Odds of 2 people having same PP

[DeadManJay]

Can anyone show me how you would calculate the chances of two people being dealt the same hand in a 10 person hold'em match?

[Cobra]

Lets use AA as an example there are 6 different combinations that the first player could have of the four aces. Once the first player has a pair of Aces there is only one combination left for the remaining two aces.

Probability of the first guy getting an ace=6/(52 choose 2)

Probability of second guy getting an ace = 1/(50 choose 2)

Probability of both getting aces=

6/(52choose 2)*1/(50 choose 2)=.000003694

Now we need to choose our players with 10 players and we choose 2 of them = (10 choose 2)=45

Multiply them together and you get

(10 c 2)*6/(52 c 2)*1/(50 c 2)=.00016622 or 1 in 6016 hands

If you have less people at the table you use the first two terms but change the player terms to the appropriate amount.

Cobra

[MickeyHoldem]

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

[Onaflag]

Why do you ask? You didn't happen to have pocket 8s last night against a guy who also had pocket 8s, did you? [img]/images/graemlins/grin.gif[/img]

Onaflag..........

[BruceZ]

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

[/ QUOTE ]

Um...yes it is exactly correct. The C(10,2) pairs of players are mutually exclusive since no more than 2 players can have the same pocket pair. Therefore the exact answer is obtained by adding the probabilities for the C(10,2) pair of players (review mutually exclusive events), which is the the same as multiplying by C(10,2).

I see this is another fine example of your analysis presented here within a week. I suggest that you don't have the background to tell people they are wrong, so you might as well quit.

[BruceZ]

[ QUOTE ]
I see this is another fine example of your analysis presented here within a week. I suggest that you don't have the background to tell people they are wrong, so you might as well quit.

[/ QUOTE ]

I confused you with motorholdem. I apologize for the tone of my response.

[MickeyHoldem]

OK... for a specific pair, that answer is correct...
.00016622 or 1 in 6016 hands
however this was not the OP's question and does the response imply that any pair would be .00016622*13 ranks = .00216086
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.

This still does not answer the OP, for he asked "same hand", which may include "J3o" vs. "J3o", although I'm not sure it was his intent. I am still working on this answer, although it may be beyond me... I would like to see an answer for this question. I'm quite sure it has not been solved before!

[ QUOTE ]
I confused you with motorholdem. I apologize for the tone of my response.

[/ QUOTE ]

No need for an apology, I take no offence. For a specific pair the posted answer was correct, and my short reply was not a useful post. I apologize!

[gaming_mouse]

[ QUOTE ]
This still does not answer the OP, for he asked "same hand", which may include "J3o" vs. "J3o", although I'm not sure it was his intent. I am still working on this answer, although it may be beyond me... I would like to see an answer for this question. I'm quite sure it has not been solved before!

[/ QUOTE ]

78 pocket pair hands (type 1)
312 suited hands (type 2)
936 offsuit non-pair hands (type 3)
1326 total

The chance that the first do players are not dealt the same hand:

(78/1326)*(1321/1326) + (312/1326)*(1322/1326) + (936/1326)*(1314/1326) = 0.99268033

Now when we add in the third person, the weights become more complex, as we have to condition on pairs of type possibilities for what the first two players were dealt. There will now be 9 weights instead of 3. The complexity will grow exponentially as we add more people, but the method will work and could in theory be done by hand. Can anybody see a trick to make this easier?

The simplest way I can see to solve this problem is to run a simulation.

gm

[gaming_mouse]

[ QUOTE ]
The simplest way I can see to solve this problem is to run a simulation.

[/ QUOTE ]

I thought of a solution that should provide a very good approximation, though it's not exact. Maybe someone can try running it in Mathematica or something... doing it in Java would be a pain.

EDIT: I made a mistake the first time in the following:

(SUM over all triplets (i,j,k) such that i+j+k=10) ((13 choose i)*6^i) * ((78 choose j)*4^j) * ((78 choose k)*12^k)

Divide the above summation by:

(1326!/(2^10*1306!))

to arrive at the answer.

I think that works, but don't have time now to really look it over. This is not exact because it assumes independence among the possible hands, while in fact they are loosely dependent.

gm

EDIT: To clarify, the above will give the chance of nobody being dealt the same hand. So subtract from 1 to answer the orig question

[MickeyHoldem]

[ QUOTE ]
(SUM over all triplets (i,j,k) such that i+j+k=10) ((13 choose i)*6^i) * ((78 choose j)*4^j) * ((78 choose k)*12^k)

[/ QUOTE ]

Assuming I follow you here... there are 66 triplets such that i+j+k=10... SUM=~2.41405E+23

[ QUOTE ]
Divide the above summation by:

(1326!/(2^10*1306!))

[/ QUOTE ]
1326! / 2^10*1306 = permut(1326,20)/1024 = 2.388E+59

so....2E23/2E59 = ~1E-36 seems a little tiny

Maybe I misunderstood you... heres the summation data:

i j k SUM
0 0 10 7.60856E+19
0 1 9 8.60098E+20
0 2 8 4.25748E+21
0 3 7 1.21528E+22
0 4 6 2.21536E+22
0 5 5 2.69485E+22
0 6 4 2.21536E+22
0 7 3 1.21528E+22
0 8 2 4.25748E+21
0 9 1 8.60098E+20
0 10 0 7.60856E+19
1 0 9 1.4335E+20
1 1 8 1.43759E+21
1 2 7 6.23631E+21
1 3 6 1.53598E+22
1 4 5 2.36709E+22
1 5 4 2.36709E+22
1 6 3 1.53598E+22
1 7 2 6.23631E+21
1 8 1 1.43759E+21
1 9 0 1.4335E+20
2 0 8 1.10584E+20
2 1 7 9.71893E+20
2 2 6 3.63785E+21
2 3 5 7.5747E+21
2 4 4 9.59633E+21
2 5 3 7.5747E+21
2 6 2 3.63785E+21
2 7 1 9.71893E+20
2 8 0 1.10584E+20
3 0 7 4.56873E+19
3 1 6 3.46462E+20
3 2 5 1.09634E+21
3 3 4 1.87661E+21
3 4 3 1.87661E+21
3 5 2 1.09634E+21
3 6 1 3.46462E+20
3 7 0 4.56873E+19
4 0 6 1.11045E+19
4 1 5 7.11908E+19
4 2 4 1.85192E+20
4 3 3 2.50215E+20
4 4 2 1.85192E+20
4 5 1 7.11908E+19
4 6 0 1.11045E+19
5 0 5 1.64286E+18
5 1 4 8.65834E+18
5 2 3 1.77785E+19
5 3 2 1.77785E+19
5 4 1 8.65834E+18
5 5 0 1.64286E+18
6 0 4 1.48006E+17
6 1 3 6.15704E+17
6 2 2 9.35708E+17
6 3 1 6.15704E+17
6 4 0 1.48006E+17
7 0 3 7.89364E+15
7 1 2 2.43041E+16
7 2 1 2.43041E+16
7 3 0 7.89364E+15
8 0 2 2.33693E+14
8 1 1 4.73457E+14
8 2 0 2.33693E+14
9 0 1 3.3722E+12
9 1 0 3.3722E+12
10 0 0 17293326336

sum 2.41405E+23