Question about the odds. Bet you haven't seen such scenario before!

What are the odds of 5 people being dealt pocket pairs (AA, QQ, JJ, TT, 99), four of them hit a set, and the TT wins holding a straight?!

this is the picture.. (I got it allin on the turn.. /sigh)

I bet it is, given your screenshot, close to 100%.

Is that real money your playing for???

thats a $11 real money buy-in tournament.

Quote: SilverFace

What are the odds of 5 people being dealt pocket pairs (AA, QQ, JJ, TT, 99), four of them hit a set, and the TT wins holding a straight?!

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)

Quote: Mission146

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)

Excuse me, I screwed something up. Even using the same initial hands, the order doesn't matter, so I should have started with:


(20/52 * 3/51) * (16/50 * 3/49) * (12/48 * 3/47) * (8/46 * 3/45) * (4/44 * 3/43)

So that changes it to:

(20/52 * 3/51) * (16/50 * 3/49) * (12/48 * 3/47) * (8/46 * 3/45) * (4/44 * 3/43) * (8/42) * (6/41) * (4/40) * (2/39) * (8/38) = 1.5652984984128933e-14

0.000000000000015652984984128932 or 1/0.000000000000015652984984128932 = 1 in 63,885,578,438,485.2

So, that's a bit more likely...lol

There was a hand at a Four Queens tournament many years ago where Mike Sexton got all his money in with pocket queens against two guys, both holding pocket tens. The flop came Q9X. Sexton had flopped top set, but an 8 on the turn and a jack on the river and Sexton was knocked out of the tournament.

That looks so much like a con-hand cold-deck setup that, IRL, people would have been pulling knives and guns out at the table when the hands were revealed. I guess there's a good rationale for playing online after all.

Now here's a little food for thought. This may be comparing apples to oranges but I am going to throw this out there anyway. I read on the Wizard of Odds website on the list of blacklisted sites that one site was blacklisted in December of last year (Winaday I believe) when it was found that a player placed a single number wager on the same number on a single-zero roulette wheel 1156 times with the number never winning. It was said the odds of that happening was in the area of 1 in 57 trillion. Shouldn't this be a pause for concern?

Quote: Mission146

Quote: Mission146

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)

Excuse me, I screwed something up. Even using the same initial hands, the order doesn't matter, so I should have started with:


(20/52 * 3/51) * (16/50 * 3/49) * (12/48 * 3/47) * (8/46 * 3/45) * (4/44 * 3/43)

So that changes it to:

(20/52 * 3/51) * (16/50 * 3/49) * (12/48 * 3/47) * (8/46 * 3/45) * (4/44 * 3/43) * (8/42) * (6/41) * (4/40) * (2/39) * (8/38) = 1.5652984984128933e-14

0.000000000000015652984984128932 or 1/0.000000000000015652984984128932 = 1 in 63,885,578,438,485.2

So, that's a bit more likely...lol

A million thanks for the calculation, dear admin. :)

Quote: BIGDADDY910S

It was said the odds of that happening was in the area of 1 in 57 trillion. Shouldn't this be a pause for concern?

Maybe. But when you lose in roulette, the money goes to the house. When you lose at poker, the money goes to other players. So the other players would need to be affiliated with the casino in order for the house to benefit from this scenario.

Quote: BIGDADDY910S

Now here's a little food for thought. This may be comparing apples to oranges but I am going to throw this out there anyway. I read on the Wizard of Odds website on the list of blacklisted sites that one site was blacklisted in December of last year (Winaday I believe) when it was found that a player placed a single number wager on the same number on a single-zero roulette wheel 1156 times with the number never winning. It was said the odds of that happening was in the area of 1 in 57 trillion. Shouldn't this be a pause for concern?

Possibly, but PokerStars is the most reputable online poker site in the world, and it didn't help that Mission was off by a factor of 126. See below.

Quote: Mission146

Excuse me, I screwed something up. Even using the same initial hands, the order doesn't matter, so I should have started with:


(20/52 * 3/51) * (16/50 * 3/49) * (12/48 * 3/47) * (8/46 * 3/45) * (4/44 * 3/43)

So that changes it to:

(20/52 * 3/51) * (16/50 * 3/49) * (12/48 * 3/47) * (8/46 * 3/45) * (4/44 * 3/43) * (8/42) * (6/41) * (4/40) * (2/39) * (8/38) = 1.5652984984128933e-14

0.000000000000015652984984128932 or 1/0.000000000000015652984984128932 = 1 in 63,885,578,438,485.2

So, that's a bit more likely...lol

This number only assumes five players were dealt cards, not all nine. Multiplying by a factor of C(9,5) = 126 is needed here to choose which five players receive the cards. Doing this makes makes the result a moderately more rational sounding: 1 in 507 billion. For a six max game, then you would need to multiply by C(6,5) = 6 to create a chance of 1 in 10.65 trillion. You did a good job on everything else though. :D

But considering PokerStars has dealt about 90 billion hands of holdem by now (since everyone but me loves that game), this happening at least once isn't THAT unlikely. If half of those hands are full ring (lots of tourneys are full ring, fwiw), then the chances this hand scenario would have played out by now would roughly be (B=billion) 1 - (1 - 1/507B)^45B ~ 0.0849 ~ 8.5%

This is the rarest event I have seen posted from a PokerStars game though.

Quote: tringlomane

This number only assumes five players were dealt cards, not all nine. Multiplying by a factor of C(9,5) = 126 is needed here to choose which five players receive the cards. Doing this makes makes the result a moderately more rational sounding: 1 in 507 billion. For a six max game, then you would need to multiply by C(6,5) = 6 to create a chance of 1 in 10.65 trillion. You did a good job on everything else though. :D

Thank you for the correction, Tringlomane, especially the last sentence to salvage my ego!

When I mess up something like this, it's always forgetting something simple that I actually knew to do and just neglected. No more Math on work days!!!

Quote: beachbumbabs

That looks so much like a con-hand cold-deck setup that, IRL, people would have been pulling knives and guns out at the table when the hands were revealed. I guess there's a good rationale for playing online after all.

Kenny Rogers needs a re-write... You still have to know when to hold 'em and when to fold 'em, but you never have to run and they count your money for you!

Quote: SilverFace

What are the odds of 5 people being dealt pocket pairs (AA, QQ, JJ, TT, 99), four of them hit a set, and the TT wins holding a straight?!

this is the picture.. (I got it allin on the turn.. /sigh)

Dont put all your cards on the table at once...Next time you get some evidence of something like this, hold it back at first. Fist just ask if something like this is possible. Then once people say probably not, say you think may have happen, but don't say the name of the casino. Every one will tell you how impossible it is, and you were dreaming. EVERYONE will tell you, if it really did happen its probably a rigged casino, and don't play there anymore. Then after the story gains attention and everyone calls you a liar......BAM !! show what you have.

It seems that when people talk about rigged online casinos with no proof, at first the stories, complaints and roomers seem to last much longer and get more attention. Once you prove somethings rigged it all goes away very fast.