Read recently that the odds of flopping two pair in PLO are ~10% on a non-paired board. Wanted to extend this to the river, though not sure how to go about calculating?
There really is nothing difficult or mysterious how to perform such a calculation. Maybe this could be an opportunity for you.
What have your tried so far?
Sometimes it helps to imagine a specific example. Suppose board is AKQJT. It doesn't matter what the board ranks are since, presumably, we are ignoring straights and flushes in this calculation.
Now suppose you have four distinct ranks in your hand. You need to come up with the following two pieces of information:
(1) How many total 4-card hands are possible with four distinct ranks given that the 5-card board has five distinct ranks (e.g., AKQJT)?
(2) How many of the 4-card hands having four distinct ranks "interact" with the 5-card board having five distinct ranks (e.g., AKQJT) to make 2 pair?
Both of these questions are fairly straightforward to tackle. Feel free to give them a try and post what you have done. (Hint: it may be a good idea to break down each question into the logical "cases" that could occur.)
Good luck!
P.S.: If anybody else feels the unquenchable urge to post an answer, I would kindly suggest you use the Spoiler tag. My view is that OP should be given the "opportunity" (should s/he be so inclined) to try this with as many hints as necessary before s/he is handed the answer or the derivation of the answer.
I am thinking that maybe it would be best to just present an answer at this point. Even though it has only been two days, it seems that this was not a burning issue for OP -- either the answer or how to derive the answer. (Perhaps I am a little sensitive on this front after I spent several hours deriving answers to a question posed in another recent thread in which that OP apparently never returned.)
I will put my derivation in a Spoiler. Of course, others are free to post whatever they'd like at this point, their own derivation/answer or any other comments.
Spoiler:
Board is unpaired means that there are 5 distinct ranks on the board. Consider the deck stub from which your hand emanates. It clearly consists of 47 cards (52-5=47).
There are 8 ranks having 4 cards in the stub and 5 ranks having 3 cards in the stub (8*4 + 5*3 = 32+15 = 47).
Now let's consider all the 4-card hands you may be dealt. There are C(47,4) = 178,365 hands possible, but we are only interested in those hands having 4 distinct ranks.
It is easiest to simply tally how many 4-card hands having 4 distinct ranks have X ranks in common with a 5-card board having 5 distinct ranks, where X can take values 0,1,2,3,4.
The derivation uses combinatorial notation. C(X,Y) is the choose or combination function read "X choose Y" and gives the number of ways you can choose Y items out of a total of X items without replacement where the order of the choices is irrelevant. The formula for C(X,Y) is well-known and widely available on modern calculators and in many common computer environments.
Case 0: Number of Common Ranks = 0
Clearly this requires that all 4 of your ranks come from the 8 ranks that do not appear on the board. And the card from each rank can be any of the 4 cards available of that rank in the deck stub.
= C(8,4)*C(5,0)*C(4,1)*C(4,1)*C(4,1)*C(4,1)
= 17,920
Case 1: Number of Common Ranks = 1
Clearly this requires 3 of your ranks come from the 8 ranks that do not appear on the board (the card from each rank being any of the 4 cards available of that rank in the deck stub) and 1 of your ranks come from the 5 ranks that do appear on the board (the card being any of the 3 cards available of that rank in the deck stub).
= C(8,3)*C(5,1)*C(4,1)*C(4,1)*C(4,1)*C(3,1)
= 53,760
Case 2: Number of Common Ranks = 2
Clearly this requires 2 of your ranks come from the 8 ranks that do not appear on the board (the card from each rank being any of the 4 cards available of that rank in the deck stub) and 2 of your ranks come from the 5 ranks that do appear on the board (the card from each rank being any of the 3 cards available of that rank in the deck stub).
= C(8,2)*C(5,2)*C(4,1)*C(4,1)*C(3,1)*C(3,1)
= 40,320
Case 3: Number of Common Ranks = 3
Clearly this requires 1 of your ranks come from the 8 ranks that do not appear on the board (the card being any of the 4 cards available of that rank in the deck stub) and 3 of your ranks come from the 5 ranks that do appear on the board (the card from each rank being any of the 3 cards available of that rank in the deck stub).
= C(8,1)*C(5,3)*C(4,1)*C(3,1)*C(3,1)*C(3,1)
= 8,640
Case 4: Number of Common Ranks = 4
Clearly this requires all 4 of your ranks come from the 5 ranks that do appear on the board (the card from each rank being any of the 3 cards available of that rank in the deck stub).
= C(8,0)*C(5,4)*C(3,1)*C(3,1)*C(3,1)*C(3,1)
= 405
Putting it all Together
Cases 2, 3, and 4 give you Two Pair. So the prob of having Two Pair given that you have 4 distinct ranks and the board has 5 distinct ranks is:
I should add that often the "trick" in answering this type of question is developing the correct "logic". Once the logic is established, deriving the correct combinatorial expressions is straightforward and, as above, often quite tedious.
