We have seen the Monty Hall problem in an earlier post. This time, instead of 3, we have four doors. There is $1000 behind one door, -$1000 behind another (you lose $1000), and two other doors have nothing ($0). Like in the previous game, you choose one door, and then the game host opens a door that contains nothing. You have an option to change to one of the other closed doors now. What will you do?
No Change
In the beginning, before hosts reveals the $0 door, the probabilities are P($1000) = 1/4, P($0) = 1/2 and P(-$1000) = 1/4. The expected return is (1/4) x $1000 + (1/2) x $0 + (1/4) x -$1000 = $0. After the clue, if you still don’t want to change, this remains the case.
Change
Here, we use solution 2, the argument method, of the Monty Hall problem. Before you get the clue, the chance that you chose the $1000 door is 1/4, and that the prize was outside your choice is 1 – 1/4 = 3/4. After the clue, that probability of 3/4 sits behind two doors. In other words, if you shift, the chance of getting $1000 is 3/8. Using similar arguments, we shall see that the chance of losing became 3/8, and for $0 is 1/4. The expected return is (3/84) x $1000 + (1/2) x $0 + (3/8) x -$1000 = $0.
Will you change?
Well, it depends on your risk appetite. The chance of winning and the chance of losing have increased. But the expected returns remained the same, at zero. Or the risk has increased if you shift. If you are risk-averse, stay where you are!
