Hold'em Flush Scenario

Factual Questions
1

I have 2 clubs down in my hand, there are 2 clubs on the flop, what’s the odds at least one more club will be turned?

Let’s say there are 10 players, so 20 cards are dealt out. Assuming randomness, that means 20/52 of the 13 clubs are out there already. That’s 5 clubs total, and I have 2 of them. Now, 2 clubs are in the flop, that leaves 6 clubs in the deck of 29 remaining.

So there are 4 possibilities,

N N = no-club, no-club
N C = no-club, club
C N = club, no-club
C C = club, club

N N = 23/29 X 22/28 = 0.62
N C = 23/29 X 6/28 = 0.17
C N = 6/29 X 23/28 = 0.17
C C = 6/29 X 5/28 = 0.04

So, the odds that AT LEAST one club will be rolled is 0.32, or about 1 chance in 3.

Is this correct?

2

Whoops, almost made it all the way through without a dumb mistake.

Should be:

So, the odds that AT LEAST one club will be rolled is 0.38, or just a little bit better than 1 chance in 3.

3

It’s 35%. Here’s an Out Chart

4

BTW, the way you calculated the odds is a little off. It makes no difference how many players are playing. Since you don’t know their hands, it makes no sense to use the probably distribution of the suits in figuring out our figure. Just pretend it’s you vs. the rest of the deck.

What you do is basically (like you did) calculate the odds of your card NOT hitting twice in a row, and then subtract that from 1.

The formula would be this:

1 - [ (47-OUTS)/47 * (46-OUTS)/46 ]

Since there are 47 cards remaining on the first draw, there are 38/47 possibilities that will not improbe your hand. On the second draw, there are 37/46 possibilities that will not improve your hand. Subtract that from 1, and you get 35%.

5

I’ll assume you mean exactly 2 clubs on the flop (and not 3, since then you’d already have the flush).

You only have knowledge of 5 cards (your two plus the three in the flop), so each of the 47 remaining cards has an equal chance of being dealt, among those are the 9 remaining clubs.

38 of the remaining 47 are not clubs. There are (47 choose 2) = 1081 different two card combinations that can come up next. (38 choose 2) = 703 of these contain no clubs, so 1081 - 703 = 378 possible pairs contain at least one club.

So the probability of getting the flush is 378/1081 = 34.9676…%

6

Great stuff, thanks.