Actual flush draw odds

[theflyingcow]

As we all know, chasing flush draws is a typical situation that comes up in most holdem games. Say we have 9 [img]/images/graemlins/spade.gif[/img] T [img]/images/graemlins/spade.gif[/img] and the flop comes x [img]/images/graemlins/spade.gif[/img] x [img]/images/graemlins/spade.gif[/img] x [img]/images/graemlins/heart.gif[/img]. So I know that the general accepted way of calculating the odds of hitting a flush draw is to pretend that none of your other opponents have no spades giving us 9 spade outs to catch the flush. However, I personally think that this is incorrect since it is almost impossible in a 9-handed game for nobody else to have been dealt a spade. Does anyone know (or can calculate, since I'm too lazy to) the average number of an arbitrary suit (spades in our example) are dealt in an x-handed (given that we have 2 in our hand) game where 2 <= x <= 10? I really think we should probably be using these numbers to count outs. Thoughts?

[ispiked]

This is talked about a little bit in this Wikipedia article. I think what it comes down to is that you have to base your calculations on the information available to you. You could probably figure out the probability of a player holding a card of the same quit, but in the end it's only a probability and you could have guessed wrong.

[Mister Moe]

I think this is a great question, and I've seen plenty of thinking people fall into this trap. The quick answer though, is that it doesn't matter, because your opponents' cards are unknowns. The high probability that your opponents have few spades counters the low probability that your opponents are spade-heavy. Take the simplest example of a two-handed game, using your example, and simplify further by only considering the turn card. As you point out, conventional wisdom is that you have 9 outs of 47 cards left-- i.e., a your P of hitting the flush on the turn is 9/47.

Now try to figure out the probability given that there is another hand out there, i.e., another two unknown cards that may or may not be [img]/images/graemlins/spade.gif[/img]s. Given your hand of two [img]/images/graemlins/spade.gif[/img]s and a flop of two [img]/images/graemlins/spade.gif[/img]s and one non-[img]/images/graemlins/spade.gif[/img], you know there are 9 [img]/images/graemlins/spade.gif[/img]s left and 38 non-spades. Plus, there are four possible hands your opponent has:

1: both [img]/images/graemlins/spade.gif[/img]s
2: 1st card [img]/images/graemlins/spade.gif[/img], 2nd card non-spade
3: 1st card non-spade, 2nd card [img]/images/graemlins/spade.gif[/img]
4: both non-spade

Your probability of hitting the flush on the turn (the third unknown) is the sum of the probabilities of doing so given each possible opponent hand.

(In the following, the 1st and 2nd cards represent your opponent's hand and the 3rd card represents the turn.)

Case 1: 9/47 1st [img]/images/graemlins/spade.gif[/img], 8/46 2nd [img]/images/graemlins/spade.gif[/img], 7/45 3rd [img]/images/graemlins/spade.gif[/img]= 504/97290.

Case 2: 9/47 1st [img]/images/graemlins/spade.gif[/img], 38/46 2nd non-spade, 8/45 3rd [img]/images/graemlins/spade.gif[/img]= 2736/97290.

Case 3: 38/47 1st non-spade, 9/46 2nd [img]/images/graemlins/spade.gif[/img], 8/45 3rd [img]/images/graemlins/spade.gif[/img]= 2736/97290.

Case 4: 38/47 1st non-[img]/images/graemlins/spade.gif[/img], 37/46 2nd non-spade, 9/45 3rd [img]/images/graemlins/spade.gif[/img]= 12654/97290.

Add 'em up, get (504 + 2736 + 2736 + 12654) / 97290 = 18630/97290.

= 9/47 !

(proving this for an arbitrary x opponents and including the river card are exercises left for the reader) [img]/images/graemlins/laugh.gif[/img]

[GittyUP]

Heres another thread which the same questions are asked and mine and others responses.
Flush draw outs

[ghostface]

50%

[Reckless1der]

True enough, on average 4.3 spades were also distributed to the other players. It would be highly unlikely that there are still 9 spades left in the stub, and it would actually average that there 5.7 left. At the same time there are not 47 remaining cards, but really only 31 (ignoring the burn). Proportionately, it is the same difference. A spade will come by the river a little more often than once every 3 times.

[NormandySD]

If you took the ratio of spades in the deck would be the same..... therefore the outs would be the same. It is expected that the spades you are looking for are spread proportionaly throughout the players hole cards, but then you also must remove the other hole cards from the possiblities of cards to come. there wouldnt be 47 possibilites... only 35.

But yes to answer your question... the odds would be GREATLY adjusted by knowing the spades in the muck/hands vs the spades left in the deck. Just ask stew ungar.

Since there is already so much going on this type of thinking would never work. I tried doing so starting with "Ace hands" and it just didnt pan out. PPLs hands cannot be read so plainly.