What's the chance one of my opponents has a flush draw?

[M.B.E.]

Here's a problem I've been wondering about for some time, with some major implications for correct holdem strategy.

Suppose I'm in the big blind with Ah-Jd and take the flop with four opponents. The flop is Js-7s-2d. What is the probability that at least one of my opponents has a flush draw?

Notice that I haven't specified the preflop action: whether there was a raise, whether the players are in early, middle, or late position, etc. My instinct is that this shouldn't make too much of a difference to the solution.

Obviously to answer this question you'll have to make some assumptions about what hands my opponents will play and what hands they will throw away preflop. Be sure to state your assumptions. Hopefully you can come up with assumptions that would reasonably approximate a typical holdem game (let's say the Mirage 20-40 if you need a reference point) but still make the problem easy to calculate. For example, you might say "assume my opponents will play anything in Groups I through V of the Sklansky hand rankings, and nothing else". Or if that turns out to be too complicated, you might say "assume my opponents will play any pocket pair, any suited ace, any suited connector 76 or bigger, and AK, AQ, AJ, AT, and KQ offsuit, and nothing else".

By the way, my guess is that whatever reasonable assumptions you make, the answer will turn out to be between 40 and 50%. One of the reasons I ask this question is that a lot of times when we're discussing hands on another forum, it's assumed that on a two-tone flop, you're probably up against a flush draw; I'd like to know whether that's correct.

[Wake up CALL]

Do you mean a one card flush draw, a runner runner flush draw, either or both?

[M.B.E.]

[ QUOTE ]
Do you mean a one card flush draw, a runner runner flush draw, either or both?



[/ QUOTE ]
I mean the opponent's hole cards are both spades.

[Copernicus]

Without taking into account likely calling hands, the random chance of at least 1 of 4 opponents having two cards that match the suited part of a two-tone board is only about 23%. I doubt that factoring out non-calling hands would change that much, certainly nowhere near the 40% range

[BruceZ]

Without taking into account likely calling hands, the random chance of at least 1 of 4 opponents having two cards that match the suited part of a two-tone board is only about 23%.

No it's not. It is less than 11/47 * 10/46 * 4 = 20.3%.

In fact, it is exactly:

C(4,1)*C(11,2) / C(47,2) -
C(4,2)*C(11,4) / C(47,4) +
C(4,3)*C(11,6) / C(47,6) -
C(4,4)*C(11,8) / C(47,8) +
= 19.3%

Note: The first line is the same as 11/47 * 10/46 * 4.

If you want to do this for a particular group of starting hands, just replace the 11's with the actual number of possible flush cards your opponents could hold, and replace 47 with the total number of possible cards your opponents could hold. You can do this for different raising situations too. In fact, there is no reason everyone shouldn't have a good idea what these numbers are for different situations. Make a spreadsheet.

[M.B.E.]

Thanks, Bruce. That method makes sense. I appreciate it.

[jwp]

My college probability course is now 18 years in the past. Would you mind breaking down the setup of the equation you used:

[ QUOTE ]
C(4,1)*C(11,2) / C(47,2) -
C(4,2)*C(11,4) / C(47,4) +
C(4,3)*C(11,6) / C(47,6) -
C(4,4)*C(11,8) / C(47,8) +
= 19.3%


[/ QUOTE ]

I understand the notation, just not the "why".

Thanks,
jwp22

[Copernicus]

I said "about". I wasnt "about" to do the exact calculation in the middle of a tourney at 12:30 AM! The point was his estimate was way off.

[BruceZ]

Would you mind breaking down the setup of the equation you used:


Quote:
--------------------------------------------------------------------------------

C(4,1)*C(11,2) / C(47,2) -
C(4,2)*C(11,4) / C(47,4) +
C(4,3)*C(11,6) / C(47,6) -
C(4,4)*C(11,8) / C(47,8) +
= 19.3%

I'm using the inclusion-exclusion principle, which you most likely didn't cover in your probability course anyway, at least not in this detail. You can find many explanations of this by me on this forum if you do a search. Basically, the 4 lines correpond to 1,2,3,and 4 people having the flush. The first 3 lines purposely overcount the times that more people have it. For example, the first line counts the times exactly 1 opponent has it once, but it counts the times exactly 2 people have it twice, exactly 3 people 3 times, etc. Each term makes successive corrections to this, until on the 4th line the answer becomes exact. Normally this becomes very close after the first couple of terms. C(4,n) correponds to the number of players who have it. C(11,2n) correpond to the ways n players can have flush cards. C(47,2n) correponds to the ways n players can have n cards.