# Question about probabilities with playing cards

by jldibble
Tags: cards, probability, stats
 Sci Advisor PF Gold P: 1,354 This is not my field, so a more elegant solution may exist. By this procedure should give the correct result. First, the only relevant cards are the 15 dealt to the other players. There are 52C15 = 4 481 381 406 320 possible combinations. The deck itself can be seen as 13 hearts + 39 other cards. The probability that each other player has been dealt exactly one heart is the easiest to calculate. For the first player, this is 13C1 $\times$ 39C4 = 1 069 263 For the second player: 12C1 $\times$ 35C4 = 628 320 For the third player: 11C1 $\times$ 31C4 = 346 115 So the probability is 2 033 698/4 481 381 406 320 ≈ 4.5e-7, or approximately one chance out of 2 203 563. To find the possibility that each player has at least one heart is more involved. You can sum up the probability that one player has 2 hearts and the two others 1 heart + the probability that two players have 2 hearts and the other 1 heart + the probability that one player has 3 hearts and the two others 1 heart, and so on.