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 11-11-2009, 04:51 PM #1 spadebidder Actually Shows Proof     Join Date: Aug 2008 Location: This looks interesting. Posts: 7,904 31.7% chance someone at table has the same ranks? What is the chance that at a 9-player table, two players hold matching hands like AK AK? There are 13 pair ranks and 78 non-pair rank combinations you can hold ((13*12)/2), or 91 unique rank combinations possible. So we can make an approximation and assume independence and solve it using the birthday problem formula, giving the chance that no one at a 9-player table has the same rank combination as Perm(91,9) / (91^9) = 66.5%, so 33.5% chance that someone does. This is the same as doing 1 - [(91/91 * 90/91) * 89/91 * (88/91).........] for 9 players But it’s actually going to be a little less because hands are not independent and we deal to all players without replacement. Each of the 78 non-pairs can happen 16 ways. Each of the 13 pairs can happen 6 ways. If we weight this to get an average we have [(78*16)+(13*6)] / 91 = 14.57 ways each hand can happen. Same as just 1326/91. So I did an adjusted calculation where I removed 1/14.57 from the denominator for each successive player and came up with 31.7%. N = 1/14.57 1 - [90/(91-N) * 89/(91-2N) * 88/(91-3N) .....] = 31.7% I think that 31.7% is very close, but can someone show me the exact way to solve this problem? Last edited by spadebidder; 11-11-2009 at 05:01 PM.
 11-12-2009, 10:38 AM #2 spadebidder Actually Shows Proof     Join Date: Aug 2008 Location: This looks interesting. Posts: 7,904 Re: 31.7% chance someone at table has the same ranks? bump - help
 11-12-2009, 02:22 PM #3 brothertupelo banned   Join Date: Nov 2009 Posts: 490 Re: 31.7% chance someone at table has the same ranks? this is off the top of my head, so i may be wrong, but here goes heads up: if you have a pair, the odds that your opponent has the same hand is 2/50*1/49 if you don't, the odds are 6/50*3/49 since you can draw either one first there are 52*3=156 possible paired hands and 52*51-156=2469 possible unpaired hands, which means that 6.3% of the time, you'll catch a pair, and the rest, not. multiplying the respective numbers and: odds you'll both have the same pair pair is .000051578 and the odds you'll have the same unpaired is .00691. add the 2 and you have a .00695 chance of matching heads up for 9 players, you have 8+7+6...+1=36 heads up possibilities, so .00695*36=.2502, or about a 1 in 4 chance. in fact it might be exactly 1/4 due to rounding errors, but i'm not positive about that. but the answer is .25 Last edited by brothertupelo; 11-12-2009 at 02:29 PM.
 11-12-2009, 02:28 PM #4 brothertupelo banned   Join Date: Nov 2009 Posts: 490 Re: 31.7% chance someone at table has the same ranks? the problem you ran into is that you have 13 pairs and 78 unpaired, but the weights are wrong because, for instance, with an ace spades, you can match it with ah, ac, or ad, for a pair but with, for instance ace deuce, you not only have 2h, 2c, 2d, you also have the suited 2s, so the ratio of paired to unpaired isn't 1/6, it's a little less than 1/15. i made the same mistake the first time through.
 11-12-2009, 02:30 PM #5 brothertupelo banned   Join Date: Nov 2009 Posts: 490 Re: 31.7% chance someone at table has the same ranks? what brought this up, btw?
11-12-2009, 02:42 PM   #6
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Re: 31.7% chance someone at table has the same ranks?

Quote:
 Originally Posted by brothertupelo but the answer is .25 ... the problem you ran into is that you have 13 pairs and 78 unpaired, but the weights are wrong because, for instance, with an ace spades, you can match it with ah, ac, or ad, for a pair but with, for instance ace deuce, you not only have 2h, 2c, 2d, you also have the suited 2s, so the ratio of paired to unpaired isn't 1/6, it's a little less than 1/15. i made the same mistake the first time through.
From my post:
N = 1/ [(78*16)+(13*6)] / 91 = 1/14.57

I know for sure that my solution is very close to the correct answer, and that 25% is not close at all. This formula is how to solve the problem:

If N = 1/14.57 then
1 - [90/(91-N) * 89/(91-2N) * 88/(91-3N) .....extend to 8 opponents] = 31.7%

I'm just not sure that applying the weighting in that way for the lack of independence is exact or not, but it's very very close.

