Monty Hall-Souvenir Problem

Logic Level 2

Although you chose to switch doors on the Monty Hall game, you lost. However, as the first contestant to lose this way, Monty offers you a chance to play for a golden goat!

He explains that there are 6 doors in this game, 2 containing a golden goat, 2 containing a regular goat, and 2 containing, well, nothing.

You will choose 1 of the doors. Then, Monty opens 2 doors(not the one you chose), 1 has a regular goat, 1 has nothing.

So you’re left with 4 doors: your door and 3 others, 2 contain the golden goat, 1 contains a regular goat, and the other 1 contains nothing.

You may choose to keep your door or switch it to another one. However, there’s a catch. If you switch and get a regular goat, then you get nothing.

You know a golden goat costs twice as much as a regular goat. Should you switch?

For the answer, suppose that the expected value of $\frac{switching}{not switching}$ is $\frac{a}{b}$ , where a and b have no common factor except 1. What is a+b?

Bonus: If a regular goat costs $180, at what price of a golden goat will you rather switch? Give the highest possible integer price.

Assumptions:

If there are two possible doors, Monty picks randomly.
The doors are randomly arranged.

The answer is 19.

1 solution

Jerome Te
Jun 13, 2020

First, calculate the probabilities of getting each of the prizes when switching/not, then calculate the expected value.

Let doors 1 and 2 contain a golden goat, 3 and 4 a regular goat, 5 and 6 nothing, and that 3 and 5 were opened.(Note: the doors are not necessarily in ascending order. They are randomly assigned.)

If you chose door 1, then the probability that 3 and 5 were opened is $\frac{1}{2*2}$ = $\frac{1}{4}$ .

If you chose door 4, then the probability is $\frac{1}{2}$ .

Because of this, the door you chose is more likely 4 or 6, but the probability of each of the outcomes {golden goat, regular goat, nothing} are $\frac{1}{3}$ because there are 2 doors with a golden goat.

If you switch, there is a 2- $\frac{1}{3}$ = $\frac{5}{3}$ divided by 3= $\frac{5}{9}$ chance of containing a golden goat. That means there’s a $\frac{4}{9}$ chance of nothing.

Assuming regular goats cost $180, If you don’t switch, your expected value is $\frac{1}{3}$ $360+ $\frac{1}{3}$ $180=$180. If you switch, your expected value is $\frac{5}{9}$ *$360=$200

$\frac{\$200}{\$180}$ = $\frac{10}{9}$

10+9=19, so that’s the answer.

Bonus: Solve the equation $\frac{1}{3}$ $x+ $\frac{1}{3}$ $180= $\frac{5}{9}$ *$x and we get the price where it doesn’t really matter.

$\frac{1}{3}$ $x+ $\frac{1}{3}$ $180= $\frac{5}{9}$ *$x

$\frac{1}{3}$ $180= $\frac{2}{9}$ $x

$60= $\frac{2}{9}$ *$x

$30= $\frac{1}{9}$ *$x

$270=$x

Therefore, subtracting $1, we get $269.

If I have forgotten any assumption, please tell me.

Jerome Te - 12 months ago

Monty Hall-Souvenir Problem

The answer is 19.

1 solution

0 pending reports