(3) players hit (3) different sets on flop (10 handed)

I just busted out in a tourny when i held middle set of (3) that all hit on the flop, each player holding a pp. Anyone know the odds of this happening??

Also, earlier this week, I busted out when another player flopped a flush, just as I had. Anyone know the odds on (2) players both flopping a flush 10 handed? Thanks.

[BBD]

Try posting this in the probability forum.

i had this happen to in an SNG, flop 974, i had 77, opponents had 99 and 44.

[BBD]

ok, not 100% sure about this but I think the math for all three players flopping a set would look like this:
2C1 * 2C1 * 2C1 / 46C3 that is 2 outs in the deck for player 1's PP with 1 on the flop * 2 outs in the deck for player 2's PP with 1 on the flop * 2 outs in the deck for player 3's PP with 1 on the flop / the total 46 card deck to be drawn from 3 times.

2C1 = 2
46C3 = (46)(45)(44) / (3)(2) reduces to (46)(15)(22)= 15180
2 * 2 * 2 / 15180 = 8 / 15180 = 0.00052 or 0.052%

thus all three players will make a set on the flop 0.05% of the time.

like I said though, you might want to post this in probabiltiy as my math might be incorrect.

[BBD]

As to your question about 2 players flopping a flush. I think the math would look like:

9C3 / 48C3 that is 9 of you and your opponent's suit remaining in the deck (as you both hold a suited hand) three of which will be on the flop / the total 48 card deck drawn from 3 times
9C3 = (9)(8)(7) / (3)(2) reduces to (3)(4)(7) = 84
48C3 = (48)(47)(46) / (3)(2) reduces to (16)(47)(23) = 17296
84/ 17296 = 0.00485 or 0.485%

thus you and your opponent will both flop a flush 0.48% of the time you both hold a hand of the same suit.

again, you may want to post this in probability to make sure the math is right but it should be something like this.

[ QUOTE ]
As to your question about 2 players flopping a flush. I think the math would look like:

9C3 / 48C3 that is 9 of you and your opponent's suit remaining in the deck (as you both hold a suited hand) three of which will be on the flop / the total 48 card deck drawn from 3 times
9C3 = (9)(8)(7) / (3)(2) reduces to (3)(4)(7) = 84
48C3 = (48)(47)(46) / (3)(2) reduces to (16)(47)(23) = 17296
84/ 17296 = 0.00485 or 0.485%

thus you and your opponent will both flop a flush 0.48% of the time you both hold a hand of the same suit.

again, you may want to post this in probability to make sure the math is right but it should be something like this.

[/ QUOTE ]

THANKS for the reply and your time - I did post it over there as well. Any thoughts on factoring in the odds that two players would even hold the same suited cards before the flop, then factoring that in figure to the above numbers to get odds on it happening before any cards are dealt, or is that getting to crazy??

And also for the "(3) Sets" question, could you factor in the odds that even (3) players would hold pocket pairs before the flop, and determine odds that this scenario would happen before any cards are dealt?

Thanks again.

[PeteGI]

A friend of mine was playing a cash game on pokerstars where all three people to the flop hit a flush. He was holding the king, and happened to have enough to create a side pot and break even. Not sure on the odds though.

[BBD]

Hi again DJ. No problem at all. Ok, this is a bit more difficult and I hope I am not giving you the wrong info. But I think the math to figure you AND your opponent holding a suited hand(of the same suit) would look like:
13C4 / 52C4 that is 13 of your suit in the deck before any cards are dealt, of which 4 will get dealt out(2 to you and 2 to your opponent) / the entire deck of 52 cards, drawn from 4 times(2 hole cards to you and 2 hole cards to your opponent)
13C4 = (13)(12)(11)(10) / (4)(3)(2) reduces to (13)(11)(5) = 715
52C4 = (52)(51)(50)(49) / (4)(3)(2) reduces to (13)(17)(25)(49) = 270725
715 / 270725 = 0.00264 or 0.26%

thus you and your opponent will hold a suited hand(of the same suit) 0.26% of the time.

Now, with reguards to your question about you and 2 of your opponents all holding PPs.
To calculate:
4C2 * 4C2 * 4C2 / 52C6 that is 4 of any given rank in the deck with 2 of that rank dealt out to a player * 4 of another given rank with 2 of that rank dealt out to another player * 4 of another given rank with 2 of that rank dealt out to the other player / the entire 52 card deck drawn from 6 times(2 hole cards to each of the three players involved)
4C2 = (4)(3) / (2) reduces to (2)(3) = 6
52C6 = (52)(51)(50)(49)(48)(47) / (6)(5)(4)(3)(2) reduces to (4)(10)(13)(17)(49)(47) = 20358520
6 * 6 * 6 / 20358520 = 216 / 20358520 = 0.0000106 or 0.001%

thus 3 players will be dealt different PPs 0.001% of the time.

again, I certainly hope my math is on the up and up. I'm not really a math guy, I just dabble. hope this helps out!

cheers,
Ben

[Phogster]

hey hey, no insightful analysis allowed in this forum!

There was a post somehere on 2+2 where someone posted a screen cap of 3 sets on the flop. He was outdone by someone else posting a screen cap of 4 people with pocket pairs hitting their sets. I have no idea where it was though.