How to compute probability of winning The producer places at random a valuable prize behind one of the three doors, numbered 1 to 3, and nothing behind the remaining two and each door is equally likely to hide the prize. If you have selected a door that hides a prize, the host will always open the smaller-numbered one of the remaining doors and then will always offer you the possibility of switching doors. Otherwise, with probability 0.5 the host will open immediately the door you have selected i.e. revealing that you have lost, and with probability 0.5 he will open the other door that does not hide the prize. In the latter case he will always give you the opportunity of switching doors. In the end, the door you select will be opened and if there's a prize, you win it.
If initially you select door 3 and if the host opens door 2 you switch, otherwise you stick with door 3, what is your probability of winning the prize?

If the prize was behind door $1$, you lose or win with probability $\frac12$ each.
If the prize was behind door $2$, you lose, since you never switch to door $2$.
If the prize was behind door $3$, you win, since the host opens door $1$ and you stick with door $3$.
Thus you win with probability $\frac13\left(\frac12+0+1\right)=\frac12$.
