Max amount of different bridge distributions?

[raul_l]

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?

 

[Tide]

Can you clarify that? The probability that all players will receive the same cards is 0.

 

[raul_l]

What I meant was, that all players would get the same distribution of cards, that they have already had in a previous game.

 

[Tide]

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.

 

[raul_l]

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.

 

[Tide]

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!