Max amount of different bridge distributions?

AI Thread Summary
The discussion centers on calculating the probability of all players in bridge receiving the same distribution of cards as in a previous game. One participant suggests using the formula C(52, 13)^4/52! for four players, while another proposes a different approach with C(13, 52)C(13, 39)C(13, 26)C(13, 13)/52!. The latter highlights that once the first player receives their specific cards, the chances for the remaining players increase, ultimately suggesting that the fourth player's probability becomes 1 after the first three have received theirs. The conversation emphasizes the complexity of calculating card distributions and probabilities in bridge. Overall, the participants find the topic interesting and engage in exploring different mathematical perspectives.
raul_l
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I was wondering about this when playing bridge with my friends. I understand, that one player can be dealt C^{13}_{52} different distributions of the cards. But how to calculate the probability, that all players will get the same cards?
 
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Can you clarify that? The probability that all players will receive the same cards is 0.
 
What I meant was, that all players would get the same distribution of cards, that they have already had in a previous game.
 
I don't play bridge so I don't know some details but I think it would be

\frac {C(52, 13)^4}{52!}

if there are four players. You are distributing 52 cards into 52 positions of which you reserve a specific set of 13 cards to each of 4 groups (players) in the 52 positions.
 
Interesting.
But how about \frac{C^{13}_{52}C^{13}_{39}C^{13}_{26}C^{13}_{13}}{52!} ? Because after the first player has got the right cards (of a previously played game), the other players have a bigger chance of getting their cards. And after three players have been distributed the right cards, the probability of the fourth player getting the right cards should be 1.
 
raul_l said:
Interesting.
But how about \frac{C^{13}_{52}C^{13}_{39}C^{13}_{26}C^{13}_{13}}{52!} ? Because after the first player has got the right cards (of a previously played game), the other players have a bigger chance of getting their cards. And after three players have been distributed the right cards, the probability of the fourth player getting the right cards should be 1.

I like yours better!
 
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