Can anyone calculate these odds?

Evil Genius

Was playing in an MTT tournament on a site that i am doubting the randomness of the random number generator. Anyway....


6 Handed

Player 1: 6,6
Player 2: 10,10
Player 3: Q,Q

Flop.......10,Q,6

Player 1 was me btw

Just wondering what the odds of 3 people having pocket pairs and the board hitting them all.

Marq

I go roughly 20,505,600 to one. but I'm not very good at these things. DEV!!!! your expertise is required.

gerire

Its going to be quiet around here for a while Dev is on holiday i think

Marq

no worries, long boring bus journies light a fire under my ass to work out obscure poker probabilities. I arrived at 7.8 million to one:

started calculating odds after the first three cards are dealt, and didn't factor in the odds on them all being different ranks, as it makes little difference:

(49/3)*(48/3)*(47/3)*(46/6)*(45/4)*(44/2). roughly 7,768,787:1, although we await DeV's confirmation.

melachi

How about me almost getting dealt an ace high straight in pocket pairs playing this free roll.

got AA, followed by KK, QQ and JJ.
Won all ins on aces and kings. Was one of these online freerolls where people aren't afraid of going all in. Finally on the queens someone calls me with an ace high saying "no way not again", he hits his ace and i lose some of my stack... next I'm dealt JJ I go all in once again cause it's the only way to go!
Was very early in the tournament and any significant lead wouldn't even half guarantee you to finish in the money.

Maybe their rigged high card dealin to get rake contributors got messed up j/k

Thraktor

Your 7 million to one is well off, that's the possiblity of getting a very specific selection of 6 cards dealt in a row (as in, saying that the cards are going to be dealt as 6 of hearts, 10 of diamonds, etc. and then them being dealt in that exact order), and even then, I don't quite think the numbers are correct. In order to get the actual odds of this sort of thing happening (each person getting dealt a pair simultaneously) you'd need to be pokin' around at that "nCr" button on your calculator quite a bit, and I'm far too lazy to be getting out the old Maths textbooks to look up the correct procedure.

Of course, in terms of your particular experience, there's every chance it's crooked, as if you get a good hand, you'll be willing to bet more, and the higher the betting, the more money they rake in off you. It's also much easier for internet casinos to get away with this sort of thing, as no-one in their right mind is going to go to the trouble of filing an international lawsuit over a few euros you feel you've been cheated out of. One of the main reasons I steer clear of internet gambling.

Marq

look at the maths again, you may have been too lazy to do it, for some reason I wasn't. everyone gets three cards, they're different. the first person then has a three in 49 chance of getting a pair in his hand. the second person 3 in 48, and so on. on the flop, the first card can be any one of six out of 46 to give one person trips, the second card can then be any one of four out of 45 to give the second person trips and then there are only two cards in the deck out of 44 that will give the third set of trips. Multiply all these probabilites together and you get the correct odds, as above.

Don't mind me, I'm just trying to claw back some credibility after going out on the bubble in the BSOP yet again this month.

PHB

Well imagein theres 3 people, first to all get pocket pairs

(52/52 * 3/51) * ( 50/50 * 3/49 ) * (48/48 * 3/47)

Then the flop coming down

6/46 * 4/45 * 2/44

Giving you a 1/8254335

But that doesnt include a **** load of stuff, like 7 other players not having certain cards so I'd say it goes higher again

Ta marq

Marq

first parts right, but cos it doesn't matter which order the cards come down in on the flop, first card is 6/46 then 4/45, then 2/44. ok, this thread has become too anal. Hyzepher, feel like closing another one?

quozl

it should be closed because people aren't agreeing with you? How can a thread about working out mathematical odds be too anal? Either someone is correct, or they're not and should be corrected.

Greg

bohsman

Ok this happened in a cash game in the fitz, i got dealt 33 the sb got dealt 66 and utg got dealt 55.
flop comes 356 SB (bertie) bets utg raises, I fold Bertie thinks about it for ages realises utg wasnt bb and goes all in, anyway the point im making is that the fitz must be fixed

Marq

Originally posted by quozl
it should be closed because people aren't agreeing with you? How can a thread about working out mathematical odds be too anal? Either someone is correct, or they're not and should be corrected.

you're right on the ball there greg, I'm asking for a thread to be closed because I can't take the emotional strain of someone questioning the accuracy of my mathematical calculations. you see, my methods of calculating poker odds fill me with pride and define my character, when someone questions them, my very validity as a person is also called into question. I can't take the constant barrage of conflicting methodology, so again, I must ask: please close this thread.

Or, maybe I'm just having a laugh at my own expense. I wouldn't get your knickers in a twist over it either way.

Hectorjelly

Originally posted by bohsman
Ok this happened in a cash game in the fitz, i got dealt 33 the sb got dealt 66 and utg got dealt 55.
flop comes 356 SB (bertie) bets utg raises, I fold Bertie thinks about it for ages realises utg wasnt bb and goes all in, anyway the point im making is that the fitz must be fixed

Why did you fold? You're supposed to lose money in set over set situations.

Hyzepher

With a flop like that and betting so hard it's the straight youare worried about no a higher set.

Hyzepher

tricky D

52!/(9!*43!) is the probability of getting any 9 chosen cards in any exact order,

then you work out the order combos, the first round of dealing you've 1 in 4, next round 1 in 3, then 1 in 2 (1 in 24 for the combos), for each of 3 combos,

so I reckon its 52!/(24 * 24 * 24 * 43!* 9!)
= 266136 to 1 and a bit
tho are the other 3 unshown hands are included in the 43! bit, have to find and dust off my 2nd/3rd yr stats book and check up on that