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A poker hand consists of 5 cards randomly dealt from a standard deck of cards without replacement. What is the probability that you're dealt a hand that contains exactly one pair of red Queens with all other cards being black and distinct (so the pair of Queens is the only pair in the hand)?

My understanding is that there are two options for the first card (the first red queen), one option for the second card (the second red queen), 24 options for the third card (the black cards minus the queens), 22 options for the fourth card (now also minus the third card and the card that would be its pair), and 20 options for the fifth card (now also minus the fourth card and the card that would be its pair).

I’m new to combinatorics and don’t know if this assumption is correct or exactly what to do with it to find the probability.

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    $\begingroup$ Well, there is only the one pair of red queens so forget them. Now you just have to choose three ranks from the $12$ possible ranks (assuming you are excluding the black queens) and then choose spade or clubs for each of those. $\endgroup$ – lulu Apr 20 at 15:24
  • $\begingroup$ Should have said: your calculation is close but not correct, because you fail to account for the fact that permuting the order of your selection doesn't change the hand. Thus, fo example, you get $2$ ways to choose the pair of red queens when, of course, there's really just $1$. As there are exactly $6$ ways to permute the selection of black cards, your answer is off by a multiplicative factor of $2\times 6=12$. $\endgroup$ – lulu Apr 20 at 15:35
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For typesetting, please use MathJax. $\endgroup$ – dantopa Apr 20 at 15:37
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Your approach is promising, but it needs some rethinking. You are going to evaluate the probability by dividing "number of successful hands" over "total number of possible hands". But then you must decide if you are counting hands with order or not. That is, whether you want to count the hands "2,3,4,5,6" and "6,5,4,3,2" of diamonds as a single one or two. It's ok to do it in either way (the difference will cancel out in the division) but you must take a conscious decision and stick to it.

Your counting mixes both strategies. If you want to count without order, then you should reason thus:

To count the ways of extracting the two red queens: there is a single such pair, hence the count is $1$. For the rest, we need to extract three cards, excluding all queens and red cards. Then, we'd have $\binom{25}{3}$ alternatives... except for the restriction that the ranks must be distinct. Then, scrap that, let's think it in this other way: we have $12$ different ranks (queen is forbidden) from which we want to extract $3$: this gives $\binom{12}{3}$ combinations. And each one of these cards can be of two possible (black) suits. Hence the total number is $1 \times \binom{12}{3} \times 2^3$. This, we'll divide by the total number of hands (without order!), hence $p=\binom{12}{3} 2^3/\binom{52}{5}=\frac{22}{32487}=6.772 \cdot 10^{-4}$

With order: the queen of diamonds can go into 5 positions and the other red queen in 4 positions. For the rest we have: in the first free position we can place $24$ cards, in the next $22$ , in the last $20$. Hence the number is $5 \times 4 \times 24 \times 22\times \ 20=211200$, and the probability is $p=211200/( 52 \times 51 \times 50 \times 49 \times 48)=\frac{22}{32487}$

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