3 Handed-We get AA, KK, QQ...What are the odds?

[Steelerman]

3-handed in a SNG. I hold KK and opponents QQ & AA. This one's gotta be way up there.

[ScottHoward]

it is way up there:
100%

[Steelerman]

Good useful info bro, anybody else?

[BruceZ]

[ QUOTE ]
3-handed in a SNG. I hold KK and opponents QQ & AA. This one's gotta be way up there.

[/ QUOTE ]

100% that it happened, assuming you're not lying. The odds against these 3 hands occuring on a given hand before it happened:

18*12*6/C(52,2)/C(50,2)/C(48,2) =~ 1,413,785-to-1.

[Steelerman]

[ QUOTE ]
[ QUOTE ]
3-handed in a SNG. I hold KK and opponents QQ & AA. This one's gotta be way up there.

[/ QUOTE ]

100% that it happened, assuming you're not lying. The odds against these 3 hands occuring on a given hand before it happened:

18*12*6/C(52,2)/C(50,2)/C(48,2) =~ 1,413,785-to-1.

[/ QUOTE ]

Lol. No, not lying. Just curious. Thanks for the info.

PartyPoker $11 Regular Tournament, Big Blind is t600 (3 handed) Converter on pregopoker.com

Button (t5850)
Hero (t7940)
BB (t6210)

Preflop: Hero is in SB with K[img]/images/graemlins/heart.gif[/img] K[img]/images/graemlins/diamond.gif[/img]
<font color="red">Button raises to t1800</font>, <font color="red">Hero raises to t7940 (All-in)</font>, BB calls t6210 (All-in), Button calls t5850 (All-in)

Flop: (t22700) 6[img]/images/graemlins/club.gif[/img] T[img]/images/graemlins/heart.gif[/img] 5[img]/images/graemlins/diamond.gif[/img] (3 players)


Turn: (t22700) 8[img]/images/graemlins/heart.gif[/img] (3 players)


River: (t22700) 2[img]/images/graemlins/heart.gif[/img] (3 players)


Results in gray below:
<font color="#f7f7f7">Button has Qs, Qc (a pair of queens)</font>
<font color="#f7f7f7">Hero has Kh, Kd (a pair of kings)</font>
<font color="#f7f7f7">BB has Ac, As (a pair of aces)</font>

[AlienBoy]

[ QUOTE ]
[ QUOTE ]
3-handed in a SNG. I hold KK and opponents QQ &amp; AA. This one's gotta be way up there.

[/ QUOTE ]

100% that it happened, assuming you're not lying. The odds against these 3 hands occuring on a given hand before it happened:

18*12*6/C(52,2)/C(50,2)/C(48,2) =~ 1,413,785-to-1.

[/ QUOTE ]


I believe him - I was just in a ring game with QQ - villains had KK and JJ. Same probability for that combo.

BTW: I came up with a different solution, though I suspect my math is wrong - Did I make an error?


(4/52)*(3/51)*(4/50)*(3/49)*(4/48)*(3/47) = 8482716 to 1


No?

AB

[BruceZ]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
3-handed in a SNG. I hold KK and opponents QQ &amp; AA. This one's gotta be way up there.

[/ QUOTE ]

100% that it happened, assuming you're not lying. The odds against these 3 hands occuring on a given hand before it happened:

18*12*6/C(52,2)/C(50,2)/C(48,2) =~ 1,413,785-to-1.

[/ QUOTE ]


I believe him - I was just in a ring game with QQ - villains had KK and JJ. Same probability for that combo.

BTW: I came up with a different solution, though I suspect my math is wrong - Did I make an error?


(4/52)*(3/51)*(4/50)*(3/49)*(4/48)*(3/47) = 8482716 to 1


No?

AB

[/ QUOTE ]

You have computed the probability that a particular player gets AA, a particular player gets KK, and a particular player gets QQ. If you multiply this by 3! = 6, you will get my answer, which is the probability of these 3 hands being dealt, regardless of who got which hands.

[AlienBoy]

[ QUOTE ]
You have computed the probability that a particular player gets AA, a particular player gets KK, and a particular player gets QQ. If you multiply this by 3! = 6, you will get my answer, which is the probability of these 3 hands being dealt, regardless of who got which hands.

[/ QUOTE ]

Divide by six you mean?

I think I'm unsure of how dividing by 3! brings us this answer - and if we were to seek one specific player, and the other players non-specific, would that be divide by 2! ??

Can you point me in the direction of the proof?


Thanks!

AB

[BruceZ]

[ QUOTE ]
[ QUOTE ]
You have computed the probability that a particular player gets AA, a particular player gets KK, and a particular player gets QQ. If you multiply this by 3! = 6, you will get my answer, which is the probability of these 3 hands being dealt, regardless of who got which hands.

[/ QUOTE ]

Divide by six you mean?

[/ QUOTE ]

No, I mean multiply the probability by 6. There are 6 ways to assign the 3 hands AA,KK,QQ to each of the 3 players, and you are computing the probability of one particular assignment, so you need to multiply your probability by 6 to make it 6 times larger or more likely, which in turn makes your odds-to-1 smaller. Like this:

Your calculation:

(4/52)*(3/51)*(4/50)*(3/49)*(4/48)*(3/47) =~ 8482716 to 1


Correct exact calculation:

6*(4/52)*(3/51)*(4/50)*(3/49)*(4/48)*(3/47) =~ 1413785 to 1


My exact calculation (equivalent to above):

18*12*6/C(52,2)/C(50,2)/C(48,2) =~ 1,413,785-to-1


[ QUOTE ]
I think I'm unsure of how dividing by 3! brings us this answer

[/ QUOTE ]

The exact answer is obtained by multiplying your probability by 6 because the number of ways to assign the 3 hands AA,KK,QQ to 3 players is 3! = 6. This is the same as the number of ways to order 3 things. Count them as: 3 hands that player 1 can have, times 2 remaining hands that player 2 can have, times 1 remaining hand that player 1 can have. Here are the 6:

Player1,Player2,Player3:
AA,KK,QQ
AA,QQ,KK
KK,AA,QQ
QQ,AA,KK
KK,QQ,AA
QQ,KK,AA


[ QUOTE ]
- and if we were to seek one specific player, and the other players non-specific, would that be divide by 2! ??

[/ QUOTE ]

If we wanted the probability that a specific player, say player 1, gets a specific hand, say AA, while the other 2 players get the other 2 hands in either order, then there would be 2 ways to do that, and we would multiply your probability by 2.


[ QUOTE ]
Can you point me in the direction of the proof?

[/ QUOTE ]

This is very basic, but let me know if you need anything else.

[AlienBoy]

Got it - explanation makes it more clear now. I was confused for a bit since I was working with percentages instead of directly with odds.

But if I wanted to find the probability of 3 players in a 9 handed game getting the above pairs, it would *not* be:


9! = 362880

362880*(4/52)*(3/51)*(4/50)*(3/49)*(4/48)*(3/47) =~ 1413785 to 1


Right? Instead I have to figure out how many ways 3 hands can be distributed amongst 9 players? Or is 3! correct regardless of the number of players in the hand, since we are only concerned about the 3 players who get one of the 3 pairs?


AB