# Card Game Bridge Probability

I'm trying to self-educated myself and I bought a probability book, which has this interesting question. It says not to look at any resources before you try it, but you may use a calculator.

In the card game bridge, each of $4$ players is dealt a hand of $13$ of the $52$ cards. What is the probability that each player receives exactly one Ace?

I immediately thought that this was long-winded, but then I thought that it could be $\dfrac{1}{13} \times \dfrac{1}{13} \times \dfrac{1}{13} \times \dfrac{1}{13}$, although this in probably not correct. The reason I thought that was because $\dfrac{4}{52} \times \dfrac{3}{39} \times \dfrac{2}{26} \times \dfrac{1}{13}$, bu this is a very simple solution. Any help? Thanks a lot.

• Hint: What makes the problem a little tricky is that each time you place an Ace, you reduce the number of available slots in that hand (making it a little more likely the next Ace will go to a different hand). Think about it this way...the first Ace goes wherever it likes. Then the second Ace can go to any of 51 remaining slots, 39 of which are in "new" hands. And so on.
– lulu
Jul 19 '15 at 15:51

We need an ace in each hand of 13, how the rest of the cards go doesn't matter !

The first ace has to be in some group, each of the other aces have to fall in a different group, so the 2nd ace has 39 permissible slots out of 51, and so on.

Thus Pr = $\dfrac{39}{51}\cdot\dfrac{26}{50}\cdot\dfrac{13}{49} =\dfrac{2197}{20825}$

Place the cards on spots, numbered as $1,2,\dots,52$.

Only distinguish aces from non-aces.

Every $4$ element subset of $\{1,\dots,52\}$ has equal probability to contain the $4$ aces.

Let's say that one player receives the cards on spots $1,\dots,13$, another receives the cards on spots $14,\dots26$, et cetera.

There are $\binom{52}{4}$ ways to place the $4$ aces.

There are $13^4$ ways to place the $4$ aces in such a way that the spots $1,\dots,13$ contain one ace, the spots $14,\dots,26$ contain one ace, the spots $27,\dots,39$ contain one ace and the spots $40,\dots,52$ contain one ace.

This gives probability: $$\frac{13^4}{\binom{52}{4}}$$

• Another approach (that gets the same answer) is to place the aces ($13^4 \times 4!$), then place the remaining cards ($48!$), then divide by the total number of sequences ($52!$). Jul 19 '15 at 16:48
• @trueblueanil: This answer agrees with yours, doesn't it? (Athough yours is easier to follow.) Jul 19 '15 at 17:39
• @TonyK: My answer agrees only with those of David Quinn, and Daniel V in the comments above. Jul 19 '15 at 18:37
• According to my calculations, $\frac{13^4}{\binom{52}{4}}$ is equal to $\dfrac{2197}{20825}$. Jul 19 '15 at 21:11
• Sorry, I guess I must have typed in a wrong number. Yes, all answers agree. Jul 20 '15 at 2:45

It's probably easiest to do this using permutations and combinations. Assuming the hands or players are distinguishable, you can distribute the four aces amongst the players in $4!$ ways.

The first player can have a choice of $\binom{48}{12}$ other cards

Likewise the second player can have a choice of $\binom{36}{12}$ other cards

The third player has a choice of $\binom{24}{12}$ other cards, while the fourth player must automatically have the remaining cards.

Multiply these together to get the total number of hands each containing just one ace.

To get the required probability you need to divide by the unrestricted total number of sets of four hands, which is $$\binom{52}{13}\times\binom{39}{13}\times\binom{26}{13}$$

• fact(4)*comb(48,12)*comb(36,12)*comb(24,12) / (comb(52,13)*comb(39,13)*comb(26,13)) is also 2197/20825 Oct 27 '19 at 23:18