Low Board Blues: Part II

Brian Alspach

Poker Digest Vol. 1, No. 5, September 25 - October 8, 1998

Lookin' at big cards, just one way to go
Lookin' at big cards, just one way to go
So when the board has fallen
Don't wanna see no low

A few days later I ran into Bib again. ``Hey, professor, remember last Thursday when you asked me to try to work out the chances of a low showing given that I have four big cards? Well, I think I have solved it. How 'bout checkin' it?''

``Sure, Bib. Let's go over what you did. Actually, the reason I mentioned this problem to you is that I wanted to introduce you to something called conditional probability. Last time we met we worked out the probability that an observer on the rail will see a board in Omaha that allows a low. That might be of some interest to somebody, but if you are in the game, you are more interested in the probability of a low board given the cards in your hand. This is what is meant by conditional probability, namely, the probability of some event given some partial information. In this case the partial information is that your hand has four high cards.''

I continued, ``So let's take a look at what you have done. Even better, why don't you explain your solution?''

I thought I detected a little pride on his part as he started his explanation. After all, he claimed he was non-mathematical and there is a certain joy in discovering that one has underestimated one's abilities. Bib Started, ``The first thing I did was to calculate the total number of boards. After I did it, I remembered that we had made that calculation during our first session. At least it gave me something I could check. Anyway, the total number of boards is C(48,5) which is 1,712,304 possible boards from the remaining 48 cards.

I decided to follow what we did last time as an outline. So I broke the count into boards which preclude--I like that word, professor--a low by examining the number of big cards. The board may contain five big cards, like 9-10-J-Q-K, or four big cards and a low card, or three big cards and two low cards, and so on. Since I am using the word `or', I am going to add the numbers I get. How am I doin' so far?''

``Perfectly, so far,'' I replied.

``If the board contains five big cards, then we have chosen five cards from the 16 big cards remaining after removing the four big cards in my hand. This is $C(16,5)=16\cdot 15\cdot 14\cdot 13\cdot 12/120$ which is 4,368. Next, the board may contain four big cards and one low card. The four big cards can be chosen in $C(16,4)=16\cdot 15\cdot 14\cdot 13/24 = 1,820$ ways. The single low card can be any of 32 cards. I have to multiply this by 1,820 because I am choosing four big cards AND one low card and I remember that `and' means multiply. Anyway, we obtain 58,240 such boards.

Next I want to count the number of boards with three big cards and two low cards. It will be the product of the number of ways of choosing three big cards from the 16 available and two low cards from the 32 available. This is the product of C(16,3) = 560 and C(32,2) = 496. Their product is 277,760. All of the preceding boards preclude a low hand because there are at most two low cards in the board. This is very much like what we did before, right?''

``That is certainly true, Bib.''

``Next I consider boards with two big cards and three low cards. The two big cards can be chosen in C(16,2) = 120 ways. That leaves three low cards and a low is possible if they have three different ranks. So a low is impossible if the three low cards form trips or there is one pair. For trips there are eight possible ranks and four ways of choosing trips. If 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 $1,376\cdot 120 = 165,120$ boards with precisely three low cards which preclude a low.

When there are four low cards on board, if there's trips or two pair there can't be a low.''

``What about quads, Bib?''

``Damn, I forgot them!''

``It is so easy to forget a subcase in these kinds of counting problems, Bib. It is also easy to make an arithmetic error. Don't worry about it because it won't take much of an adjustment to include them since there are only eight ways of getting low quads.''

``I suppose you're right. I just don't like making errors of any kind. So I'll continue. As far as trips are concerned, 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 1,904 ways that four low cards can occur and yet preclude any low. Oh, yeah, I have to add eight more for the quads giving me 1,912 altogether. For each of them we can choose any of 16 high cards so that the number of boards with four low cards and no low possible is $16\cdot 1,912 = 30,592$.

The final situation is having five low cards and no possible low. 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 quads 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 537,648 boards which do not allow a low given that my hand has four high cards. We obtain the probability that no low is possible by dividing that by the total number of boards which, as we saw earlier, is 1,712,304. This gives us a probability of 0.314. So roughly two-thirds of the time a low is possible when you hold four big cards. To be precise, the probability is 0.686 that a low is possible.''

``Very impressive, Bib.''

``Thanks, professor, but there's one thing which doesn't make sense to me.''

``What's that?''

``Well ... last time you showed me the probability of a low board showing for an observor on the rail is 0.601, and I've worked out the probability of a low board is 0.686 when I have four big cards. That's a 14% increase and I don't even understand why they are different at all.''

``There is an explanation, Bib, but let's save it for another time.''

Next time we shall consider hands containing low cards.

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