Three suited on the river, how often is there a flush?

[bisonbison]

I'm not one of the paranoid types who thinks that every time there's a flush on the board that someone has it, but I do wonder what the actual frequency is, and I don't know how to do the math. Here's a ton of flush related questions. If anyone feels like answering any of these, I'd be forever grateful.

Assuming 10 players who will automatically go to showdown, if there's a three-flush on the board, how often will someone/anyone have a flush? For 2-9 players?

If I'm heads-up and don't have the flush, how often will my opponent have a flush?

If I have 32s and it's 10 to the river, how often will someone have a higher flush? For 2-9 players?

[sthief09]

Here's the probability that a random opponent has a flush:
C(10,2)/C(47,2) = 4.16% = 23-1

2-card combinations out of 10 clubs. On the denominator, we take out your 2 cards (assuming neither is a club) and the 3 clubs on the board, to get 47 cards.

This probably isn't right, but I'll take a stab at the chance someone has the flush. I think it's 1-(.9584)^n.

1 4.16% 23-1
2 8.15% 11-1
3 11.98% 7-1
4 15.64% 5-1
5 19.15% 4-1
6 22.52% 3-1
7 25.74% 3-1
8 28.83% 2-1
9 31.80% 2-1

[Henke]

Brian Alspach has written an article about loosing flushes that can be found here. (http://www.math.sfu.ca/~alspach/mag57/) The computations can be found here. (http://www.math.sfu.ca/~alspach/comp37/)

[alThor]

Let's look at the easiest one.

[ QUOTE ]
Assuming 10 players who will automatically go to showdown, if there's a three-flush on the board, how often will someone/anyone have a flush?

[/ QUOTE ]

Let's do P(no one has flush) with 10 random hands at showdown, and exactly three hearts on the board. If no one has a flush, then each player has either zero or one heart in his hand. To get an exact answer to the question, we have to add up all the combinations of possibilities for how many of the ten players have zero and/or one heart in his hand.

Example: get the probability that no one has a flush because no one has even a single heart in his hand. This is the easiest, it's just 37/47 * 36/46 * ... * 18/28 = .001629. I.e., all 20 cards in the hands are all non-hearts.

Example: get the probability that no one has a flush because nine people have not a single heart, but a tenth player has one heart. If we knew that seat#1 held the one heart, the probability would be 2*(37/47 * 36/46 * ... * 19/29 * 10/28). Since we don't know who holds the heart, we multiply the probability by 10 choose 1 (=10), and get around 1.81%.

There's probably an elegant formula that does this more quickly, but I used Excel and these kinds of formulas to get P(no flush) = .63438. So there's a 37% chance of at least one flush.

Unfortunately I don't have time to do all the other exercises you posed. But this methodology should work. Good luck.

al'Thor

[k000k]

[ QUOTE ]
Here's the probability that a random opponent has a flush:
C(10,2)/C(47,2) = 4.16% = 23-1

2-card combinations out of 10 clubs. On the denominator, we take out your 2 cards (assuming neither is a club) and the 3 clubs on the board, to get 47 cards.

1 4.16% 23-1
2 8.15% 11-1
3 11.98% 7-1
4 15.64% 5-1
5 19.15% 4-1
6 22.52% 3-1
7 25.74% 3-1
8 28.83% 2-1
9 31.80% 2-1


[/ QUOTE ]

Why put your 2 cards back in the deck for him to draw from? His 2 cards MUST be from the 45 remaining cards.. (We are talking about after the river is down right?) You know he can't have any of the 5 board cards, and he surely doesn't have one of your hole cards. So I'm not understanding the 47 there...

Your numbers say: "For if you get a completely suited flop where you hold none of that suit, what is the probability that an opponent FLOPPED a flush?"

I think it's C(10,2)/C(45,2) = 45/990 = .5 = 19:1. The odds he got his flush are slightly better than you think.

[ QUOTE ]

I think it's 1-(.9584)^n


[/ QUOTE ]
I say 1-(.95)^n,

And similarly for bisonbison's numbers for if you hold ONE of the suit, your opponent has a flush more often than that..

[sthief09]

Yeah, you're right. I wasn't considering that he other 2 board cards can't be clubs and that he can't have them. Good catch...

[k000k]

Here's a list of pct's/odds:

If you hold NONE of that suit:
C(10,2)/C(45,2) = 45/990 = .5 = 19:1

1 .05 19:1
2 .0975 9.3:1
3 .1426 6.0:1
4 .1855 4.4:1
5 .2262 3.4:1
6 .2649 2.8:1
7 .3017 2.3:1
8 .3366 2.0:1
9 .3698 1.7:1


If you hold ONE of that suit:
C(9,2)/C(45,2) = 36/990 = .5 = 27.5:1

1 .0364 27.5:1
2 .0714 14.0:1
3 .1052 8.5:1
4 .1377 6.3:1
5 .1691 4.9:1
6 .1993 4.0:1
7 .2284 3.4:1
8 .2565 2.9:1
9 .2835 2.5:1

If you hold TWO of that suit:
C(8,2)/C(45,2) = 28/990 = .5 = 34.4:1
1 .0283 34.4:1
2 .0558 16.9:1
3 .0825 11.1:1
4 .1084 8.2:1
5 .1336 6.5:1
6 .1581 5.3:1
7 .1812 4.5:1
8 .2051 3.9:1
9 .2276 3.4:1

[k000k]

woops, small error on the

C(10,2)/C(45,2) = 45/990 = .5 = 19:1
C(9,2)/C(45,2) = 36/990 = .5 = 27.5:1
C(8,2)/C(45,2) = 28/990 = .5 = 34.4:1

They arent all = to .5... It's .05, .0364, and .0283.....

The charts are correct tho..

[z32fanatic]

excellent post guys. I too am always scared of a flush whenever there are 3 suited cards on the board. Now I know that when there are 3 suited cards, there is only a 1/3 chance that anyone would've had the flush, and a percentage of those people may have folded preflop. This will probably make me, and other people tons more money.