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_. We are able to determine by its use the probabilities of A by dividing the smallest numbers, and by observing that the probability or the function which expresses it is equal to unity when the player A does not lack a single point, or when the first index is zero, and that this function becomes zero with the second index. Supposing thus that the player A lacks only one point, we find that his probability is ½, ¾, ⅞, etc., according as B lacks one point, two, three, etc. Generally it is then unity less the power of ½, equal to the number of points which B lacks. We will suppose then that the player A lacks two points and his probability will be found equal to ¼, ½, 11/16, etc., according as B lacks one point, two points, three points, etc. We will suppose again that the player A lacks three points, and so on.