Odds of holding a pocket pair in Omaha?

Hello everyone. I'm a new fish to both the forum, and to Omaha in general. I've been playing NLH for several years and the incorporation of Omaha has been both fun and challenging. I ran across the question in the thread title the other day and have tried to calculate those odds, and have now come up with three answers. Surely at least two of them are wrong! :)

To make a long post somewhat shorter, I believe the answer is 36.25%. My rationale is below:

1. Pick a card, any card (let's say an ace)

2. The second card dealt has a 3/51 chance of pairing your ace, or about 5.88%.

3. If you didn't pair up (let's say you got a king), you have a 6/50 chance of pairing your ace or your king, or exactly 12%.

4. If you still haven't paired up (let's say you got a queen for your third card), you have a 9/49 chance of pairing either your ace, king, or queen on your last card, or about 18.37%.

Since we don't need all of these events to come true, simply one of them, the odds should be the sum of the probabilities, or about 36.25%.

I think I'm right. I'm willing to be wrong. I just want to know for sure, because this is now a brain splinter.

Hello and welcome. Usually it's easier to work out how often it won't happen and then use 1-X. In this case to miss you need to get four different ranks.

Total number of hands = 52*51*50*49/4/3/2/1 = 270725
Total number of hands with different ranks = 52*48*44*40/4/3/2/1 = 183040
Total number of hands with a pair = 87685 (32.39%)

I haven't checked but suspect in your case in later stages you need to multiply 6/50, 9/49 etc. by the probability of getting past the earlier stages. So instead of 6/50 becoming 12% you also need to multiply by 48/51 (to get a slightly smaller number).

As an exercise, it might be easier to see what's going on, you could try similar logic by working out the chances of throwing a die and getting a 6 sometime in four rolls. For instance at each stage it's 1/6 you'll throw a six. But clearly you don't just add up the 1/6ths (otherwise after six rolls you're guaranteed to roll one!).

Quote: charliepatrick

Hello and welcome. Usually it's easier to work out how often it won't happen and then use 1-X. In this case to miss you need to get four different ranks.

Total number of hands = 52*51*50*49/4/3/2/1 = 270725
Total number of hands with different ranks = 52*48*44*40/4/3/2/1 = 183040
Total number of hands with a pair = 87685 (32.39%)

I haven't checked but suspect in your case in later stages you need to multiply 6/50, 9/49 etc. by the probability of getting past the earlier stages. So instead of 6/50 becoming 12% you also need to multiply by 48/51 (to get a slightly smaller number).

As an exercise, it might be easier to see what's going on, you could try similar logic by working out the chances of throwing a die and getting a 6 sometime in four rolls. For instance at each stage it's 1/6 you'll throw a six. But clearly you don't just add up the 1/6ths (otherwise after six rolls you're guaranteed to roll one!).

Agree with this. Just noting that you are excluding from your hands that have three of a kind and four of a kind. I can’t tell from the OP how he wanted to treat those hands. In Omaha a three of a kind and four of a kind are both trash starting hands.

I calculated it differently, but got the same result. This way makes more sense to me.

No need to calculate the first card. It's any card. 52/52.
48/51 - 48 cards out of 51 remaining do not match your first card.
44/50 - 44 cards out of 50 don't match either of the two cards you already have.
40/49 - 40 cards out of 49 don't match and of the three you have.

(48/51) * (44/50) * (40/49) = 0.67611

1 - ( (48/51)*(44/50)*(40/49) ) = 0.32389 = ~32.4%


Side note: The /4/3/2/1 part in the other responses is optional. It reduces the number of hands by ignoring the order that you get dealt them. The math works out either way since you're doing that division on both parts. (Simple algebra there.) And besides, /1 is totally unnecessary! 🤪

My thanks to all of you for your replies. The 1-x method makes a lot of sense. For the purposes of the exercise, the difference between having a hand with exactly one pair (99xx) and a hand with trips (999x), two pair (9988) or quads (9999) is pretty small, but obviously affects the outcome. Intuitively, I understand the 1-x method better, and it seems to make more sense than my method in the OP.