WPT Post-Producer seeks help
 

Howdy Poker/Computer Experts!

My name is Dan Abrams and I'm the Post Producer for the World Poker Tour. I also wrote&produced the 2000 World Series of Poker for the Discovery Channel(when Ferguson won). However another Dan Abrams wrote a book on poker with Epstein.

We here at the WPT like putting interesting factoids & trivia in the bumpers (as the show goes to commercial).

We have high standards and have to make 13 episodes and 18 bumpers per. So we need some help.

We want to supply accurate stats on the following questions:

1) If you have pocket kings, what are the odds you'll run into pocket aces in a 9 handed game? (And in a 4 handed game?). Is it just the negative parlay of 1:220 to the 8th power (and thus about 1/25)?

2) If you have pocket queens, what are the odds you'll run into pocket aces or pocket kings in a 9 handed game? (and in a 4 handed game?)

3) If you have pocket jacks, what are the odds you'll run into pocket aces or pocket kings or pocket queens in a 9 handed game? (and in a 4 handed game?)

4) If you have Ace-King, what are the odds you'll run into pocket aces or pocket kings, nine handed? And 4 handed?

5) If you have Ace-Queen, what are the odds you'll run into pocket aces or pocket kings or pocket queens, nine handed? And 4 handed?

Thank you for your gracious efforts and consideration.

Sincerely,

Dan Abrams

Re: WPT Post-Producer seeks help
 

1) If you have pocket kings, what are the odds you'll run into pocket aces in a 9 handed game? (And in a 4 handed game?). Is it just the negative parlay of 1:220 to the 8th power (and thus about 1/25)?

Your approximation using 220:1 would actually give 1-(220/221)^8 = 1/28. You really want to use 204:1 since you having kings means there are only 50*49/2 = 1225 remaining hands possible, so the odds of AA are 1225/6 = 204:1. Then 1-(203/204)^8 = 1 in 25.9. The exact answer does turn out to be 1 in 25.5 as you said. The exact answer is more difficult since the player's hands are not independent, that is, certain hands make AA more or less likely. We must also consider the possibility of more than one player having AA. That could be ignored in this problem because it is small, but we cannot ignore it in some of your other problems. This is how the exact answer must be computed:

[ 6*8*P(48,14)/2^7 - C(8,2)*P(46,12)/2^6 ] / [ P(50,16)/2^8 ]
= 1 in 25.5.

The denominator is the number of ways to deal hands to the remaining 8 players. It is the number of ways to deal 16 cards out of 50, and then we divide by 2^8 since we don't care about the order of the 2 cards in each player's hand. The first term in the numerator says there are 6 ways a player can have AA, there are 8 players that can have AA, and there are P(48,17)/2^7 ways to deal the remaining hands to 7 players. This would double count cases where two players get AA, so the second term in the numerator subtracts this off. This is so small it could be ignored in this problem.

For 4 players this becomes:

[ 6*3*P(48,4)/2^2 - C(3,2)*P(46,2)/2^1 ] / [ P(50,6)/2^3 ] = 1 in 68.1


4) AK vs. AA or KK

This problem is similar to 1 except there are now 9 ways that two players could have AA or KK. Note that no more than two players can have these hands, and one must have AA while the other has KK. There are still 6 ways for a player to have AA or KK.

9 handed:

[ 6*8*P(48,14)/2^7 - 9*C(8,2)*P(46,12)/2^6) / [ P(50,16)/2^8 ]
= 1 in 25.6

4 handed:

[ 6*3*P(48,4)/2^2 - 9*C(3,2)*P(46,2)/2^1 ] / [ P(50,6)/2^3 ]
= 1 in 68.1


The other problems are done similarly, but they will be more time consuming because there are more ways that multiple players can have better hands that we must consider. This means more terms to be added together. The worst case is problem 3 where up to 6 players could have a better hand. This would require 6 terms to be computed, however, some of these could be ignored. Basically you compute terms until they become sufficiently small. The second term is subtracted, the 3rd is added, and so on alternately. This is called the inclusion-exclusion principle.

For Dan
 

Hi Dan:

I didn't read all of BruceZ's post, but let me assure you that he knows his stuff.

For everyone else, David and I met Dan Abrahms a couple of years ago when he interviewed us for the WSOP show that was shown on The Travel Channel. He did a great job then, and I'm sure he'll be top notch for the WPT.

