[sdw]

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Heads up, the chance the other player has 2 cards of the same suit is

p = (11/50) * (10/49) = 4.490 %
or once in every 22.3 attempts.

Versus N opponents, an approximation that at least one player has 2 cards of the same suit is

P(N) = 1 - (1-p)^N

for N = 9 this is ~ 33.9 %

As approximations go, its not too bad.


Since the hands are not completely independent, I believe the second formula is only correct if

You keep your two cards (leaving 50 in the deck)
The other N guys randomly select 2 cards from the remaining 50.
After each opponent selects 2 cards, he puts them back in the deck and you reshuffle for the next guy.

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My probability is a little fuzzy, but doesn't P(N) = 1 - (1-p)^N calculate the probability that exactly one opponent has 2 cards of the same suit?

If you wanted one or more players isn't it p * N?


Here is another interesting one:

Heads up, the chance the other player has 2 cards of the suit when 3 of the 5 cards on the board are the suit and your holding 2 cards of the suit is:

p = (8/45) * (7/44) = 2.83%