Last edited by spadebidder; 11-12-2009 at 02:50 PM.

 11-12-2009, 02:46 PM #7 brothertupelo banned   Join Date: Nov 2009 Posts: 490 Re: 31.7% chance someone at table has the same ranks? yeah, you're right. i realized my mistake. i didn't take into account the fact that suited unpaired is worth more than unsuited unpaired. subtract out suited unpaired from my work and make a separate category for it, and run it again.
11-12-2009, 08:10 PM   #8
BruceZ
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Re: 31.7% chance someone at table has the same ranks?

Quote:
 Originally Posted by spadebidder I know for sure that my solution [31.7%] is very close to the correct answer, and that 25% is not close at all.
That's funny, because I know that the correct answer cannot possibly be more than 25.1%, and that is trivial to show.

The probability that 2 specific players have the same hand is

(16/17)*9/C(50,2) + (1/17)*1/C(50,2)

That is the probability that the first guy has a non-pair, 16/17 = C(13,2)*4*4/C(52,2), times the probability that the second guy has the same non-pair, of which there are 3*3=9 remaining out of C(50,2) total hands. The probability that the first guy has a pair is 1/17, times the probability that the second guy has the same pair, of which there is 1 remaining out of C(50,2) total hands.

Now for an upper bound on the probability that at least 2 players share the same hand at a 9-player table, multiply the above by the total number of pairs of players C(9,2). This an upper bound since it over counts the cases where more than one pair of opponents have the same hand.

P(at least 2 players of 9 have same hand) <

C(9,2) * [ (16/17)*9/C(50,2) + (1/17)*1/C(50,2) ]

=~ 25.1%

Now with a fair amount of work, I can show with the incluson-exclusion principle that the exact answer to within 0.1% is actually 24.9%, which is quite close to 25% . I will put that in a separate post. So the exact answer is very close to the mutually-exclusive approximation.

11-12-2009, 10:30 PM   #9
BruceZ
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Re: 31.7% chance someone at table has the same ranks?

Quote:
 Originally Posted by brothertupelo this is off the top of my head, so i may be wrong, but here goes heads up: if you have a pair, the odds that your opponent has the same hand is 2/50*1/49 if you don't, the odds are 6/50*3/49 since you can draw either one first there are 52*3=156 possible paired hands and 52*51-156=2469 possible unpaired hands, which means that 6.3% of the time, you'll catch a pair, and the rest, not. multiplying the respective numbers and: odds you'll both have the same pair pair is .000051578 and the odds you'll have the same unpaired is .00691. add the 2 and you have a .00695 chance of matching heads up for 9 players, you have 8+7+6...+1=36 heads up possibilities, so .00695*36=.2502, or about a 1 in 4 chance. in fact it might be exactly 1/4 due to rounding errors, but i'm not positive about that. but the answer is .25

This is essentially the same as my upper bound calculation, though you do not state that the multiply by C(9,2)=36 is an approximation. You ignored the pairs in the end which makes a small difference.

The problem is not with the suited non-pairs as spadebidder wanted to count those the same as the unsuited. If the suited non-pairs are counted separately, then the upper bound of the probability is even lower since an increase in the number of hands decreases the probability that 2 players have the same hand. I got 16.8% for the upper bound in that case.

Last edited by BruceZ; 11-12-2009 at 10:36 PM.

11-12-2009, 10:33 PM   #10
BruceZ
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Re: 31.7% chance someone at table has the same ranks?

This equation is still under construction. I will post a note later when it is complete.

Quote:
 Originally Posted by spadebidder What is the chance that at a 9-player table, two players hold matching hands like AK AK?

You can get as accurate as you like with the inclusion-exclusion principle. I will compute it to within 0.1%. I have aleady shown in a separate post that the upper bound is 25.1%.

Although this looks like alot of work, there is really alot of repetition of simple terms, and the whole thing can be done quite quickly.

