Poker Digest Vol. 1, No. 6, October 9 - 22, 1998
I got an ace-deuce, other two are high
I got an ace-deuce, other two are high
Now if the flop don't hit me
Gonna fold 'em and sigh
Two articles back I showed Bib Ladder how to determine the probability of a low board occurring in Omaha high-low given that no cards in any of the players' hands are known. In the last article, Bib Ladder himself worked out the probability of a low board in Omaha high-low given that a player holds four high cards in her hand. Later I asked Bib to try working out for himself the probabilities of a player making a low given that the player holds precisely two or three low cards. I finally heard from him two weeks later via e-mail (he is slowly mastering aspects of his recently purchased computer). What follows is essentially what he gave to me.
Since the probability of making a low is the same for any hand having two low cards and two high cards, we assume the two low cards are A-2. This is the most interesting situation because it probably is the only such two-card low hand many people would play.
If the board has exactly three low cards and two high cards, then a low is possible only if the three low cards are of distinct ranks chosen from 3-8. There are C(6,3) = 20 possible choices for the three distinct ranks. There are four choices for a card of each rank giving 43 = 64 ways of choosing cards of these three ranks. Finally, there are C(18,2) = 153 ways of choosing the two high cards. Multiplying these numbers yields 195,840 possible boards.
If the board has four low cards and one high card, there are three different patterns under which the player can make a low: All four cards have distinct ranks from the range 3-8, the four cards are from the range 3-8 and there is exactly one pair, or three of the cards have distinct ranks from the range 3-8 and the remaining card pairs either the ace or deuce. The first can happen in C(6,4) = 15 ways for the four ranks multiplied by 44 = 256 for the choices of cards of the chosen ranks yielding 3,840. In the second case, there are six choices for the rank which will be paired, C(4,2) = 6 choices for pairs of that rank, C(5,2) = 10 choices for the two ranks which will not be paired, and 42 = 16 for the choices of the two cards. We multiply these numbers to obtain 5,760. The last situation can happen in C(6,3) = 20 multiplied by 43 = 64 multiplied by six ways because of choosing three ranks from 3-8, choosing the three cards, and choosing a duplicate of an ace or deuce. This yields 7,680. For each of these we multiply by 18 since there are that many high cards from which to choose. This yields boards with four low cards that will produce a low given that the player has two distinct low cards.
Finally, we consider boards with five low cards. The number of subcases is increasing and making the calculation tedious, but let's stick with it. First, allfive of the ranks may be distinct and from the range 3-8. This can happen in ways. Another possibility is that all five of the cards may be from 3-8 with one pair. There are six choices for the rank of the pair and C(4,2) = 6 pairs of that rank. This gives 36 ways to get the pair. There are C(5,3) = 10 choices for the three ranks of the non-paired cards multiplied by 43 = 64 for choosing the three cards. Multiplying yields 23,040. The other subcases produce similar computations. If there are five cards from the range 3-8, and three-of-a-kind present along with two distinct ranks, this can occur in ways. If there are five cards from the range 3-8 and two pairs present, this can happen in C(6,2) = 15 ways for choosing the ranks of the two pairs times ways of choosing the two pairs times 16 for the number of choices for the one remaining card. This product is 8,640.
There may be four cards from the range 3-8 and one card duplicating an ace or a deuce. If the four cards from the range 3-8 have distinct ranks, this can occur in C(6,4) = 15 ways for the choices of the four ranks times 44 = 256 for the choices of the cards. This product is 3,840. If there is a pair amongst the four cards, this can happen in ways. There are six choices for the card duplicating the ace or deuce so there are 6(3,840 + 5,760) = 57,600 boards with five low cards and four cards in the range 3-8 making a low for a hand with an A-2 and two big cards.
The last case is if there are three cards from the range 3-8 and two cards duplicating ace and deuce. The three cards from the range 3-8 must have distinct ranks. The number of ways of achieving this is C(6,3) = 20 for the three ranks times 43 = 64 for the choices of cards of the given ranks. There are C(6,2) = 15 choices of the two cards duplicating A,2 yielding boards.
Adding these values gives 625,344 boards which will yield a low to a player who holds two small cards and two big cards in Omaha high-low. The total number of possible boards is 1,712,304 and thus the probability of making a low is 0.365.
We see that we expect to make a low more than one-third of the time if we play a hand with two low cards of different ranks and two big cards. A natural question to ask is how often we should expect to make a nut low. Let's consider this when the two low cards are A-2.
If the board has exactly three low cards and we make a low, then it will be a nut low. As we saw above, this happens for 195,840 boards. If the board has exactly four low cards from the range 3-8 and we make a low, it must be a nut low. This happens for 18(3,840 + 5,760) = 172,800 boards. Bad things can begin to happen at this point. In the case there are four low cards with three distinct ranks x,y,z from the range 3-8 and a low card that duplicates either the ace or deuce, you will not have the nut low unless x,y,z is 3,4,5. The latter happens in ways.
Even more of the possibilities with five low cards have the nut low counterfeited when your A-2 is duplicated. If all five low cards are in the range 3-8, then you will have a nut low. From above we have 6,144 + 23,040 + 3,840 + 8,640 = 41,664 boards of this form. Now suppose that four of the low cards are from the range 3-8 with distinct ranks and the other low card duplicates your ace or deuce. The only way you have the nut low is if the 4 ranks from the range 3-8 have the form 3,4,5,y. There are three choices for y, four choices for each of the four ranks and six choices for the duplicating card. This yields only boards that guarantee nut low for your A-2. In the case that there is a pair amongst the four cards in the range 3-8, the only three ranks that guarantee nut low are 3,4,5. There are three choices for which of the ranks has a pair, six choices for the pair, 42 choices for the other two cards, and six choices for the duplicating card. This product gives another 1,728 boards.
Finally, if there are only three cards from the range 3-8 and the other two cards duplicate A-2, the only nut low situation is when the ranks of the three cards in the 3-8 range are 3,4,5. So the total number of boards is .
Summing the previous numbers we obtain 424,512 boards which give the holder of A-2 and two big cards a nut low. This means the probability of achieving a nut low is 0.248.
Bib also produced the following table of useful probabilities. We use H to denote a big card (9,10,J,Q,K), the column headed `low' is the probability of the hand making a low, and the column headed `nut low' is the probability of the hand making a nut low.