
#1
Feb1313, 06:10 PM

P: 43

Let's say that I'm playing a card game with 3 friends. The game involves a full deck of regular playing cards (52 cards, 4 suits, etc.) For this game, each player is dealt 5 cards. The cards are dealt out 5 cards at a time and in a clockwise direction starting with the person to the left of the dealer (I don't think the manner in which the cards are dealt will be relevant).
Now my question is this; how can I find the probability that each other player, besides myself, has been dealt at least ONE heart. And then, how could I find the probability that each player, besides myself, has been dealt ONLY ONE heart. Thanks for the help. 



#2
Feb1813, 04:04 PM

Sci Advisor
PF Gold
P: 1,108

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 [itex]\times[/itex] 39C4 = 1 069 263 For the second player: 12C1 [itex]\times[/itex] 35C4 = 628 320 For the third player: 11C1 [itex]\times[/itex] 31C4 = 346 115 So the probability is 2 033 698/4 481 381 406 320 ≈ 4.5e7, 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. 



#3
Feb1813, 04:35 PM

P: 325

Or are you asking this question without having looked at our hand? 


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