Odds of 2 people having same PP

[gaming_mouse]

Yeah,

Mickey. The answer is not right. I'm working it myself too. I think there's a way to make it work tho.... I'll post back in a few.

[gaming_mouse]

[ QUOTE ]
Yeah,

Mickey. The answer is not right. I'm working it myself too. I think there's a way to make it work tho.... I'll post back in a few.

[/ QUOTE ]

Okay, I think I got it.

The divisor should be: 52!/(32!*2^10)

Also, you need to add the multinomial coefficient 10Ci,j,k into each term of the sum.

This results in the answer:
0.0046903

which seems reasonable.

A simulation could confirm....

EDIT: Forget it. This would mean that it is almost certain that it would happen. I forgot about the subtraction from 1.....

[BruceZ]

[ QUOTE ]
OK... for a specific pair, that answer is correct...
.00016622 or 1 in 6016 hands
however this was not the OP's question

[/ QUOTE ]

The title of the OPs question is "Odds of 2 People Having Same PP". "PP" stands for pocket pair. Cobra stated that he was solving for a specific pair AA. Your response to his post was:

[ QUOTE ]
This is a very close approximation (for a specific pair), however it is not correct.

[/ QUOTE ]

That was what I was responding to. His solution for a specific pair was not an approximation. It was exact and correct.


[ QUOTE ]
and does the response imply that any pair would be .00016622*13 ranks = .00216086

[/ QUOTE ]

No, it certainly doesn't imply that. We cannot just multiply by 13 here because the 13 ranks are not mutually exclusive. There can be 2-5 ranks shared between 2-5 pairs of players. This is different from the case where we had C(10,2) mutually exclusive and disjoint pairs of players that can share a specific pocket pair. Do you understand the difference?


[ QUOTE ]
This is not correct... as by inclusion-exclusion "I" get
0.00215899941535865 for any pair. The solution is lengthy, but I would gladly post it for you if you would like.

[/ QUOTE ]

I get a slightly different answer. Let me know where we differ, and I will attempt to explain the difference.

C(10,2)*13*6/C(52,2)/C(50,2) -

C(10,4)*C(4,2)*13*12*6^2/2! /C(52,2)/C(50,2)/C(48,2)/C(46,2) +

C(10,6)*C(6,3)*13*12*11*6^3/3! /C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2) -

C(10,8)*C(8,4)*13*12*11*10*6^4/4! /C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)/C(40,2)/C(38,2) +

C(10,10)*C(10,5)*13*12*11*10*9*6^5/5! /C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)/C(40,2)/C(38,2)/C(36,2)/C(34,2)

= 0.00215899865959

[MickeyHoldem]

Here's my calculations: SHORT VERSION

no duplicate pair = 1 - p(i) + p(ii) - p(iii) + p(iv) - p(v)

p(i) = c(13,1) * 3 * c(48,16) * 15!! / (c(52,20) * 19!!)
p(ii) = c(13,2) * 3^2 * c(44,12) * 11!! / (c(52,20) * 19!!)
p(iii) = c(13,3) * 3^3 * c(40,8) * 7!! / (c(52,20) * 19!!)
p(iv) = c(13,4) * 3^4 * c(36,4) * 3 / (c(52,20) * 19!!)
p(v) = c(13,5) * 3^5 / (c(52,20) * 19!!)

no duplicate pair = 0.99784100058464

atleast 1 duplicate PP = 1 - .99784100058464 = .00215899941536

---- for any lurkers -----
for odd n
n!! = n * n-2 * n-4 * ... * 3 * 1
19!! = ways to arrange 20 cards into 10 hands

LONG VERSION

Same PP

Most pairs of PP = 5

How many deals contain 5 pairs of PP?
There are 13 ranks... we need 5 and each rank can be arranged into PP 3 ways, so...
D5 = 13 choose 5 * 3^5 = 312741

How many deals contain exactly 4 pairs of PP?
There are 13 ranks... we need 4, each rank can be arranged 3 ways
Now the other 2 hands need 4 cards from 36 remaining, and are arranged 3 ways
d4 = 13c4 * 3^4 * 36c4 * 3 = 10234449225
however this will include hands in which there are 5 pairs of PP(PoPP) so we need to subtract some hands, but how many? d4 contains 5 pairs of PP (5 choose 4) times, so...
D4 = 10234449225 - 5 * 312741 = 10232885520

How many deals contain exactly 3 PoPP?
13 choose 3 * 3^3 times the number of ways to arrange the (40 choose 8) cards into 4 hands. How many ways are there to arrange 8 cards into 4 hands... 7!! = 7 * 5 * 3 * 1 = 105, so...
d3 = 13c3 * 27 * 40c8 * 105 = 62355087644850
however this counts the 4 and 5 pair hands, so we need to subtract D4 and D5. d3 contains 4 PoPP (4choose3) times and 5 PoPP (5choose3) times, so...
D3 = 62355087644850 - (4c3 * 10232885520) - (5c3 * 312741) =
62314152975360

