Probability of completing a flush draw
I've been thinking about this for the last few days and wondered what others thought about this.
The situation in question is when you have 2 suited cards (let's say As 9s) and there are 2 of your suit on the flop (let's say 3s Ks 8d).
Everything I have read about this has suggested that you have 9 outs to make your flush on the turn card, giving you a 9/47 = 19% chance.
While in the best case it is true that no one else would hold any of your suit, I think it is not accurate in the average case.
In a 10-handed ring game, there are 20 cards dealt to the players. On average, there will be 20 cards / 4 suits = 5 cards of each suit dealt out. Since you hold 2 spades, on average there will be 3 spades dealt out to the other players.
So, on average, you really only have 9 - 3 = 6 outs to make your flush on the turn card, giving you a 6/47 = 13% chance.
Does this make sense to anyone else? Or am I breaking correct probability calculations with my "average" thinking?
It just seems to me that having 9 outs for the flush is really the best case, where in poker we really want to rely on the average case. And I'm trying to figure out what that average case is.
Has anyone seen this covered elsewhere?
Thanks.
Re: Probability of completing a flush draw
You overthought it and came to the wrong conclusion. There are 47 cards you haven't seen, including your opponents' hands; 9 of those are spades. That's 9 outs in 47 cards. If you wanted to go about it the way you are proposing, you would account for not just the missing spades but also the missing cards your opponents hold, so your outs are 6 in 29, which works out to be basically the same percentage and therefore gives you the same calling odds.
