**Brian Alspach**

**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
4^{3} = 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 4^{4} = 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 4^{2} = 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 4^{3} = 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 4^{3} = 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 4^{4} = 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 4^{3} = 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, 4^{2} 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.

hand | low | nut low |

A,2,H,H | 0.365 | 0.248 |

A,3,H,H | 0.365 | 0.138 |

2,3,H,H | 0.365 | 0.138 |

A,2,3,H | 0.495 | 0.431 |

A,2,4,H | 0.495 | 0.323 |

A,2,5,H | 0.495 | 0.263 |

2,3,4,H | 0.495 | 0.225 |

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