Flush on Stud - Help

[kav]

Hello.

I'm trying to calculate the probability of making a flush in stud when my first three cards are of the same suit. Suppose I have no opponents, otherwise I'll go crazy [img]/images/graemlins/smile.gif[/img]

Is this right?

(10/49 * (9/48 + 9/47 + 9/46)) + (10/48 * (9/47 + 9/46)) + (10/47 * 9/46)

I tried to explain what I did by words, but I couldn't cause my english is kinda poor [img]/images/graemlins/blush.gif[/img], the numbers speak for themselves.

Cheers

[pzhon]

[ QUOTE ]
Is this right?

(10/49 * (9/48 + 9/47 + 9/46)) + (10/48 * (9/47 + 9/46)) + (10/47 * 9/46)

[/ QUOTE ]
No. Every term which relies on not getting a flush card is missing a factor.

So, for example, the 10/49(...+9/47+...) is supposed to represent picking up the 4th card to your flush on 4th street and making the flush on 6th street. For that to happen, you need to miss on 5th street, so you need to multiply by 39/48. Similarly, the 10/47*9/46 term represents picking up the 4th card on 6th street and making the flush on the river. For that to happen, you had to miss on 4th and 5th streets, so you need to multiply by 39/49 and 38/48.

[bigpooch]

The chances of NOT getting a flush are [C(39,4)+10 C(39,3)]
over C(49,4) or 173641/211876 so the chances of making a
flush are about 0.180459325.

[tshort]

4 cards left to be dealt You want to find the probability that two, three, or four of them are of the desired suit. It is easier to take 1 - the probability none or one of them are of the desired suit.

Prob 0: C(39,4) / C(49,4) = .3882

Choose 4 cards from 39 non-desired suit cards. Divide by total possibilities.

Prob 1: (C(10,1) * C(39,3)) / C(49,4) = .4313

Choose 1 card from 10 desired suit cards. Choose 3 cards from 39 non-desired suit cards.

1 - .3882 - .4313 = .18046

Alternatively:

Prob 2: (C(10,2) * C(39,2)) / C(49,4) = .1574

Prob 3: (C(10,3) * C(39,1)) / C(49,4) = .0221

Prob 4: C(10,4) / C(49,4) = .0010

.1574 + .0221 + .0010 = .1805

[kav]

Guys, thanks A LOT for the help. I knew I was missing something but I couldn't see what.

I tried again after reading your post's and I *think* I got it right.

(10/49 * (9/48 + (39/48 * 9/47) + (39/48 * 38/47 * 9/46)) +
39/49 * (10/48 * (9/47 + (38/47 * 9/46)) +
39/49 * 38/48 * (10/47 * 9/46)

It's big [img]/images/graemlins/laugh.gif[/img].

I've never studied probability so I don't know what these C(X, Y) mean or how to calculate it.

Thanks again, I really appreciate! [img]/images/graemlins/smile.gif[/img] [img]/images/graemlins/smile.gif[/img]