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I was reading Introduction to Probability, 2nd Edition, and the following question appears as exercise $60$ in the first chapter:

A well-shuffled 52-card deck is dealt to $4$ players. Find the probability that each of the players gets an ace.

I have some issues with the answer from the manual, the size of sample space is clearly:

$$\frac{52!}{13!13!13!13!}$$

But then the solution states that:

There are $4!$ different ways of distributing the $4$ aces to the $4$ players.

Why did we consider the order of the aces here? when we use the multinomial for calculating the size of the sample space we ignore the order as far as I know and so we should do when we calculate the number of ways to distribute the $4$ aces.

So is the order between groups is considered in the multinomial? even if the type of the aces is not relevant here (because we care about the existence of an ace not about its type)?

RobPratt
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Kais Hasan
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2 Answers2

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The count $\dfrac{52!}{13!13!13!13!}$ ignores the order of cards within each hand. It doesn't ignore which hand is given to which player.

So if you have a situation where each player has an ace, but the players with the ace of hearts and ace of spades exchange their hands, then you're in a different one among the $\frac{52!}{13!13!13!13!}$ deals -- so you also need to count them as two different instances of "deals where each player gets an ace".

Troposphere
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I like to think of the multinomial coefficient as number of ways of selecting people for labeled groups, Thus if groups A, B, C, D, are to have $4,3,2,1$ people respectively, we get $\dbinom{10}{4}\dbinom63\dbinom32\dbinom11\equiv \dbinom{10!}{4,3,2,1}\equiv \dfrac{10!}{4!3!2!1!}$

For your problem, cards can be thought of as people, thus distributing the aces and non-aces to A,B,C,D

answer will be $\dfrac{\dbinom{4}{1,1,1,1}\dbinom{48}{12,12,12,12}}{\dbinom{52}{13,13,13,13}}$

true blue anil
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