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View Full Version : Probability of an overpair with # opponents -- Whats the Formula?

etgryphon
12-15-2004, 02:30 PM
Hey all I try to do a search on this, but I don't know what the title would be...

Here is my stab at it and tell me if it is off...
P = probablity
x = Rank of your PP

This is finding the probability that your opponents DO NOT have a higher PP that you...

P = product of [((13-x)*4/(50-2n)*(((50-2n)-3)/((50-2n)-1))) + (((x-1)*4)/(50-2n))] where n = 0...# of opponents

I hope this is clear and makes sense to you all...or if someone has already written this then please point me in the right direction...

Thanks,

-Gryph

gaming_mouse
12-15-2004, 06:16 PM
Gryph,

It's not clear at all what you are asking. Does PP mean pocket pair? Or just the highest card in your hand? Please clarify.

gm

etgryphon
12-15-2004, 08:17 PM
Sorry PP stands for Pocket Pair...

I'm look to answer the question if you are dealt JJ what is the probability that one of your opponents has QQ-AA if you have 3 opponents or 4 or n-number of opponents

-Gryph

gaming_mouse
12-15-2004, 08:33 PM
Okay.

There are 6 ways to make QQ. Same for KK-AA. So 18 hands total that you are worried about. There are 50 cards total left. So against 3 opponents, the chance that at least one person has QQ-AA given that you have JJ is:

Note:
50 choose 2 = 1225
48 choose 2 = 1128
46 choose 2 = 1035
44 choose 3 = 946

1 - ((1225-18)/1225)*((1128-18)/1128)*((1035-18)/1035)= 0.047

So about 5% with 3 oppos. For 4 oppos, it's about 6.5%:

1 - ((1225-18)/1225)*((1128-18)/1128)*((1035-18)/1035)*((946-18)/946)= .065

The same method can be used to solve for any number of players.

gm

gm

etgryphon
12-16-2004, 10:30 AM
GM:

Thanks a lot...I think I need to simplify my thinking in these math problems...

-Gryph

Cobra
12-19-2004, 02:51 PM
I did figure out an excel computation for at least one person holding a higher pocket pair than you have. I used inclusion-exclusion with three terms to get an approximation.

N = Number of opponents
X = Number of pair above yours(You have JJ, X = 3)

First term - 6X/C(50,2)
Second term - (36*Power(X,2)-30x)/c(50,2)/c(48/2)
Third term -

(216*Power(X,3)-540*Power(X,2)+324x)/c(50,2)/c(48,2)/c(46,2)

Take the first term and multiply by # opponent combo's, etc.

Term 1 * c(N,1) - Term 2 * c(N,2) + Term 3 * c(N,3)

This will give you the proboability that at least one person will hold a pocket pair higher than you with N number of opponents.

memphis57
12-19-2004, 10:23 PM
Seems to me you should base the calc on the number of players who saw the deal times the probability that they would still be in the hand, rather than just the number of current opponents. Thus, in the case where you hold JJ, say its on the turn, and say it's a full 10-man table, you would want to use 9.9 or 10 for your number of opponents since it's extremely unlikely anybody who got that deal would have dropped out of the hand.

BruceZ
12-25-2004, 01:50 AM
[ QUOTE ]
Okay.

There are 6 ways to make QQ. Same for KK-AA. So 18 hands total that you are worried about. There are 50 cards total left. So against 3 opponents, the chance that at least one person has QQ-AA given that you have JJ is:

Note:
50 choose 2 = 1225
48 choose 2 = 1128
46 choose 2 = 1035
44 choose 3 = 946

1 - ((1225-18)/1225)*((1128-18)/1128)*((1035-18)/1035)= 0.047

So about 5% with 3 oppos. For 4 oppos, it's about 6.5%:

1 - ((1225-18)/1225)*((1128-18)/1128)*((1035-18)/1035)*((946-18)/946)= .065

The same method can be used to solve for any number of players.

gm

gm

[/ QUOTE ]

This is incorrect. The second term is not (1128-18)/1128 because that assumes that the first player does not have an A,K, or Q. The same is true for the remaining terms. The problem cannot be done easily this way. Cobra has done this problem exactly right with inclusion-exclusion. The answer comes out to 12.6%.

This is the same as the mistake you made in this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=1404106&amp;page=&amp;view=&amp;s b=5&amp;o=&amp;vc=1).

gaming_mouse
12-25-2004, 05:39 PM
Bruce,

Yep. You're exactly right. It was a bad error too, as the answer doesn't even provide a good approximation.

Thanks for the correction,
gm

BruceZ
12-26-2004, 03:01 PM
[ QUOTE ]
Bruce,

Yep. You're exactly right. It was a bad error too, as the answer doesn't even provide a good approximation.

