hi there
first of all let me start by saying hello to all the forum members i;m new in the forum so go easy on me ok
I saw the craziest thing ever last night in an MTT on PokerStars.
Four Players called all-in. Hole cards:
AA
KK
QQ
JJ
I also had a pair (which I folded), 55.
Board ends up with an A, K, and 5. There would have been three sets had I been in to go along with the four top pocket pairs. I've never seen anything like it. Anyone care to calculate the odds for it?
I saw the craziest thing ever last night in an MTT on PokerStars.
Four Players called all-in. Hole cards:
AA
KK
JJ
I also had a pair (which I folded), 55.
Board ends up with an A, K, and 5. There would have been three sets had I been in to go along with the four top pocket pairs. I've never seen anything like it. Anyone care to calculate the odds for it?
Comments
Rest assured, the odds are so astronomical that you will never see it again in your life. Unless you are on pokerstars. That site is rigged.
You will see it at least 3 or 4 more times
Love to. The odds are 1:0 for, since this an event which has occurred.
Unless you happened to have initiated a statistical experiment before the hand occurred which was monitoring for the outcome of a flop with two flopped sets and a folded hand also flopping a set, computing the probability/odds of this event is a meaningless exercise.
Observer bias will naturally lead to reports of a lot of strange looking outcomes which are a priori improbable but ex post certainties. This seems to be yet another ex post certainty.
Hey, a new initialism! YAEPC
Mais non. I also have a YAEPC I'd like to share.
I once saw a flop come down K64 and flop 3 sets for 3 active players. Unfortunately for me, I had the 66 and the KK thought that limping in behind 3 other limpers was a cute play.
I suppose that by some of the "logic" in this thread, this proves that Brantford is rigged.
ScottyZ
How many Queens?
Someone in early pos. raises to 20 folds to me i have AA so i raise and the BB Caps it.
Flop comes K-4-9. Early pos. bets. I raise. BB re-raise. Early Pos Caps.
Turn comes a 2. Not sure of the suits but there is no Flush draw.
Early bets. I raise, I'm now all in. BB re-raises and Early caps it.
River a is a 4. Early and BB go to the max and early reveals...
POCKET 4's!
BB had pocket cowboys.
And all i had is this bad story to tell.
NL Tournament (Live ay my house)
6 Players left.
EP Raises with KK
Fold
MP Calls QQ
Fold
SB Calls AA
BB (me with short stack) goes all-in with JJ.
EP calls
MP Re-raises all-in
SB Goes all in.
EP folds.
Flops comes rag,rag,rag
Turn K
River Q
So, I guess my house is rigged too.
My point is you are only likely to see a scenario like this once in your life. Maybe twice - randomness happens.
Let's get Mickey working on the odds. When you factor in the concept that the hands would have to be distributed in a certain order to get everyone to go all-in before the flop, I'd have to expect you'd have to play millions of hands before you see it again.
OK.... lets see.... what are the odds of 4 players getting AA, KK, QQ & JJ, and getting it all-in pre-flop?
odds of getting AA = 6/(52c2)
odds of getting KK = 6/(50c2)
odds of getting QQ = 6/(48c2)
odds of getting JJ = 6/(46c2)
now there are 9 players at the table so we have choices on who gets these 4 hands = (9c4)
= 6/(52c2) * 6/(50/2) * 6/(48c2) * 6/(46c2) * (9c4)
Now this is for a specific order.... will only one order work for getting all the players all-in? There are 4!=24 ways to order these hands but would JJ call 3 previous all-ins, would QQ even.... if you answer yes then order doesn't matter and we can add the term 4! to the above equation giving us....
= 6/(52c2) * 6/(50/2) * 6/(48c2) * 6/(46c2) * (9c4) * 4!
Now this will include hands where other pocket pairs get dealt to some other players at the table. These could include other pocket AAs, or QQs, etc. Another term must be added to the above. This term would be the odds that 5 player not get dealt any pocket pair from a deck of 44 cards (missing AAKKQQJJ). This is not a trivial problem.... Lets call this term Q.
So finally we have....
= 6/(52c2) * 6/(50/2) * 6/(48c2) * 6/(46c2) * (9c4) * 4! * Q
= 0.00000206660619132731 * Q
Now Q<1 but not terribly so... let's say it's ~.9 so...
= 0.00000206660619132731 * .9
= 0.0000018599455721945800 or about 1 in 537650*
*remember we don't care about order here.... if order matters then this becomes more rare...up to 24 times more rare if we want only 1 order.
Anyone want to try and solve actual Q?
I would say order of the hands is important. Factoring in that plus the 'Q factor', I'd say we are looking at millions of hands. So Doyle and the hard core players would see this a few times perhaps but me, once.
AA vs KK vs AK vs TT - 4 all ins
flop goes QJT - nut straight for AK
turn no help
river T- Quads
UnbelievableÂ
i had AK
English is a wonderful language in that there are multiple ways to parse any given sentence. When I say "three of whom had" and then list three hands, I think it's pretty clear that I'm delineating the three hands in question. It is of course possible to parse the sentence in a nonsensical manner (as is possible for many sentences), because you're trolling or want to score board l33tnezz points or whatever, and that's your prerogative.
TrollyZ
He's a lot of things and let me tell you a really crappy moderator, like beyond crappy
we've been signing peititions to get him out of here, but to no avail,
maybe if we wrote l33tnezz on them they'd get more flashy and more people would pay attention, whoops I mean more whoms
Hope you have better luck getting him bounced then us,
Cheers
P.
You are soooooo
Did you mean to say all_aces instead of ScottyZ?
Man that guy burns me up.
ScottyZ