Each of these six hypotheses is possible and they have respectively for probabilities:
p1, p2, p3, p4, p5, p6.
The sum of these six numbers equals 1; but this is all we know of them; these six probabilities depend naturally upon the habits of the player which we do not know.
At the second shuffle and the following, this will recommence, and under the same conditions; I mean that p4 for example represents always the probability that the three cards which occupied after the nth shuffle and before the n + 1th the places 123, occupy the places 321 after the n + 1th shuffle. And this remains true whatever be the number n, since the habits of the player and his way of shuffling remain the same.
But if the number of shuffles is very great, the cards which before the first shuffle occupied the places 123 may, after the last shuffle, occupy the places
123, 231, 312, 321, 132, 213
and the probability of these six hypotheses will be sensibly the same and equal to 1/6; and this will be true whatever be the numbers p1 ... p6 which we do not know. The great number of shuffles, that is to say the complexity of the causes, has produced uniformity.
This would apply without change if there were more than three cards, but even with three cards the demonstration would be complicated; let it suffice to give it for only two cards. Then we have only two possibilities 12, 21 with the probabilities p1 and p2 = 1 − p1.
Suppose n shuffles and suppose I win one franc if the cards are finally in the initial order and lose one if they are finally inverted. Then, my mathematical expectation will be (p1 − p2)n.
The difference p1 − p2 is certainly less than 1; so that if n is very great my expectation will be zero; we need not learn p1 and p2 to be aware that the game is equitable.