Best wishes,
Mason

More answers
 

Dan,

Awhile ago on this board we computed the exact probability of someone having JJ,QQ,KK,AA when we hold TT in a 10 handed game. This came out to 16.5%. We also showed that using the "negative parlay" approximation as you call it, we would get 16.3%. For this reason, I feel that if we use this approximation for all of your problems, we should be within a few tenths of the exact answer. For all practical poker purposes, this would always be good enough; however, for your purposes of putting the numbers on the screen, you may still want to get the exact answers the long way just to be very precise and to make sure that some of them aren't off by more than we expect. Note that even if you round to the nearest whole number you could be rounding the wrong way.

If you want to go the exact route, I can help you with that. For now, here are the results of the approximation which should be very close. I am just subtracting the better hands from 1225 to get the fraction rather than using 204:1, etc, it's the same thing.

2) QQ vs. KK,AA

9 handed: 1-(1213/1225)^8 = 1 in 13.2
4 handed: 1-(1213/1225)^3 = 1 in 34.4

3) JJ vs. QQ,KK,AA

9 handed: 1-(1207/1225)^8 = 1 in 9.0
4 handed: 1-(1207/1225)^3 = 1 in 23.0

5) AQ vs. AA,KK,QQ

9 handed: 1-(1213/1225)^8 = 1 in 13.2
4 handed: 1-(1213/1225)^3 = 1 in 34.4

Note this is just the same as problem 2 since there are again 12 better hands.

You already have my exact answers to problems 1 and 4 in the above post, and they are close to what we would get with the approximate method.

-Bruce

Final Exact Answers
 

Dan,

I've computed all the exact answers accurate to the tenths decimal place (and in most cases better). This required going out to 3 terms in each case. As expected, they only changed by a few tenths from the approximate values given earlier. Since they are all in agreement with the approximate results, we can have confidence in this computation, and it would be better to use these more accurate numbers. If you want the the Excel spreadsheet I made which does this, PM me with your email. Here are the exact results:

1) KK vs. AA

9 handed: 1 in 25.6
4 handed: 1 in 68.1

2) QQ vs. AA,KK

9 handed: 1 in 13.0
4 handed: 1 in 34.2

3) JJ vs. AA,KK,QQ

9 handed: 1 in 8.9
4 handed: 1 in 22.9

4) AK vs. AA,KK

9 handed: 1 in 25.8
4 handed: 1 in 68.2

5) AQ vs. AA,KK,QQ

9 handed: 1 in 13.1
4 handed: 1 in 34.3


I made a minor correction to the formulas given in my first post for cases 1 and 4. In case 1 there are 6 ways 2 players can have AA, and in case 4 there are 18 ways 2 players can have AA or KK, not 9 ways. Here are the corrected formulas:

1) KK vs. AA

9 handed:

[ 6*8*P(48,14)/2^7 - 6*C(8,2)*P(46,12)/2^6 ] / [ P(50,16)/2^8 ] = 1 in 25.6.

4 handed:

[ 6*3*P(48,4)/2^2 - 6*C(3,2)*P(46,2)/2^1 ] / [ P(50,6)/2^3 ] = 1 in 68.1


4) AK vs. AA,KK

9 handed:

[ 6*8*P(48,14)/2^7 - 18*C(8,2)*P(46,12)/2^6) / [ P(50,16)/2^8 ] = 1 in 25.8

4 handed:

[ 6*3*P(48,4)/2^2 - 18*C(3,2)*P(46,2)/2^1 ] / [ P(50,6)/2^3 ] = 1 in 68.2

The other formulas are a little more complex, so I won't give them here, but they are on the spreadsheet.

-Bruce

Re: Final Exact Answers
 

Bruce,

Wow, you really do know your 'stuff' ! Listen, I have a probability question which has been bothering me for a couple days, and everyone I ask gives me different answers and I don't think I can calculate it so here goes: What is the probability of another player getting the same pocket pair dealt as you in the same hand (say, pocket 99's "Gretzky") in an 11-handed game. Note, that in Toronto here we play 11 players.

Thanks if you can.

Re: Final Exact Answers
 

That's an easy one. The probability of a particular player having your pair is 1/1225, so the probability that one of the other 10 players has it is 10/1225. The reason we can just multiply by 10 here is because it is impossible for more than one other player to have your pair.

Re: Final Exact Answers
 

Bruce,

Thanks for the quick answer...okay, but...

.. since it'st 220:1 to get the specific pair to start, wouldn't it be the product of 220 * (10/1225) ?

Re: Final Exact Answers
 

I was assuming you already had the pair. Before the deal, the probability that you and someone else will get a pair of 9s is (1/221)(10/1225). The probability of you getting any pair and someone else getting the same pair is (13*6/1326)(10/1225).

Re: Final Exact Answers
 

Yep, ok got it....thanks Bruce !