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WPTDan
02-18-2003, 03:57 PM
Howdy Poker/Computer Experts!

My name is Dan Abrams and I'm the Post Producer for the World Poker Tour. I also wrote&amp;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 &amp; 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

BruceZ
02-18-2003, 06:33 PM
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.

Mason Malmuth
02-18-2003, 11:29 PM
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

BruceZ
02-19-2003, 08:59 AM
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

BruceZ
02-20-2003, 09:46 AM
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

Karatitis
02-20-2003, 06:39 PM
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.

BruceZ
02-20-2003, 06:53 PM
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.

Karatitis
02-20-2003, 07:08 PM
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) ?

BruceZ
02-20-2003, 07:37 PM
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).

Karatitis
02-20-2003, 07:54 PM
Yep, ok got it....thanks Bruce !

WPTDan
02-21-2003, 07:57 PM
Dear Bruce,

Thank you. We value and appreciate your gracious efforts.
I'm impressed with your scholarship and dedication.
I hope to see you at the final table and thus get a chance to interview you.

stay groovy,
Dan

WPTDan
02-21-2003, 08:01 PM
Dear Mason,

Thanks for the endorsements (of Bruce's scholarship and my experience).

I've always loved twoplustwo!!!

Stay groovy,
Dan Abrams

gisborne
02-23-2003, 10:17 PM
"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?"

It might be a better factoid to include running into AK in this scenario.

Rick Nebiolo
02-24-2003, 03:22 AM
Dan,

You came to the right place for answers to your odds questions. I'll second that DanZ is tops as he has straightened me out now and then.

Anyway, I stopped by the Commerce tonight to watch about an hour or so of the final two WPT \$10K no limit tables. Damn if I had to take a leak at the wrong time so I missed Men getting busted.

Watching tonight leads me to another odds question: What are the odds/chances that the majority of the players who make the final table will dress half decently?

Regards,

Rick

Rick Nebiolo
02-24-2003, 03:34 AM

AceHigh
02-26-2003, 09:37 PM
"other 10 players has it is 10/1225"

Hate to nitpick, but I will.

Won't this be 9/1225 at a ten-handed table?

BruceZ
02-27-2003, 12:35 AM
Yes, but the question concerned an 11-handed game. So what's the nitpick?

AceHigh
02-27-2003, 08:50 AM
OK, I guess I misread it. /forums/images/icons/tongue.gif

02-28-2003, 03:24 AM
I get another answer for how often someone has A's when a particular player
has K's.

Here's how I got it:

P(no one has A's)

=P(no other A's are out) + P(one A is out) +P(two A's are out)((no one has
AA (given only two A's are out) ) +P( 3 A's are out)(no one has AA)
+P(4A's are out) (no one has AA)

When i say n A's are out i mean the total numbers of A's held by the active
players.

So this is (for 9 players)=

(32C4)/ (50C4) + (32C3)(18C1)/(50C4)
+ 32C2(18C2)/(50C4) [(9C2)2^2/(18C2)]
+ 32C1(18C3)/(50C4) [(9C3)2^3/(18C3)]
+(18C4)/(50C4) [(9C4)2^4/(18C4)]

= .9560...

So the probability someone has A's is .0439...

or the odds someone has it is 21.76 to 1.

BruceZ
02-28-2003, 08:17 AM
P(no one has A's)

=P(no other A's are out) + P(one A is out) +P(two A's are out)((no one has
AA (given only two A's are out) ) +P( 3 A's are out)(no one has AA)
+P(4A's are out) (no one has AA)

This approach will work, but your evaluation of all the terms is wrong. I have evaluated your terms correctly, and I get exactly the same answer I reported earlier.

For example, the probability of no aces being dealt is C(46,16)/C(50,16) = 0.201. That is, there are C(50,16) ways to choose the remaining cards for the 8 players, and C(46,16) of these have no ace. It is not C(32,4)/C(50,4) = 0.156 as you said. Apparently you are counting the number of ways the 4 aces can be arranged in the undealt cards (which would be C(34,4) by the way) divided by the number of ways they can be in 50 cards. This doesn't work since it doesn't count all the combinations generated by the other cards. You can't just focus on aces. If you want to focus on the undealt cards, you can say there are C(50,34) total ways to choose them, and there are C(46,30) ways to make sure they have all the aces, so C(46,30)/C(50,36) = 0.201 as before. Your other terms have similar problems.

