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dabluebery
05-14-2004, 10:15 AM
Hi. I know that when you hold 2 suited cards in holdem, there's about an 11% chance of flopping at least two of your suit, for a flush draw. I know this because I've read it everywhere.

But HOW can I calculate this myself? I'm fuzzy on my combinations and permutations, and I can't find an explicit example to guide me anywhere.

I know there are 50 cards left, and you have to choose three of them, so the denominator should be 50c3 which is 19,600 possible ways to arrange the flop, but after that, I'm lost. How many different arrangements of the flop have two of my suit? There are 11 left.

Anyway, I'd LOVE some detailed equations explaining how I can do this myself. Thanks.

Rob

Johan
05-14-2004, 10:37 AM
19600

possibilities first 2 your suit 11/50*10/49= 4.49%

card 2 3 your suit 11/49 * 10/48 = 4.68%

card 1 3 your suit 11/50 * 10/48 = 4.54%
Total = 13.71 %

So a little higher i guess... Includes 3 your suit flops so should be about accurate
Johan
PS: 11/50 * 10/49 * 9/48 = .84% 13.71%-.84% = 12.87% so maybe i am missing something here, or the 11% is not completely accurate

PairTheBoard
05-14-2004, 01:09 PM
The number of ways the two cards of your suit can be dealt is 11c2 = 55. The number of ways the offsuit card can be dealt is 39. So your answer is:

55*39 / 50c3 = 10.94%

PairTheBoard

dabluebery
05-14-2004, 03:26 PM
THANKS. That's perfect. I'm so stupid.

Rob

steveyz
05-14-2004, 03:41 PM
[ QUOTE ]
19600

possibilities first 2 your suit 11/50*10/49= 4.49%

card 2 3 your suit 11/49 * 10/48 = 4.68%

card 1 3 your suit 11/50 * 10/48 = 4.54%
Total = 13.71 %

So a little higher i guess... Includes 3 your suit flops so should be about accurate
Johan
PS: 11/50 * 10/49 * 9/48 = .84% 13.71%-.84% = 12.87% so maybe i am missing something here, or the 11% is not completely accurate

[/ QUOTE ]

In your calculation, you need to include the percentage that the 3rd card is NOT one of your suit. I.e.

Probability that ONLY card #1 & #2 are your suit:

11/50 * 10/49 * 39/48 = 3.65%

This is equivalent to the probability that only #1 & #3, or #2 and #3 are your suit, so the probably is 3 * 3.65 = 10.94% ~ 11%

brassnuts
05-20-2004, 04:40 AM
I thought I was good at math until now. I cannot figure out why there are only 19600 variations on a flop with 50 cards left in the deck. I played around with the numbers and found out that 50×49×8=19600. I'm just wondering why 50c3 (and what does 'c' stand for?) is not 50×49×48=116700?

tech
05-20-2004, 05:00 AM
50c3 has a mathematical meaning called a combination. You have 50 possible cards, and you are choosing 3 of them. The order you choose them doesn't matter (i.e., doesn't matter which of the three cards you pull first).

The formula is as follows:

mCn = m! / (m-n)! / n!

So, 50c3 = 50! / 47! / 3! = 50*49*48/6 = 19600

The ! is for factorial.

BugsBunny
05-20-2004, 07:04 AM
you transposed some digits. 50 * 49 * 48 = 117600

This gives the number of permutations rather than combinations. With permutations order is significant. With combinations order is not significant.

Assume a deck with 3 cards:
ABC

using your calculation you would come up with 3 * 2 *1 = 6 possible flops. And you'd be right if order nattered:
ABC
ACB
BAC
BCA
CAB
CBA

but order doesn't matter. All those flops are equivalent for purposes of evaluating a hand so the number of combinations is 1.

The general formula for permutations with n objects taken in groups of r is:
n! / (n-r)! where ! is a factorial symbol (and 0! = 1). So in the case of a flop you would have:
50!/(50-3)! = 50!/47!

This can be rewritten as:
50*49*48*47!/47!

The 47! in the denominator and numerator cancel each other out leaving your original calculation.

The general formula for combinations on the other hand is:
n!/((n-r)!r!)

so again using the flop as an example we have:

50!/((50-3)!*3!) = 50*49*48*47!/(47!*3!) = 50*49*48/3*2*1 = 117600/6 = 19600

It's easier to just use the appropriate keys on the calculator /images/graemlins/smile.gif

Hope this helps.

brassnuts
05-20-2004, 04:42 PM
Thanks Bugs and Tech... it was late last night when I saw the post and shoulda have realized that I was calculating permutations when the order didn't matter. And, it makes perfect sense to me now that the number of combinations would be the number of permutations divided by x = number of selectees factored, because x would in fact be the number of different ways these same cards could be arranged. Awesome, thanks again.

EDIT: BTW, anyone know where the factorial button is on a TI-86?

Randy Burgess
05-21-2004, 08:20 AM
A little obsessive-compulsive but does lay out doing combos pretty well. Or you can get ahold of any basic probability text. I have one published by Dover that I like.