And I got it in on the turn..
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Well, it is clearly possible.
In the ONLY live tournament I ever played in, at the table behind me 4 aces lost to a royal flush.
What are the odds?
Who knows?
There are 8.06582E+67
different ways to arrange 52 cards.
This is just one of them.
Several of them could produce this exact scenario:
What I REALLY want to know is which one of the other players folded pocket kings?
Cheers,
--VS
Originally posted by Gifl
@VorpalF2F WSOP ME ?
Oh how I wish!No, it was my niece's grad class put on the $50 BI event.
It took the tournament organizer several minutes to get everyone settled back down.He did it by announcing the next blind level.
Cheers,
--VS
[Link edited by HollyMichelle as it connects to another forum and against our rules]
I will quote the post for you so everyone can see it
In terms of the initial hands, I'm going to use those specific pairs, but I'm going to say that it doesn't matter which player gets what hands.
(4/52 * 3/51) * (4/50 * 3/49) * (4/48 * 3/47) * (4/46 * 3/45) * (4/44 * 3/43) ----
With that out of the way, we're looking for four of those players to hit a set. Because we do not know what cards the other players had, we do not remove them from the calculation:
(4/52 * 3/51) * (4/50 * 3/49) * (4/48 * 3/47) * (4/46 * 3/45) * (4/44 * 3/43) * (8/42) * (6/41) * (4/40) * (2/39)
NOTE:
This does not actually reflect the probability of this happening, in general, but the probability of it happening exactly the way it happened. IOW, four of the players hit their sets prior to the card that made the straight coming out, although, my calculation in such that which set happens 1st-4th is irrelevant. If you also want the sets to come out in the specific order, then just change the first number in the last four parentheses to, "2."
Finally, we need the straight card for the ten out of 38 cards remaining. This card could have actually been a King or an Eight and the hand result would have been the same. Interestingly, the player with the Tens was about 21% to win the hand going into the River card!
(4/52 * 3/51) * (4/50 * 3/49) * (4/48 * 3/47) * (4/46 * 3/45) * (4/44 * 3/43) * (8/42) * (6/41) * (4/40) * (2/39) * (8/38) = 1.3044154153440777e-16
0.00000000000000013044154153440776 or 1/0.00000000000000013044154153440776 = 1 in 7,666,269,412,618,226
I do hope you meant those specific hands and not just any such result. The reason is that, obviously, there are other scenarios in which five players could have an inside pair, four of them could make a set, but the probability of the other inside pair making a straight is zero. (ex. AA, KK, 22, 33, 77)
Dear HollyMichelle
Thank you for editing my post.
Sorry for neglecting the rules. It was unawares.
Regards
Seraj
Of course it is possible......
Why wouldn't it be?
On WSOP Quad aces ran into a royal flush, it happens.
Its easier to believe in Santa and other "mystical beings" rather than this. 5 times pocket pairs which translate into 4 sets which make possible for a fifth hand make a straight on river.
Jep this falls into category Royal over poker... dear god.
Originally posted by JediMasterC
Its easier to believe in Santa and other "mystical beings" rather than this. 5 times pocket pairs which translate into 4 sets which make possible for a fifth hand make a straight on river.
I AM Santa Claus -- so I guess it is possible to have 5 PP and a straight on the river.
Peace,
--VS
The nines should've been the ones to make the straight and everybody would be complaining about riggedness
(52/52 * 3/51) = 1 pocket pair or a, which says you start off with any card in the deck, then multiply by the odds of pairing that card
a * (48/50 * 3/49) = 2 different pocket pairs or b, which says you start off with 1 card different from the first hand (meaning there's 2 that can't hit)
b * (44/48 * 3/47) = 3 different pocket pairs or c, which says that you start off with 1 card different in the first two hands (meaning there's 4 that can't hit)
c * (40/46 * 3/45) = 4 different pocket pairs or d, which says you start off with 1 card different from the first three hands (meaning there's 6 cards in the deck that can't hit)
d * (36/44 * 3/44) = 5 different pocket pairs or e, which says you start off with 1 card different from the first four hands (meaning there's 8 cards in the deck that can't hit)
The total so far is (52*48*44*40*36*3*3*3*3*3)/(52*52*50*49*48*47*46*45*44) or 486/17214925
Next, take the total number of possible combinations the board can have, which is 43*42*41*40*39.
Next, say cards a, b, c, and d must show up on the board, and any other card in the deck that isn't a, b, c, d, or e. Since there's 2 of each card left and 43 cards in the deck, we say the possible number of combinations would then be 2*2*2*2*(43-10).
So the odds of 4/5 hands hitting the board are (2*2*2*2*(43-10))/(43*42*41*40*39) = 11/2406495
And the odds of this situation happening, where the specific cards don't matter, only the fact that everyone has a pocket pair and 4 of them made a set are:
(486/17214925) * (11/2406495) = 1782/13809210312625 or about 1 in 7.75 billion.
i hate to rain on this parade but i just want to point out that the odds of this hand being dealt are the same as any other hand... 50/50 either it is or it isnt
I don't play MTTs so I might be wrong but since it doesnt say the MTT name on top,
dude were you playing NL 6k cash games?
EDIT: Everyone way too short for it to be a cash game, has to be a donkament. Funny its not showing the name of the MTT on top, thought Stars always did.
Maybe not for the replays, I don't know.
To badgerer: Shuffle a deck of 52 cards until it's completely random and you don't know the order.
Is the top card of the deck the 2 of spades, followed by the 3 of diamonds, followed by the 6 of clubs?
If not, shuffle again, and keep count of how many times you shuffle until you get these 3 cards in that exact order.
Also keep count of how many times there is a spade, diamond, and club within those cards, in no particular order. They don't have to be a 2, 3, or 6, just a combination of those 3 suits in any order within the top 3 cards.
Post the results on this thread.
(Note: If the cards don't show up in that exact order, then just keep trying. It should only take about 20 days straight worth of shuffling, assuming you don't get up to eat, sleep, or use the bathroom. During that time you'll also notice that the suits I mentioned will show up in no particular order with no particular number attached to them approximately 31,000 times. Hopefully when it finally shows up you'll realize that it wasn't a 50/50 shot.)