It continues to embarrass me that ultimately I was only "convinced" that the calculated answer really was right, and not some kind of plausible-sounding sleight-of-hand, when I confirmed that it was commonly believed by the right people.

Comment author:MikeDobbs
25 March 2013 05:29:23PM
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One of my favorites for exactly that reason- if you don't mind, let me take a stab at convincing you absent "the right people agreeing."

The trick is that once Monty removes one door from the contest you are left with a binary decision. Now to understand why the probability differs from our "gut" feeling of 50/50 you must notice that switching amounts to winning IF your original choice was wrong, and loosing IF your original choice was correct (of course staying with your original choice results in winning if you were right and loosing if you were wrong).

So, consider the probability that you original guess was correct. Clearly this is 1/3. That means the probability of your original choice being incorrect is 2/3. And there's the rub. If you will initially guess the wrong door 2/3 of the time, then that means that when you are faced with the option to switch doors you're original choice will be wrong 2/3 of the time, and switching would result in you switching to the correct door. Only 1/3 of the time will your original choice be correct, making switching a loosing strategy.

It becomes more clear if you begin with 10 doors. In this modified Monty Hall problem, you pick a door, then Monty opens 8 doors, leaving only your original choice and on other (one of which contains the prize money). In this case your original choice will be incorrect 9/10 times, which means when faced with the option to switch, switching will result in a win 9/10 times, as opposed to staying with your original choice, which will result in a win only 1/9 times.

(nods) Yah, I'm familiar with the argument. And like a lot of plausible-sounding-but-false arguments, it sounds reasonable enough each step of the way until the absurd conclusion, which I then want to reject. :-)

Not that I actually doubt the conclusion, you understand.

Of course, I've no doubt that with sufficient repeated exposure this particular problem will start to seem intuitive. I'm not sure how valuable that is.

Mostly, I think that the right response to this sort of counterintuitivity is to get seriously clear in my head the relationship between justified confidence and observed frequency. Which I've never taken the time to do.

## Comments (33)

BestI remember how my roommates and I drew a game tree for the Monty Hall problem, assigned probabilities to outcomes, and lo, it was convincing.

(nods)

It continues to embarrass me that ultimately I was only "convinced" that the calculated answer really was right, and not some kind of plausible-sounding sleight-of-hand, when I confirmed that it was commonly believed by the right people.

*0 points [-]One of my favorites for exactly that reason- if you don't mind, let me take a stab at convincing you absent "the right people agreeing."

The trick is that once Monty removes one door from the contest you are left with a binary decision. Now to understand why the probability differs from our "gut" feeling of 50/50 you must notice that switching amounts to winning IF your original choice was wrong, and loosing IF your original choice was correct (of course staying with your original choice results in winning if you were right and loosing if you were wrong).

So, consider the probability that you original guess was correct. Clearly this is 1/3. That means the probability of your original choice being incorrect is 2/3. And there's the rub. If you will initially guess the wrong door 2/3 of the time, then that means that when you are faced with the option to switch doors you're original choice will be wrong 2/3 of the time, and switching would result in you switching to the correct door. Only 1/3 of the time will your original choice be correct, making switching a loosing strategy.

It becomes more clear if you begin with 10 doors. In this modified Monty Hall problem, you pick a door, then Monty opens 8 doors, leaving only your original choice and on other (one of which contains the prize money). In this case your original choice will be incorrect 9/10 times, which means when faced with the option to switch, switching will result in a win 9/10 times, as opposed to staying with your original choice, which will result in a win only 1/9 times.

*1 point [-](nods) Yah, I'm familiar with the argument. And like a lot of plausible-sounding-but-false arguments, it sounds reasonable enough each step of the way until the absurd conclusion, which I then want to reject. :-)

Not that I actually doubt the conclusion, you understand.

Of course, I've no doubt that with sufficient repeated exposure this particular problem will start to seem intuitive. I'm not sure how valuable that is.

Mostly, I think that the right response to this sort of counterintuitivity is to get seriously clear in my head the relationship between justified confidence and observed frequency. Which I've never taken the time to do.