Recall the Monty Hall Problem: \"On the show “Let’s Make a Deal”, hosted by Mont
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Question
Recall the Monty Hall Problem:
"On the show “Let’s Make a Deal”, hosted by Monty Hall, there are three doors. There is a prize behind one of the doors and goats (?!) behind the other two. The contestant chooses a door. Then the host opens a different door behind which there is always a goat. The contestant is then given a choice to either switch doors or to stay put. The contestant wins the prize if and only if the contestant chooses the door with the prize behind it. Is it to the contestant’s benefit to switch doors?"
Analysis of this problem shows that if the contestant switches doors after Monty (the announcer) revealed a goat then they would win with probability 2/3.
Now we have a new contestant, Jeremy, who instead decides whether to switch or not by consulting a BIASED crystal ball, (where the probability that Jeremy switches is x). Assume that Monty acts like normal and that Jeremy chooses doors uniformly at random.
(a) Compute, in terms of x, the probability that Jeremy wins.
(b) What is the smallest value of x that still gives Jeremy a 50% or better chance of winning?
(c) Right before Jeremy went on stage, Thomas tampered with the crystal ball and sets p to be 1/5. Since Thomas is backstage, he has no idea what Jeremy does. All Thomas knows is that Jeremy loses in the end. What is the probability that Jeremy did not switch doors when Monty revealed a goat?
Explanation / Answer
a) probabilty Jeremy wins =he switches and wins+does not switch and wins =x*(2/3)+(1-x)*(1/3)=(1/3)(1+x)
b) (1/3)(1+x)>0.5
1+x>1.5
x>0.5
c)probability of losing =switch and loses+does not switch and loses =(1/5)*(1/3)+(4/5)*(2/3)=9/15
probability that Jeremy did not switch doors when Monty revealed a goat given loses
=P(does not switch and loses)/P(loses) =(4/5)*(2/3)/(9/15)=8/9
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