I'll try an alternative counting method and see if I get the same answer as one of these other folks. Their two answers agree to about 7 significant figures, 0.00000012583437773342733107186438969809670960643871... vs 0.00000012583439386511151359617331968028508948288906.... Anyway, here's my approach. First of all, you could have only one suit: the probability of that happening is 4 possible hands out of (52 C 13). If that's the case, then the probability that partner has only two of the other three suits is, let's see, three ways to choose which two suits, times (26 C 13) ways of choosing the cards, out of (39 C 13) possible hands. But that overcounts the ways where partner has only one suit, counting each twice (say, if you have only clubs and partner has only spades, that's counted as one of the ways with hearts and spades and as one of the ways with diamonds and spades). So subtract the ways that partner has only one suit, 3 ways. On the other hand, you might have two suits: (4 C 2) ways to choose which two suits, times (26 C 13) ways to choose your cards, minus the two ways where you only have one suit, out of (52 C 13). Then partner has to have the other two suits, (26 C 13) out of (39 C 13) being that probability. So my final calculation looks like 4 / (52 C 13) * (3 * (26 C 13) - 3)/(39 C 13) + (4 C 2) * ((26 C 13) - 2) / (52 C 13) * (26 C 13) / (39 C 13) 54086240179999/429820857815004205200 is then my answer. --Joshua Zucker