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Q15E

Expert-verifiedFound in: Page 477

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

** **

**Question: In this exercise we will use Bayes' theorem to solve the Monty Hall puzzle (Example 10 in Section 7.1). Recall that in this puzzle you are asked to select one of three doors to open. There is a large prize behind one of the three doors and the other two doors are losers. After you select a door, Monty Hall opens one of the two doors you did not select that he knows is a losing door, selecting at random if both are losing doors. Monty asks you whether you would like to switch doors. Suppose that the three doors in the puzzle are labeled 1, 2, and 3. Let W be the random variable whose value is the number of the winning door: assume that \(p\left( {W = k} \right) = \frac{1}{3}\) for k=1,2,3. Let M denote the random variable whose value is the number of the door that Monty opens. Suppose you choose door i**

**a) What is the probability that you will win the prize if the game ends without Monty asking you whether you want to change doors?**

**b) Find \(p\left( {M = j\left| {W = k} \right.} \right)\) for j=1,2,3 and k= 1.2.3**

**e) Use Bayes' theorem to find \(p\left( {M = j\left| {M = k} \right.} \right)\) where i and j and are distinct values d) Explain why the answer to part (c) tells you whether you should change doors when Monty gives you the chance to do so, **

**Answer:**

a) The probability that you will win the prize if the game ends without Monty asking you whether you want to change door or not is 1/3

b) \(p\left( {M = j\left| {W = k} \right.} \right) = 1\) if i, j, k are distinct

\(p\left( {M = j\left| {W = k} \right.} \right) = 0\) if j=k or j=i

\(p\left( {M = j\left| {W = k} \right.} \right) = \frac{1}{2}\) if i=k and \(j \ne i\)

c) \(p\left( {W = 3\left| {M = 2} \right.} \right) = \frac{2}{3}\)

d) Hence you should change doors

Monty Hall puzzle

\(p\left( {\frac{{{F_j}}}{E}} \right) = \frac{{p\left( {\frac{{{E_1}}}{{{F_j}}}} \right)p\left( {{F_j}} \right)}}{{\sum\limits_j^n {p\left( {\frac{E}{{{F_j}}}} \right)p\left( {{F_j}} \right)} }}\)

\(p\left( {\frac{E}{{{E_1}}}} \right) = \frac{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right)}}{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right) + p\left( {\frac{{{E_2}}}{F}} \right)p\left( F \right)}}\)

Consider the following event

W = number of winning door

M = number of door Monty opens

because the winning door was chosen uniformly at random, your chance of winning, p(W=i), is 1/3, no matter which door you choose

The probability that you will win the prize if the game ends without Monty asking you whether you want to change door or not is 1/3

If you have chosen other than winning door, then Monty opens the other non-winning door.

If you have chosen the winning door, then Monty opens the other two doors with equal likelihood.

Thus

\(p\left( {M = j\left| {W = k} \right.} \right) = 1\) if i, j, k are distinct

\(p\left( {M = j\left| {W = k} \right.} \right) = 0\) if j=k or j=i

\(p\left( {M = j\left| {W = k} \right.} \right) = \frac{1}{2}\) if i=k and \(j \ne i\)

Without loss of generality assume that i=1, j=2 and k=3

By Bayes’ Theorem

\(p\left( {W = 3\left| {M = 2} \right.} \right) = \frac{{p\left( {M = 2\left| {W = 3} \right.} \right)p\left( {W = 3} \right)}}{{p\left( {M = 2\left| {W = 3} \right.} \right)p\left( {W = 3} \right) + p\left( {M = 2\left| {W = 2} \right.} \right)p\left( {W = 2} \right) + p\left( {M = 2\left| {W = 1} \right.} \right)p\left( {W = 1} \right)}}\)

\( = \frac{{1 \cdot \left( {\frac{1}{3}} \right)}}{{1 \cdot \left( {\frac{1}{3}} \right) + \left( {\frac{1}{2}} \right) \cdot \left( {\frac{1}{3}} \right) + 0 \cdot \left( {\frac{1}{3}} \right)}}\)

\( = \frac{1}{{\left( {\frac{3}{2}} \right)}}\)

\(p\left( {W = 3\left| {M = 2} \right.} \right) = \frac{2}{3}\)

You should change doors because you have 2/3 chance of winning by switching.

(A similar calculation to that part(c) shows that your chance of winning if you do not switch door is 1/3)

Hence you should change doors

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