What are the odds of 4 of a kind and royal flush in same game?

Assuming that they are the only two hands being dealt, that each player gets five cards each only, and that there is no switching (because otherwise the maths gets too complicated): * The odds of the first player getting a royal flush are four in 311875200, or one in 78 million (1/77968800 to be precise). This comes from the fact that there are 311875200 possible hands for the first player (52 x 51 x 50 x 49 x 48), four of which may be a royal flush. * There are also 36 possible straight flushes (10 per suit, but one of each of those is a royal flush as well). For the second player, this number is reduced to 32 possible straight flushes (because four of those 36 use at least one card that has already been given to the first player). * However there are fewer possible hands for the second player; about 184 million in total. This gives the second player a 32 in 184072680, or one in 5752271.25 (1/5.7 million). * The probability of the first player getting a royal flush and the second player getting a straight flush, in the same deal, then, is 1 in 448497686637000; one in every 448 trillion deals or so. However this does not cover the chances of the first player having the straights and the second getting the royals, so: * The chance of the first player getting a straight flush is 36 out of 311875200, or one in 8.7 million (1/8663200). The chance of the second player getting the royals after that is four in 184072680, or one in 46018170 (1/46 million).* * Multiplying these together, we get the chance of the first getting the straights and the second having the royals; it comes out as about one of every 399 trillion deals or so (actually 1 in 398664610344000). * Adding these probabilities together, we get the chance of one player having a straight flush and one having a royal flush; this number is 847162296981000 in 178800155483325020773128000000 or one in every 211 trillion (1/211057734888000) hands. *See discussion.