistoles.

Next, suppose that the first player has gained two points and the second player none, and that they are about to play for a point; the condition then is that, if the first player gain this point, he secures the game and takes the 64 pistoles, and, if the second player gain this point, then the players will be in the situation already examined, in which the first player is entitled to 48 pistoles and the second to 16 pistoles. Thus if they do not wish to play, the first player would say to the second, ``If I gain the point I gain 64 pistoles; if I lose it, I am entitled to 48 pistoles. Give me then the 48 pistoles of which I am certain, and divide the other 16 equally, since our chances of gaining the point are equal.\'\' Thus the first player will have 56 pistoles and the second player 8 pistoles.

Finally, suppose that the first player has gained one point and the second player none. If they proceed to play for a point, the condition is that, if the first player gain it, the players will be in the situation first examined, in which the first player is entitled to 56 pistoles; if the first player lose the point, each player has then a point, and each is entitled to 32 pistoles. Thus, if they do not wish to play, the first player would say to the second, ``Give me the 32 pistoles of which I am certain, and divide the remainder of the 56 pistoles equally, that is divide 24 pistoles equally.\'\' Thus the first player will have the sum of 32 and 12 pistoles, that is, 44 pistoles, and consequently the second will have 20 pistoles.