Two-Pair probability

• July 5th 2011, 03:51 PM
arcketer
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!
• July 5th 2011, 04:28 PM
Plato
Re: Two-Pair probability
Quote:

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!

$\dfrac{\binom{13}{2}\binom{4}{2}\binom{4}{2}(44)}{ \binom{52}{5}}$ does that give the correct answer?

If so WHY? If not WHY NOT?
• July 5th 2011, 04:31 PM
Also sprach Zarathustra
Re: Two-Pair probability
Quote:

Originally Posted by Plato
$\binom{13}{2}\binom{4}{2}\binom{4}{2}(44)$ does that give the correct answer?

If so WHY? If not WHY NOT?

LOL! (Giggle)
• July 5th 2011, 04:45 PM
Soroban
Re: Two-Pair probability
Hello, arcketer!

Quote:

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.

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

For the first card, there are 52 choices.. . $P(\text{any}) = \tfrac{52}{52}$
Suppose it is a Jack.

The second must be another Jack. . $P(\text{another Jack}) = \tfrac{3}{51}$

The third card must not be a Jack: . $P(\text{another value}) = \tfrac{48}{50}$
Suppose it is a 7.

The fourth card must another 7. . $P(\text{another 7}) = \tfrac{3}{49}$

The fifth card must not be a J or 7. . $P(\text{other}) = \tfrac{44}{48}$

All this is correct.

You had ordered the five cards, so to un-order them,
. . you multiplied by: ${5\choose2,2,1} = 30$ . . . . correct

But this last factor is the number of ordered partitions.

It considers $\{JJ|77|3\}$ to be different from $\{77|JJ|3\}$

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

• July 5th 2011, 04:59 PM
arcketer
Re: Two-Pair probability
Edit: NVM!
• July 5th 2011, 05:49 PM
Plato
Re: Two-Pair probability
Quote:

Originally Posted by arcketer
Edit: NVM!

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.
• July 5th 2011, 06:10 PM
arcketer
Re: Two-Pair probability
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!
• July 5th 2011, 08:04 PM
arcketer
Re: Two-Pair probability
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!