- #1
morrowcosom
- 54
- 0
I run into the dilemma of having a great hand and then a possible flush hits, so I tried to calculate the probability of a single opponent having a flush when the board has 3 suited cards by the river.
I used the old 1-(unhelpful cards/unseen cards) trick, but my answer seems too high, and I hope it is!
3 suited cards on board:
1- [(36/46)*(36/45)]
=38%
Three of the flush cards are already taken, so 10 are left to be helpful, 46 unseen cards remain, they must have one of these as well as another helpful card, so 9 helpful cards remain out of the 45 unseen.
Thanks for the help, I hate this situation.
I used the old 1-(unhelpful cards/unseen cards) trick, but my answer seems too high, and I hope it is!
3 suited cards on board:
1- [(36/46)*(36/45)]
=38%
Three of the flush cards are already taken, so 10 are left to be helpful, 46 unseen cards remain, they must have one of these as well as another helpful card, so 9 helpful cards remain out of the 45 unseen.
Thanks for the help, I hate this situation.