Flush Probability

[kyro]

I just started playing NL hold-em at PartyPoker today and I must say I'm not happy. Can someone tell me the odds that if you flop a flush, someone else will also hold a flush? I'd do it out but I'm lazy and figure someone might know it off the top of their heads. Thanks

[Mike Haven]

please tell us the real reason for your post

[LongOdds]

It depends on how many players are at the table, but there should be only 8 cards of that suit left versus 39 of the other three suits when you subtract the 5 known cards. The odds depend on how many players, but regardless, they aren't good.

[kyro]

heh, OK. I don't remember the exact cards or anything, but I'll try and reenact it as best as I can. (While we're at it, is there some sort of program everyone has that helps them list their hands? I've heard of one but I don't know the specifics)

OK, it's a 10-person PL game at Party. First hand of the sitting I get dealt 87 of clubs. I check my BB post. There are about 6 people who stay in, no raises. Flop comes 3 clubs, one of which is an A. Since I only put in 15 bucks, there were a couple of raises and I decided I had to call the all-in raise. The next two cards are blanks, no pairs, no clubs. One guy shows Kc and something else. Second guy shows T2c. I was just curious what the odds that if I flopped a flush, that someone else flopped one along with me. Thanks.

[BruceZ]

[ QUOTE ]
I just started playing NL hold-em at PartyPoker today and I must say I'm not happy. Can someone tell me the odds that if you flop a flush, someone else will also hold a flush? I'd do it out but I'm lazy and figure someone might know it off the top of their heads. Thanks

[/ QUOTE ]

Of course it depends on how many players see the flop. With N opponents on the flop, the probability that someone flopped a flush with you, assuming random hands, by the inclusion-exclusion principle (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383&Forum=probability&Words=inclusion-exclusion&Match=Entire% 20Phrase&Searchpage=0&Limit=25&Old=1year&Main=4169 81&Search=true#Post417383) is:

C(N,1)*C(8,2) / C(47,2) -
C(N,2)*C(8,4) / C(47,4) +
C(N,3)*C(8,6) / C(47,6) -
C(N,4)*C(8,8) / C(47,8)

Where we take C(N,k) = 0 for N < k, so we never have more terms than we have opponents. This evaluates to:

<font class="small">Code:</font><hr /><pre>
# Opp. P(flush)
1 2.6%
2 5.1%
3 7.7%
4 10.1%
5 12.6%
6 15.0%
7 17.3%
8 19.6%
</pre><hr />

This is roughly 2.5% per opponent.

Example 3 in the inclusion-exclusion post I linked to above contains a similar problem, so refer to it for details.

Problems similar to this one have been asked a lot, but I question the value of the answers since in reality you would want to consider the hands that the opponents would actually see the flop with, and not random hands. It does have value if the opponents you are considering are forced to play any hand, because they are in a blind or forced to go all-in.

This problem gets asked in many variations. Sometimes we hold flush cards, sometimes not. Sometimes 3 suited cards flop, and we want to know the probability that someone flopped a flush. Sometimes 2 suited cards flop, and we want to know the probability that a draw is out. These are all simple variations on the same formula, so I have tablulated all of these possibilities once and for all in one place. Each of these cases is characterized by the combined number of exposed suited cards on the flop and in our hand (from 2 to 5). The table gives the probability that at least one other opponent holds 2 flush cards, assuming random hands. So the first column corresponds to the case above since there are 3 suited cards on the flop, and 2 in our hand for a total of 5.

<font class="small">Code:</font><hr /><pre>
exposed flush cards
# opp. 5 4 3 2
1 2.6% 3.3% 4.2% 5.1%
2 5.1% 6.6% 8.2% 10.0%
3 7.7% 9.8% 12.1% 14.7%
4 10.1% 12.9% 16.0% 19.3%
5 12.6% 16.0% 19.7% 23.6%
6 15.0% 18.9% 23.2% 27.8%
7 17.3% 21.9% 26.7% 31.9%
8 19.6% 24.7% 30.1% 35.8%
9 21.9% 27.5% 33.4% 39.5%
10 24.2% 30.2% 36.6% 43.1%
</pre><hr />