What are the odds of getting a full house in 7 card poker if your first 3 cards are queens?

Answer 1

If our first three cards are queens, our question really is, what is the probability that i get a pair given four cards out of 49? Since we already have three queens, we don't want to consider the case where we can get a pair of queens, being as this is clearly impossible. There are four of each card in every suit, so the number of ways we can get any pair is 4C2*12*4C2, or 432. The total number of ways we can get any two cards is 49C2, or 1176. Therefore, our probability that we get a pair in any two places of those four cards is 432/1176, or 18/49. However, we aren't done yet because we have to make sure that we don't get another queen or two more of the same previous two cards that we drew on the remaining two cards, because either case would result in four of a kind. Therefore, we need to multiply 18/49 by the odds that neither of the two previous instances occur. The probability that they don't occur is 1-(probability that they do occur). So, the number of ways to get the queen is 2C1*1 or 2, and the number of ways to draw the same two cards is 2C2, or 1. Our total number of ways to draw 2 cards from 47 is 47C2, or 1081. So, the odds that we get either the other queen or the other two cards is 3/1081. To get the probability that our cards do not result in either of those two instances is 1-3/1081, or 1078/1081. So, our final probability is 18/49*1078/1081, or 396/1081, giving you an approximate 36.6 percent chance of getting a full house.