probabilty of ATC opponent flopping flush draw
 

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?

Re: probabilty of ATC opponent flopping flush draw
 

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.

Re: probabilty of ATC opponent flopping flush draw
 

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.

Re: probabilty of ATC opponent flopping flush draw
 

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.

Re: probabilty of ATC opponent flopping flush draw
 

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.

Re: probabilty of ATC opponent flopping flush draw
 

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.

Re: probabilty of ATC opponent flopping flush draw
 

[ 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.