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Letsmakeadeal
 

Letsmakeadeal

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    Letsmakeadeal Letsmakeadeal Presentation Transcript

    • Let’s Make a Deal February 21, 2012
    • We Begin With player faces a choice of 3 doors–behind one is the prize after initial guess, Monty Hall reveals a Zonk! behind one of two remaining doors contestant can switch curtains after the reveal question: does it pay to switch?
    • Further ... Let G be the door the prize is behind ∈ (1, 2, 3) Let C be the first choice of the contestant ∈ (1, 2, 3) Let M be the door that Monty reveals ∈ (1, 2, 3) Without loss of generality, let the ratio of winning by switching to not switching be represented by this situation: P(G = 3|M = 2, C = 1) , (1) P(G = 1|M = 2, C = 1)
    • Rewriting ... P(W |S) P(G = 3|M = 2, C = 1) = P(W |N) P(G = 1|M = 2, C = 1) P(M=2,C =1|G =3)∗P(G =3) P(M=2,C =1) = P(M=2,C =1|G =1)∗P(G =1) P(M=2,C =1) P(M = 2, C = 1|G = 3) ∗ P(G = 3) = P(M = 2, C = 1|G = 1) ∗ P(G = 1) P(M = 2|C = 1, G = 3) ∗ P(C = 1|G = 3) = (2) P(M = 2|C = 1, G = 1) ∗ P(C = 1|G = 1)
    • Simplifying P(W |S) P(M = 2|C = 1, G = 3) = (3) P(W |N) P(M = 2|C = 1, G = 1) numerator = 1: Monty must choose door 2 if contestant chooses 1 and prize is behind 3 denominator depends on Monty’s behavior; if he chooses randomly, then denominator= 1/2, ratio = 2–as in the simpler calculation, switching is about twice as likely to win as not we don’t know how Monty chooses, but if there is any randomness to Monty’s choice, denominator <1–better off switching