Wrong with the solution to the Monty Hall problem

combinatoricsmonty-hallprobabilitypuzzle

I'm trying to develop an intuitive sense of why the suggestion of the Monty Hall problem is that you should switch doors when an informed host opens one of the two dummy doors.

So, I'm trying to think of this without using any formulas or rigorous proofs by breaking the problem down into three steps:

  1. the first choice of door
  2. the door opened
  3. the final choice of door

And just tabulating how many times a switch results in a win, and how many times a nonswitch results in a win. This is the result:

Car Goat Goat
S
O
S
Car Goat Goat
S
O
S
Car Goat Goat
S
O
S
Car Goat Goat
S
O
S
Car Goat Goat
S
O
S
Car Goat Goat
S
O
S
Car Goat Goat
S
O
S
Car Goat Goat
S
O
S

This is all eight outcomes that satisfy the condition:

  1. the participant can choose any of the three doors originally at random
  2. the host can only subsequently open a door that the participant did not choose
  3. the host can only subsequently open a door that hides a goat
  4. the participant is free to choose either of the remaining doors

From counting the results we see that there are four scenarios in which switching the door will result in a win, and four scenarios in which not switching the door will result in a win.

I am not debating what the correct answer to this problem is. I know that the correct answer has been both proven and demonstrated by simulation, and that the solution states that you double your chances of winning by choosing to switch doors. I simply seek to understand what is wrong about my assumptions that's leading me astray. Thanks!

Best Answer

The first four scenarios each have probability $\frac13\times \frac12 \times \frac12 =\frac1{12}$ if the decisions are random within the constraints

The last four scenarios each have probability $\frac13\times 1 \times \frac12 =\frac1{6}$ if the decisions are random within the constraints

So they are not equally likely, though combined they give a probability of winning the car of $\frac12$ for somebody who switches or not at random.

For a determined switcher, the first and third scenarios each have probability $\frac13\times \frac12 \times 1 =\frac1{6}$, making the probability of not winning the car $\frac13$. For the same determined switcher, the fifth and seventh scenarios each have probability $\frac13\times 1 \times 1 =\frac1{3}$, making the probability of winning the car $\frac23$.

Related Question