Quote:
Originally Posted by whosnext
Edit to add: wait a minute, there is another way to have quads over quads in PLO, isn't there (board could be a "full house")??
You're right, that would use two hole cards if the player had a kicker higher than the pair on board. The boat couldn't be XXXAA or AAAKK. The kicker can't tie the pair because the villain needs quads of that rank.
This is a tedious case to calculate because there are many (110?) different probabilities to add. The board is xxxrr (where r is a rank from 1 to 12) and there is a different probability for each r and for x>r vs x<r.
The player with quad x's needs a playable kicker. The # of playable kickers (k) varies with r and r's relation to x.
If x<r then k=4(13-r) else k=4(12-r)
The # of valid ways to group the cards (p) depends on the # of playable kickers dealt (j).
P(quads>quads | xxxrr) = sum of 15 * C(k,j) * C(52-k, 5-j) * (p/35) / C(47,8) from j=1 to 5
j=1 → p = 6
j=2 → p = 9
j>2 → p = 10
P(board xxxrr such that x>r) = 24(13-r) / C(52,5)
P(board xxxrr such that x<r) = 24(r-1) / C(52,5)
For r=12 we only consider x<r.
I'll compute it another time.
Last edited by heehaww; 07-24-2019 at 10:46 AM.