Max amount of different bridge distributions?

In summary, the conversation discusses the probability of all players in a game of bridge receiving the same distribution of cards that they had in a previous game. One suggestion is to use the formula \frac{C(52, 13)^4}{52!} if there are four players, while another suggestion is \frac{C^{13}_{52}C^{13}_{39}C^{13}_{26}C^{13}_{13}}{52!}. Both formulas take into account the reserved set of 13 cards for each player and the distribution of 52 cards into 52 positions. However, the latter formula has a higher probability for the fourth player to receive the correct cards after the first three players have already received them
  • #1
raul_l
105
0
I was wondering about this when playing bridge with my friends. I understand, that one player can be dealt [tex] C^{13}_{52} [/tex] different distributions of the cards. But how to calculate the probability, that all players will get the same cards?
 
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  • #2
Can you clarify that? The probability that all players will receive the same cards is 0.
 
  • #3
What I meant was, that all players would get the same distribution of cards, that they have already had in a previous game.
 
  • #4
I don't play bridge so I don't know some details but I think it would be

[tex]\frac {C(52, 13)^4}{52!}[/tex]

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
Interesting.
But how about [tex] \frac{C^{13}_{52}C^{13}_{39}C^{13}_{26}C^{13}_{13}}{52!} [/tex] ? 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
raul_l said:
Interesting.
But how about [tex] \frac{C^{13}_{52}C^{13}_{39}C^{13}_{26}C^{13}_{13}}{52!} [/tex] ? 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!
 

1. What is the maximum amount of different bridge distributions?

The maximum amount of different bridge distributions is 2,598,960. This is known as the number of possible hands in bridge, also referred to as the "bridge factor".

2. How is the maximum amount of different bridge distributions calculated?

The maximum amount of different bridge distributions is calculated using the mathematical formula nCr = n! / (r! * (n-r)!), where n is the total number of cards in a deck (52) and r is the number of cards in a hand (13).

3. Why is the maximum amount of different bridge distributions important?

The maximum amount of different bridge distributions is important because it represents the complexity and variety of possible hands in bridge. It also plays a role in probability calculations and strategy decisions during gameplay.

4. Is it possible to encounter all 2,598,960 different bridge distributions in a game?

No, it is highly unlikely to encounter all 2,598,960 different bridge distributions in a game. The number of possible hands far exceeds the number of games played in a lifetime.

5. Can the maximum amount of different bridge distributions be surpassed?

No, the maximum amount of different bridge distributions cannot be surpassed as it is the total number of possible hands in bridge. However, variations of the game such as duplicate bridge can increase the number of possible hands within a specific round of play.

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