You are playing a TV game show involving three doors; behind one of them is a pr

Question

You are playing a TV game show involving three doors; behind one of them is a prize. You picked door 1 and the host opened door 3, showing that the prize was not behind it. You are then offered a cash prize instead of the chance to win whatever is behind the door 1. You saw that the (conditional) probability that the prize is behind door 1 is now 1/3 . Show that if the prize has a value of V , then your expected gain in staying 3 with door 1 is 1/3V. You are playing a TV game show involving three doors; behind one of them is a prize. You picked door 1 and the host opened door 3, showing that the prize was not behind it. You are then offered a cash prize instead of the chance to win whatever is behind the door 1. You saw that the (conditional) probability that the prize is behind door 1 is now 1/3 . Show that if the prize has a value of V , then your expected gain in staying 3 with door 1 is 1/3V. You are playing a TV game show involving three doors; behind one of them is a prize. You picked door 1 and the host opened door 3, showing that the prize was not behind it. You are then offered a cash prize instead of the chance to win whatever is behind the door 1. You saw that the (conditional) probability that the prize is behind door 1 is now 1/3 . Show that if the prize has a value of V , then your expected gain in staying 3 with door 1 is 1/3V.

Explanation / Answer

Initially prize is in one of the three doors, so probability that each door having a prize is,

P(D1) = 1/3, P(D2) = 1/3, P(D3) = 1/3,

Now conditioned that door 3 (D3) is having no prize, so the prize would be either in door 1 (D1) or door 2 (D2),

Hence probability that door 1 having prize is P(D1 | D3=0) = 1/3 * 1/2
similarly probability that door 2 is having prize is P( D2 | D3=0) = 1/3 * 1/2
where P(D3 | D3 = 0) = 1/3 * 0 = 0

Hence expected gain for value V would be,

E(gain) = V * P(D1 | D3=0) + V * P(D2 | D3=0)
E(gain) = V * 1/3 * 1/2 + V * 1/3 * 1/2
E(gain) = V * 1/3 * ( 1/2 + 1/2 )
E(gain) = 1/3 * V

Hence expected gain in staying 3 with door 1 is 1/3*V

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