Dodgeball

The number of ways of choosing which four of the twenty-four players are left on the playing field is $\dbinom{24}{4}$ . The number of ways of making that choice by taking three players from Team A and one player from Team B is $\dbinom{12}{3} \dbinom{12}{1}$ , which is the same as the number of ways of taking one player from Team A and three players from Team B.

So the desired probability is

$\dfrac{2 \dbinom{12}{3} \dbinom{12}{1}}{\dbinom{24}{4}} = \boxed{\dfrac{80}{161}}$