Monty Hall Problem with a twist // Conditional Probabilities: I am getting hung

Question

Monty Hall Problem with a twist // Conditional Probabilities:

I am getting hung up how to handle the fact the host chooses two doors, and am further unsure of how to apply the discrete expected value by conditioning equation of E[X] = E[E[XIY]] = sum x*P(X=x)...

Game show with five doors (A, B, C, D and E). There is a prize behind one of the doors and a nothing behind the other four doors. You pick Door A and the host shows you that there is nothing behind Door B and Door C. Should you switch to another door (D or E) or stay with Door A? What is the new expected value of Doors A and D (given that you initially picked Door A and then the host opened B & C)?

Explanation / Answer

The best way to do monty hall problems or its variations is to make cases and solve it. Here as we have 5 doors and the prize could be behind any of the 5 doors, therefore there could be 5 possible cases here. These can put as:

Now as all of the 5 cases above are equally likely, that is the prize could be behind any of the 5 doors, therefore the required probability that the person wins if he switches the door is computed as:

= (1/5)*(0 + 0.5 + 0.5 + 0.5 + 0.5) = 2/5 = 0.4

Also Probability of winning if switched is computed as:

= (1/5)(1 + 0 + 0 + 0 + 0 ) = 0.2 < 0.4

Therefore there is a better chance of winning if the door is switched.

Door Chosen Prize Behind A Prize Behind B Prize Behind C Prize Behind D Prize Behind E Door Revealed by host Winning probability if switched Winning probability if not switched A Yes No No No No B and C 0 1 A No Yes No No No C and D 0.5 0 A No No Yes No No B and D 0.5 0 A No No No Yes No B and C 0.5 0 A No No No No Yes B and C 0.5 0

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