Probability of pocket pairs in 10-handed game

[bigpooch]

[ QUOTE ]

If you scroll all the way down to the bottom of the page , it tells you the probability that a larger pair is out given that you hold pocket pair X . His numbers are for n=10 so if you hold pocket deuces , then the probability someone holds a higher pocket pair is =~ 0.3269 + 0.0818 +0.01186=~ 0.42056


[/ QUOTE ]

I'll be the nit here and say that the table on that page of
Professor Alspach's site is incomplete (it only has columns
for "one rank", "two ranks" and "three ranks" although it's
true that the contribution for higher ranks will be small).

Also, the bottom entries are 0.3269, 0.08186 and 0.01186 and
the sum of these is 0.42062, which only represnts a lower
bound.

I'll post a calculation using the inclusion-exclusion
method, but I'll probably be a bit lazy and round off
to the nearest 10^(-10), instead of expressing the answer as
an exact fraction.

[bigpooch]

Multiplying (3/51) = 1/17 by 10 is just the start. You
have to also consider when two or more players have a
pocket pair.

There is a useful idea called the "inclusion-exclusion"
principle. Suppose the n events A[1], A[2], ..., A[n] are
given and let

P(S) denote the probability of S,
S*T denote the intersection of sets S and T,
S+T dentoe the union of the sets S and T,
and sigma(conditions; expression) denote summation notation

Then,

P(S[1]+S[2]+...+S[n]) =
sigma(1<=j<=n; P(S[j]) )
-sigma(1<=j<k<=n; P(S[j]*S[k]) )
+sigma(1<=j<k<l<=n; P(S[j]*S[k]*S[l]) )
-...
+[(-1)^(n-1)]P(S[1]*S[2]*...*S[n])

You could also find this on various websites, perhaps if
this seems unclear. The idea is you simply subtract off
all "double counting" errors (you double counted when two
players had a pocket pair) but then add back "triple
counting" adjustments, etc.


For the problem in question, the probability that the first
player gets a pocket pair is the same as that for each of
the other players, if there are no other conditions (if we
don't know anything about what other players hold).

Similarly, for any two different sets of n players, the
probability that each set of players all have pocket pairs
is identical (doesn't matter which players are selected)
where 2<=n<=10.

Let A[j] be the event that the jth player has a pocket pair
and let C(m,n) be the number of combinations of m objects
taken n at a time (which is simply m!/[(n!)(m-n)!] ). Note
that there are C(10,n) ways of choosing n players at a table
of ten players.

Thus, the probability that at least one player has a pocket
pair in ten-handed hold'em is:

10*P(A[1])
-C(10,2)*P(A[1]*A[2])
+C(10,3)*P(A[1]*A[2]*A[3])
-C(10,4)*P(A[1]*A[2]*A[3]*A[4])
+C(10,5)*P(A[1]*A[2]*A[3]*A[4]*A[5])
-C(10,6)*P(A[1]*A[2]*A[3]*A[4]*A[5]*A[6])
+C(10,7)*P(A[1]*A[2]*A[3]*A[4]*A[5]*A[6]*A[7])
-C(10,8)*P(A[1]*A[2]*A[3]*A[4]*A[5]*A[6]*A[7]*A[8])
+C(10,9)*P(A[1]*A[2]*A[3]*A[4]*A[5]*A[6]*A[7]*A[8]*A[9])
-P(A[1]*A[2]*A[3]*A[4]*A[5]*A[6]*A[7]*A[8]*A[9]*A[10])

The first term is just (1/17)10 = 0.588235294118

The second term (negative) is:
-45(1/17)((12*6+1)/C(50,2)) = (-45/17)(73/1225) = 657/4165
= -0.157743097239.

The first two terms sum to 0.430492196879

The third term is:
120(1/17)[(1/1225)(72/1128)+(72/1225)(68/1128)]
=4968/195755 = 0.025378662103

The first three terms sum to 0.455870858982

The fourth term (negative) is:
-210(1/17)[(1/1225)(72/1128)(67/1035)+
(72/1225){(2/1128)(67/1035)+(66/1128)(63/1035)}
=-6590808/2431277100 = -8718/3215975
=-0.002710841969

The sum of the first four terms is 0.453160017013

The fifth term is:
252(1/17)[(1/1225)(72/1128){(1/1035)(66/946)+
(66/1035)(62/946)}
+(72/1225){(2/1128)*[(1/1035)(66/946)+(66/1035)(62/946)]
+(66/1128)*[(3/1035)(62/946)+(60/1035)(58/946)]}
=4616377920/(17*1225*1128*1035*946)
=0.000200713119

The sum of the first five terms is 0.453360730132

The sixth term (negative) is:
-210(1/17)[(1/1225)(72/1128){(1/1035)(66/946)(61/861)
+(66/1035){(2/946)(61/861)+(60/946)(57/861)} }
+(72/1225){(2/1128)*[(1/1035)(66/946)(61/861)
+(66/1035){(2/946)(61/861)+(60/946)(57/861)}]
+(66/1128)*[(3/1035){(2/946)(61/861)+(60/946)(57/861)}
+(60/1035){(4/946)(57/861)+(54/946)(53/861)}]}]
=-206075469600/[17*1225*1128*1035*946*861]
= -0.000010406329

The sum of the first six terms is 0.453350323893

From the Bonferroni inequalities, the probability is
strictly between the last two sums, and since the seventh
term will be small (anyone dare compute this?), the
probability is approximately 0.45335.

Anyone care to check this?

[jay_shark]

Hey Pooch , you must have had fun working out the sixth term :P

Everything looks right which is consistent with the formula I've provided . To be honest with you , I don't really like the way Brian computed the probability of multiple pocket pairs because all of that work isn't even necessary and it's much better to use the inclusion/exclusion like you did .

I haven't checked all the little details , but did you compute the probability that only distinct ranks of pairs are out ?

[bigpooch]

In my calculations, if you check carefully, they include any
identical pairs. Well, I thought I could continue, but I
didn't think there would be much value to know this to more
than five decimal places!