The reader is advised that the results in this article and the preceding article are nonsense and should be ignored. See the article Low Board Blues: Reprise and Coda for corrections.

Low Board Blues: Part V

Brian Alspach

Poker Digest Vol. 1, No. 8, November 6 - 19, 1998

It's nearly mornin', gotta leave real soon
It's nearly mornin', gotta leave real soon
But man I love this high-low
Be back this afternoon

It will be helpful to have Part IV of this series since this is a direct continuation of that article. I shall use some of the notation and terminology from Part IV. Recall that we are discussing the probabilities of having one or more players also having a nut low given that your hand is A,2,H,H and the board has a form such that another player must have A,2 in her hand in order to have a nut low. I then promised to say more about determining why the probability of one other player also having a nut low is 0.298, the probability of two other players also having nut lows is 0.024, and the probability of three other players also having nut lows is 0.0006.

We are assuming the game is 10-handed. Recall that the number of semi-deals is L = 27,033,832,321,979,553,040,334,558,256,562,500. Let's examine the simplest situation first, namely, the probability that three other players also have nut lows. This means that the semi-deal must contain the remaining 3 aces and 3 deuces. Furthermore, the partition of the 36 cards must put an A,2 in 3 separate hands. There are 6 different ways the 3 deuces can be paired with the 3 aces. You can picture this by thinking of the 3 aces as place holders and allowing all 6 permutations of the 3 deuces into the place holders. There are 30 cards remaining to be distributed. The first A,2 can be completed in C(30,2) = 435 ways, the second A,2 can be completed in C(28,2) = 378 ways, and the last A,2 can be completed in C(26,2) = 325 ways. Since the A,2 pairs are distinct, the total number of completions is the product of the preceding numbers. Finally, we partition the remaining 24 cards into 6 hands of 4 cards each. This can be done in 4,509,264,634,875 ways as seen in Part IV. We multiply the latter number by $6\times 435\times 378\times 325$ obtaining 1,445,843,848,629,367,687,500 ways for 3 other hands to also have a nut low. Dividing this by L yields the probability 0.00059. This tells us that the chance of having the low portion of the pot quartered is negligible when the nut low requires two particular low cards. It is not worth worrying about.

The above probabilities tell us it is about 40 times more likely for your nut low being compromised twice than thrice. Being compromised twice is going to happen about once every 40 times you have the nut low in the circumstances under discussion. This is no longer insignificant. Let's see what is involved in obtaining this result.

The semi-deal A,2 patterns allowing two other people to have nut lows are A,A,2,2; A,A,2,2,2; A,A,A,2,2; and A,A,A,2,2,2. Consider each separately remembering that the numbers for A,A,2,2,2 and A,A,A,2,2 are the same by interchanging the role of A and 2. There are 9 ways to have A,A,2,2 in the semi-deal. There are 2 ways to split them into two subsets of A,2 each. We can complete the first to a hand in C(32,2) = 496 ways, and the second in C(30,2) = 435 ways. The remaining 28 cards can be partitioned into 13,189,599,057,009,375 hands. Then there are $9\times 496\times
435$ times the latter number ways for two people to have A,2 from a set of 36 fixed cards which have 2 aces and 2 deuces. This number is 51,224,182,065,726,169,500,000. We multiply the latter number by C(42,32) to get the number of semi-deals having A,A,2,2 with an A,2 in two different hands. This number is 75,373,462,748,285,396,252,981,923,500,000.

We then do a similar count for a semi-deal containing A,A,2,2,2. It is slightly more complicated because we can have either A,2 in two separate hands, or we can have A,2 in one hand and A,2,2 in another hand. There are 3 ways to choose A,A,2,2,2 to be in the semi-deal. There are 6 ways of choosing the A,2 pairs to be in separate hands. The first A,2 can be completed to a hand in C(31,2) ways (we are insisting here that the remaining deuce is in neither hand containing an ace), the second A,2 can be completed to a hand in C(29,2) ways, and the remaining 28 cards are partitioned into 7 hands of 4 cards each. Multiplying all these numbers yields 14,940,386,435,836,799,437,500.

If we require A,2 in one hand and A,2,2 in another hand, there are 3 ways to split the aces and deuces in this way. The A,2 hand can be completed in C(31,2) ways and the A,2,2 can be completed in 29 ways. Again, the remaining 28 cards are partitoned into 7 hands of 4 cards each. The product of these numbers is 533,585,229,851,314,265,625.

The two numbers above are added and then multiplied by 2 to account for the semi-deals having A,A,A,2,2 (with the roles of A and 2 interchanged) which is multiplied by C(42,31) because this is the number of ways of completing A,A,2,2,2 to 36 cards. The final result is 132,474,570,890, 925,847,959,786,000,000 ways of getting two other hands holding A,2 with the semi-deal containing either A,A,2,2,2 or A,A,A,2,2.

Similar considerations are used when the semi-deal contains A,A,A,2,2,2. There are now a few more subcases because we can have A,2 in each of two hands with the remaining A and 2 in separate hands, or we could have A,A,2 in one hand and A,2 in another hand, and so on. There turn out to be 447,672,687,838,301,141,381,347,182,000,000 ways of two other hands having A,2 when the semi-deal contains A,A,A,2,2,2. Adding this number to the previous two numbers and dividing by L yields 0.0242 as the probability of having two other hands with nut low.

The calculations involved in determining the probability of having one other hand with a nut low follow the same pattern as what has been discussed here. There are more cases because we allow only one ace and deuce to appear together in any of the possible semi-deals. As mentioned in Part IV, this probability is 0.298. This means the probability is about 0.32 that there is at least one other nut low against you. So the odds are about 2 to 1 against at least one other nut low.

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