Low Board Blues: Part I

Brian Alspach

Poker Digest Vol. 1, No. 4, September 11 - 14, 1998

Woke up this mornin' Omaha on my mind
Woke up this mornin' Omaha on my mind
Don't wanna play no low hands
'Less I'm in the big blind

As I pulled into the parking lot of the local cardroom, I stopped singing my rendition of the Low Board Blues. When I entered the room, I saw Bib Ladder sitting by himself drinking a cup of coffee so I asked if I could join him. He invited me to take the chair opposite him.

``Actually, professor, I'm glad ya dropped in. I bought a computer three weeks ago and have discovered the poker newsgroup. They've been droppin' a lot of numbers about the chances of making low hands in Omaha high-low. Since I understood your earlier explanation about the number of three-card poker hands, I was hopin' you could show me how to get these numbers.''

``Sure, Bib. I'll retrieve my ubiquitous pad of paper and see what we can come up with. The counting is a little more sophisticated for this problem, but as long as you keep common sense and the basic principles I mentioned earlier in mind, we should be alright. Something else we shall develop is that the numbers change according to your viewpoint; more about that later.''

I continued, ``Let's assume that you are on the rail watching the boards appear and you ask yourself what the chances are that a low will not be possible. Let's first calculate the total number of boards. We are choosing 5 cards from 52. As we saw the last time we talked, this number is C(52,5). Evaluating this yields $52\cdot 51\cdot 50\cdot 49\cdot 48$ divided by $5\cdot 4\cdot 3\cdot 2\cdot 1=120$. After doing the arithmetic, we find that there are 2,598,960 possible boards.

The question now is how many of those boards preclude a low. The board may contain five high cards, such as K-Q-J-10-9, or four high cards and a low card, or three high cards and two low cards, and so on. Since I am using the word or, we have to add the numbers we get. If the board contains five high cards, then we have chosen five cards from the 20 high cards in the deck. This is $C(20,5)=20\cdot 19\cdot 18\cdot
17\cdot 16/120$ which is 15,504. Next, the board may contain four high cards and one low card. The four high cards can be chosen in $C(20,4)=20\cdot 19\cdot 18\cdot 17/24 = 4,845$ ways. The single low card can be any of 32 cards. We have to multiply this by 4,845 because we are choosing 4 high cards AND one low card - remember that and means multiply. Anyway, we obtain 155,040 such boards. Is this alright so far?''

``Yeah, professor, I understand so far. Continue.''

``Next we want to count the number of boards with three high cards and two low cards. It will be the product of the number of ways of choosing three high cards from the 20 available and two low cards from the 32 available. This is the product of C(20,3) and C(32,2). We easily calculate that C(20,3)= 1,140 and C(32,2)= 496. Their product is 565,440. All of the preceding boards preclude a low hand because there are at most two low cards in the board. Things get a little stickier now because three or more low cards will be in the board. Just keep the common sense here and we shall prevail.

We now move to boards with two high cards and three low cards. The two high cards can be chosen in C(20,2)=190 ways. The problem is that even though three low cards are present, a low may be precluded because at least two of the low cards form a pair. So we have to break this into subcases. If the three low cards form trips, this can happen in $8\cdot 4=32$ ways because there are eight possible ranks for low cards and four ways of choosing trips. The other way of counterfeiting a low is that exactly two of the low cards form a pair. There are eight choices for the rank that is paired, seven choices for the rank that is not paired, C(4,2)= 6 ways of choosing the pair and C(4,1)=4 ways of choosing the card that is not paired. This gives us $8\cdot 7\cdot 6\cdot 4=1,344$. We do not have to divide by anything in the last product because we have counted any such scheme exactly once. Altogether we have 1,344+32= 1,376 ways that three low cards do not allow a low. So there are $1,376\cdot 190=261,440$ boards with precisely three low cards which preclude a low. Are you still with me, Bib?''

``Let me think for a minute,'' he replied. After a short silence he nodded for me to continue.

``The next situation is four low cards on board that preclude a low. How can that happen?''

Bib immediately replied, ``Well, if there's quads, trips or two pair there can't be a low.''

``That's right. Since you said or that means we have to add those three counts together. There are precisely eight ways to get quads. There are eight ranks to choose for the trips, seven choices for the rank of the other low card, four ways of choosing trips and four choices for the other card. That gives us $8\cdot 7\cdot 4\cdot
4=896$ ways. If there are two pair, there are C(8,2)=28 ways to choose the two ranks to be paired, and each pair can be chosen in six ways. This gives $28\cdot 36=1,008$ ways for two pair to occur. Adding them together gives 1,912 ways that four low cards can occur and yet preclude any low. For each of them we can choose any of 20 high cards so that the number of boards with 4 low cards and no low possible is $20\cdot

The final situation is having five low cards and no possible low. Using the same reasoning you just demonstrated, Bib, we must have either four-of-a-kind or a full house on board. There are eight choices for the rank of the card in the quartet and seven choices for the rank of the remaining card. The latter card can then be any of four yielding $8\cdot 7\cdot 4=
224$ boards with four-of-a-kind. For full houses, we have eight choices for the rank of the trips, seven choices for the rank of the pair, four choices for the trips and six choices for the pair. Multiplying them gives 1,344 such boards.

Now we add all of the above boards together to obtain 1,037,232 boards which do not allow a low. We obtain the probability that no low is possible by dividing that by the total number of boards which, as we saw earlier, is 2,598,960. This gives us a probability of 0.399. So roughly 60 percent of the time an observer on the rail will see that a low is possible.

Oh, I see my seat is open. Let's continue this conversation later, Bib. Why don't you see if you can determine the probability of a low being possible given that you hold exactly four high cards. That may get you singing the blues too.''

As I headed towards my seat, Bib gave me a quizzical look as he tried to understand what that last comment was about. Next issue: The Low Board Blues: Part II.

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