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DISCLAIMER: this is a question already on the site, but I am seeking to gain clarity on a different approach to this problem.

Question is the probability of a full house (i.e., $3$ of a kind + $2$ of another kind) when drawing 5 cards at random from a standard deck of $52$ cards.

Solution 1) $$P(Full\ House) = \frac{13 \cdot 4C3 \cdot 12 \cdot 4C2}{52C5}$$

Solution 2) $$P(Full\ House) = \frac{52}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48}$$

Solution 1 provides the correct answer; however, solution 2 is incorrect by a factor of $10$. I believe I might need to introduce $5C2\ or\ 5C3 = 10$ into solution 2 to correct for the discrepancy, but I am unsure as to what the reasoning behind this might be.

What is the reason for the difference between solution 1 versus solution 2?

user946947
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    Possible duplicate: Probability of getting a full house – KingLogic Jul 03 '21 at 23:17
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    You aren't accounting for different ways to pick up the same hand in the second case. The order of the $3$ of a kind and $2$ of a kind doesn't matter – Rushabh Mehta Jul 03 '21 at 23:18
  • @DonThousand If that is the case, then wouldn't I multiply by $5!$ to account for ordering? I feel like I am misinterpreting your statement since ordering should not matter IMO – user946947 Jul 04 '21 at 00:51
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    The only permutations that are different are those where the position of the pair items are different i.e. (1,2),(1,3),....(4,5). There are ten of these. Interchanging pair items in the same position is already accounted for. Similarly for triples. – herb steinberg Jul 04 '21 at 01:59

1 Answers1

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Your instinct is right about why the second answer is correct: it's missing a $_5 C_2$ (or equivalently, a $_5 C_3$). In the comments, you asked an insightful question: why should you multiply by that and not ($5!$)?

Running with the $5!$ instinct for a second, let's consider what it might represent: it would be the number of ways to rearrange all cards. So, for instance, your full house might be $K \diamondsuit, Q \diamondsuit, 5 \diamondsuit, 3 \clubsuit, 7 \clubsuit$. There are indeed $5!$ ways to rearrange those, such as:

$$3 \clubsuit, K \diamondsuit, 5 \diamondsuit, Q \diamondsuit, 7 \clubsuit$$ $$K \diamondsuit, 7 \clubsuit, Q \diamondsuit, 3 \clubsuit, 5 \diamondsuit$$ $$Q \diamondsuit, 5 \diamondsuit, K \diamondsuit, 7 \clubsuit, 3 \clubsuit$$

...but let's take a closer look at that last rearrangement. Notice that it has the same form as the "original" deal -- that is, $\diamondsuit, \diamondsuit, \diamondsuit, \clubsuit, \clubsuit$. That means we had already implicitly counted this case in the calculation $$\frac{52}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48},$$ so we've now overcounted it.

The picture is a little clearer now though, because we can think carefully about exactly what we've overcounted. Specifically, we've overcounted all the $\diamondsuit \diamondsuit \diamondsuit \clubsuit \clubsuit$ configurations. There are $3! \cdot 2!$ of these, corresponding to the $3!$ and $2!$ ways to rearrange the diamonds and clubs, respectively. So, the correct multiplier wasn't $5!$, but $\frac{5!}{3! \cdot 2!}$, or $_5 C_3 =$ $_5 C_2$.

Aaron Montgomery
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