Calculating percentage of better hand.

Question

For example, I have two cards and there are two other players. They have both 20% chance of having a better hand then me, how much chance is it in total that either one of them has a better hand then mine?

Thanks!

Answer 1

It means you have a hand that's worse EITHER than the first OR than a second, so don't count twice:

$$ P(A \cup B) = P(A) +P(B) - P(A \cap B) $$

If these probabilities are independent, you have $P( A \cap B) = P(A)P(B)$.