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Old 12-29-2005, 03:01 PM
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Default probability of getting flush with suited hand

Hey guys,

I'm ready Ed Miller's Getting Started in Hold 'Em and he says in there that if you have a suited hand, the probability of you completeing the flush is 6%. "A board with three of your suit will come 6% of the time." (p. 69) I am having a little trouble figuring out how he computed this...could someone help me?

BT
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Old 12-29-2005, 03:38 PM
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Default Re: probability of getting flush with suited hand

he is referring to the probability of getting exactly three cards of your suit on the board. If you include the chance that there are four or five of your suit on the board the chances of making a flush are ~8.4%.

I did the math by taking the probability of getting exactly three of your suit on the board times the ways of getting two more cards on the board [C(11,3)*C(47,2)]/total number of possible boards C(50,5). This computation includes the times that the other cards are of your suit. To exclude those times you would figure the numerator by saying [C(11,3)*C(39,2)]. This yeilds 5.77%, which rounds to the 6% he reported.

If my calculations are off, one of the more probability savvy posters will correct me, I'm sure.
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Old 12-29-2005, 07:45 PM
AaronBrown AaronBrown is offline
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Default Re: probability of getting flush with suited hand

ChicagoVince's method works pretty well, here is the exact way to do it. You have to figure the cases of getting exactly three, four and five suited cards separately.

Three suited cards: C(11,3)*C(39,2) = 122,265
Four suited cards: C(11,4)*C(39,1) = 12,870
Five suited cards: C(11,5)*C(39,0)= 462

To get the probabilities, you have to divide by C(50,5) = 2,118,760. You get 5.77%, 0.61% and 0.02% respectively, for a total probability of 6.40%.
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Old 12-31-2005, 04:49 AM
TheProdigy TheProdigy is offline
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Default Re: probability of getting flush with suited hand

Hey,

I'm just wondering, when you use C(39,2), C(39,1), etc on your three examples, what is the reasons for using those numbers?

Still trying to learn Probabilities,
Prodigy
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  #5  
Old 12-31-2005, 04:59 AM
BruceZ BruceZ is offline
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Default Re: probability of getting flush with suited hand

[ QUOTE ]
Hey,

I'm just wondering, when you use C(39,2), C(39,1), etc on your three examples, what is the reasons for using those numbers?

[/ QUOTE ]

There are 39 non-flush cards and 11 flush cards remaining. C(39,2) is the number of combinations of 2 non-flush cards, and C(11,3) is the number of combinations of 3 flush cards, so C(11,3)*C(39,2) is the number of ways to make a flush with exactly 3 flush cards and 2 non-flush cards on the board.

C(39,1) = 39 is the number of non-flush cards, and C(11,4) is the number of combinations of 4 flush cards, so C(11,4)*C(39,1) is the number of ways to make a flush with exactly 4 flush cards and 1 non-flush card on the board.
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Old 12-31-2005, 06:40 AM
Mike Haven Mike Haven is offline
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Default Re: probability of getting flush with suited hand

For anyone who can't remember why, for example,
C(11,3)*C(39,2) should equal 122,265 this might help:

C(n,r) is often written as nCr

and it equals (n!)/[(n-r)!r!]

(note: n! is spoken as "n factorial" and means multiplying all the numbers from 1 up to n together; so if n = 4, n factorial would mean 1*2*3*4 or 1x2x3x4 or 4x3x2x1 or 24)

which for 11C3 means (11x10x9x8x7x6x5x4x3x2x1)/[(11-3)!3!]

or (11x10x9x8x7x6x5x4x3x2x1)/(8x7x6x5x4x3x2x1)x(3x2x1)

(note: you can simply cross off the (8x7x6x5x4x3x2x1) from the end of the (11x10x9x8x7x6x5x4x3x2x1) without having to multiply them both out, which will give

or (11x10x9)/(3x2x1)

or 165

and for 39C2 means (39x38x...)/[(39-2)!2!]

or (39x38x...)/[(37x36x...)x(2x1)]

or (39x38)/(2x1)

or 741

so 11C3*39C2

or C(11,3)*C(39,2)

or C(11,3) x C(39,2)

or 165 x 741

equals 122,265.

I hope is useful to someone, and if you save these calculations they might make the next combinations thread a little easier to understand.
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  #7  
Old 12-31-2005, 08:41 AM
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Default Re: probability of getting flush with suited hand

As a quick analysis, may I use outs and Bayes Theorem?

38:9 multiplied by 37:9, 1/5 x 1/5 leaving me an answer of about 4%?

Cordially,

mfp
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