Basketball Riddle

You have a basketball hoop and someone says that you can play 1 of 2 games.

Game #1: You get one shot to make the hoop.
Game #2: You get three shots and you have to make 2 of 3.

If 'p' is the probability of making a particular shot, for which values of p should you pick one game or the other?

Solution #1: The probability of winning Game #1 = p.

Probability of winning Game #2 = P(2,3) + P(3,3)
where,
P(2,3) = probability to make 2 of 3 shots, and
P(3,3) = probability to make 3 of 3 shots.

2 of 3 shots can be taken in C(3,2) ways: (YYN, YNY, NYY). Consider YYN, since these shots are independent, we can simply multiply their probabilities to find the probability of getting YYN.

Therefore, P(YYN) = p*p*(1-p).
It's obvious that the probability of other sequences(YNY, NYY) is also p*p*(1-p). So the total probability:
P(2,3) = 3*p*p*(1-p).

3 of 3 shots can be tken in one way, therefor:
P(3,3) = p*p*p.

So, probability of winning Game #2 = P(2,3) +P(3,3) = 3*p*p - 2*p*p*p.

Now, if P(Game #1) > P(Game #2):

p  > 3*p*p - 2*p*p*p,

on solving this equation, we get,

p < .5

So, play Game #1 when 'p' < .5 otherwise play Game #2.