probabilty of ATC opponent flopping flush draw

[helter skelter]

I guess this is the forum to post this in. I was about to post in poker theory when I saw this forum.

I am trying to determine the probability of an ATC limper flopping a flush draw, given that I have one of his suit

I take out my 2 cards and opponents 2 cards from the 52 card deck to get 48

There are 10 flush cards left in the deck, he needs to flop 2 out of three on the flop

If I use the poker odds calculator, plugging in 10,2,3,48, I get 0.098866790 probability

If I work backwards after the flop:

I use binomial calculator plugging in 47 (subtracting my cards and flop), 2, I get 1081 possible hands he could have started with

then I go to a hand chart and count the possible hands he could have with 2 flush cards and they total 45

so 45/1081 = 0.041628122

I was expecting these 2 numbers to be the same, so what am I missing here?

[DarkMagus]

Er, your post isn't very clear.

The first number you calculated is, I think, the odds of him flopping a flush draw given that he has two suited cards. The second number appears to be the odds of him having a flush draw given that there are two suited cards on the flop.

Why would you expect those numbers to be the same? They're completely different events.

[helter skelter]

Ok, well I will try to clarify it better.

I am trying to evaluate playing a certain marginal hand against ATC opponents. I want to estimate implied odds and weigh it against the cost of entering the pot, so I am trying to estimate the probabilty of certain situations arising, for example, I flop 2 pair while my opponent flops a flush draw. There are many other situations that might occur with my marginal hand, and I want to evaluate each one.

My marginal hand is offsuit, so first I wanted to find the probability that I will flop 2 pair at the same time that my opponent flops a flush draw given that I have one of his suit. I figured I would multiply the probability that I flop 2 pair with the probability that he will flop a flush draw, using the poker odds calculator.

I then realized that implied odds will be different if his flush draw has an overcard. Although I will win slightly less often in that case, my opponent might be willing to call a much larger bet with that flush draw, so the implied odds could be significantly higher.

I don't know how to model that using the poker odds calculator, so I set up a scenario with a given flop and then went and counted the hands he could have using a hand chart to get a separate count for the 2 cases. Before even separating the cases, though, I noticed that it showed that the percentage of flush draw hands he could have was a much lower percentage out of the possible hands using this method, I wanted to be sure I use the correct method/procedure before I continue, so that is why I am asking this question.

[DarkMagus]

Well, it's a weird question but I guess if you really want the answer the easiest way would be through combinations.

With 48 cards remaining and 10 cards of his suit, there are 10*9 / 2 = 45 combinations of two cards on the flop that could hit. Multiply this by the remaining 38 non-suited cards and you get 1710 possible flops where he has a flush draw. Divide this by 48c3 = 17296 total possible flops, and you get 0.0989 probability.

I'm not really sure why you stipulated that you have one card of his suit, but that's the answer assuming such a condition.

[helter skelter]

Thanks. The numbers make sense.

I was just having trouble wrapping my brain around why the 2 events were so different, where post-flop, the probability that my opponent has a flush draw was less than half the pre-flop probability that he would flop a flush draw.

I had to think about it a couple hours to get the light bulb to turn on. It is not glowing brightly, but at least I can see shadows.

[DarkMagus]

Call the event that he has two spades A, and the event that the flop contains two spades B. The probability of A, given that B has occured, is different from the probability of B given that A has occured.

Maybe it would be a little easier to think of a more extreme example.

Suppose on the board there a 4 card straight flush - say, 5s6s7s8s. What is the probability that he has a straight flush if he has random cards? Not too bad, since he only needs the 9s or the 4s. It actually works out to be 8.5%.

Now, if we know that he has the 9s or 4s preflop, what is the probability of him hitting a straight flush? I won't bother calculating it, but I think you'll agree it should be much less than 8.5%.

You can see that the same concept applies in your flush draw example. If you want to see some math on the subject, look up Baye's Theorem.

[helter skelter]

[ QUOTE ]
The probability of A, given that B has occured, is different from the probability of B given that A has occured.

[/ QUOTE ]

Yeah, I vaguely remember that from my long ago statistics class. I will have to do a review of statistics and poker math before I try to solve any more such problems.