Here is the proper evaluation of your terms:

1/C(50,16)*[
C(46,16) +
4*C(46,15) +
6*C(46,14)*[ 1 - 8/C(16,2) ] +
4*C(46,13)*[ 1 - 3*8/C(16,2) ] +
C(46,12)*[1 - ( 6*8/C(16,2) - C(8,2)/C(16,4) ) ] = 1 in 25.6.

For the last 3 terms, I used the trick of mulitiplying the probability of a single AA by 8 players. This works when there is no possibility of 2 players having AA. For the last term 2 players can have AA, so the probability of all pairwise AA is subtracted off. This produces exactly the same answer as the exact method I used earlier to all decimal places. This method would become cumbersome for the other problems.

Incidentally, when I evaluate the (incorrect) expression you gave, I get 1 in 7.3, not 1 in 21.76.

BTW, another distinguished poster has PMed me that he has repeated all my calculations by the exact method and got the same answers I did. I won't mention his name in case he doesn't want to take credit/responsibility /forums/images/icons/smile.gif

BruceZ
02-28-2003, 08:53 AM
I take it back about the evaluation of your (incorrect) expression, it does come out to 1 in 22.76, but it is incorrect nonetheless.

irchans
02-28-2003, 08:55 AM

<pre><font class="small">code:</font><hr>
P(no one has A's)
= P(no other A's are out) +
P(1 A is out) +
P(2 A's are out)*((no one has AA (given only two A's are out) ))+
P(3 A's are out)(no one has AA)+
P(4 A's are out)(no one has AA)
</pre><hr>

Let d = c[50,16] and c[x,y] = x! /y! /(x-y)! .

Then, for a 9 handed game (8 opponents), I get

<pre><font class="small">code:</font><hr>
P(no one has A's) = c[46,16] / d = .2014
P(1 A is out) = c[4,1] c[46,15] / d = .4157
P(2 A's are out) = c[4,3] c[46,13] / d = .2923
P(3 A's are out) = c[4,3] c[46,13] / d = .0827
P(4 A's are out) = c[4,4] c[46,12] / d = .0079
</pre><hr>

and

<pre><font class="small">code:</font><hr>
P(no one has AA given only 2 A's are out) = 14/15
P(no one has AA given only 3 A's are out) = 4/5
P(no one has AA given only 4 A's are out) = 8/13.
</pre><hr>

Plugging those values into your formula gives

<pre><font class="small">code:</font><hr>
P(no one has A's) = .2014 + .4157 + .2923 * 14/15 +
.0827*4/5 + .0079* 8/13 = .960938

P(someone has A's) = 1 - .960938 = .039062

1/.039062 = 25.6

</pre><hr>

03-01-2003, 01:06 AM
"This approach will work, but your evaluation of all the terms is wrong. I have evaluated your terms correctly, and I get exactly the same

I did notice one problem I assumed there was 9 *other players* not 8.

" For example, the probability of no aces being dealt is C(46,16)/C(50,16) = 0.201. That is, there are C(50,16) ways to choose the remaining
cards for the 8 players, and C(46,16) of these have no ace. It is not C(32,4)/C(50,4) = 0.156 as you said. Apparently you are counting the
number of ways the 4 aces can be arranged in the undealt cards (which would be C(34,4) by the way) divided by the number of ways they
can be in 50 cards. This doesn't work since it doesn't count all the combinations generated by the other cards. You can't just focus on aces.
If you want to focus on the undealt cards, you can say there are C(50,34) total ways to choose them, and there are C(46,30) ways to make
sure they have all the aces, so C(46,30)/C(50,36) = 0.201 as before. Your other terms have similar problems. "

I am almost certain what I said is correct (which you say is incorrect) that is If one takes out two K's,

the probability no A's are out if two cards are dealt to 9 other players is the probability no A's are out is the probability the 4 A's are distribulted amoung the other 32 ((52-20)) cards. And thats 32C4/(50C4). Though this should equal (as you said) 46C18/ 50C18.

Ill look at it again.

