wrote application which simulates set or better on flop

[SuitedKingAce]

Hello all,

I wrote a little application (I am an computer-science student, so it's a good exercise) which enumerates all tuples of seven cards, of which the first two cards (the player's cards) have equal rank:

Ah Ak 5h 8c 3d 9h 3s
6h 6d 7c 2d Kd 3s Td
6h 6d 7c 2d Kd 3s 6c
6h 6d 7c 6c Kd 3s 7h
...

According to my application, there are 3494569 tuples like that.

Then, I enumerated all tuples of seven cards, of which the first two cards (the player's cards) have equal rank AND of which the flop (third, fourth and fifth card) has at least one card of the same rank as the rank of the player's cards:

6h 6d 7c 6c Kd 3s 7h
6h 6d 7c 6c 7h 3s Kd
...

(in other words, the second set is a subset of the first)

According to my application, there are 542906 tuples like that.

So, this might lead to the following conclusion: the cance of flopping set or better is 542906/3494569 = 15.53%.

However! We all know that the chance of flop a set or better is 1-[(48/50)*(47/49)*(46/48)] = 11.86%.

I'm pretty confident that my application is correct, as it's very straightforward.

Does anybody have an idea why I'm getting this result?

[AaronBrown]

This doesn't make sense to me.

There are 13 pairs for the pocket cards, and 6 ways to get each. Since the order of the board matters, there are 50 x 49 x 48 x 47 x 46 = 254,251,200 boards. Overall, that's 19,831,593,600 tuples, over 5,000 times your value. So I'm not sure what you're doing in your enumerations.

Suppose you start with a single pair, say AA (obviously it doesn't matter which pair). With 50 cards, there are C(50,3) = 19,600 possible flops. C(48,3) = 17,296 of these have no Aces. 2*C(48,2) = 2,256 of these have one Ace. 48 of these have two Aces. (2,256 + 48)/19,600 = 0.1176 (not 0.1186). Any enumeration you do should have results proportional to these numbers.

[DarkMagus]

So you want us to debug your program without being able to see the source code?