I'm having a bit of trouble understanding this particular version of the Monty Hall problem.
Say that there are 5 doors, and I pay \$100 to switch doors, with a potential reward of \$1,000. If I switch doors, what is my expected winning value? In class, my professor said that the probability of us switching and winning (net win = \$900) is (4/15). Further, the probability of us switching and not winning is (11/15).
Can anybody offer any insight into why these are our probabilities?
Assuming there is only one prize the probability of being right on the first pick is: $1/5$.
Thus the expectation given staying is $\mathsf E(W\mid S^\complement) = \tfrac 15\cdot 1000= 200$
Assuming the host always reveals a no-prize, then the probability of picking a prize given switching is $4/15$, because you can be right first time and certainly loose if you switch, or be wrong first time and then the prize is equally likely to be behind one from the three remaining doors. $$\tfrac 15\cdot 0+\tfrac 45\cdot \tfrac 13=\tfrac 4 {15}$$
The probability of losing given you switched is thus $11/15$.
Thus the expectation given switching is: $\mathsf E(W\mid S) = \frac 4{15}\times 900 - \frac {11}{15}\times 100 = \frac {500}3 = 166.\dot 6$
In conclusion, you are more likely to win if you switch, however it costs more than it is worth.