Two players flopping a flush - Gambling and Probability

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Two players flopping a flush

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07-05-2009 , 11:56 PM

T0ken

enthusiast

Join Date: Apr 2009 Posts: 54

I'm trying to figure out the odds of three of the same suit on the flop giving two players a flush.

If someone could answer this for me and possible explain how to arrive at the answer, I'd appreciate it.

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07-06-2009 , 09:57 AM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 5,038

If you assume two players already have suited cards of the same suit, there are 9 cards left of that suit with 48 cards left in the deck. Therefore, both flopping a flush has probability

9/48 x 8/47 x 7/46 = 0.004857.

If you are asking what is the probability that two specific players get the same suited cards and then flop a flush, the result is a lot lower. This implies that of the 7 cards dealt - two to each player and three on the flop all are the same suit. This can happen with probability 4*C(13,7)/C(52,7) =0.000051.

For n players, a first order approximation is 1- (1-0.000051)^(n-1)

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07-06-2009 , 03:45 PM

T0ken

enthusiast

Join Date: Apr 2009 Posts: 54

Thanks for the reply. I'm not sure if that answers my question or not. Let me try and word it differently.

Say I have some mid or lower suited connectors and a couple of players see the flop. I flop a flush and a situation arises where I have to call an AI. Mathematically speaking only, what are the odds that I'm up against another flush?

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07-06-2009 , 03:51 PM

jmark

Pooh-Bah

Join Date: Dec 2003 Posts: 3,790

There are 8 of that suit left and 47 unknown cards. Each opponent has an (8/47)*(7/46) chance of having the flush as well.

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07-06-2009 , 05:41 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 5,038

Quote:

Originally Posted by T0ken

Okay - that's a different question

For 1 or 2 opponents to have a flush under the stated condition, then

Case 1:A and B each have 2 flush cards

Case 2 A has 2 flush cards and B doesn't
or A doesn't have 2 flush cards and B does

The probability the first player has two flush cards is 8*7/(47*46) = 0.025902, say P1

The probabilty the second has 2 flush cards given the first has is (6*5)/(45*44)=0.015152, say P2

For Case 1, the probability is
P1*P2= 0.000392

For Case 2, the probability is
2*[P1* (1-P2)]=.051019

Therefore, the total probability is .0514, or odds against of 18.5 to 1

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07-08-2009 , 02:19 PM

porquinho

adept

Join Date: Dec 2008 Posts: 709

Where I play (brick and mortar), on sunday regular tournament, 3 people flopped a flush draw, and it hit on turn. One with 78, one with A9 and one witl KJ.

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07-19-2009 , 09:19 PM

sld007

stranger

Join Date: May 2006 Posts: 1

That cannot be right. If he odds of one person flopping a flush is approximately 118 to 1 then how can the odds of two people flopping a flush only be 18.5 to 1?????

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07-19-2009 , 09:55 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 5,038

Quote:

Originally Posted by sld007

That cannot be right. If he odds of one person flopping a flush is approximately 118 to 1 then how can the odds of two people flopping a flush only be 18.5 to 1?????

You are comparing two different things. The 18.5 to 1 was answering OP's question of one or two other players having a flush GIVEN that OP had flopped a flush. In other words, the flop is three cards of the same suit matching OP's suited cards.

The 118 to 1 you quoted would be the odds against OP flopping a flush or the odds against a flop of three cards that match OP's suited cards.
'

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