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View Full Version : Probability of flopping 1 card flush draw?

Mark1808
04-13-2005, 04:28 PM
I come up with a number I think is too high and I think it has to do with inclusion - exclusion.

Probability you hold two diamonds, that you will flop 2 diamonds =

P(1st and 2nd cards are diamonds)= 11/50 * 10/49 = 4.49% or
P (2nd &amp; 3rd cards are diamonds) = 11/49 * 10/48 = 4.68% or
P (1st &amp; 3rd card are diamonds) = 11/50 * 10/48 = 4.58%

P(of floppin 1 card draw) = 4.49 + 4.68 + 4.58 = 13.75%

Then probaility of flopping flush = 11/50 * 10/49 * 9 /48 = .84%

I'm wondering if I should be using the choose function to calculate these?

David K.
04-13-2005, 05:00 PM
You're adding the probability that the 3 of them are diamonds 3 times:

P(1st and 2nd cards are diamonds, 3rd is or is not a diamond)+
P(2nd &amp; 3rd cards are diamonds, 3rd is or is not a diamond)+
P(1st &amp; 3rd cards are diamonds, 3rd is or is not a diamond)

That's your mistake. You probably have the right result if you subtract 2x0.84 from your result.

The proper way to calculate those odds are with combinations (Cn,p), the easy way is by subtracting the odds of NOT making your flush to 1.

RedeemerKing
04-13-2005, 05:02 PM
I think its around 10% so u aint that far off

gaming_mouse
04-13-2005, 06:06 PM
[ QUOTE ]
I come up with a number I think is too high and I think it has to do with inclusion - exclusion.

Probability you hold two diamonds, that you will flop 2 diamonds =

P(1st and 2nd cards are diamonds)= 11/50 * 10/49 = 4.49% or
P (2nd &amp; 3rd cards are diamonds) = 11/49 * 10/48 = 4.68% or
P (1st &amp; 3rd card are diamonds) = 11/50 * 10/48 = 4.58%

P(of floppin 1 card draw) = 4.49 + 4.68 + 4.58 = 13.75%

Then probaility of flopping flush = 11/50 * 10/49 * 9 /48 = .84%

I'm wondering if I should be using the choose function to calculate these?

[/ QUOTE ]

The chance of flopping EXACTLY two diamonds:

(11 choose 2)*39/(50 choose 3)= .109

To get this same answer using your method, you'd need to do:

P(1st diamond, 2nd diamond, 3rd NOT diamond) = 11/50*10/49*39/48

Alter your second and third calculations similarly. Then the two answers will agree.

Mark1808
04-13-2005, 08:20 PM
Thank you so much, I'm getting someplace. I think I understand the choose method now. 11 chose 2 gives the number of ways you can get 2 diamonds from remaining 11 diamonds, multiply by the number of non-diamonds (each non-diamond can match up with each 2 card diamond combo) divided by 50 chose 3 (the number of possible flops).

So the possibility of flopping a flush would be (11 choose 3) / (50 choose 3) = .84% Damn!! Thank you Gambling Mouse!

Now the odds of hitting the flush with two cards to come would be p(diamond turn &amp; blank card river) + P(blank card turn &amp; diamond card river) + P(Diamond card turn &amp; river)

= 9/46 * 37/45 + 37/46 * 9/45 + 9/46 + 8/45 = 35.65%?

Can you do that with the choose method too? I am off I get
(9*38)+ (9*8) / (47 choose 2)= 38%

gaming_mouse
04-13-2005, 11:15 PM
[ QUOTE ]

= 9/46 * 37/45 + 37/46 * 9/45 + 9/46 + 8/45 = 35.65%?

Can you do that with the choose method too? I am off I get
(9*38)+ (9*8) / (47 choose 2)= 38%

[/ QUOTE ]

Sure. First fix your method:

9/47 * 38/46 + 38/47 * 9/46 + 9/47 * 8/46 = 0.349676226

Using the counting method:

(9*38 + (9 choose 2))/(47 choose 2) = 0.349676226

And finally, taking the complement of NOT getting any more of your suit:

1 - (38/47)*(37/46) = 0.349676226

They all agree.