I am not understanding this problem:

In a deck of 52 cards, of 13 ranks, and 4 suits, how many different 5 card hand can we get such that, there is always exactly one pair.

There is a similar example here . But I don't understand the logic behind it.

My thoughts are:

assume we have the hand ${a_{1},a_{2},b,c,d}$ where $a_{1}$ and $a_{2}$ are two cards of the same rank and $b,c,d$ are distinct cards from each other and $a$.

So my logic:

there are $\binom{52}{1}$ ways to pick $a_{1}$ and $ \binom{3}{1}$ ways to pick $a_{2}$.

and then $ \binom{50}{1}$ for $b$, $\binom{49}{1}$ for $c$, and $ \binom{48}{1}$ for $d$

which gives us $\binom{52}{1}\binom{3}{1}\binom{50}{1}\binom{49}{1}\binom{48}{1}$ = $18345600$ ways to get a pair...

...which is the wrong answer.

Where have I gone wrong?

  • $\begingroup$ Are you sure that you aren't getting 3 of a kind or 2 pair with that hand? $\endgroup$ – JB King Oct 28 '14 at 3:16
  • $\begingroup$ Related: a neat probability question: a player announces either "I have an Ace" or "I have the Ace of Spades" . In which situation is he more likely to have a pair of aces? (originally seen in Marylin vos Savant's column) $\endgroup$ – Carl Witthoft Oct 28 '14 at 12:35

There are $\binom{13}{1}$ ways to pick the kind we have a pair in. For each such way, we have $\binom{4}{2}$ ways to pick the actual cards.

For each of these ways, there are $\binom{12}{3}$ ways to pick the kinds we will have one each of. For each of these kinds, there are $\binom{4}{1}$ ways to pick the actual cards, for a total of $\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3$ ways.

Remark: One of common problems in counting is inadvertent double (or multiple) counting. For example, your $\binom{52}{1}\binom{3}{1}$ includes the Ace of $\spadesuit$ together with the Ace of $\heartsuit$. But it also counts the Ace of $\heartsuit$ together with the Ace of $\spadesuit$. These give the same one-pair hand.

  • $\begingroup$ okay, that makes more sense, thank you for clarifying. $\endgroup$ – JLL Oct 28 '14 at 3:22
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    $\begingroup$ You are welcome. Counting is a bit tricky at first, but after the first few throroughly analyzed examples it gets much easier. $\endgroup$ – André Nicolas Oct 28 '14 at 3:26
  • $\begingroup$ Fortunately, there is no need to account for flush or straight once we have a pair. $\endgroup$ – Hagen von Eitzen Oct 28 '14 at 14:54
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    $\begingroup$ @AndréNicolas why last item has power of 3? $\endgroup$ – zeleniy Mar 14 '17 at 10:44

Hint: When you pick $b,c,d$ you need to make sure they don't match the pair, so there are less than $50$ choices for $b$.

  • $\begingroup$ Ahh yes touche, forgot about that. Is my first assumption C(52,1) wrong? $\endgroup$ – JLL Oct 28 '14 at 3:20
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    $\begingroup$ As stated, no, but your whole calculation assumes that the pair are the first two cards you draw. You need to multiply by $5 \choose 2$ to select the two cards that are the pair. So your approach would be $52$ (choose the first card of the pair) times $3$ (choose the second card of the pair) times 48 (choose the third card-can't match the pair) times 44 (can't match the pair or the first odd card) times 40 times $5!/2$ ways of ordering the cards (the divide by 2 because reversing the pair doesn't change anything). $\endgroup$ – Ross Millikan Oct 28 '14 at 3:47

there are $52\choose 1$ ways to pick $a_1$ and $3\choose 1$ ways to pick $a_2$ .

So far so good, but you have over counted.   For example: you have counted picking a diamond in for $a_1$ and a heart for $a_2$ as well as picking a heart for $a_1$ and a diamond for $a_2$, but both are the same selection.   So, as order does not matter, you need to divide the count by the $2!$ ways to rearrange those two cards.

Alternatively I'd suggest counting ${13\choose 1}{4\choose 2}$ ways to pick the one face and two suits of the paired cards.

and then $50\choose 1$ for b , $49\choose 1$ for c , and $48\choose 1$ for d

You have to avoid picking cards that have the same face as a, or each other. You want only the one pair. So by your way you need to count $48\choose 1$ ways for b, $44\choose 1$ for c, and then $40 \choose 1$ for d, and likewise deal with overcounting the arrangement of three singletons by dividing by $3!$.

I'd count the way to pick three different faces and a suit for each as: ${12\choose 3}{4\choose 1}^3$

So your result should be $$\frac 1 {2! 3!}{52\choose 1}{3\choose 1}{48\choose 1}{44\choose 1}{40\choose 1} = {13\choose 1}{4\choose 2}{12\choose 3}{4\choose 1}^3 $$

  • $\begingroup$ There are two ways to choose the paired cards but there are also 3! ways to choose the other three cards, so you must also divide by 3!: $$(\frac{1}{3!}) (\frac{1}{2!}) {52 \choose 1} {3 \choose 1} {48 \choose 1} {44 \choose 1} {40 \choose 1} $$ $\endgroup$ – Michael Dec 8 '19 at 18:47
  • $\begingroup$ Ah, yes, of course. Thank you. @Michael. $\endgroup$ – Graham Kemp Dec 9 '19 at 2:36

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