Suppose that a deck of 52 cards containing four aces is shuffled thoroughly and the cards are then distributed among four players so that each player receives 13 cards. Determine the probability that each player will receive one ace.
The answer to this is given as$$\frac{13^4}{\binom {52}4}$$
My doubt is the following:
- The book justifies ${\binom {52}{4}}$ as number of possible different combinations of the four positions in the deck occupied by 4 aces. That sounds like a case of arrangements to me, so shouldn't we think about permutations and not combinations if we are concerned about how the aces are to be arranged in the deck ?.
- Shouldn't the denominator be ${\binom {52}{13}}$ since you are choosing 13 cards for 4 people.
3 Answers
$\begingroup$There are $$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$$ ways to distribute $13$ cards to each of four people.
There are $4!$ ways to distribute the aces so that each person receives one and $$\binom{48}{12}\binom{36}{12}\binom{24}{12}\binom{12}{12}$$ ways to distribute the remaining cards so that each person receives twelve of them. Hence, the desired probability is \begin{align*} \frac{4!\dbinom{48}{12}\dbinom{36}{12}\dbinom{24}{12}\dbinom{12}{12}}{\dbinom{52}{13}\dbinom{39}{13}\dbinom{26}{13}\dbinom{13}{13}} & = \frac{4! \cdot \dfrac{48!}{12!36!} \cdot \dfrac{36!}{12!24!} \cdot \dfrac{24!}{12!12!} \cdot \dfrac{12!}{12!0!}}{\dfrac{52!}{13!39!} \cdot \dfrac{39!}{13!26!} \cdot \dfrac{26!}{13!13!} \cdot \dfrac{13!}{13!0!}}\\[2mm] & = \frac{4! \cdot \dfrac{48!}{12!12!12!12!}}{\dfrac{52!}{13!13!13!13!}}\\[2mm] & = \frac{4!48!}{12!12!12!12!} \cdot \frac{13!13!13!13!}{52!}\\[2mm] & = \frac{4!48!13^4}{52!}\\[2mm] & = \frac{13^4}{\dfrac{52!}{4!48!}}\\[2mm] & = \frac{13^4}{\dbinom{52}{4}} \end{align*}
Let's compare this solution with the approach of your author. As you stated, there are $\binom{52}{4}$ ways to choose the four positions occupied by the aces in the deck. Since each person receives $13$ cards, there are $13$ possible places for the position of the ace in each person's hand. Hence, the desired probability is $$\frac{13^4}{\dbinom{52}{4}}$$
$\endgroup$ 2 $\begingroup$All we really care about is the placement for the aces; not where the other cards may be in the deck, nor even the suits of the aces.
So let us take 52 blank cards and 4 stickers with 'ace' written on them. To 'deal the cards', place the blank cards in for lines of 13, then unbiasedly select four from them and put a sticker on each.
How may ways are there to stick the aces on 4 different cards in the deck of 52? That is a selection of 4 from 52. $$\binom {52}4$$
How many ways are there to do this so that each ace is stuck on one card in each from the four lines of 13? That is a selection of 1 from the first line, 1 from the next line, and so forth. $$\binom{13}1\binom{13}1\binom{13}1\binom{13}1$$
Divide and calculate.$$\dfrac{13^4}{\dbinom{52}{4}}$$
$\endgroup$ $\begingroup$Informal analysis: we can re-imagine the process of the deal as one in which there are 52 slots, 13 assigned to each hand and the cards are distributed from the top unifomly to the remaining slots. We put the four aces on the top of the deck. The first ace can be dealt to any of the four players. The second ace can be dealt to any of the remaining 39 slots. The third ace can be dealt to any of the remaining 26 slots. The last ace can be dealt to 13 slots. So odds of this happening are:\begin{equation*} {52 \over 52} \times {39 \over 51} \times {26 \over 50} \times {13 \over 49} \approx 10.5\% \end{equation*}Formal approach: first, we have to identify our probability space. Using the multiinomial coefficient we can count the total number of atomic events as $|\Omega| = {52 \choose 13,13,13,13}$. We consider how remaining cards are dealts, 12 to each of the remaining players, and each player has one ace. So $|N_{\text{one ace each}}| = 4! \cdot {48 \choose 12, 12, 12, 12}$. So we can compute:\begin{align} \Pr(N_{\text{one ace each}}) &= \frac{|N_{\text{one ace each}}|}{|\Omega|} \\ &= {4! \cdot {48 \choose 12,12,12,12} \over {52 \choose 13,13,13,13}}\\ &= {{52 \times 39 \times 26 \times 13} \over {52 \times 51 \times 50 \times 49}}\\ &\approx 10.5\% \end{align}
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