How 21 got the Monty Hall problem wrong

Last night I watched the movie 21, based on the book Bringing Down the House, about the MIT blackjack team. There is a scene in the movie in which math professor Mickey Rosa poses the Monty Hall problem to a student, Ben Campbell. Predictably enough Ben is able to figure out the correct strategy and realizes that switching doors yields a 2/3s probability of winning the car (which is roughly the same chance you’ll predict the entire plot of the movie within five minutes). The problem is, he’s wrong.

At first the error is subtle. There is not enough information about the behavior of the host to determine whether the best strategy is to switch or not. The problem is not constrained enough. This is made glaringly obvious when Mickey Rosa asks “how do you know he’s not trying to play a trick on you?” If there is the possibility that the host is trying to play a trick on you, then there is the possibility that the host only allows you to switch if you have already chosen the car. In that case, switching doors would result in a 0% chance of winning the car.

Only when you constrain the problem does it make sense to switch doors. If (and only if) the host always opens a door, the door always reveals a goat, and the host always offers the player the option of switching does switching result in winning the car 2/3s of the time. Wikipedia has a good analysis of this constrained version using Bayesian products. (The short answer, the host is essentially giving you the option of switching from your initial choice of one door, to the two remaining doors and if the car is behind one of those two, you win).

Of course if the host only gives you the option to switch when you’ve chosen the car it won’t take long for people to catch on, and everyone will win all the time. It makes more sense for the host to offer a switch when you’ve selected the car with probability p, and offer a switch when you’ve selected a goat with probability q. Thus the host opens the door with probability: $p \times \frac13 + q \times \frac23$ .

Borrowing from the analysis on the wikipedia page, we see that:

$P(H_{ij}|C_k) = \begin{cases}0 \text{ if } i=j\\0 \text{ if } j=k\\\displaystyle\frac{p}{2} \text{ if } i=k\\q \text{ if } i \neq k \text{ and } j \neq k\end{cases}$

We can now use this new piecewise function to determine the probability of winning the car by switching:

$P(C_3|H_{12}) = \displaystyle\frac{P(H_{12}|C_3)P(C_3)}{P(H_{12})}\\P(H_{12}|C_3)P(C_3) = q \times \frac13\\P(H_{12}) = P(H_{12} \cap C_1) + P(H_{12} \cap C_2) + P(H_{12} \cap C_3)\\ = P(H_{12}|C_1)P(C_1) + P(H_{12}|C_1)P(C_2) + P(H_{12}|C_1)P(C_3)\\ = 0 \times \frac13 + \frac{p}{2} \times \frac13 + q \times \frac13 = \displaystyle\frac{p + 2q}{6}$

So:

$P(C_3|H_{12}) = \displaystyle\frac{2q}{p+2q}$

If we let p = 1 and q = 1 (in other words, the host always opens a door) then the probability of winning the car by switching is 2/3s, which is what we should expect.

Update:
It makes the most sense for the host to always offer the switch when you’ve chosen the car (p=1), and to offer the switch when you’ve selected the goat with probability q. By solving the inequality:

$\displaystyle\frac{2q}{1+2q} > \frac12$

We find that we should switch when q > 1/2.

This entry was posted on Thursday, May 7th, 2009 at 3:00 pm and is filed under math. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

4 Responses to “How 21 got the Monty Hall problem wrong”

Matt
May 8th, 2009 00:29
1

“Of course if the host only gives you the option to switch when you’ve chosen the car it won’t take long for people to catch on, and everyone will win all the time.”

Unless I have misunderstood you, that is not correct. If the host always gives you the option to switch when you originally chose the car and never lets you switch when you pick a goat then you’d only win 1/3 of the time, assuming you knew what he was up to. You’d win every time you originally chose a car because you would reject his offer of switching. However, you’d lose the other 2/3 of the time when you choose a goat and are not allowed to switch.
Niki
May 8th, 2009 09:06
2

Yes, of course you are right. What I meant to say was that you’d win every time the host offered a switch.
Randy
June 16th, 2009 13:04
3

Awesome. I like your mathematical insight on this problem, but it would be far for me to explain it to most people. Anyway, you reminded me of Ben Campbell. So, when I was watching the movie, I was really thinking it was you.

I hope you get that scholarship.
Andy Bates
March 7th, 2010 05:08
4

Although the probably can’t be determined to be exactly 2/3, the best option is always to switch, and here’s why:

Maybe the host always offers you the option to switch. Maybe he opens a door at random. Maybe he only offers you the option when you’ve chosen a car, and maybe only when you’ve chosen a goat. In the absence of that specific information, you have to make a best guess. If he always opens a door no matter what, then your odds of success when switching are 2/3. If he always opens a door at random, your odds are 1/2 when switching. If he only offers you a switch when you’ve chosen the car, your odds are 0; if he only offers you a switch when you’ve chosen the goat, your odds are 1/1.

Looking at all of those options, you are better off switching every time, unless you have specific information that he only offers you the chance to switch when you’ve chosen the car (which you don’t have). So, switch. Every time.

a+bi

What does aploosby mean?

How 21 got the Monty Hall problem wrong

4 Responses to “How 21 got the Monty Hall problem wrong”

Leave a Reply