Monty Hall problem- Is it worth changing your mind?

Probability Level 2

Monty Hall used to host a program called "Let's Make A Deal" and one of the "deals" went as follows. You are shown 3 doors and are told that behind one door is a new car yet behind each of the other doors is a gag prize - a goat. You make your pick then Monty Hall opens a door you haven't chosen and shows that behind it is a goat. He then asks you if you want to change your selection. Should you change your mind, stick with your original choice or does it really matter?

Stick with your original choice Always change your choice It makes no difference

2 solutions

Robert Gagnon
May 15, 2016

You will win 2 out of 3 times if you ALWAYS change your choice. Let's take 3 cases and in all three cases your choice is ALWAYS door 1. CASE 1 - car is behind door #1 and since you will always change your mind, you always end up with a goat. CASE 2 - car is behind door #2. Monty Hall has no choice and must show you the goat behind door #3. You change your choice from door #1 to door #2 and win the car. CASE 3 - car is behind door #3. Monty can only show you the goat behind door #2. You change your choice from door #1 to door #3 and win the car. As can be seen ALWAYS changing your mind, allows you to win the car 2 out of 3 times.

Hobart Pao
May 20, 2016

Another way to think about it is by grouping.

At first, the chance of winning is 1/3 and chance of losing is 2/3.

Let's say you have doors A, B, C and you choose A. Let's say B is the losing door that is shown to you. Then you can group B with C. Initially before the info was given, you know that there is 1/3 chance of A being your winner and 2/3 chance of it not (so the 2/3 is distributed over B and C). Given the new information about door B not being the winner, then the 2/3 chance of A NOT being the winner is distributed over C only. Thus, you should switch from A to C because A still only has 1/3 chance of being the winner.

For me, it was easier to envision with 5 doors instead (A, B, C, D, E).

Let's say you choose A. You have 1/5 chance of A being winner and 4/5 chance of it not being the winner. Then, you are given the new information that door B is not the winner. That 4/5 chance of A NOT being the winner becomes equally distributed over doors C, D, and E so instead of C, D, and E only each having a 1/5 chance of being winners, they now each have a 4/15 chance of being the winner, which is greater than the 1/5 chance of A, so then it would be better to switch from A to C, D, or E.

Monty Hall problem- Is it worth changing your mind?

2 solutions

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