The 2 pair Question

[slimeywater]

2 hands delt out of a deck of cards (52 cards)
5 cards in each hand
if boths hands contain 2 pair
what hand does player A require to on average be beating player Bs hand

[BadPenguin]

I believe there are exactly 1584 ways to make a given two-pair hand of "X's over Y's" (XXYYZ, in other words). If you crank this out for all possible 2-pair hands you get the following numbers for each possible higher pair of the two:

1584 // 33xxy
3168 // 44xxy
4752 // 55xxy
6336 // 66xxy
7920 // 77xxy
9504 // 88xxy
11088 // 99xxy
12672 // TTxxy
14256 // JJxxy
15840 // QQxxy
17424 // KKxxy
19008 // AAxxy

This makes 123,552 total 2-pair hands, so the median hand would be hand 61,776 out of these. Assuming I walked through the numbers correctly I end up with the upper half of 2-pair hands starting with JJ552 and the lower half ending with JJ44A.

Since the original question asked what hand player 1 needs to expect to be beating player 2, I guess the answer would be JJ552, in which case player 1 would be the favorite to have the best hand by a tiny margin.

I think this is a useful thing to know, particularly in 5-card draw where it's not an uncommon situation to have 2-pair after the draw and be fairly sure that your opponent was drawing to 2-pair. Jacks-up sounds like a strong hand, but when you factor in the small chance that your opponent improved to a full house then you're more likely to be behind in this situation than not.

[slimeywater]

hey bad peguin
yes jj552 would be the average 2 pair hand
but if u held a lower jacks up hand
you would have a better two pair on average because you have taken 2 of the jacks away from ur opponents possible hand
i have a feeling he answer is JJ223

[AaronBrown]

Okay, now that BadPenguin has narrowed it down to JJxxy we can reduce the number of possible two-pair hands for your opponent form 123,552 to 75,592. If we assume that y is 10 or lower, 37,332 of the hands are headed by A's, K's or Q's and beat all your J's. 36,369 of the hands are headed by 10's or lower, and lose to all your J's. 1,891 of the hands are headed by J's. We need to win 1,427 of them and lose 464 to make our hand exactly an even bet.

Both hands contain a pair of Jacks, so let's just remove the Jacks from the deck and deal three card hands. Both hands hold a pair that's 10 or lower. If you hold J's and 8's, there are 364 ways to lose to J 10 or J 9. If you hold JJ 88 2, there are another 40 ways for you to lose, for a total of 404.

So JJ 88 2 is a slight favorite (37,853 to 37,736 with 3 ties). However, now it gets complicated. JJ 77 A is a bigger favorite, although a weaker hand, because holding the A reduces the chance of your J's losing. You recall our initial assumption that xxy were all 10 or lower. By the time we get down to JJ 77 10, we are a significant underdog. So I think the best you can do is JJ 88 2.

[BadPenguin]

You're right, I forgot about taking that into account. I'm pretty sure you're answer is correct - any jacks-up will give you a better than average 2-pair hand.

I made a rough calculation that JJ223 should be the better 2-pair hand about 51% of the time or so.

Thanks for the interesting problem!

[Siegmund]

To a first approximation, the posters so far are right that jacks up is the middle of the range.

However, so far everyone has oversimplified. [Edit: AaronBrown was very warm.] Not just the size of your underpair, but also the size of your kicker makes a huge difference: JJ22K runs into kings up only half as often as JJ223 does.

There are 123552 two-pair hands, as previously noted. Once Player A has been dealt two pair, there remain just 75595 ways for Player B to be dealt two pair from a deck than contains 4 each of 10 ranks, 3 of one rank, and 2 of 2 ranks.

For any given hand, we can calculate what the 37798th opposing hand is, a make a list of whether it's better or worse than our starting hands.

We find that we have an above-average hand if we have:
JJ22 through JJ77 with a Q/K/A kicker
JJ88 with a 9/T/Q/K/A kicker
JJ99+ with any kicker.

(And no, I don't know of a fast way to calculate this; I just wrote a short program to count up the number of ways to get each possible villain's hand and stop when it got to halfway.)