I’m sure many of you will be familiar with the Monty Hall Problem, especially those of you who’ve read “The Curious Incident of the Dog in the Night-time” but I was taken aback at the strength of feeling it provoked in a recent discussion at work. Essentially the problem is thus:
You are the contestant in a game show with the chance to win a car. The car is behind one of three doors. Behind each of the other two doors is a goat. You get to select a door. The host then opens one of the remaining doors to reveal a goat and eliminates that door from the game. You then have the option to stick or switch to the remaining door. It’s as simple as that. Are you better off sticking or switching.
In the discussion at work (which occupied a very long email thread) there were two rival camps:
- it makes no difference, it’s a 50:50 chance and anyone that tells you any different is a gullible fool (or words to that effect)
- you should always switch as the odds shift in your favour if you do (from 0.333 to 0.666 in fact)
This became a very emotional debate and I was surprised it took so long before someone said “why don’t we just try it out”. Get a friend, get three cups and a pea and play the game. Or you could model it in software exactly as I did. In fact, there’s a very neat model here that will run a simulation in your browser and has a great explanation of what’s going on.
In fact, there’s a very good way to see through this problem which a number of people in the discussion pointed out. Imagine that instead of 3 doors it’s a lottery game where there are one million tickets and only one prize. You pick a random ticket. The “host” then systematically destroys 999,998 tickets so you’re left with the option of sticking with your original ticket (one in a million chance of winning) or switching (.999999 chance of winning). Not a difficult choice!
The good news is that everyone kissed and made up so all is once again well on the Microsoft Campus…