re: Odds of 2 people making a Flush at the Same time

[OOgie]

Scenario: You get in cheap in late position (game type unimportant) with suited connectors. (87s). You hit the flop with middle pair and have a flush draw. (Ts-8o-3s). You call. The turn comes Qs....you've made your flush. What are the chances that your opponent has also made a flush/ a higher flush ?

In trying to work this out....
1. 13 suited of which 5 suited are accounted for, leaving 8s. of the 46 remaining cards at the turn.
2. Since the odds of any one player starting off with 2 suited cards is about 3:1, then (2.5) opponents at this table of 10 should be holding suited cards. Here is where I get lost. But of the four suits there is now less than a 25% chance the opponent has the same suit since you already hold two of those cards.
3. Trying to put the early opponent on a hand doesn't help too much either since he could just as easily have an overpair, two overcards, a straight draw, a straight flush draw, or the nut flush draw.
4. In this case scenario .... so what are the chances your opponent holds AXs, KXs,JXs, TXs, or 9Xs ??? to have you beat.

Any insight into this problem would be much appreciated. Two opponents to a flush seems to happen more often than one would think. But then again so do quads.

[the_hawk]

This is not really a probability question. More precisely, you could answer it as such, ascribing random hands to the field, but that really wouldn't give you any useful information. I'm afraid the unhelpful answer is: "it depends, but not very often".

[janeg]

I'll take a stab at this (hopefully Bugs .. where is he?? or fep will jump in if I'm wrong).

Since you have 2 cards in the suit, there are only 11 remaining cards in the suit to choose from. C(11,2) = 55, 55/1,326 = 4% times 8 opponents = 32% chance another player has 2 cards in the same suit.

Of that, there are 6 cards higher than yours. C(6.1) = 6/55 = 10%. So there's a 10% chance that if another player did receive 2 cards in the same suit, one of them is higher than your 8. So roughly 3.2% chance some else was dealt a higher hand in the same suit.

Most books recommend discounting 1 out for the possibility your flush will be beat by a higher flush.

Edit: Aarrrgghhh ... I'm pretty sure I've goofed up on determining how many of the 55 possible hands could have cards higher than 8. And I'm not convinced the following is right either

There are 6 cards higher than yours, there's 4 ways to make a specific suited hand so 6 x 4 = 24. 24/55 = 43% x 33% = 14% chance someone else holds a higher flush hand?

[nsidestrate]

I'll get you started, but Hawk is right that it is a difficult problem to approach from a statistical basis.

The odds of any individual player having two cards of the same suit as you pre-flop are calculated as ((11/50)*(10/49)) or about 4.5% of the time. If you face nine other players, the odds are about about 40% that they hold one of those flush draws. However, most of those combinations are junk like J4s which any reasonable player will have folded.

There are 2,450 two card combinations and 110 suited cards that don't include the two cards you hold. 90 of these combinations are higher than your flush, so there is only a 3.7% chance that someone was dealt a higher flush draw than you. The only flush cards that you beat are 62/52/42/32/54/53/52/43/42/32.

Even that exaggerates the risk, since hands like Q2s beat you, but are unlikely to be played. If we restrict the list to A2s+, K2s+, QTs+, J9s+ and T9s+, we are looking at 49 combinations or 2% risk per person. Even this exaggerates the risk. If you restrict them to KTs+ instead of K2s+ the risk drops to 1.5% per player in the pot or about 13.5% for 9 players.

[nsidestrate]

[janeg]

Ahhh... thanks nside, that looks more reasonable. Knew I had it wrong some how but wasn't sure where

[OOgie]

I appreciate your replies and as always thank-you very much.

thought I would attack the problem 'Backwards."

1. 1,326 different ways to be dealt any two hands
(52 x 51 divided by 2)

2. Of the hands that can beat me.
AXs = 7 hands (AKs...... A2s) - (AQs, ATs,A8s,A7s,A3s)
KXs= 6
JXs- 5
9Xs=4

(7+6+5+4)

Hand totals = 22. 22/1,325 = about 1.6% of possible hands can beat my 8 hi flush

And

of the total possible hands {variations of the remaining 8 suited cards} making it (AXs......42s) = 28/1,325 or a little over 2%


So if I'm correct (?) when you've hit your flush, there is about a 50:1 chance that your opponent, especially if he is a Flush Chaser also hit.

[janeg]

You still have to multiply 1.6% by your 8 opponents since the odds are the same for everyone at the table. 1.6% x 8 = 12.8% or roughly 7:1 someone will have a higher flush.

Usually what you have to worry about is someone sticking around with A-K-Q in the same suit hoping to river you.

[nsidestrate]

[OOgie]

janeg, thanks for the Icing on the Solution.

2% or 1 out chance that someone else also holds two of that same suit, - Times the number of opponents at the table (minus 2% for each player who mucked suited garbage ).

Thank-you All for helping on this one.