[gaming_mouse]

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Finally, I realize this is unrealistic, but assume that each third card is equally likely (2's as likely as A's; as if opponents were playing random hands).

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Given this assumption, which you correctly note is unrealistic, the calculation is fairly easy. Say you are against three players. There are two T's left in the deck, out of a total of 47 unseen cards (we substract the 3 cards on the board, as well as the 2 cards in your hand). We'll now calculate that chance that NONE of your opponents has a T.

The chance that your first oppo does not hold a T is:

(45 choose 2)/(47 choose 2)

Given that your first oppo does not hold a T, the chance that your second oppo does not hold T is:

(43 choose 2)/(45 choose 2)

And the third:

(41 choose 2)/ (43 choose 2)

Cancelling terms, we get (for the chance that no oppo holds a T):

(41 choose 2)/(47 choose 2)= .7585

So there is about a 24% chance that at least 1 of 3 opponents holds a T. The calculation is similar for more players.

gm