Calculating Runer-Runner In Omaha

[EasilyFound]

Trying to calculate the probability of hitting a nut low by hitting the turn and the flop. You start w/A-2-3-X and the flop it K-10-8. To hit a nut low, I need to hit two cards cards below and 8 but no more than one of my three low cards. I'm not sure how to calculate that. I know how to figure runner runner nut low if I have A-2-x-x, then the equation is 20/45 x 16/44 = 16.2%. But I think the equation is different if you add another low card to my hand. Need a math wizard for this, I think.

Thanks.

[BruceZ]

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Trying to calculate the probability of hitting a nut low by hitting the turn and the flop. You start w/A-2-3-X and the flop it K-10-8. To hit a nut low, I need to hit two cards cards below and 8 but no more than one of my three low cards. I'm not sure how to calculate that. I know how to figure runner runner nut low if I have A-2-x-x, then the equation is 20/45 x 16/44 = 16.2%. But I think the equation is different if you add another low card to my hand. Need a math wizard for this, I think.

Thanks.

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P(4-7 on turn)*P(A-7 on river - turn rank) + P(A-3 on turn)*P(4-7 on river) =

(16/45 * 21/44) + (9/45 * 16/44) =~ 24.2%.

[EasilyFound]

Thanks. Same idea for calculating w/starting hand of A-2-3-4?

[AaronBrown]

No. Now there are 24 cards that can help you. If you get one of the 12 that pair you (A, 2, 3 or 4), there are 22 cards that can go with it. If you get one of the other 12 (5, 6 or 7), there are now 21. That comes out to a 26% chance.

[BruceZ]

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Thanks. Same idea for calculating w/starting hand of A-2-3-4?

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With A234, you can pair 2 of your hole cards. Any 2 board cards from A-7 will make your hand, as long as the board doesn't pair.

Note that there are 12 remaining cards A-4, and 12 remaining cards 5-7.

P(5-7)*P(A-7 minus 4 of turn rank) + P(A-4)*P(A-7 minus 3 of turn rank) =

(12/45 * 20/44) + (12/45 * 21/44)= 24.8%.

This isn't much better than A23x which was 24.2%. The only difference is that it allows the exact combinations A2, A3, and 23. On the other hand, A23x is much better than A2xx which was 16.2%, because A23x allows the A or 2 to pair.

[BruceZ]

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No. Now there are 24 cards that can help you. If you get one of the 12 that pair you (A, 2, 3 or 4), there are 22 cards that can go with it.

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There are 21 cards that go with it, not 22. A-4 (12-3=9) + 5-7 (12) = 21.


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If you get one of the other 12 (5, 6 or 7), there are now 21.

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There are now 20, not 21. A-4(12) + 5-7(12-4=8) = 20.


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That comes out to a 26% chance.

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24.8%. Only 0.6% better than A23x. I didn't believe it myself at first, but that's because it only allows for additional specific 2 card combinations which have relatively low probability. We also lose all the combinations with one of the fours.

[EasilyFound]

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[ QUOTE ]
Thanks. Same idea for calculating w/starting hand of A-2-3-4?

[/ QUOTE ]

With A234, you can pair 2 of your hole cards. Any 2 board cards from A-7 will make your hand, as long as the board doesn't pair.

Note that there are 12 remaining cards A-4, and 12 remaining cards 5-7.

P(5-7)*P(A-7 minus 4 of turn rank) + P(A-4)*P(A-7 minus 3 of turn rank) =

(12/45 * 20/44) + (12/45 * 21/44)= 24.8%.

This isn't much better than A23x which was 24.2%. The only difference is that it allows the exact combinations A2, A3, and 23. On the other hand, A23x is much better than A2xx which was 16.2%, because A23x allows the A or 2 to pair.

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Thanks. I feel proud. After my post, I did the math using the formula, exactly as above, and came out with the same answer. Not much difference between A234 and A23.

[AaronBrown]

You are correct. I forgot to subtract the card you just drew.

In addition to the reason you mention, the case is not much better because you're now holding one more of the cards that could otherwise help you.