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$4$ cards are drawn from a pack without replacement. What is the probability of getting all $4$ from different suits?

Here's how I tried to solve:

For the first draw, we have $52$ cards, and we have to pick one suit. So, probability for this is $\frac{13}{52}$.

For the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\frac{13}{51}$.

Similarly, the third and fourth draw have probabilities $\frac{13}{50}$ and $\frac{13}{49}$ respectively.

Since the draws are independent, the total probability becomes $$\frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} \times \frac{13}{49}$$

But my book says the answer is $\frac{{13\choose 1} \times {13 \choose 1} \times {13\choose1} \times {13\choose1}}{52 \choose 4}$.

My answer differs by a factor of $4!$. What did I do wrong?

Lhachimi
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Pratyush Yadav
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2 Answers2

39

The following uses a slightly different idea that is close to yours.

It does not matter what the first card is. Whatever it is, the probability the next is of a different suit is $\frac{39}{51}$.

Given the first two cards were of different suits, the probability the next draw is of a new suit is $\frac{26}{50}$. And given the first three were of different suits, the probability the fourth is of a new suit is $\frac{13}{49}$.

Thus our probability is $\frac{39}{51}\cdot\frac{26}{50}\cdot \frac{13}{49}$. If you like symmetry, and who doesn't, you may want to put a $\frac{52}{52}$ in front of the expression.

Remarks: $1.$ You calculated the probability that we get the suits in a specific order, say $\heartsuit,\spadesuit,\diamondsuit,\clubsuit$. But there are $4!$ orders in which the suits could come. That accounts for your being off by a factor of $4!$.

$2.$ The book solution is based on a somewhat different idea. Just look at the final hand we end up with, not the order in which we got the cards. There are $\binom{52}{4}$ equally likely hands. Now we count the favourables. The number of hands with exactly one spade, one diamond, one heart, and one club is $13^4$. For the spade can be chosen in $13$ ways, and for each choice the heart can be chosen in $13$ ways, and so on.

André Nicolas
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  • Can anyone explain why it is only $\frac{39}{51}.\frac{26}{50}.\frac{13}{49}$, but not multiplied by $4!$?...please – Ankita Pal Oct 09 '20 at 01:34
14

You have to multiply your answer with $4!$, because there are $4!$ ways in which you can choose the order of the $4$ suits.

Stan
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