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I suck way too much at math for this, but I just HAVE to know how often we expect this to happen...
Three way to the flop. All three of us have pocket pairs. All three of us flop sets. One guy (the one who flopped top set, so it doesnt really matter) turns quads and so we have quads over boat over boat.
That has to be insanely rare.
Full HH: http://www.flopturnriver.com/phpBB2/...oker-48251.htm
lol rigged!!!!11!!1!shiftone!!!
The probability of the whole hand happening like this depends on the number of players being dealt in (more players makes it more likely for there to be pocket pairs out there, and thus the whole probability increases). So I am going to calculate for the game being 5-handed.Quote:
Originally Posted by euphoricism
The chance of getting dealt a pocket pair of a specific rank is equal to the chance of being dealt any card of that rank and then another card of that rank: P(pp) = 4/52 * 3/51 = 1/221 (= 220:1).Quote:
Originally Posted by euphoricism
Lets calculate the probability of three specific people each being dealt a pocket pair, P(tsp).
The first guy has 13 ranks to choose from, so the probability that he gets dealt a pocket pair is 13*P(pp).
The second guy has a different pocket pair, so he has only 12 ranks to choose from, so the probability is 12*P(pp).
The third guy has only 11 ranks to choose from, or 11*P(pp).
So the chance of three different pocket pairs is P(3sp) = 13*12*11*P(pp)^3 = 132/830297 (~ 6289:1).
With 5 people playing though, there are a number of possibly ways to choose 3 specific people. 5 choose 3 = 5!/(3!)(5-3)! = 10 ways.
So the probability that any 3 people out of the 5 each get dealt a different pocket pair is P(3ap) = 10*P(3sp) = 10 * 132/830297 = 1320/830297 (~ 628:1).
First lets figure out the probability that the first player hit his set on the first card of the flop, second player on the second card, and third player on the third card.Quote:
Originally Posted by euphoricism
The probability that the first player hit on the first card is 2/46.
Second player's probability is 2/45.
Third player's probability is 2/44.
So the probability that the sets hit in that particular order is P(3ss) = 2^3/44*45*46 = 1/11385 (= 11384:1).
But the sets can happen in any order. With three cards the number of possible orders is 6 (123,132,213,231,312,321). So the probability of 3 sets being flopped (where order doesn't matter) is P(3as) = 6 * P(3ss) = 2/3795 (~ 1897 : 1).
So the probability of 3 people out of 5 each having a pocket pair and each flopping a set is P(3ap)*P(3as) = 1320/830297 * 2/3795 = 16/19096831 (~ 1193551 : 1).
This is easy. It's the chance that one specific player (the one with top set) hits his last card of rank. This is 1/43.Quote:
Originally Posted by euphoricism
So the chance of the whole sequence is 16/19096831 * 1/43 = 16/821163733 (~ 51322732 : 1).
More rare than fifty million to 1.Quote:
Originally Posted by euphoricism
you make it look easy
More likely than hitting the lottery.
Holy shit Seq. You are THE MAN.
PM me your stars account, for your free $10 hidden bonus :D
Quote:
Originally Posted by thenonsequitur
I'm so bad at math thats the only part I can understand.
I kind of wanted to figure it out for my own sake too, just to know, so a bonus isn't really necessary. But if you want to send me $10, I won't turn down the offer :) . My Stars name is JujuJuice.Quote:
Originally Posted by euphoricism
anybody read the thread on 2p2 (boosted J trip report I think from PCA) where 3-way flop everyone flops a set and goes bananas. Dood with bottom set (777 on 7-q-j board) folds face up (twincaracas from pokerstars 25-50) after push-call in front of him.
Turn is a 7 obv
ditto that, thx nonQuote:
Originally Posted by gabe
talk about the right play at the wrong time...Quote:
Originally Posted by Genitruc