How many ACES?
 

I sit down in the main game and watch pocket Aces take down a nice pot.I comment:nice to see the best hand win..Next hand I got pocket ACES&win..Next hand nothing..Next hand I gots ACES,so does other guy who wins with a flush[4 hearts on board]..NEXT HAND I gots pocket ACES & win..pocket Aces 6 times in 6 hands What the chances?I got em 3 times& lose money...WHAT A COUNTRY! [img]/forums/images/icons/confused.gif[/img]

Re: How many ACES?
 

what kind of deck were u playing with?

If you got a pair of aces six times in six hands in holdem -- the apriori probability is about one in 116507435287321 (or 1 in 527182965101 after first getting a pair of wired aces). That is remarkable. Losing like you did is very possible and things like that happen to every poker player if you play a lot.

Re: How many ACES?
 

what kind of deck were u playing with?

If you got a pair of aces six times in six hands in holdem -- the apriori probability is about one in 116507435287321 (or 1 in 527182965101 after first getting a pair of wired aces). That is remarkable. Losing like you did is very possible and things like that happen to every poker player if you play a lot.

Re: How many ACES?
 

> If you got a pair of aces six times in six hands in holdem -- the apriori probability is about one in 116507435287321 ...

I would have thought that the probability of getting aces six times in a row would be (12/1326)^6 ~= 1/(1.8 10^12) or about 1 in two trillion.

( [img]/forums/images/icons/cool.gif[/img] Informative Note -->> two trillion is approximately the number of cubic millimeters of water in the nearest four person hot tub.)

Getting the 5 ace pairs out of 6 would be 6*(12/1326)^5*(1-12/1326) which is approximately 1 in 3 billion.

Re: How many ACES?
 

"I would have thought that the probability of getting aces six times in a row would be (12/1326)^6 ~= 1/(1.8 10^12) or about 1 in two trillion. "
Before the fact (apriori): the probability of getting wired aces : 4/52 * 3/51 = 1/221,
and
(1/221)^6 = 1 / (116,507,435,287,321) =
1/(1.16507435287321 * 10^14)

oops : 1/221 correct!
 

Re: How many ACES?
 

Read the question again guys. 6 pairs of aces, for the entire table, in 6 hands. One hand no aces, another 2 sets of rockets. Although this probability is small, it's a lot larger than the numbers you guys are throwing around.

Re: How many ACES?
 

We all added wrong #1st hand A'swin#2nd My A's win#3NONE#4th handMy A's lose to A's#5my A's win..In 5 hands thier were 5 pair of Ace's...WOW! [img]/forums/images/icons/crazy.gif[/img]

Neat Trick
 

With 10 players, the probability of at least 6 AA in 6 rounds with at least one round having 2 AA is 1 in 24.8 million. First I’ll do this the long way, and then I’ll show how to get the same answer by a neat trick.

P(1) = probability of getting AT LEAST 1 AA in a round
P(2) = probability of getting 2 AA in a round
P(n,k) is permutations of n things taken k at a time = n!/k!

P(2) = [6*P(10,2)*P(48,16)/2^8] / [P(52,20)/2^10] = .00033

This is (6 ways to get 2 AA * ways to choose 2 players * ways to deal remaining hands)/ways to deal total hands

P(1) = [6*10*P(50,18)/2^9] / [P(52,20)/2^10] – P(2) = .0449188

This is (6 ways to get AA * 10 players * ways to deal remaining hands)/total ways to deal hands minus P(2) by inclusion-exclusion.

Now here’s a trick to get P(1) and P(2) directly. Note that the odds of one player having aces is 1/221, so 10/221 would count all double aces twice, so we have:

P(1) = 10/221 – P(2)

Now if one player has aces, the probability of any other particular player having the other aces is 1/1225, so the chance that one of 9 players has the other aces is 9/1225 since it is not possible for more than one other players to have aces. Thus from conditional probability we have:

9/1225 = P(1)/P(2)

Solving these last two equations gives P(1) = .0449188 and P(2) = .00033.

Now we want P(2) in 1 of the 6 hands and P(1) in 4 of the 5 remaining hands:

6P(2)*5P(1)^4 = 4.03 x 10^-8 or 1 in 24.8 million.

While an individual would not expect to see this result in his lifetime, it is expected that some posters here would see it periodically.

The poster has corrected this to 5 AA in 5 hands with 2 on one hand, so this is:

5P(2)*4P(1)^3 = 6 x 10^-7 or 1 in 1.7 million.

typo: P(n,k) = n!/(n-k)!