03-01-2003, 03:03 AM
Poster: BruceZ
Subject: Re: WPT Post-Producer seeks help

" I take it back about the evaluation of your (incorrect) expression, it does come out to 1 in 22.76, but it is incorrect nonetheless. "

I don't think so. I noted before that this is for a ten handed game and I think everyone else did it for a ninehanded game.

I did redo it for a nine handed game and got 24.60 to 1, not the number
you stated.

BruceZ
03-01-2003, 05:12 AM
Ahhh, if you were evaluating a 10-handed game, then I agree with your answer completely. I get the same answer you got for a 10-handed game using both my method and your method. The evaluation of all your terms is correct for a 10-handed game. The original poster was asking about a 9-handed game.

I did redo it for a nine handed game and got 24.60 to 1, not the number
you stated.

That is the number I've been stating all along. I said 1 in 25.6, which is 24.6 to 1.

All the exact answers I posted for WPTDan are accurate. These anwers have been obtained 2 different ways by 2 people,and now 3 different ways and 3 different people for the KK vs. AA problem.

BruceZ
03-01-2003, 05:21 AM
I said:
" For example, the probability of no aces being dealt is C(46,16)/C(50,16) = 0.201. That is, there are C(50,16) ways to choose the remaining
cards for the 8 players, and C(46,16) of these have no ace. It is not C(32,4)/C(50,4) = 0.156 as you said. Apparently you are counting the
number of ways the 4 aces can be arranged in the undealt cards (which would be C(34,4) by the way) divided by the number of ways they
can be in 50 cards. This doesn't work since it doesn't count all the combinations generated by the other cards. You can't just focus on aces.
If you want to focus on the undealt cards, you can say there are C(50,34) total ways to choose them, and there are C(46,30) ways to make
sure they have all the aces, so C(46,30)/C(50,36) = 0.201 as before. Your other terms have similar problems. "

You said:

I am almost certain what I said is correct (which you say is incorrect) that is If one takes out two K's,

the probability no A's are out if two cards are dealt to 9 other players is the probability no A's are out is the probability the 4 A's are distribulted amoung the other 32 ((52-20)) cards. And thats 32C4/(50C4). Though this should equal (as you said) 46C18/ 50C18.

Agreed. Disregard my comments about focusing on the aces, your way is fine, that wasn't the problem. The only issue here is 9-handed vs. 10-handed. Your evaluation is correct for 10-handed, and can be easily adjusted for 9-handed. WPTDan was interested in 9-handed, so that is what I answered. I get the same anwer as you for 10-handed by my method. I also get the same answer as I reported originally for WPTDan for 9-handed using your method (as did irchans).

So comparing our methods:

10 handed:
P(no aces out given we have KK) =
C(46,18)/C(50,18) my way focusing on dealt cards
= C(46,28)/C(50,32) my way focusing on undealt cards
= C(32,4)/C(50,4) your way focusing on only aces

P(no AA given 2 aces out) =
1 - 9/C(18,2) my method

9 handed:
P(no aces out given we have KK) =
C(46,16)/C(50,16) my way focusing on dealt cards
= C(46,30)/C(50,34) my way focusing on undealt cards
= C(34,4)/C(50,4) your way focusing on only aces

P(no AA given 2 aces out) =
1 - 8/C(16,2) my method

And you used your way for a compact evaluation of no one having AA for the other numbers of aces being out as well. This provides an advantage in certain cases. I will add the counting method you used here to my bag of tools.

irchans
03-05-2003, 04:40 PM
Bruce wrote
"
P(no aces out given we have KK) =
C(46,18)/C(50,18) my way focusing on dealt cards
= C(46,28)/C(50,32) my way focusing on undealt cards
= C(32,4)/C(50,4) your way focusing on only aces
"

I was surprized that there were 3 very different ways to compute that probability. I noticed that you can also get the same number with

(*)

P(no aces out given we have KK) = C(78,28)/C(78,32).

Questions:

1) I have no idea why (*) is true. Can anyone figure out a probabalistic reason for (*) ?

2) Are there any other positive integers m,n,q, and q such that

P(no aces out given we have KK) = C(m,n)/C(p,q) ?

My guess is that there are no soultions where m &gt; 78.

Easy E
03-06-2003, 03:10 PM
or a syndication fee, or sumtin'....