Is this correct? (Pocket pair v. Pocket Pair)

[tpir90036]

I am dealt a pocket pair... what are the odds that *any* of my opponents also has a pocket pair?

My answer:
there are (C(6,2) * 12 + 1) pocket pair combinations left out of C(50,2) possible hands. or, 73/1225 for each person.

for N opponents we get:
1 - [1 - (73/1225)]^N
i.e. one minus the odds of *none* of my opponets having a PP.

alternate answer:
or is it ok to just use this?
N * (73/1225)

it would seem that the first answer is more correct as the 2nd one begins to diverge as N gets large. i.e. if we pick N to be 20 opponents the 2nd solution gives an answer that is over 100%!

is the first answer correct? is the 2nd one just a reasonable approximaton for normal sized N? am i missing something?

thanks in advance,
-tpir

[BruceZ]

[ QUOTE ]
there are (C(6,2) * 12 + 1) pocket pair combinations left out of C(50,2) possible hands. or, 73/1225 for each person

[/ QUOTE ]

I assume the C(6,2) is a typo. It should be C(4,2) = 6, and the rest is correct.

[ QUOTE ]
for N opponents we get:
1 - [1 - (73/1225)]^N
i.e. one minus the odds of *none* of my opponets having a PP.

alternate answer:
or is it ok to just use this?
N * (73/1225)


[/ QUOTE ]

Both of these methods are approximations. The first is an approximation because the hands are not independent (the event of one player holding a pair changes the probablity of another holding a pair). The second is an approximation because the hands are not mutually exclusive. That is, more than one opponent can hold a pair, and the approximation counts these more than once. These approximations are often very good, and the second approximation is the starting point to the exact answer by the inclusion-exclusion principle. (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=417383&page=&view=&sb =5&o=&vc=1)

This thread (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Number=396447&page=&view=&sb =5&o=) discusses the exact solution to the problem of determining if any player holds a pair.