Cards were dealt, the lowest chipped player called ALL IN, the rest all followed. Everyone was ALL IN.
When the dealer got everyone to turn their cards over we had the following:-
1 with pocket 10's
1 with pocket 4's
1 with pocket 6's
1 with pocket Q's
1 with pocket Aces
With the flop NOT ONE PERSON GOT TRIPS.... the winner was the 10's with a straight. Two others got paid because they had the highest chips. Two were gone.
I would like to know what the odds were with 5 plays and 52 cards ALL GETTING A PAIR. Can you help. It was amazing.
Quote: mopnz2000...the odds were with 5 plays and 52 cards ALL GETTING A PAIR...
I get about 1 in 1,493,840.
I have heard stories of dealers stacking a deck so the game ends quickly.
Confirmed, but that assumes that the pairs are of different ranks. You didn't allow for the case that the pairs can also be of the same rank.Quote: ChesterDogI get about 1 in 1,493,840.
Allowing players to have pairs of the same rank as each other for their pairs gives 1 in 1,268,018.
--Ms. D.
Sounds like a home game or league game.Quote: mopnz2000Because it was a final table we had a dealer.
Odd stuff happens more in home and league games simply because there's less at stake so more people tend to stay in the hand.
Also, they tend to have less skilled players.
A couple years ago, my wife was in league game with a similar hand where there were two all-ins and one caller. All three had pocket pairs. The flop gave all three a set, and was suited. The next two cards matched the suit, resulting in a flush on board. Chop pot.
Quote: DJTeddyBearA couple years ago, my wife was in league game with a similar hand where there were two all-ins and one caller. All three had pocket pairs. The flop gave all three a set, and was suited. The next two cards matched the suit, resulting in a flush on board. Chop pot.
And none of the three had a matching suit higher than a community card?
EDIT, if it gave all three a set, they couldn't have a matching suit. :-s D'oh. But thats so rare!
Quote: DorothyGaleConfirmed, but that assumes that the pairs are of different ranks. You didn't allow for the case that the pairs can also be of the same rank.
Allowing players to have pairs of the same rank as each other for their pairs gives 1 in 1,268,018.
--Ms. D.
Thanks; I did miss that. But when I try to get your value of 1 in 1,268,018, I get 1 in 1,255,523. For two players having identical ranks, I get 1 in 8,066,737, and for two pairs of two players, I get 1 in 322,669,492.
I made a small boo boo.Quote: ChesterDogThanks; I did miss that. But when I try to get your value of 1 in 1,268,018, I get 1 in 1,255,523.
--Ms. D.
Quote: buzzpaffWhat I find amazing is that 4 other players would call the all in player, Can Ii assume this game was not for money ??
I wondered the same thing. I assume the 4s and 6s belonged to two of the smaller chip-stacks at the table. The aces and queens would have called just about any all-in. I don't know what to say about the 10s. Maybe that player felt priced-in after the two people in front of him went all-in?
As I mentioned before, because of the comment "We were down to 5 players and had a dealer..." lead me to believe that it was a home game or a league where there isn't normally a dealer at every table.
That being the case, I assume that the players simply wanted to end it already. Whether this was a verbal decision, or just happened that way, is irrelevant.
The dealer runs the league, several clubs in the area are involved, league members around the 500 mark so his reputation would have been in tatters if he had split the cards, I can vouch for his honesty. It was not the first hand at the last table because 9 players ended up on that table but only 5 were left.
I suppose I am still gob smacked that this came up that is why I was so interested.
Anyway, your response great, and thanks for the odds.