1,10-11. The Monty Hall 3-door puzzle

Monty Hall 3-door Puzzle

You have a chance to win a large prize in a game show hosted by Monty. The prize is behind one of 3 closed doors, and the other two doors are losers, with a goat behind each.

• You are asked to select (but not open) one of the doors.
• Then Monty, opens one of the doors he knows is a losing door, selecting randomly if both remaining doors are losers.
• Then asks you if you want to switch doors.

Which strategy is the best?

• Switch doors.
• Stick to original selection.

title: Not in exam!

You selected a door, let’s say ${\text{door}}_1$. Then Monty opened another door:

As $\frac{1}{3}+\frac{1}{3}=\frac{2}{3}>\frac{1}{3}=\frac{1}{6}+\frac{1}{6}$, hence you have a better chance switching from ${\text{door}}_1$ to the door Monty did not open.

Using Bayes’ Theorem

• You select ${\text{door}}_1$.
• Monty opens another, ${\text{door}}_2$.
• We label the remaining door as ${\text{door}}_3$.

Consider the following:

• $W_1:{\text{door}}_1\text{ is the winning door}$
• $W_2:{\text{door}}_2\text{ is the winning door}$
• $W_3:{\text{door}}_3\text{ is the winning door}$
• $M_2:\text{Monty opened }{\text{door}}_2$

We want to know, $\boxed{p({\bar{W}}_1|M_2)}$:

• If $p({\bar{W}}_1|M_2)>\frac{1}{2}$, we should switch doors.
• If $p({\bar{W}}_1|M_2)<\frac{1}{2}$, we should stick to our door.
• If $p({\bar{W}}_1|M_2)=\frac{1}{2}$, it doesn’t matter.

We need to know:

• $p(W_1)=p(W_2)=p(W_3)=\frac{1}{3}$
• $p({\bar{W}}_1)=\frac{2}{3}$
• $p(M_2|W_1)=\frac{1}{2}$ If your selected door is the winning one, then Monty randomly opens one of the other two doors. We also know that $p(M_2|W_3)=1$ because if ${\text{door}}_3$ is the winning door and you selected the first one, then Monty must open the second. (as he never opens the winning door) $$\begin{array}{rl}p(M_2|W_1)& =\frac{p(M_2\cap {\bar{W}}_1)}{p({\bar{W}}_1)}\\ & =\frac{p(M_2\cap (W_2\cup W_3))}{\frac{2}{3}}\\ & =\frac{p((M_2\cap W_2)\cup (M_2\cap W_3))}{\frac{2}{3}}\\ & =\frac{p(M_2\cap W_3)}{\frac{2}{3}}\\ & =\frac{p(M_2|W_3)\cdot p(W_3)}{\frac{2}{3}}\\ & =\frac{1\cdot \frac{1}{3}}{\frac{2}{3}}\\ & =\frac{1}{2}\end{array}$$
• Hence, $p(M_2|{\bar{W}}_1)=\frac{1}{2}$

We can now apply Bayes’ theorem:

$$\begin{array}{rl}p({\bar{W}}_1|M_2)& =\frac{p(M_2|{\bar{W}}_1)\cdot p({\bar{W}}_1)}{p(M_2|{\bar{W}}_1)\cdot p({\bar{W}}_1)+p(M_2|W_1)\cdot p(W_1)}\\ & =\frac{\frac{1}{2}\cdot \frac{2}{3}}{\frac{1}{2}\cdot \frac{2}{3}+\frac{1}{2}\cdot \frac{1}{3}}\\ & =\frac{\frac{2}{6}}{\frac{3}{6}}\\ & =\frac{2}{3}\end{array}$$

As $p({\bar{W}}_1|M_2)=\frac{2}{3}>\frac{1}{2}$, you have a better chance switching.