It's very easy to compute the wrong answer, and a relative pain in the butt to compute the correct one:


1-

[(50*49/2)-24]/(50*49/2)*

[(48*47/2)-24]/(48*47/2)*

[(46*45/2)-24]/(46*45/2)*

[(44*43/2)-24]/(44*43/2)*

[(42*41/2)-24]/(42*41/2)*

[(40*39/2)-24]/(40*39/2)*

[(38*37/2)-24]/(38*37/2)*

[(36*35/2)-24]/(36*35/2)*

[(34*33/2)-24]/(34*33/2)


= 22%.


The denominator will come up in other problems with 9 opponents, so it is worth remembering its value(2.2575x10^26).


Note that the odds of a given player having one of these hands is 24/1326=1.8%, but if the first 9 players do not have one of these hands, the chance that the final player has it is 4.3% or more than twice as great.