By Bayes's Theorem, we have PrOoF APRon Proy PrOOF rICProl cy PrOf We assume that the prize is randomly placed behind the curtains We compute all of the conditional probabilities for event O, as we'll need them to compute PrO PrO A)= 1/2, if the prize is behind a, Monty Hall can open either b or c PrOB 0, if the prize is behind b, Monty Hall cannot open curtain b Prole) 1, if the prize is behind c, Monty Hall can only open curtain b To compute Prof, we note that O=On A)U(On B)U(OnC) and A,OnB, and OnC are mutually exclusive events as the prize can be behind one curtain only. So, Prof= PronA+PronBPrloncy PrlAPr(| A)+ Pr( B+PrICPrlolCy /3)·(1/2)+(1/3)·0+(1/3)·1 Therefore. we have Pr(AJO- PriA)PrIoJA 1/3)·(1/2) /2 Furthermore, we have. PrIolo)= Pr PrICProlcy PrOf (1/3)·1 In other words, if we switch we double our chances of winning. We could have also arrived at this answer by noting PrBlo)=0 and therefore PrCO=1-Pr(Alof=1-1/3=2/3 Generalized Monty Hall Problem With the same analysis, we can show that if there are n curtains and the emcee opens n-2 of them after the contestant's initial choice, the contestant's chances of winning will be 1/n if she decides to stay with the initial choice and(n-1)/n if she switches. In other words, the contestant's chance of winning goes up by a factor of n-1 in case of switchinBy Bayes’s Theorem, we have: P r{A | O} = P r{A}P r{O | A} P r{O} P r{C | O} = P r{C}P r{O | C} P r{O} We assume that the prize is randomly placed behind the curtains: P r{A} = P r{B} = P r{C} = 1/3 We compute all of the conditional probabilities for event O, as we’ll need them to compute P r{O}: P r{O | A} = 1/2, if the prize is behind a, Monty Hall can open either b or c P r{O | B} = 0, if the prize is behind b, Monty Hall cannot open curtain b P r{O | C} = 1, if the prize is behind c, Monty Hall can only open curtain b To compute P r{O}, we note that O = (O ∩ A) ∪ (O ∩ B) ∪ (O ∩ C) and O ∩ A, O ∩ B, and O ∩ C are mutually exclusive events as the prize can be behind one curtain only. So, P r{O} = P r{O ∩ A} + P r{O ∩ B}P r{O ∩ C} = P r{A}P r{O | A} + P r{B}P r{O | B} + P r{C}P r{O | C} = (1/3) · (1/2) + (1/3) · 0 + (1/3) · 1 = 1/2. Therefore, we have: P r{A | O} = P r{A}P r{O | A} P r{O} = (1/3) · (1/2) 1/2 = 1/3. Furthermore, we have: P r{C | O} = P r{C}P r{O | C} P r{O} = (1/3) · 1 1/2 = 2/3. In other words, if we switch we double our chances of winning. We could have also arrived at this answer by noting P r{B | O} = 0 and therefore P r{C | O} = 1 − P r{A | O} = 1 − 1/3 = 2/3. Generalized Monty Hall Problem With the same analysis, we can show that if there are n curtains and the emcee opens n−2 of them after the contestant’s initial choice, the contestant’s chances of winning will be 1/n if she decides to stay with the initial choice and (n−1)/n if she switches. In other words, the contestant’s chance of winning goes up by a factor of n − 1 in case of switching. 2