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Subject:Re: Are you a "Mental Gymnast"? From:SteveFJong -at- aol -dot- com To:"TECHWR-L" <techwr-l -at- lists -dot- raycomm -dot- com> Date:Wed, 22 Jan 2003 13:22:15 -0500
Eric,
First off, the "Monty Hall" problem is an example of a mathematical illusion, akin to an optical illusion. Apparently, even PhD's in mathematics don't see this correctly, and get all bent out of shape arguing. Writers tend to be more prone to math blindness than the average, so what chance do we stand? It's a more fair question to ask a programmer 8^) And that's about as close to technical writing as we'll get here.
That said, the key to understanding the problem is to know two things. First, there are two sets: The set of doors you pick (one member) and the set of doors you didn't pick (two members). Your odds of guessing right are one in three, and therefore your odds of guessing wrong are two in three. Another way to express it is that if you get to switch to the other set, your odds of winning become two in three.
The other thing to remember is that Monty knows where the prize is, and can always pick a door that (a) you didn't pick and (b) has Rice-a-Roni behind it. His opening the door does not change the odds that you guessed right in the first place; in fact, his opening a door doesn't change *your* odds at all. If you switch, you'll be right two times out of three.
This is entirely counterintuitive for most people, but another example can make things clearer. Let's increase the number of doors to 100; Monty still knows where the prize is. You pick a door at random. He can actually now open 98 doors, all losers, leaving you with the same stay-or-switch choice. In this case, if you stick with your choice, you will win 1 time in 100; but if you switch, you will win 99 times out of 100!
If you can write a game simulation in any language, you can run it a couple of thousand times, and you'll see immediately that in the three-door case, the winning percentage converges on 33% for staying and 67% for switching.
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