basic odds question - AA 3 times in a row...

whiskeytown · #1 09-23-2005, 10:58 PM

what are the odds of getting AA 3 times in a row? - I wanna say it's 1/224 times 3, or 1/672 - but I'm not sure - been a while since I did this kind of math...

happened to me on Full Tilt, BTW, and by the 3rd time I had busted out - lost the last two - LOL

RB

BruceZ · #2 09-23-2005, 11:11 PM

[ QUOTE ]
what are the odds of getting AA 3 times in a row? - I wanna say it's 1/224 times 3, or 1/672 - but I'm not sure - been a while since I did this kind of math...

happened to me on Full Tilt, BTW, and by the 3rd time I had busted out - lost the last two - LOL

RB

[/ QUOTE ]

(1/221)^3 = 10,793,860-to-1.

UATrewqaz · #3 09-24-2005, 01:05 AM

He's correct and the best way to interpret the result is as follows

If you got 10,793,861 people together and dealt each of them 3 hands then statisically 1 person will receive pocket AA for all three deals (over hte long run).

Neil Stevens · #4 09-24-2005, 01:43 AM

There is no one "correct" way to interpret probabilities, and your example doesn't even work.

Assuming all the 10,793,861 people's hands are being dealt from different decks, there's actually a 37% chance that someone won't get aces all three times, by my estimate:

(10,793,860/10,793,861)^10,793,861 = about 37%

So either the interpretation is wrong, or the math is wrong, and I trust the math in this case.

BruceZ · #5 09-24-2005, 03:11 AM

[ QUOTE ]
There is no one "correct" way to interpret probabilities, and your example doesn't even work.

Assuming all the 10,793,861 people's hands are being dealt from different decks, there's actually a 37% chance that someone won't get aces all three times, by my estimate:

(10,793,860/10,793,861)^10,793,861 = about 37%

So either the interpretation is wrong, or the math is wrong, and I trust the math in this case.

[/ QUOTE ]

He said in the long run, meaning that if we repeated this experiment a large number of times, each time dealing 3 hands each to 10,793,861 players, then on average 1 player would get 3 AA. That is, the expected value of the number of players getting 3 AA is 1. This is correct.

Another correct interpretation is that if we were to deal 3 hands to each player until some player gets 3 AA, then on average we would need to deal hands to 10,793,861 players. That is, the expected value of the number of players dealt hands before a player gets 3 AA is 10,793,861.

whiskeytown · #6 09-24-2005, 09:10 AM

the point is....

it's pretty far out there to get three times in a row (of course, it happened online) -

and now what are my odds of being all in the last two times and losing HU to one opponent who always had the worst hand when the money went in the middle ? [img]/images/graemlins/grin.gif[/img]

RB

Chillout · #7 09-28-2005, 12:42 PM

This happened to me too once, btw [img]/images/graemlins/smile.gif[/img] On partypoker about a year ago. Quite hilarious. Don't remember if I won or lost tho, but I think i'd remembered if I lost all 3 :P

Vetstadium · #8 09-28-2005, 01:18 PM

I was in a tourn at Taj last year and someone got it 3 times in a row doubled up twice and got sucked out on once.

Luzion · #9 09-28-2005, 01:30 PM

[ QUOTE ]
[ QUOTE ]
what are the odds of getting AA 3 times in a row? - I wanna say it's 1/224 times 3, or 1/672 - but I'm not sure - been a while since I did this kind of math...

happened to me on Full Tilt, BTW, and by the 3rd time I had busted out - lost the last two - LOL

RB

[/ QUOTE ]

(1/221)^3 = 10,793,860-to-1.

[/ QUOTE ]

Bruce... its only (1/221)^3 if you have exactly 3 trials, which is not very useful to know at all. What if you had 1,000 trials, as in you played 1000 hands. Can you calculate the odds on getting AA 3x in a row within those 1000 hands?

Thanks for any information.

BruceZ · #10 09-28-2005, 08:13 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
what are the odds of getting AA 3 times in a row? - I wanna say it's 1/224 times 3, or 1/672 - but I'm not sure - been a while since I did this kind of math...

happened to me on Full Tilt, BTW, and by the 3rd time I had busted out - lost the last two - LOL

RB

[/ QUOTE ]

(1/221)^3 = 10,793,860-to-1.

[/ QUOTE ]

Bruce... its only (1/221)^3 if you have exactly 3 trials, which is not very useful to know at all. What if you had 1,000 trials, as in you played 1000 hands. Can you calculate the odds on getting AA 3x in a row within those 1000 hands?

Thanks for any information.

[/ QUOTE ]

(1/221)^3 is the probability that a hand will start (or end) a sequence of 3 AA in a row. This is what people normally mean by the probability of it happening if they don't specify some number of hands. If you want the probability of it happening in 1000 hands, use this method:

Let P(n) be the probability of a sequence of 3 AA in a row by the nth hand. Then:

P(1) = 0
P(2) = 0
P(3) = (1/221)^3
P(4) = P(3) + (220/221)*(1/221)^3
P(n) = P(n-1) + [1 - P(n-4)]*(220/221)*(1/221)^3 for n > 4

P(1000) =~ 1 in 10,865.

That is, the probability that it happens in n hands is the probability that it happens in n-1 hands plus the probability that it happens first on the nth hand. To happen first on the nth hand, it must not happen by hand n-4, then we must not get AA on hand n-3 with probability 220/221, followed by 3 pairs in a row with probability (1/221)^3. This can be computed in a single column in Excel.

This can also be done with matrix multiplication as a Markov process.

Of course you can appromimate this result by simply multiplying (1/221)^3 by 998 trials. It's actually less than 998 trials, because some trials last more than 1 hand. The average length of a trial is 1 + 1/221 + 1/221^2, so the average number of trials is 998/(1 + 1/221 + 1/221^2), which when multiplied by that useless number (1/221)^3 gives 1 in 10,865, same as the exact answer to the nearest integer. This method is an approximation in general because it could happen more than once, but the probability of that here is small.