Thanks for the correction,
gm

[/ QUOTE ]

The 12.6% was for 9 opponents. For 3 opponents it is 4.36%, and for 4 opponents it is 5.77%. So your results of 5% and 6.5% weren't that far off, but it would actually be more accurate to just use independence since 1 - (1207/1225)^N gives 4.34% and 5.75%.

The advantage of inclusion-exclusion is that you always have a bound on the error. Although the terms can sometimes become complicated to compute for multiple opponents, it converges so fast that it is rarely necessary to compute more than 2 or 3 terms for accuracy to within 0.1% regardless of the number of opponents.

Cobra
12-26-2004, 03:25 PM
Hopefully someone can answer this. When using inclusion-exculsion, like in this example. Is the first term minus the second term the exact probability that exactly one person will be dealt it. The second term would be the exact probability that two people are dealt it. And the third term the exact probability that three or more people are dealt it. Is this how it works?

Thanks,

Cobra

BruceZ
12-26-2004, 05:26 PM
[ QUOTE ]
Hopefully someone can answer this. When using inclusion-exclusion, like in this example. Is the first term minus the second term the exact probability that exactly one person will be dealt it. The second term would be the exact probability that two people are dealt it. And the third term the exact probability that three or more people are dealt it. Is this how it works?

Thanks,

Cobra

[/ QUOTE ]

Not exactly. The first term is P(opponent 1 has it) + P(opponent 2 has it) + ... P(opponent 9 has it). That is why we multiply by 9. We are computing the probability of a union of 9 events. Now remember that when these are not mutually exclusive, this will count the times that 2 opponents have it twice, 3 opponents 3 times, etc. The second term of inclusion-exclusion is the sum of the probabilities for every pair of players having it. It will count the times that 3 players have it 3 times, the times 4 players have it C(4,2) = 6 times, etc. The third term is the sum of probabilities for every combination of 3 opponents having it, etc. We actually want to count each of these just once. Summing these with coefficients of +/- 1 gives the probability of 1 or more opponents having it. For an explanation of why this is so, see this post (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=865342&amp;page=&amp;view=&amp;sb =5&amp;o=&amp;vc=1). The first part of this post derives inclusion-exclusion for the probability of at least 1. The second part shows how you would modify this to get the probability of exactly 1, exactly 2, etc. as you are asking about. It turns out that to get exactly 1, the coefficients become 1, -2, +3, -4,...etc. For exactly 2, they become 1, -3, +6, -10, etc. In general, for exactly N, the coefficients turn out to be the same as the Nth diagonal of Pascal's triangle, if you know what that is. This thing:

<font class="small">Code:</font><hr /><pre>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
...
</pre><hr />

Let me know if you have any more questions about this. It's a rather difficult thing to explain in this forum, and you will have to do some careful thinking about it, along with reading the posts that I linked to.

BruceZ
12-26-2004, 06:33 PM
My above post might take you a little far afield, so this might answer your specific question more directly:

When we subtract the second term from the first, at that point we have counted exactly once all the deals for which exactly 1 player has it (good), and we have also counted exactly once all the deals that exactly 2 players have it (good), but it will count the deals where exactly 3 players have it 3 - 3 = 0 times. That is because the first term counted these 3 times, and the second term counted these C(3,2) = 3 times, and when we subtracted we ended up with 0. It also counts the deals where exactly 4 players have it 4 - C(4,2) = 4 - 6 = -2 times (that's negative 2 times). Remember that the goal is to ultimately count each of these exactly 1 time, and we have so far accomplished this for 1 player and 2 players, but not 3, 4 or more players.

Now we ADD the 3rd term which will count the deals with 3 players having it once, so 0 + 1 = 1 (good), but this will count deals where 4 players have it 4 times, 5 players C(5,3) = 10 times, etc. So now we are good up to 3 players, but 4 players has become -2 + 4 = 2. Since we want this to be 1, you can see that subtracting the 4th term will achieve this. So we are finding that this pattern of alternating +/- 1 times each term seems to be the magic formula for "fixing" each count to be exactly 1, and you can convince yourself that this will continue to work for all the terms. By the time we add all 9 terms, we will have counted each deal with 1 or more players having it exactly once.

gaming_mouse
12-26-2004, 06:56 PM
Bruce,

Thanks for that link. Interesting stuff....

gm

Cobra
12-26-2004, 07:19 PM
Bruce,

Thanks for the link. I will try to work throught this and see if it makes sense to me. I appreciate the help.

