AA v KK back to back hands odds?

[Kevmath]

Just happened in a 6.50 turbo I was playing on stars, first time it was 8 handed, very next hand it was 7 handed.

Kevin...

P.S. AA won both times, imagine that!

[Zapped]

I'm glad there's no answer here because I wanted to try to think about this one. I'm a n00b here so I welcome corrections to my flawed logic.

First off, I'll calculate the probability for any AA vs KK matchup in the pocket with 8 players. There are 52C16 possible two-card hands for 8 players. The player with AA chose 2 cards from the 4 aces, the player with KK chose 2 cards from the 4 kings. The remaining six players had 48 cards from which they chose a total of 12 cards. If I did my math correctly:

P(aa-v-kk-p8) = (4 choose 2) (4 choose 2) (48 choose 12) / (52 choose 16)
= 0.24202

btw, "(4 choose 2) (4 choose 2) (48 choose 12) / (52 choose 16)" can be entered directly into a google search to double-check my answer.

Using same logic:

P(aa-v-kk-p7) = (4 choose 2) (4 choose 2) (48 choose 10) / (52 choose 14)
= 0.13311


The probability of having this twice in a row is 1 minus the products of the probabilities of NOT having it twice in a row.

P(overall) = 1 - [1 - P(aa-v-kk-p8)] * [1 - P(aa-v-kk-p7)]
= 1 - (1-0.24202)(1-0.13311)
= 0.34292

or about 1.9-to-1 against.

This seems too likely, so I'll leave it to the forum demigods to correct any errors in my reasoning /images/graemlins/smile.gif

[RiverTheNuts]

Here's my crack at it:

((4/52)*(3/51)*(4/50)*(3/49))^2

[RiverTheNuts]

Edit:

Make that 8*7*((8/52)*(3/51)*(4/50)*(3/49))^2

Conditional probability, D given C given B given A

Since there are 8 and 7 people in the hands, youll get 8 and 7 cracks at someone starting this off.

The first card to a person must be an ace or a king, so 8/52. The second must be the same as the first... 3/51 ... if these are true, then someone else must get the A or K that first person didn't... 4/50, and then a second one, 3/49. Multiply all that junk together, and then square it since you want back to back.

I get 1 in 9,088,793 times will back to back hands have AA and KK against each other with 8 dealt in the first time and 7 the second time

I know just about nothing about prob. and stats. but this seems pretty intuitive to me, someone help us out here

[BruceZ]

[ QUOTE ]
Edit:

Make that 8*7*((8/52)*(3/51)*(4/50)*(3/49))^2

Conditional probability, D given C given B given A

Since there are 8 and 7 people in the hands, youll get 8 and 7 cracks at someone starting this off.

The first card to a person must be an ace or a king, so 8/52. The second must be the same as the first... 3/51 ... if these are true, then someone else must get the A or K that first person didn't... 4/50, and then a second one, 3/49. Multiply all that junk together, and then square it since you want back to back.

I get 1 in 9,088,793 times will back to back hands have AA and KK against each other with 8 dealt in the first time and 7 the second time

I know just about nothing about prob. and stats. but this seems pretty intuitive to me, someone help us out here

[/ QUOTE ]

The only thing really wrong is that the 8*7 should be C(8,2)*C(7,2) = (8*7/2)*(7*6/2). That is because there are C(8,2) ways to choose the 2 players out of 8 to have AA or KK, and C(7,2) ways to choose the 2 players out of 7.

It is actually a slight approximation to just multiply by C(8,2) and C(7,2), because it is possible that 3 or 4 players can have AA or KK, and we are counting the 3 player cases 3 times, and the 4 player cases C(4,2) = 6 times. To account for this, we would need to compute additional terms as per the inclusion-exclusion principle (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383). In this case the difference is so small that it is not worthwhile, unless you just wanted practice using the inclusion-exclusion principle. With just 1 term the answer comes out to 1 in 865,599, and the exact answer by inclusion-exclusion comes out to 1 in 874,093.

[BruceZ]

[ QUOTE ]
I'm glad there's no answer here because I wanted to try to think about this one. I'm a n00b here so I welcome corrections to my flawed logic.

First off, I'll calculate the probability for any AA vs KK matchup in the pocket with 8 players. There are 52C16 possible two-card hands for 8 players. The player with AA chose 2 cards from the 4 aces, the player with KK chose 2 cards from the 4 kings. The remaining six players had 48 cards from which they chose a total of 12 cards. If I did my math correctly:

P(aa-v-kk-p8) = (4 choose 2) (4 choose 2) (48 choose 12) / (52 choose 16)
= 0.24202

[/ QUOTE ]

This isn't right. The denominator should count the total number of ways to deal the hands, not just the number of ways to choose 16 cards. You can just count the total number of ways to deal 2 hands, so the denominator would be C(52,2)*C(50,2). Then since this is order dependent, the numerator would be C(4,2)*C(4,2)*2 since either player could have the AA, while the other has KK. Then we just multiply this by C(8,2) ways to choose the 2 players out of 8. For the 7 player case, multiply by C(7,2). See my response to RiverTheNuts for more about this. Also see his answer for an alternative form that uses fractions instead of combinations.


[ QUOTE ]

The probability of having this twice in a row is 1 minus the products of the probabilities of NOT having it twice in a row.


P(overall) = 1 - [1 - P(aa-v-kk-p8)] * [1 - P(aa-v-kk-p7)]
= 1 - (1-0.24202)(1-0.13311)
= 0.34292

or about 1.9-to-1 against.

[/ QUOTE ]

This isn't right. Just multiply the probabilities of having it on each hand to get the probabilties of having it on both hands, since the hands are independent. You multiplied the probabilites of NOT having it on each hand, which gives you the probability of not having it on either hand, and then subtracted from 1, which gives you the probability that you DO have it on at least 1 hand. This is not what you wanted. See the difference?

[BruceZ]

[ QUOTE ]
It is actually a slight approximation to just multiply by C(8,2) and C(7,2), because it is possible that 3 or 4 players can have AA or KK, and we are counting the 3 player cases 3 times, and the 4 player cases C(4,2) = 6 times. To account for this, we would need to compute additional terms as per the inclusion-exclusion principle (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383). In this case the difference is so small that it is not worthwhile, unless you just wanted practice using the inclusion-exclusion principle. With just 1 term the answer comes out to 1 in 865,599, and the exact answer by inclusion-exclusion comes out to 1 in 874,093.

[/ QUOTE ]

The exact answer actually involves a "custom" modification to inclusion-exclusion that I have not run into before. The problem is that (AA KK AA) and (AA KK KK) get counted twice when we count (AA KK), so we subtract off the number of threesomes. Now the (AA AA KK KK) got counted 4 times when we counted the (AA KK), but when we counted the threesomes, they got counted twice since there are 2 ways to make (AA AA KK) or (AA KK KK) from (AA AA KK KK). So when we subtracted off the threesomes, we reduced the foursomes from 4 to 2. Since the goal is to count everything 1 time, we need to subtract the foursomes, so we get this interesting result that we have to subtract two consecutive terms instead of alternating addition and subtraction. The exact answer comes out to 1 in 874,112.