Odds Question

[Hybrid_11]

I had an argument with a buddy about how to calculate odds in a situation. Were assuming im going after a backdoor flush after the flop. I have two suited cards in my hand and there is one on the flop. Now to calculate odds do u take the 47 of the unseen cards or only the 29 remaining cards to be possibly dealt out on the turn?

[sinfulslick18]

where are you getting 29 from...your opponents hold cards?

-sinful

[jdl22]

Try the probability forum next time.

There are 47 cards left in the deck. Out of those two are taken for the turn and river. That makes 47 choose 2 or 47*46/2=1081 different turn/river combinations (in this case order doesn't matter). You have 10 flush cards and you need both of them in there so there are 10 choose 2 or 10*9/2=45 different combinations that win.

Hence the probability of hitting is 45/1081 ~ .04. So you have about a 4% chance of hitting. Odds against are 96:4 or 24:1.