Cobra

BruceZ
12-26-2004, 08:05 PM
I wanted to make sure you saw this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=1439096&amp;page=0&amp;view=e xpanded&amp;sb=5&amp;o=14&amp;fpart=1) that I just put up, which might answer your question more directly.

Cobra
12-26-2004, 11:01 PM
I did see that, thanks.

Cobra

Cobra
12-26-2004, 11:43 PM
After reading through your other post I think I understand it but I want to make sure.

Exactly one = 1*term 1 - 2*term 2 + 3*term 3 - 4*term 4

Exactly two = 1*term 2 - 3*term 3 + 6*term 4

Three or more = 1*term 3 - 4*term 4

Am I correct that in finding exactly two you start with the second term and then apply the coefficients. Once again thanks for your time.

Cobra

BruceZ
12-27-2004, 01:13 AM
[ QUOTE ]
After reading through your other post I think I understand it but I want to make sure.

Exactly one = 1*term 1 - 2*term 2 + 3*term 3 - 4*term 4

[/ QUOTE ]

Correct.

[ QUOTE ]
Exactly two = 1*term 2 - 3*term 3 + 6*term 4

[/ QUOTE ]

Correct.

[ QUOTE ]
Three or more = 1*term 3 - 4*term 4

[/ QUOTE ]

No, this should be 1*term 3 - 3*term 4 + 6*term 5 - ...

Term 3 counts 4 players 4 times, so we have to subtract 3 times term 4 to make this 4-3 = 1. If you subtract 4*term 4 you are making this 0, which would be correct if you were doing exactly 3. Notice that the coefficients for 3 or more should be the same as for exactly 2. In general, exactly N will have the same coefficients as N+1 or more (applied to shifted terms of course).

[ QUOTE ]
Am I correct that in finding exactly two you start with the second term and then apply the coefficients.

[/ QUOTE ]

Yes.

Normally we just want 1 or more, and we never solve explicitly for exactly anything, we just combine the terms +/-1. That is basic inclusion-exclusion. Using coefficients other than +/-1 is a modified form used for exactly N, or N or more for N &gt; 1.

Of course to be exact you would have to evaluate all the terms until you run out. I only stopped at 4 here for convenience, and because the terms get so small. Normally we only need 2 or 3 terms to be as accurate as we need.

Cobra
12-27-2004, 09:33 AM
Thanks again Bruce. I thought about it again last night and realized I subtracted out too many of the fourth term. I believe I understand now. Thanks for the help.

Cobra

gamble4pro
12-28-2004, 10:19 AM
If:
n = number of your opponents
p = number of pairs (as value type) higher than the one you hold, then the probability for at least one opponent to hold a higher pair is:
P = 6*n*p / 1225 – n*(n – 1)*p*(6*p – 1) / 460600 + n*(n – 1)*(n – 2)*p*(6*p – 1)*(6*p – 2) / 1430163000. This is a +/- 8% approximation (not all terms shown). This result is from "Texas Hold'em Odds", by Catalin Barboianu (www.probability.go.ro/page8.html). So your result should be 1 minus this one. Seems to be pretty much different.

Marm
12-29-2004, 06:20 PM
Without Reading all the other posts, this could have been covered, I worked out a simple formula that you can use. There is a slight fudge factor to allow you to do the math in your head:

7n - xn/2

Where N is the number of players and X is the value of your pair (A=14, K=13, etc... )

This gives you a % chance of there being an overpair in the number of players specified.

EX. You have JJ (11) with 6 players remaining; 6x7 - (11x6)/2

42- 33 = 9%

Of course this ignore the probability of others having PP's too. It just gives you the % chance of at least one player having a greater PP.

Marm
01-03-2005, 01:07 AM
I have posted that formula on other forums, usually got positive feedback for it, but since this is THE place for this type of question, just curious on what you guys thought of it.

Cobra
01-03-2005, 01:49 AM
Marm,

I have also created a formula but it is not something you could use in a game. It is referenced in the above thread on 12/19. I compared my numbers to your's and I believe your quick formula gives a reasonable result. First to make sure I did it correctly in your formula I assumed you meant number of opponents, not players for N. I looked at 9,5, and 2 opponents with me holding a Q,8, or 3.

I list your probablility first and then mine.

.........QQ...........88...........33

9-----.09/.086 .27/.237 .495/.393

5-----.05/.048 .15/.139 .275/.242

2-----.02/.0195 .06/.058 .11/.1048

As you can see your formula becomes very accurate with less opponents, or a higher starting pair. It is inaccurate with a lot of opponents and a lower pair to start with.
I hope this helps.

Cobra

Marm
01-04-2005, 02:26 AM
Very much so, thanks.