Odds of two people flopping a flush in a 4 handed game ?

[theredpill5]

I just got knocked out of a tourney by some guy who kept flopping flushes. He flopped two flushes in 3 hands. The first one, I also flopped a flush but a 9 high one. He had a queen high. I shouldn't have let him complete the SB. He was completing too many SB's. Then he flopped another one 2 hands later.

This game only had 4 players in it.

[codewarrior]

1.25/10,000 <-- (EDIT: wrong, this is heads-up

1.6/100,000 four-handed assuming neither of the two other players holds a flush card)

[Slim Pickens]

Nice catch. /images/graemlins/smirk.gif

[codewarrior]

Wow, you saw the first reply before I deleted it, forgetting to calculate the odds of two players holding four cards of the same suit? /images/graemlins/shocked.gif

[Slim Pickens]

Yeah. I got 1.5/10,000 but realized it's a little lower than that since the other two can be holding some of the suited cards too. Then I saw your edited post. I'm no statistician.

Slim

[rickr]

I've told you before Red, about 90%, lol. Least it is for me.

Later,
Rick

[codewarrior]

Odds of 2 players holding 4 flush cards:

(13/52)*(12/51)*(11/50)*(10/49) = .00264

There are 48 cards remaining, but the two other opponents have four of them, so 44 cards remaining. Now, assuming neither of those two hold any of the remaining 9 flush cards, the odds of a suited flop are:

(9/44)*(8/43)*(7/42) = .00632

The probability of both of these happening is:
(.00632)*(.00264) = .000016 or 1.6/100,000

[Benholio]

[ QUOTE ]
Odds of 2 players holding 4 flush cards:

(13/52)*(12/51)*(11/50)*(10/49) = .00264

There are 48 cards remaining, but the two other opponents have four of them, so 44 cards remaining. Now, assuming neither of those two hold any of the remaining 9 flush cards, the odds of a suited flop are:

(9/44)*(8/43)*(7/42) = .00632

The probability of both of these happening is:
(.00632)*(.00264) = .000016 or 1.6/100,000

[/ QUOTE ]

Don't forget, there are 6 different pairs of players that could be holding the suited hands, and 4 different suits in which they could be suited.

[Slim Pickens]

[ QUOTE ]
Odds of 2 players holding 4 flush cards:

(13/52)*(12/51)*(11/50)*(10/49) = .00264

[/ QUOTE ]
...but they can be any one of four suits.

Odds of 2 players holding 4 flush cards:

(12/51)*(11/50)*(10/49) = .0106

...and any one of your three opponents can be holding them, so it's close enough to multiply by three. Right?
=0.0317

Now we have 48 unseen cards, 9 of which are of the correct suit, so the flop comes suited 9/48*8/47*7/46=0.00486

(0.00486)*(0.0317)=1.5/10,000?

Anyway, I'm not good at this stuff but that's what I get.

Slim

[theredpill5]

is that 1.5 % ?

or 1.5 in 10,000

so .00015 % ?

Geez, is it .00015 % ..oh my god. I knew it was pretty high.

I'm having crazy bad beat day today.

AA beaten by 99 just now

Also this happened. Heads up

I have K J

Flop: K 7 7

guy raised preflop. I called.

I check so he would hopefully bluff at it.

He pushes all-in

I call

Turn: 6
river: 6

He shows A6

I'm like WTF. I can see him bluffing at it but why push all-in on it with ace high ? GEEZ

[Slim Pickens]

1.5/10,000=1.5e-4=0.015%

No promises that number is correct.

[codewarrior]

You lost me. I thought this was 4-handed situation specific. I added the condition that the other two players don't hold flush cards to the board.

What did I miss? I did this while I was on the phone.

[codewarrior]

Odds of two players flopping a flush: I.E. they both hold two cards of the same suit, and three more of that suit on the board.

[codewarrior]

Your calculation is correct for two heads-up players. That's the mistake I made the first time, unless, as I've already said, I'm missing something.

[Benholio]

Your calculation would be true for a specific 2 players and a specific suit. Like say, player 1 and 3 both flop a clubs flush. But what about player 2 and 3 both flopping a hearts flush? Or player 1 and 4 both flopping a diamonds flush?

[codewarrior]

You are of course, correct.

I read it as "I flopped a flush, what are the odds my opponent did too?"

I think we are both right, but with different starting conditions. Mine promotes better paranoia!!!!!

[Slim Pickens]

I read it as "I flopped a flush. How often did any one of my three opponents flop the same flush?"

[codewarrior]

This is good reasoning also. Isn't this fun?

In summary - not too god-damn often to fear the flush playing that short!!!

[Benholio]

(nitpick)
Well, if you already know hero has flopped a flush, you only need to calculate the odds that someone else has 2 of the remaining 8 cards of that suit.
(/nitpick)

[Slim Pickens]

Um... uh... oh yeah... huh... That's, like, totally right Beavis.

[gasgod]

Doesn't this belong on the "I am such a whiner" forum?


GG

[Slim Pickens]

How about this?

1-[1-(9/47)*(8/46)]^3=0.0966 = 10%

[Benholio]

[ QUOTE ]
How about this?

1-[1-(9/47)*(8/46)]^3=0.0966 = 10%

[/ QUOTE ]

Well, there are only 8 left of the suit, so I think 8/47 and 7/46 would do it.

[Slim Pickens]

No way man. 13-5=9. End of story. It's crap like this that makes me glad I'm not a engineering undergrad anymore. Feh.

1-[1-(8/47)*(7/46)]^3=0.0757 = 7.6%

...and that's why I stay away from the stats forum.

Slim