How rare is this to happen/odds?

Justwar · #1 07-16-2007, 02:38 PM

I recieved pocket aces back to back hands, and faced pocket aces both hands but against different people. Lost the first hand, won the 2nd hand. What are the odds of that?

btmagnetw · #2 07-16-2007, 03:04 PM

why didn't you throw in the details of suits, or what seat you were in, or what color shirt you were wearing?

sixhigh · #3 07-17-2007, 10:14 AM

Aced back to back - approx. every 50k hands.

timotheeeee · #4 07-17-2007, 05:27 PM

Not really sure, but I think the calculation is something like (4/52)(3/51)(2/50)(1/49) all multiplied by 2 (for occuring twice), but I'm not sure.

timotheeeee · #5 07-17-2007, 05:29 PM

actually I'm dumb. Don't listen to me.

timotheeeee · #6 07-17-2007, 05:38 PM

[ QUOTE ]
Not really sure, but I think the calculation is something like (4/52)(3/51)(2/50)(1/49) all multiplied by 2 (for occuring twice), but I'm not sure.

[/ QUOTE ]

Maybe it's [(4/52)(3/51)(2/50)(1/49)] [(4/52)(3/51)(2/50)(1/49)]

I have no experience with this. I need to shut up.

Camp11 · #7 07-18-2007, 12:58 AM

I had AAs back to back today. I was wondering what the odds were myself. Didnt face off against any AA though.

qpw · #8 07-18-2007, 05:06 AM

[ QUOTE ]
I recieved pocket aces back to back hands, and faced pocket aces both hands but against different people. Lost the first hand, won the 2nd hand. What are the odds of that?

[/ QUOTE ]

One would expect it to happen once in approximately 136,000 million pairs of hands.

So I think you can claim to have experienced a genuine oddity!

Edit: That does not take into account your winning/losing or the fact that you were playing different opponents.

sixhigh · #9 07-18-2007, 05:43 AM

To answer you question correctly: if you have AA, then there's only one possible combination of AA left in the deck (from 1225 remaining hands). Assuming you are at a 10-player table the odds of someone holding AA when you hold AA are roughly 1-(1-1/1225)^(9) ~ 1/136. The probability of you holding AA is 1/221. And the probability of not tying is 1/23.

So the probability of holding AA in a particular hand, facing AA and not tying is 1/221 * 1/136 * 1/23 ~ 1/690,000. The probability of this happening back to back is (1/690k)^2 ~ 1/480,000,000,000.

qpw · #10 07-18-2007, 06:29 AM

[ QUOTE ]
To answer you question correctly: if you have AA, then there's only one possible combination of AA left in the deck (from 1225 remaining hands). Assuming you are at a 10-player table the odds of someone holding AA when you hold AA are roughly 1-(1-1/1225)^(9) ~ 1/136. The probability of you holding AA is 1/221. And the probability of not tying is 1/23.

So the probability of holding AA in a particular hand, facing AA and not tying is 1/221 * 1/136 * 1/23 ~ 1/690,000. The probability of this happening back to back is (1/690k)^2 ~ 1/480,000,000,000.

[/ QUOTE ]
Indeed. My answer assumed heads up. The number of players involved is obviously significant.