The question is to show that the probability that each of the four players in a game of bridge receives one ace is $$ \frac{24 \cdot 48! \cdot13^4}{52!}$$ My explanation so far is that there are $4!$ ways to arrange the 4 aces, $48!$ ways to arrange the other cards, and since each arrangement is equally likely we divide by $52!$. I believe the $13^4$ represents the number of arrangements to distribute 4 aces among 13 cards, but I don't see why we must multiply by this value as well?


Imagine $4$ groups of $13$ spots for the $4$ hands

An ace can be placed in each group in any of $13$ places, hence $13^4$
The aces' suits can be distributed between groups in $4!$ ways, hence $24$
The remaining cards can be placed in 48! ways,
and 52! is the unrestricted ways of placing the cards
thus (putting the terms in the order you have), $Pr = \dfrac{24\cdot48!\cdot13^4}{52!}$


There is a much simpler way to get the same result.

We need an ace in each group 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 spots out of $51,$ and so on

thus $Pr = \dfrac{39}{51}\cdot\dfrac{26}{50}\cdot\dfrac{13}{49}$


There are two approaches to card-game questions like these. In the first approach, we temporarily assume that order of cards within the hand matters. In doing so, we treat our sample space as all ways of arranging the fifty-two cards in a row.

Approaching like this, we have the following steps for multiplication principle:

  • Pick who gets which ace: $4!$ ways
  • Arrange the remaining $48$ cards in a row and give them to the players (12 to north, the next twelve to east, the next twelve to south, etc...): $48!$ ways
  • Pick where in the hand the ace goes for each player: $13^4$ ways

Note that the final step was necessary in order to have what we count cover all possible ways that the 52 cards be dealt such that every player receives at least one ace. If we hadn't multiplied by $13^4$, it would have been as though we only considered the ace being the first card in each players' hand. Since we are considering order important, the hand $A\spadesuit 2\spadesuit 3\spadesuit\dots$ is a different outcome than $2\spadesuit A\spadesuit 3\spadesuit\dots$

Since there are $52!$ number of equiprobable ways to deal the cards (where order matters) the probability is as given:


My preference is instead to work in the situation that order doesn't matter.

Here, we break apart as multiplication principle:

  • Choose which player gets which ace: $4!$ ways
  • Choose twelve additional cards for each player: $\binom{48}{12,12,12,12}=\frac{48!}{12!12!12!12!}$ number of ways

There are a total of $\binom{52}{13,13,13,13}=\frac{52!}{13!13!13!13!}$ number of deals (where order within each hand doesn't matter) for a probability then of:


which after a short amount of manipulation you see is precisely the same answer as before.


There are $4!$ ways to distribute the aces, then $\frac{48!}{12!^4}$ ways to distribute the other $48$ cards, $12$ to a player.

Without worrying about the aces, there are $\frac{52!}{13!^4}$ ways to distribute $52$ cards, $13$ to a player.

Therefore, the probability is $$ \frac{4!\frac{48!}{12!^4}}{\frac{52!}{13!^4}}=\frac{13^4}{\binom{52}{4}}\doteq0.1054982 $$

  • $\begingroup$ Being a bridge player, I am very interested by this answer. My question is : does this result means that if I have one ace and my partner has one ace, the probability that one of the opponents has the two other aces is 90% ? Or, is such a statement totally wrong ? $\endgroup$ – Claude Leibovici Jan 8 '16 at 4:54
  • $\begingroup$ If you have one ace and your partner has one ace, then the other two players have two aces. It should be close to $50\%$ that one has both, but not quite. $\endgroup$ – robjohn Jan 8 '16 at 8:11
  • $\begingroup$ To count the number of ways that one of them does not have both aces, note that there are $2!$ ways to distribute the aces, then $\frac{24!}{12!^2}$ ways to distribute the other $24$ cards, $12$ to a player. Without worrying about the aces, there are $\frac{26!}{13!^2}$ ways to distribute $26$ cards, $13$ to a player. Therefore, the probability is $$\frac{2!\frac{24!}{12!^2}}{\frac{26!}{13!^2}} =\frac{13^2}{\binom{26}{2}} =0.52$$ So the probability that one of them has both aces is $0.48$. $\endgroup$ – robjohn Jan 8 '16 at 8:11
  • $\begingroup$ Thanks for tha answer ! I am real bad with probabilities and I totally misunterpreted the result of $1-0.105$. Shame on me !! Cheers $\endgroup$ – Claude Leibovici Jan 8 '16 at 8:28

Since the order doesn't matter, let's make hands one by one.

  1. For the first, there are 4 Aces and I need to choose 1 to give to this person, $\binom{4}{1}$. Then I need to choose 12 cards from the 48 non-Aces to complete this hand, $\binom{48}{12}$. Thus the number of ways to make this hand is $$\binom{4}{1}\binom{48}{12}.$$

  2. I already gave the first person an Ace, so I have three left and I need to choose one to give to this person. I also used up 12 non Aces on the last person, so there are 36 non Aces left and I need to choose 12. The number of ways to make this hand is thus, $$\binom{3}{1}\binom{36}{12}.$$

  3. and 4. follow the same logic and so I have $$\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}$$ ways to make hands 3 and 4.

Finally, all the possible ways to make 4 13-card hands are $$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13} = \binom{52}{13,13,13,13} = \frac{52!}{13!\,13!\,13!\,13!}.$$

Lastly, putting it all together, the probability of each person getting an Ace is thus \begin{align*}\frac{\binom{4}{1}\binom{48}{12}\binom{3}{1}\binom{36}{12}\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}}{\binom{52}{13,13,13,13}} &= \frac{4!}{1!3!}\cdot\frac{48!}{12!36!}\cdot\frac{3!}{1!2!}\cdot\frac{36!}{12!24!}\cdot\frac{2!}{1!1!}\\ &\qquad\times\frac{24!}{12!12!}\frac{1!}{1!0!}\cdot\frac{12!}{12!0!}\cdot\frac{13!\,13!\,13!\,13!}{52!}\\ &=\frac{24 \cdot 48! \cdot13^4}{52!}. \end{align*}


After the cards have been shuffled and cut, there are $\binom{52}4$ equally likely possibilities for the set of four positions in the deck occupied by the aces. Among those $\binom{52}4$ sets, there are $13^4$ which result in each player getting an ace; namely, make one of the $13$ cards to be dealt to South an ace, and the same fo West, North, and East. So the probability is $$\frac{13^4}{\binom{52}4}=\frac{4!\cdot48!\cdot13^4}{52!}=\frac{2197}{20825}\approx.1055$$


Suppose you are arranging the cards in rows of 13. Each row must have an ace; if we place the aces as the first cards, there are 4! ways to arrange them. Then, there are 48! ways to arrange the remaining 48 cards. However, this restricts the aces to being the first card; they can, in fact, be in any of 13 positions - for each of the 4 players. So, multiply by 13^4 to account for the 4 aces being in any of the 13 positions.


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