Odds on hitting 4 flush?

[carpe2k]

I realize this has probably been asked before, but as the search function is worthless, I will ask anyway.
My pocket aces have been beaten by players catching a 4 flush in the last two tournaments. What are the odds of catching a flush if you have one to the suit?
Thanks in advance. /images/graemlins/confused.gif

[closer2313]

Should be

[C(12,4) * C(36,1) / C(48,5)] * 100 ~ 1.04%

[Luzion]

[ QUOTE ]
Should be

[C(12,4) * C(36,1) / C(48,5)] * 100 ~ 1.04%

[/ QUOTE ]

Hes asking what is the probability of hitting a 4 flush if you only hold one of the suited cards. It should be

[ C(12,4) * C(38,1) ] / C(50,5) = 0.0089 = 0.89%

[closer2313]

I think the first part of his statement was implying what were the odds of them hitting a 4 flush versus his pocket aces. If that was the question then my answer is right.

If he wants to know what the probability of hitting a 4 flush if he only knows his hand, then your answer is right.

But actually, both answers need to be multiplied by 2 becuase you can make 2 flushes. so

If you hold AA and your opponent has 2 offsuit cards of different suits from your AA, then the probability of him making a 4 flush is

[C(12,4) * C(36,1) / C(48,5)* 2] * 100 ~ 2.08%

If you hold 2 offsuit cards then the probability of hitting a 4 flush is

[ C(12,4) * C(38,1) ] / C(50,5) * 2 = 0.0178 = 1.78%

[Luzion]

Good point with the * 2 since he is holding two offsuit cards that can each make a 4-flush.

[Cobra]

There is some confusion of what information you are looking for. Lets assume you have aces, the flop comes with two to a suit and you hold one of that suit. The following is the probability that if one of your opponents has two of the suit in question he will hit a flush.

You: /images/graemlins/spade.gif /images/graemlins/diamond.gif
Flop: /images/graemlins/spade.gif /images/graemlins/spade.gif /images/graemlins/heart.gif
Opponent: /images/graemlins/spade.gif /images/graemlins/spade.gif

In this situation there would be 45 unknown cards eight are spades and 37 are blanks.

Probability of ending up with 3 /images/graemlins/spade.gif= 8*37/(45c2)=29.9%

Probability of ending up with 4 /images/graemlins/spade.gif=(8c2)/(45c2)=2.8%

Note that if you end up with four spades your opponent would loose. So the chance of you loosing to a flush draw is 29.9%.

Cobra

[Schizo]

what dies C(12,4) mean?

[benneh]

12 choose 4, or its just another way of showing a math equation.

in plain math, it translates into 12! / ((12-4)! * 4!)