3 years ago
Statistics question about a game of bridge.
There are 4 players, 2 teams of two players each, and a standard 52 card deck, and each player gets 13 cards.
The number of bridge games is equal to (52C13)*(39C13)*(26C13)*(13C13 = 1) according to an answer key I have.
In another question, we had to find the probability of a single player getting all the spades, which is 4*(39C13)*(26C13)*(13C13 = 1) divided by answer from previous question. We multiply by four because we can choose between 4 players for getting all the spades.
Here's my question - why don't we account for there being 4 players in the total number of games possible like we did in the problem about the spades? Combinations inherently don't care about order, but doesn't it create for a different game from the point of view of a specific player if the cards were in different hands? I've confused myself a little bit here, but just want to know why we multiplied by 4 in one of those and not the other.
The number of bridge games is equal to (52C13)*(39C13)*(26C13)*(13C13 = 1) according to an answer key I have.
In another question, we had to find the probability of a single player getting all the spades, which is 4*(39C13)*(26C13)*(13C13 = 1) divided by answer from previous question. We multiply by four because we can choose between 4 players for getting all the spades.
Here's my question - why don't we account for there being 4 players in the total number of games possible like we did in the problem about the spades? Combinations inherently don't care about order, but doesn't it create for a different game from the point of view of a specific player if the cards were in different hands? I've confused myself a little bit here, but just want to know why we multiplied by 4 in one of those and not the other.
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M$1 Answer
In the first case, you *are* taking into account rotations of the hands around the table. Start with player 1 (is that East? Not a bridge player). If you pull 13 cards out of the deck randomly for that player, there are 52C13 possible hands (which are really *all* the possible 13 card hands for a 52 card deck). Then move to player 2 and pull 13 cards from the 39 remaining; that's the 39C13. And proceed for the last 2 players. If you think about it a little, you should see that we've counted every possible game, including "rotations", where the hands are all the same but rotated around the board.
In the second case, you basically pull all the spades out of the deck, leaving a deck of 39. Then hand that hand to player 1, leaving 39C13 for player 2, 26C13 for player 3, and 1 for player 4. But players 2-4 never got the all-spades hand when counting those up, so you have to hand the all-spades hand to the other players in turn, and count the same set of possibilities for the others. Thus the 4.
Does that make sense?
In the second case, you basically pull all the spades out of the deck, leaving a deck of 39. Then hand that hand to player 1, leaving 39C13 for player 2, 26C13 for player 3, and 1 for player 4. But players 2-4 never got the all-spades hand when counting those up, so you have to hand the all-spades hand to the other players in turn, and count the same set of possibilities for the others. Thus the 4.
Does that make sense?
You can leave an optional "tip" with Mahalo's virtual currency, Mahalo Dollars. If you are asking a difficult question that might require some research, or if you'd like a wide variety of feedback, a higher tip often leads to more answers to your question.
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