P(at least 2 players with same hand) =

P(at least 2 players with same pair) +
P(at least 2 players with same non-pair) -
P(at least 2 players with same pair AND 2 players with same non-pair)

P(at least 2 players with same pair) =
C(9,2)*P(2 specific players with same pair) -
C(9,4)*P(2 specific pairs of players which each share a pair) +
...

= C(9,2)*13*6 /C(52,2)/C(50,2) -
C(9,4)*13*6*12*6 /C(52,2)/C(50,2)/C(48,2)/C(46,2)

=~ 0.173%

P(at least 2 players with same non-pair) =

C(9,2)*P(2 specific players with same non-pair) -

C(9,3)*P(3 specific players with same non-pair) +

C(9,4)*[ P(4 specific players with same non-pair) + P(2 specific pairs of players which each share a non-pair) ] -

...

= C(9,2)*C(13,2)*4*4*3*3 /C(52,2)/C(50,2) -

C(9,3)*C(13,2)*4*4*3*3*2*2 /C(52,2)/C(50,2)/C(48,2) +

C(9,4)*[ C(13,2)*4*4*3*3*2*2*1*1 +
C(4,2)*C(13,2)*4*4*C(12,2)*4*4 ] /C(52,2)/C(50,2)/C(48,2)/C(46,2)

=~ 24.7%

P(at least 2 players with same pair AND 2 players with same non-pair) =

C(9,4)*P(2 specific players with same pair)*P(2 specific players with same non-pair)

-

...

= C(9,4)*C(4,2)*13*6*C(12,2)*4*4 /C(52,2)/C(50,2)/C(48,2)/C(46,2)

=~ 0.003%

This last term is neglible compared to the others.

Last edited by BruceZ; 11-13-2009 at 11:23 AM.

 11-12-2009, 10:41 PM #11 spadebidder Actually Shows Proof     Join Date: Aug 2008 Location: This looks interesting. Posts: 7,904 Re: 31.7% chance someone at table has the same ranks? You're right, I found the mistake in my method too. If it comes out close I'll show it. Thanks for the work, I really needed to get the answer to this.
11-13-2009, 02:01 AM   #12
brothertupelo
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Re: 31.7% chance someone at table has the same ranks?

Quote:
 Originally Posted by BruceZ You ignored the pairs in the end which makes a small difference. .
no, i didn't. i added that in. it's just so small as to be not really noticable.

11-13-2009, 03:15 AM   #13
BruceZ
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Re: 31.7% chance someone at table has the same ranks?

Quote:
 Originally Posted by brothertupelo no, i didn't. i added that in. it's just so small as to be not really noticable.
Ah, you did add it. However,

Quote:
 if you have a pair, the odds that your opponent has the same hand is 2/50*1/49 if you don't, the odds are 6/50*3/49 since you can draw either one first there are 52*3=156 possible paired hands and 52*51-156=2469 possible unpaired hands, which means that 6.3% of the time, you'll catch a pair, and the rest, not. multiplying the respective numbers and: odds you'll both have the same pair pair is .000051578 and the odds you'll have the same unpaired is .00691. add the 2 and you have a .00695 chance of matching heads up for 9 players, you have 8+7+6...+1=36 heads up possibilities, so .00695*36=.2502, or about a 1 in 4 chance.
52*51-156 is actually 2496, not 2469. You are counting each 2 card combination twice, which is OK since the factor of 2 comes out in the divide, but you took 156/2469 = 6.3% which should be 156 / (156 + 2496) = 1/17. This is the probability of being dealt a pair, and 16/17 is the probability of being dealt a non-pair. It looks like you actually used 16/17 to get the .00691, but 6.3% to get the .000051578 which should actually be .000048019. Then the sum is 0.006963, and 36*0.006963 =~ 0.2507.

Last edited by BruceZ; 11-13-2009 at 05:50 AM.

 11-13-2009, 11:21 AM #14 BruceZ Carpal \'Tunnel     Join Date: Sep 2002 Posts: 11,877 Re: 31.7% chance someone at table has the same ranks? This is not yet finished. Since we are counting pairs of players with the same hand, we need a modified form of the inclusion-exclusion principle which puts some coefficients in front of some of those terms. I did a simulation which suggests the answer is 22.8%. I'm working to square this with the equation. The upper bound of 25.1% is still correct. Last edited by BruceZ; 11-13-2009 at 02:41 PM.

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