In my continuing quest to teach my daughter math concepts via poker, we stumbled upon one of the most basic questions in playing Omaha Hi/Lo.

How often will a low be "available" on a random 5-card board? (Assume nothing about anybody's hole cards.)

Of course, a low is available if there are at least three distinct ranks on board 8 or below (ace is low for this).

By working through seven cases of how many low ranks (and dupes) appear on board, I am able to tally that there are a total of 1,561,728 boards out of a grand total of 2,598,960 boards for which a low is available, or 60%.

I imagine that there has to be an easier way to derive this other than going through seven combinatoric cases.

What is the easiest way to derive this? Is this an application of the Principle of Inclusion-Exclusion (PIE)?

If I am totally missing an easy way to do this or am being an idiot in some way, please be gentle.

Thanks much.

Your number is verified here:

https://en.wikipedia.org/wiki/Poker_...maha)#The_flop

Sounds like you used the same method.

Thanks for the reply.

By the way, my daughter initially thought the numerator was:

C(8,3) * (C4,1)^3 * C(49,2)

but figured out that could not be correct since that gives a number greater than the denominator!

^^^

C(5,3) * C(8,3) * 4^3 / C(52,3) -
5*[3*C(8,4)*4^4 + 3*C(8,3)*6*4^2] / C(52,4) +
[6*C(8,5)*4^5 + 3*4*C(8,4)*6*4^3 + 3*C(8,3)(4*6^2 + 4^3)] / C(52,5)

= 664/1105 or about 60.09%

The PIE coefs greater than 1 are bolded.

That, divided by C(52,5), is equivalent to my first line. It quadruple-counts the 4 low-card unpaired case and double-counts the paired case. It counts the 5 low-card unpaired case 10 times, the paired case 7 times, the double-paired case 4 times, and the tripped case 3 times.


Edit: Granted, I still had to account for 7 cases, so the non-PIE approach may have been better.

Let's see:

C(8,3)*64 * C(20, 2) +
C(8,4)*4^4 * 20 +
C(8,5)*4^5 +
3*C(8,3)*(6*16*20 + 4*36 + 64) +
4*C(8,4)*6*64
/ C(52,5)

Yeah that's better because you don't have to figure out coefficients. With PIE I didn't have to calculate the coefs for the unpaired cases, but for the other cases I did.

In general, to know in advance whether PIE is better than the additive approach, one should determine whether PIE will reduce the number of cases to count.

For instance, suppose the question is, what's the probability that at least one player gets dealt AA in 6max?

PIE is preferred there because it saves us from having to tally cases where one player gets AA and another player gets Ax (where x =/= A). We don't have to because those cases don't get overcounted by any PIE step.

By contrast, with the plo8 problem, each type of case is overcounted by the first PIE step and thus has to be addressed.

So next time, I'll know not to use PIE for something like this, and you'll know whether the additive approach was more effort than necessary.

Thanks much, as always.