Problem: You have two options

1: One successful shot at the hoop
2: 2 of 3 successful shots at the hoop

Let P be the probability of making a successful hit, for what value of P will you choose option 2 over 1.

Solution:

If P is probability for option 1

Option 2, we can have 2 cases of success, one all 3 shots are successful – P*P*P

or 2 Hits and 1 Miss (HHM, HMH, MHH) 3 cases, each with probability of P*P*(1-P) i.e. 1-P for one miss. as there are 3 cases, total probability is 3*P*P*(1-p). Combining both result for option 2, P*P*P + 3*P*P*(1-P).

To prefer option 2,

P*P*P + 3*P*P*(1-P)>P
P*P+3P(1-P)>1

Can be solved to quadratic equation

-2P^2+3P-1>0
or
(2P-1)(P-1)>0

P>1 not possible so
P>1/2 or .5

Choose option 2 when probability of hitting hoop is more than 0.5

Monty Hall problem is an interesting problem that plays with probability.

Problem: There are three doors, say A, B and C. Behind one of the doors there is a prize and two are empty. You have to make a choice and try to win the prize. You choose a door say A, now host opens the door C, which is empty. The question is, should you switch the doors?

Solution: According to one solution proposed, Switching doors will increase probability of winning. This is the explanation behind that

Initially all 3 doors has probability of 1/3 of winning. So when we choose door A, we had probability of winning – 1/3. Now Doors B & C have collective probability of winning as 2/3. So when we know C is empty, probability of prize behind B is 2/3(?).

Frankly the argument does not look very strong, for me the new probability should be 1/2 for each door.

Read more here- http://en.wikipedia.org/wiki/Monty_Hall_problem