re: Marilyn vos Savant and the history of the Montel Hall question

quote:

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If Door 1 contains the car:

1) Guest picks Door 1, host opens Door 2. If guest switches, lose. If guest stays, win.

2) Guest picks Door 1, host opens Door 3. Switch: lose. Stay: win.

3) Guest picks Door 2, host opens Door 3. Switch: win. Stay: lose.

4) Guest picks Door 3, host opens Door 2. Switch: win. Stay: lose.

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Your error is assuming that all 4 of these scenarios have an equal chance of occurring. I think the problem is better to frame from the original choice. Since the scenarios would play themselves out the same way for all three doors chosen, we will just assume door 1 is chosen.


Choose Door 1 and Switch: 2 wins; 1 loss

1.) 1/3 chance car is behind door one (either door can be open; it doesn't matter). Lose

2.) Car is behind door two (door three is open). Win

3.) Car is behind door 3 (door two is open). Win

Choose door 1 and Stay: 1 win; 2 loses

1.) 1/3 chance car is behind door one (either door can be open; it doesn't matter). Win

2.) Car is behind door two (door three is open). Lose

3.) Car is behind door 3 (door two is open). Lose


Basically, choosing the correct door on one's first choice is still 1/3 and it being behind one of the other two doors is 2/3. When you switch, you are basically choosing the other 2 doors (even though we know one doesn't have the car, as shown).

I think a common error is that there are technically 4 possibilities in the above scenarios, 2 of those possibilities occur when you choose the correct door immediately; however, those 2 possibilities can only add up to 1/3. For example, if you chose the correct door (1/3 chance), and each door then has a is a 1/2 chance of being opened, then choosing the correct door and door 2 being open is actually 1/6 (1/3*1/2); same for door 3. In other words, the odds either door 2 or door 3 is chosen must add up to 1/3.