Three players of supposed equal ability play together on the following conditions: that one of the first two players who beats his adversary plays the third, and if he beats him the game is finished. If he is beaten, the victor plays against the second until one of the players has defeated consecutively the two others, which ends the game. The probability is demanded that the game will be finished in a certain number _n_ of plays. Let us find the probability that it will end precisely at the _n_th play. For that the player who wins ought to enter the game at the play _n_ - 1 and win it thus at the following play. But if in place of winning the play _n_ - 1 he should be beaten by his adversary who had just beaten the other player, the game would end at this play. Thus the probability that one of the players will enter the game at the play _n_ - 1 and will win it is equal to the probability that the game will end precisely with this play; and as this player ought to win the following play in order that the game may be finished at the _n_th play, the probability of this last case will be only one half of the preceding one. This probability is evidently a function of the number _n_; this function is then equal to the half of the same function when _n_ is diminished by unity. This equality forms one of those equations called _ordinary finite differential equations_.
We may easily determine by its use the probability that the game will end precisely at a certain play. It is evident that the play cannot end sooner than at the second play; and for this it is necessary that that one of the first two players who has beaten his adversary should beat at the second play the third player; the probability that the game will end at this play is ½. Hence by virtue of the preceding equation we conclude that the successive probabilities of the end of the game are ¼ for the third play, ⅛ for the fourth play, and so on; and in general ½ raised to the power _n_ - 1 for the _n_th play. The sum of all these powers of ½ 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 ½; the function sought is then equal to one half of this function in which we diminish by unity the first index plus the half of the same function in which the second variable is diminished by unity. This equality is one of those equations called _equations of partial differentials_.