How many deals contain exactly 2 PoPP?
d2 = 13c2 * 9 * 44c12 * 11!! = 153904827325019000 <--- excel rounds this off but it should not matter
Of course this number counts 3,4 and 5 PoPP hands so subtract again...
D2 = 153904827325019000 - (3c2 * 62314152975360) - (4c2 * 10232885520) - (5c2 * 312741) = 153717823465652000
(rounded)

Almost there: How many deals contain exactly 1 PP?
in one step:
D1 = 13c1 * 3 * 48c16 * 15!! - (2c1 * D2) - (3c1 * D3) - (4c1 * D2) - (5c1 * D1) = 177947147017904000000 (again rounded)


How many possible deals are there?
D = 52c20 * 19!! = 125994627894135 * 654729075 =
82492346176096200000000 again this number has been rounded off

odds of atleast 1 PP: (D1 + D2 + D3 + D4 + D5) / D = 0.00215899941535866
odds of exactly 1 PP: D1 / D = 0.00215713524062027

[MickeyHoldem]

for a loose approximation:

1/17 times PP... odds to match 1/1225
4/17 suited cards.... odds to match 9/2450 * 1/3
12/17 unsuited.... odds to match 9/2450 * 7/9

(1/17 * 1/1225 + 4/17 * 3/2450 + 12/17 * 7/2450) = 0.0023529412

so for 9 opponents:
1 - (1-0.0023529412)^9 = 0.02097825
or
9 * 0.0023529412 = 0.021176471

seems reasonable!

its late... I guarentee nothing (spelling included)

[gaming_mouse]

[ QUOTE ]
for a loose approximation:

1/17 times PP... odds to match 1/1225
4/17 suited cards.... odds to match 9/2450 * 1/3
12/17 unsuited.... odds to match 9/2450 * 7/9

(1/17 * 1/1225 + 4/17 * 3/2450 + 12/17 * 7/2450) = 0.0023529412

so for 9 opponents:
1 - (1-0.0023529412)^9 = 0.02097825
or
9 * 0.0023529412 = 0.021176471

seems reasonable!

its late... I guarentee nothing (spelling included)

[/ QUOTE ]

I like this approach. First, fixing some small errors:

1/17 times PP... odds to match 1/1225
4/17 suited cards.... odds to match 3/1225
12/17 unsuited.... odds to match 6/1225

(1/17 * 1/1225 + 4/17 * 3/1225 + 12/17 * 6/1225) = 0.00408163265

Now we account for all possible pairs of players (10 choose 2)*0.00408163265 = 0.183673469

Now we need the 2nd order term of inclusion-exclusion. There are 45 choose 2 = 990 pairs of pairs of players. (10 choose 3) = 120 of those pairs of pairs are actually made up of only 3 people (eg, players 1&2 and players 2&3). The remaining 870 will be made up of 4 non-overlapping players.


When the pair of pairs is made up of 3-players, the chance of both pairs having the same hand (ie, all 3 players having the same hand) is:

1/17 times 0 (can't have 3 people w/ same pair)
4/17 times 3/1225*2/1128
12/17 times 6/1225*4/1128

This piece of the calculation is therefore:

120*(4/17*3/1225*2/1128 + 12/17*6/1225*4/1128)= 0.00159382902

When we 4 distinct players:

1/17 times 0
4/17 times 3/1225*2/1128*1/1035
12/17 times 6/1225*4/1128*1/1035

This piece of the calc:

870*(4/17 *3/1225*2/1128*1/1035 + 12/17 * 6/1225*4/1128*1/1035) = .0000111645028

Adding these together we get 0.00160499352

Meaning that the first order approximation is accurate. But 18% seems way too high.

Mickey, can you see what's wrong with this?

[gaming_mouse]

So I finally just did a sim of 1 million trials, and got 19.2% -- surprising, though it kind of agrees w/ the last calc.


Making everything exact for my last calc (there was a small error in the unsuited case), the first term of inclusion-exclusion:

1/17 times PP... odds to match 1/1225
4/17 suited cards.... odds to match 3/1225
12/17 unsuited.... odds to match 7/1225

(10 choose 2)*(1/17 * 1/1225 + 4/17 * 3/1225 + 12/17* 7/1225) = 0.2096

Now we are just a bit too high.

There were some small errors in my 2nd term calcuation too (eg, there are 330 overlapping pairs of pairs, not 120), but even correcting for these will not make the 2nd term significant enough to account for the difference between the sim and the first term.

Anyway, this is getting much closer. I'd appreciate anyone who can find out why the first term of inc-excl is not closer to the sim results. Of course, there may be errors in the sim, but I think it's more likely the errors are here.

gm