53 replies to this topic

[livestrong]

Does anyone happen to know the odds of two players being dealt (in the same hand) AA? Thanks!!

[fryer98]

livestrong said:

Does anyone happen to know the odds of two players being dealt (in the same hand) AA? Thanks!!

1 in 48400??

[Rainman312]

I have only had AA while another player has had AA one time in my life and he won with a flush :wink:

[BeanGW]

livestrong said:

Does anyone happen to know the odds of two players being dealt (in the same hand) AA? Thanks!!

I'm totally no expert on odds, so everybody's probably gonna tell me this is wrong... but here's how I see it. You are asking what the odds are that FOUR specific cards are dealt to two players.Player 1: First Card = 4/52 in the deck are Aces * 3/52 once one is removed. = 1 in about 122 hands that you are dealt AcesPlayer 2: First Card = 2/52 left that are Aces * 1/52 once one is removed = one in 1352So odds are 1 in 1352 that if you are dealt a pocket pair, someone else will have the same one.... be them Aces or anything else. Of course my math is pretty bad, so that could be totally wrong.

[wrto4556]

Quote

odds of AA vs AA

Not much.

[Whatever]

Let's take another viewpoint now and ask the following: Suppose you are playing a 10-handed hold'em game and upon looking at your two hole cards, you find A-A. What is the probability another player also has been dealt pocket aces? This is now a question involving conditional probability and you must be careful. Even though the probability, as we just saw, for two people to be dealt pocket aces is small, the result may be quite different once the condition that someone has aces is given. The total number of completions to player semideals is given by the product of and 17!! because we are choosing 18 cards from 50 to be dealt to the other nine players, and there are 17!! ways to partition the 18 cards into nine hands of two cards each. Then set aside the remaining two aces to go into another hand. For the other eight hands, we are choosing 16 cards from 48 and partitioning the 16 cards in 15!! ways to form completions. This gives us completions to player semideals with another pair of pocket aces. Dividing the second number by the first gives us a probability of 9/1,225 that someone else also has been dealt pocket aces. Expressed as a decimal, the probability is about .00735, and expressed as an Egyptian fraction, the probability is about 1/136. Therefore, if you are dealt pocket aces, the odds against another player also having pocket aces is about 135-to-1.

[BeanGW]

Whatever said:

Let's take another viewpoint now and ask the following: Suppose you are playing a 10-handed hold'em game and upon looking at your two hole cards, you find A-A. What is the probability another player also has been dealt pocket aces? This is now a question involving conditional probability and you must be careful. Even though the probability, as we just saw, for two people to be dealt pocket aces is small, the result may be quite different once the condition that someone has aces is given. The total number of completions to player semideals is given by the product of and 17!! because we are choosing 18 cards from 50 to be dealt to the other nine players, and there are 17!! ways to partition the 18 cards into nine hands of two cards each. Then set aside the remaining two aces to go into another hand. For the other eight hands, we are choosing 16 cards from 48 and partitioning the 16 cards in 15!! ways to form completions. This gives us completions to player semideals with another pair of pocket aces. Dividing the second number by the first gives us a probability of 9/1,225 that someone else also has been dealt pocket aces. Expressed as a decimal, the probability is about .00735, and expressed as an Egyptian fraction, the probability is about 1/136. Therefore, if you are dealt pocket aces, the odds against another player also having pocket aces is about 135-to-1.

Dude... yr like... smart. "Completions to player semideals." I'm gonna start droppin that one like a bomb on the PP .50/$1 chat.

[MDXS]

I've had this happen twice. Both times in NL MTTs and both time it went all-in before the flop with a third person hanging around. The first time some guy with 42o hit two pair. The other time TT hit a set. Both sucked.

[MrNiceGuy]

In a 10-handed game, there are 45 possible combinations of two seats, so the odds of seeing AA vs. AA are 4/52*3/51*2/50*1/49*45 , so it happens 0.01662% of the time, or about once every 6017 hands.The odds of running into another AA when you hold AA in a 10-handed game are 2/50*1/49*9, so it will happen 0.735% of the time that you hold AA, or about every 136 times.

