So today I watched the Mythbusters episode that discussed the “Monty Hall paradox” - the infamous probability problem. The Monty Hall paradox is well-known enough that I’ve come across before during maths lessons in school, in probability and statistics textbooks, in lectures, on TV, and so on. This particular Mythbusters episode was testing the hypothesis that contestants who “switched” their choice were twice as likely to win.
Mythbusters is well known for “bad science” - that is (amongst other things), biased test conditions, contrived experiments, low sample sizes and so on. During the Monty Hall experiment in this Mythbusters episode, the low sample size used for the experiment annoyed me - and seeing as I didn’t have anything else to do this evening, I had a go at testing it myself.
So what is the Monty Hall paradox?
Essentially, the paradox arises in something like the following scenario; a contestant in a game show is presented with three doors. Behind two of the three doors are goats, and behind the remaining door is a car. Three doors suggest a 1 in 3 chance of selecting the car - and unless you’re a bit strange, the goal is obviously to end up with the car as a prize. The contestant selects a door.
From the 2 remaining doors (the ones that the contestant didn’t select), the game show host opens one losing door and shows the contestant. There are now only two doors in the game, and the contestant is asked if they would like to switch their choice of doors - that is, if they would change from their initial choice to the remaining closed door. Intuitively, this may seem like an entirely pointless choice, but in reality it isn’t. The truth is that switching the door selection actually doubles the contestant’s likelihood of walking (or driving, as the case may be) away with the car.
Wait, how does that work?!
The answer is counterintuititive and has to do with conditional probability. The best way to look at this is to look at the game as simply as possible. A contestant is presented with a 1 in 3 chance at a “winning” outcome (y’know, unless they really want a goat). There is then a 2 in 3 chance that they will have selected a goat from that initial choice.
The event space is reduced by one - that is, a possible losing outcome is removed. Because the probability of choosing a goat initially was 2 in 3, the contestant can now make the fairly safe assumption that behind the remaining door is the desired outcome - the car!
More formally, the contestant has a 2 in 3 chance of choosing a losing outcome initially (which means that they probably picked a goat the first time round). A losing outcome (a goat) is then removed from the event space. The contestant can now choose to change their choice from their 2 in 3 chance of losing to a 1 in 3 chance of losing (which means that they will there is a 2 in 3 chance of a winning outcome if they switch).
This idea seems confusing because it’s deceptively simple. Don’t believe me? I’ll prove it!
I’m not a mathematician by trade, so I’m probably not going to be able to make a beautifully elegant proof without some serious legwork. Instead, I hacked up a quick Python script to test the hypothesis empirically (incidentally, it’s on GitHub as a Gist). As it’s a very simple script, I ran the experiment for 1,000,000 iterations - 500,000 on games where the original selection is retained, and the other 500,000 where the original selection is changed.
From the snippet below, it becomes clear that “sticking” with the original choice only won 166,030 games of 500,000 - a win ratio of about 33.2%… not great. The “switch” option appears to be a lot more profitable, winning a total of 333,671 games out of 500,000 - with a win ratio of 66.7% overall. Weird.
# charles at nevis in ~ [23:44:04] $ python monty.py Iterations: 500000 Stick: 166030 Switch: 333671 Ratio (switch / stick): 2.010
The results are still surprising to me because they don’t really correspond with what my instinct tells me - but the results do show that when repeating the experiment over millions of iterations, a contestant is indeed about twice as likely to end the game with a winning outcome.
How about that?
As always, there’s plenty of information online about this topic. The Wikipedia page for the Monty Hall problem is actually pretty good, and gives a slightly more mathematical description of the problem than I have here.