probability of flopped full encountering backdraw straight flush
I was wondering if somebody had an idea of the probability of a live hand that happened to me yesterday.
Basically the board reads:
Kd-10h-10s 8s 7s
I held 9s6s
my opponent K-10
So do you know the probability of this occuring ?
flopped full is like 1/500 hands? =x
backdoor straight flush is like 1/10000 hands? =y
both occuring at the same time is x*y?
Thanks a lot for your feedback
According to one odds calculator I saw, the odds of flopping a full house with unpaired hole cards are 1 in 1,087.
For the backdoor straight flush, you need to hit perfect-perfect. On the flop, there are 47 unknown cards (52 in the deck, minus the 3 on flop, 2 in your hand AND 2 in your opponent's hand. We normally would not count your opponent's cards as "known" but we have to in this case because we are assuming that he flopped a full house). So, the odds on the flop are 2 in 47 that you will hit either the 8s or 7s. On the turn, you have 46 remaining unknown cards and you have to hit the single remaining straight flush card, so the odds are 1 in 46.
Therefore, the odds of hitting perfect-perfect straight flush after your opponent flopped a full house are 2/47*1/46 = 1 in 1,081.
The odds of both happening are 1/1087*1/1081 = 1 in 1,175,047
@Dap Poker Great thanks!
Ok it won't happen anymore in my life normally (at least live!)