[MailableRumble]

Not that it counts, I just went heads up with a guy in a sng on PP where we both had pocket AA. Ended up chopping. 3 hands later he got AA again and lost to a straight. Now that's bad luck.E

[Ebonwoulfe]

Rainman312 said:

I have only had AA while another player has had AA one time in my life and he won with a flush :wink:

You might have been playing vs. me, I've only had AA vs AA once and I won with a flush

[Rainman312]

I think the question was what are the odds of two people have AA in the same hand. You said the odds are 135 to 1 if you have AA that someone else has AA. You are already taking into consieration that someone already has AA so your figures are way off. The odds of having a pocket pair are 220 to 1 or something like that. So the odds of two people having AA have to be enormous

[Ebonwoulfe]

Rainman312 said:

I think the question was what are the odds of two people have AA in the same hand. You said the odds are 135 to 1 if you have AA that someone else has AA. You are already taking into consieration that someone already has AA so your figures are way off. The odds of having a pocket pair are 220 to 1 or something like that. So the odds of two people having AA have to be enormous

The odds of having AA is 220-1Odds of any PP is about 16-1Odds of AA after you have one A is 16-1 too

[Ebonwoulfe]

[size=20]BANG!Don't worry, folks, just killing a double post...

[Rainman312]

The odds of being dealt any pp could be 16 to 1. But the odds of a certain pp are 220 to 1

[SmileyHere]

MrNiceGuy is correct

the general formula is [ ( 4 choose 2) / (52 choose 2) ] * [ ( 2 choose 2 ) / ( 50 choose 2 ) ] * (P choose 2)

where P is the number of players

ofcourse you need to know how to evaluate the binomial coeifficents... and of course
(4 choose 2) = 6
(2 choose 2) = 1
(52 choose 2) = 1326
(50 choose 2) = 1225

and (P choose 2) = 45 36 28 21 15 10 6 3 2 ( P=10 to P=2 respectively)

MrNiceGuy did round a bit early however... I get 1 in 6016.11 ten handed but close enough
and for other "handedness":

10: 1 in 6,016
9: 1 in 7,520
8: 1 in 9,669
7: 1 in 12,892
6: 1 in 18,048
5: 1 in 27,073
4: 1 in 45,121
3: 1 in 90,242
2: 1 in 270,725

Now ofcourse if you are looking at AA and wondering, you do have a conditional probability, and the answer would be different

specifically [ ( 2 choose 2) / (50 choose 2) ] * [ ( P-1) choose 1 ] ... which simplifies to (p-1)/1225

for a ten top that would be 9/1225 or 1 in 136.111111

[Canary3]

weirdest bumb ever for a first post

[Pot Odds RAC]

MrNiceGuy is correct

the general formula is [ ( 4 choose 2) / (52 choose 2) ] * [ ( 2 choose 2 ) / ( 50 choose 2 ) ] * (P choose 2)

where P is the number of players

ofcourse you need to know how to evaluate the binomial coeifficents... and of course
(4 choose 2) = 6
(2 choose 2) = 1
(52 choose 2) = 1326
(50 choose 2) = 1225

and (P choose 2) = 45 36 28 21 15 10 6 3 2 ( P=10 to P=2 respectively)

MrNiceGuy did round a bit early however... I get 1 in 6016.11 ten handed but close enough
and for other "handedness":

10: 1 in 6,016
9: 1 in 7,520
8: 1 in 9,669
7: 1 in 12,892
6: 1 in 18,048
5: 1 in 27,073
4: 1 in 45,121
3: 1 in 90,242
2: 1 in 270,725

Now ofcourse if you are looking at AA and wondering, you do have a conditional probability, and the answer would be different

specifically [ ( 2 choose 2) / (50 choose 2) ] * [ ( P-1) choose 1 ] ... which simplifies to (p-1)/1225

for a ten top that would be 9/1225 or 1 in 136.111111

You bumped a two and a half year old thread for your first post?

...just to confirm what had already been posted?

[ThePhoenix88]

weirdest bumb ever for a first post