# Thread: Bridge Odds: How to calculate distribution of aces?

1. ## Bridge Odds: How to calculate distribution of aces?

Total number of ways to distribute Aces is: ${7 \choose 4}=35$

4 aces in one player's hands:
(4,0,0,0) = 4 ways

3 aces in 1 player's hands:
(3,1,0,0) = 12 ways

2 aces in 1 player's hands:
(2,2,0,0) = 6 ways
or
(2,1,1,0) = 12 ways

1 aces in each player's hand:
(1,1,1,1) = 1 way

Total is 35 ways.

But can anybody tell me how to calculate the probability of each scenario happening? Say, (2,1,1,0).

Thanks

2. Hello, chopet!

We need some serious clarification . . .

You said "bridge hands", so I assume 13 cards are dealt to each of four players.

Next, are the four players distinguishable? .Can we call them A, B, C, and D?

If so, there are: . ${\color{blue}{52\choose13,13,13,13}}$ possible ways to deal the cards.

And the problem is far more complicated than you think . . .

Four Aces in one player's hand

There are 4 choices of players to get the four Aces.
There is 1 way for him get the four Aces.
For his other 9 cards, there are: ${\color{red}{48\choose9}}$ choices.
Then the remaining cards are given to the other three players in: . ${\color{red}{39\choose13,13,13}}$ ways.

Hence, there are: . ${\color{blue}4\cdot{48\choose9}\cdot{39\choose13,1 3,13}}$ ways for one player to get the four Aces.

Three Aces in one player's hand, one Ace in another's

There are 4 choices of players to get the three Aces.
There are: ${\color{red}{4\choose3}}$ ways for him to get three Aces.
For his other 10 cards, there are: ${\color{red}{48\choose10}}$ ways.

There are 3 choices of players to get the one Ace.
There is one way for him to get the one remaining Ace.
For his other 12 cards, there are: ${\color{red}{38\choose12}}$ ways.

The remaining 26 cards are distributed to the other two players.
There are: ${\color{red}{26\choose13,13}}$ ways.

Hence, there are: . ${\color{blue}4\cdot{4\choose3}\cdot{48\choose10}\c dot3\cdot{38\choose12}\cdot{26\choose13,13}}$ ways

Get the idea?

Edit: corrected an error . . . sorry for the blunder.
.

3. Originally Posted by Soroban

Hence, there are: . ${\color{blue}4\cdot{4\choose3}\cdot{48\choose10}\c dot3\cdot{48\choose12}\cdot{26\choose13,13}}$ ways
Hi, thanks!
but for the guy with only ace. should it be:
. ${\color{blue}4\cdot{4\choose3}\cdot{48\choose10}\c dot3\cdot\color{red}{39\choose12}\cdot\color{blue} {26\choose13,13}}$

4. Hello again, chopet!

Thank you for pointing out my error . . .

After the first player gets his 13 cards (including 3 Aces),
. . there are 3 choices for the next player.

There is one way for him to get the remaining Ace.
His other 12 cards come from the remaining 38 cards.
So there are: ${\color{red}{38\choose12}}$ ways.

There are: . ${\color{blue}3\cdot{38\choose12}}$ ways for the second player.

Three Aces in one player's hand, one Ace in another's

. . ${\color{blue}4{4\choose3}{39\choose12}\cdot 3{38\choose12}\cdot{26\choose13,13}}$

5. Hi,

I was looking at this problem, and just want to clarify one more point. Say for the case (2,1,1,0), where one player has 2 aces, and 2 players have 1 ace each:

we have 4 ways to choose Player 1, and ${4 \choose2}=6$ ways to choose his 2 aces, followed by ${3 \choose2}=3$ ways to choose Player 2 and 3, and ${2 \choose1}=2$ way to choose their 1 ace:

hence giving us:
$4*6*3*2=144$

Is that what you mean?

6. ## bridge hands

here is my crack at this. there are a total of (52/13) hands equal to 52 factorial divided by the product 0f 13 factorial and 39 factorial.
the hand containing 4 aces is filled from the remaining 48 cards taken 9 at atime.
by eliminating the factorials the equation becomes

13x12x11x10 dividedby 52x51x50x49 equals .0026 the probability
0f a designated player getting the 4 aces . probability of one of the four getting them is 4times greater or ,01. four aces would appear once in every 100 deals