whats the formula to figure

[mostsmooth]

what the chances are of completing a flush with 3 suited on 4th? say you have 3 clubs, and there are no other clubs out
also, are there tables somewhere that would tell me this already?

[TomCollins]

Wow, tough question. 9 clubs left, and 46 cards left. I'm gonna go out on a limp and say the odds are 37-9.

[ohgeetee]

I think it is impossible, unless your question is asking something else. With only 3 clubs, you couldnt complete a flush on 4th street.

[mostsmooth]

[ QUOTE ]
Wow, tough question. 9 clubs left, and 46 cards left. I'm gonna go out on a limp and say the odds are 37-9.

[/ QUOTE ]
wow, really?, 37-9 to get 2 more clubs? im guessing youre wrong. and who told you it was heads up? and, that wasnt much of a formula you gave. and, wouldnt there be 10 clubs left? and, did you really mean "go out on a limp", or did you want to say "limb" like normal people do?
i know it was a tough question like you said, maybe ill make my next question easier for you
thanks for the help

[mostsmooth]

[ QUOTE ]
I think it is impossible, unless your question is asking something else. With only 3 clubs, you couldnt complete a flush on 4th street.

[/ QUOTE ]
sorry, i just assumed everybody would understand "3 suited on 4th" means you have 3 suited cards on 4th

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
I think it is impossible, unless your question is asking something else. With only 3 clubs, you couldnt complete a flush on 4th street.

[/ QUOTE ]
sorry, i just assumed everybody would understand "3 suited on 4th" means you have 3 suited cards on 4th

[/ QUOTE ]i also assumed that its obviously not a holdem question, its a stud question
/images/graemlins/grin.gif

[Shawsy]

[ QUOTE ]
what the chances are of completing a flush with 3 suited on 4th? say you have 3 clubs, and there are no other clubs out
also, are there tables somewhere that would tell me this already?

[/ QUOTE ]

I'll take a stab at this one, though I don't profess to be an expert in 7-stud.

If there are n players other than you in the game, then there are 48-2n cards that you haven't seen. In your example you said there are no clubs showing, so there are 10 left in the deck. Let me work with that for now, but you could make a more general formula too. Let me also for sake of illustration say that there are four other players in the hand with you, so n=4 and 48-2n is 40 (10 clubs and 30 non-clubs).

To flush, you need at least two clubs from the remaining 40 cards in the deck. There are C(40,3) = 9880 combinations of three remaining cards that you can get in all.
There are C(10,2) x C(30,1) = 1350 ways of getting two clubs, and C(10,3) = 120 ways of getting three clubs. Total probability of flushing with 3 on 4th street would be 1470/9880 or about 14.9%.

If you want a general formula, let n be the number of players other than you in the game (and assume that they all get to 4th street), and i be the number of clubs showing on the board, other than yours.
The probability of getting two more clubs is C(10-i,2)xC(48-2n+i,1). The probability of getting three more clubs is C(10-i,3). The total number of ways the remaining three cards could come is C(48-2n,3).
I don't know if there are any tables out there, but the probability is a function of both the number of other players in the game (really the number of face-up cards you have seen) and the number cards of your suit that are showing (and/or were shown if they have since been folded - when not all the players get to 4th street). You could probably do up a chart in MS Excel easily enough. I might give it a whirl some time.

Good luck,

[mostsmooth]

thanks, that looks like it makes sense
ill go with that /images/graemlins/cool.gif

[Shawsy]

I am going to try to attach a file. Sorry, can't figure out how to do it. PM me with an e-mail address and I can send you one.

[TomCollins]

Flame on, I deserve it (as I do most times). I had no clue he was asking a stud question. (To be honest, I don't even know what the hell he was trying to ask).