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Herb N.
09-26-2002, 05:07 PM
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! /forums/images/icons/confused.gif

09-27-2002, 11:42 AM
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.

09-27-2002, 11:46 AM
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.

irchans
09-27-2002, 12:04 PM
> 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.

( /forums/images/icons/cool.gif 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.

09-27-2002, 03:04 PM
"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)

irchans
09-27-2002, 03:48 PM

09-27-2002, 06:21 PM
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.

Herb N.
09-29-2002, 05:33 PM
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! /forums/images/icons/crazy.gif

BruceZ
09-29-2002, 05:48 PM
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 Iíll do this the long way, and then Iíll 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] Ė 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 hereís 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 Ė 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.

BruceZ
09-29-2002, 06:20 PM

MD_
10-01-2002, 09:02 PM
In a ten handed game, the odds that one or more players will have pocket aces in 6 particular hands is about 1 in 113 million. Note that it doesn't matter that they're consecutive, just they theye are particular hands.

-MD

10-02-2002, 09:53 AM
This would be P(1)^6 (see above) = .0449188^6 = 1 in 122 million.

BruceZ
10-02-2002, 09:54 AM

Herb N.
10-02-2002, 08:09 PM
Now I'm realy down in the dumps.I hit a 122 million to one shot and lost money /forums/images/icons/frown.gif Oh well you can't put a price on a good time! /forums/images/icons/wink.gif

BruceZ
10-02-2002, 09:30 PM
You only hit the 1 in 1.7 million. 1 in 122 million was for an AA in every round for 6 rounds. Note also that it is easier to get 6 AA in 6 rounds with 2 on one round (1 in 24.8 million) than it is to get at least one on every round.

Herb N.
10-03-2002, 05:13 PM
I knew I was a beter player than to lose money hitting 1:122 mil. /forums/images/icons/grin.gif