Two-Pair probability

The question is: What is the probability of drawing a two-pair in a 5 card set from a normal 52-card deck? That is, QQJJ2 or 337AA, for example.

I have my answer, but for some reason it is exactly double the correct answer. I haven't been able to figure out why. Here's my solution:

First you draw an arbitrary card. To pair it, the probability is 3/51. Next, you want to draw a different card; there are 48 such cards out of 50, so the probability is 48/50. You seek its pair, which has probability 3/49. Your final card must be different from both the originals (otherwise we'd have a full house), so it has probability 44/48.

That yields approx. 0.00316. There are precisely 5! / 2! * 2! * 1! orderings of two pair hands = 30. Thus, 30 * 0.00316 = 0.09507. But the correct answer is half that. Any pointers in the right direction would be awesome!

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Originally Posted by arcketer

The question is: What is the probability of drawing a two-pair in a 5 card set from a normal 52-card deck? That is, QQJJ2 or 337AA, for example.

I have my answer, but for some reason it is exactly double the correct answer. I haven't been able to figure out why. Here's my solution:

First you draw an arbitrary card. To pair it, the probability is 3/51. Next, you want to draw a different card; there are 48 such cards out of 50, so the probability is 48/50. You seek its pair, which has probability 3/49. Your final card must be different from both the originals (otherwise we'd have a full house), so it has probability 44/48.

That yields approx. 0.00316. There are precisely 5! / 2! * 2! * 1! orderings of two pair hands = 30. Thus, 30 * 0.00316 = 0.09507. But the correct answer is half that. Any pointers in the right direction would be awesome!

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does that give the correct answer?

If so WHY? If not WHY NOT?

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Originally Posted by Plato

does that give the correct answer?

If so WHY? If not WHY NOT?

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LOL! (Giggle)

Hello, arcketer!

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What is the probability of drawing Two Pair in a poker hand from a 52-card deck?

I have my answer, but for some reason it is exactly double the correct answer.
I haven't been able to figure out why. Here's my solution:

First you draw an arbitrary card.
To pair it, the probability is 3/51.

Next, you want to draw a different card.
There are 48 such cards out of 50, so the probability is 48/50.
You seek its pair, which has probability 3/49.

Your final card must be different from both the originals
(otherwise we'd have a full house), so it has probability 44/48.

That yields approx. 0.00316.
There are precisely 5!/(2!*2!*1!) orderings of two pair hands = 30.
Thus, 30 * 0.00316 = 0.09507.
But the correct answer is half that.

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Your reasoning is correct, but there is a subtle omission.


For the first card, there are 52 choices.. .
Suppose it is a Jack.

The second must be another Jack. .

The third card must not be a Jack: .
Suppose it is a 7.

The fourth card must another 7. .

The fifth card must not be a J or 7. .

All this is correct.


You had ordered the five cards, so to un-order them,
. . you multiplied by: . . . . correct

But this last factor is the number of ordered partitions.

It considers to be different from


And that is where the "times two" comes from.

Edit: NVM!

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Originally Posted by arcketer

Edit: NVM!

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Please explain yourself! Why "never mind"?
That is unfair to anyone reading this thread.
You may not need to post here again if you choose not to answer my question.

I'm not finished with the problem. I had typed out a response to you, but after submission read a more detailed response from Soroban, so I quickly edited away what I wrote. I haven't had a chance to look over his response yet as I just got home, but I will!

Ahh I see my mistake now, Soroban. So the general formula that I was using gives ordered partitions. Is there any such a formula for unordered partitions? It seems simple but I don't want to make a similar mistake on a similar problem later. Thank you for the help!