5.27. (The Monly Hall Problem) Monty Hall gives Dan the choice of three curtains

Question

5.27. (The Monly Hall Problem) Monty Hall gives Dan the choice of three curtains. Behind one curtain is a car, while behind the other two curtains are goats. Dan chooscs a curtain, but before it is opened, Monty Hall opens one of the other curtains and reveals a goat. IHe then offers Dan the option of keeping his original curtain or switching to the remaining closed curtain. The Monty Hall problem is to figure out Dan's best strategy: "To stick or to switch?" (a) What is the probability that Dan wins the car if he always sticks to his first choice of curtain? What is the probability that Dan wins the car if he always switches curtains? Which is his best strategy? (If the answer seems counter intuitive, suppose instead that there are 1000 curtains and that Monty Hall opens 998 goat curtains. Now what are the winning probabilities for the two strategies?) (b) Suppose that we give Monty Hall another option, namely he's allowed to force Dan to stick with his first choice of curtain. Assuming that Monty Hall dislikes giving away cars, now what is Dan's best strategy, and what is his probability of winning a car? (c) More generally, suppose that there are N curtains and M cars, and suppose that Monty Hall opens K curtains that have goats behind them. Compute the probabilities Pr(Dan wins a car | Dan sticks), Pr(Dan wins a car | Dan switches) Which is the better strategy?

Explanation / Answer

(a)

There are 3 curtains with prized car behind 1 curtain. Therefore,

P (car is behind the first curtain) = 1/3

P (the car is behind the second or third curtain) = 1 - 1/3 = 2/3

The probability that Dan's win's the car if he sticks with his choice is 1/3

P (the car is behind 2nd curtain) + P (the car is behind 3rd curtain) = 2/3

Now, according to the question, the third curtain contains a goat, therefore

P (the car is behind the 2nd curtain) + 0 = 2 /3

The probability that Dan's win's the car if he changes with his choice is 2/3

Changing the choice is the best stratergy as the probability of winning is more.

(b)

Since Monty Hall does not like giving away cars, there will be two cases.

Case 1: If the car is behind the curtain chosen by Dan, Monty Hall will give him a choice to change the curtain.

Case 2: If the car is not behind the curtain chosen by Dan, Monty hall won't give him a choice to change the curtain.

P (Case 1) = 1/3

P (Case 2) = 2/3

Dan's best stratergy is to stick with the door if Monty Hall gives him the choice to change it.

P (Winning the car) = 1/3*2/3 = 2/9

(c)

There are N curtains with prized car behind M curtains. Therefore,

P (car is behind the chosen curtain) = M/N

P (Dan wins a car | Dan sticks) = M/N

Now, Monty hall opens K doors leaving Chosen door and remaing door with the cars, that is, [N - (M+1)].

The probability picking the car by choosing different door is, hte product of not being able to choose the car before, that is, (N-1)/N and choosing it now, that is, 1/N-K-M. Therefore,

The probability picking the car by choosing different door = [(N-1)/N]*[1/(N-K-M)]

P (Dan wins a car | Dan switches) = [(N-1)/N]*[1/(N-K-M)]

The probability of switching the curtain is greater than probability of sticking.

Therefore, Dan should switch the curtain.

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