on; and in general 12 raised to the power n — 1 for the nth play. The sum of all these powers of 12 is unity less the last of these powers; it is the probability that the game will end at the latest in n plays.
Let us consider again the first problem more difficult which may be solved by probabilities and which Pascal proposed to Fermat to solve. Two players, A and B, of equal skill play together on the conditions that the one who first shall beat the other a given number of times shall win the game and shall take the sum of the stakes at the game; after some throws the players agree to quit without having finished the game: we ask in what manner the sum ought to be divided between them. It is evident that the parts ought to be proportional to the respective probabilities of winning the game. The question is reduced then to the determination of these probabilities. They depend evidently upon the number of points which each player lacks of having attained the given number. Hence the probability of A is a function of the two numbers which we will call indices. If the two players should agree to play one throw more (an agreement which does not change their condition, provided that after this new throw the division is always made proportionally to the new probabilities of winning the game), then either A would win this throw and in that case the number of points which he lacks would be diminished by unity, or the player B would win it and in that case the number of points lacking to this last player would be less by unity. But the probability of each of these cases is 12; the function sought is then equal to one half of this function in which we diminish by unity the first
