# Max amount of different bridge distributions?

1. May 11, 2006

### 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?

2. May 11, 2006

### Tide

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

3. May 11, 2006

### 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.

4. May 11, 2006

### 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.

5. May 12, 2006

### 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.

6. May 12, 2006

### Tide

I like yours better!!