There are a lot of people arguing that Monte not knowing still makes it worth switching. It doesn't. Here's why. (From here on I'm spelling it Monty.) If Monty knows, the possibilities are:
1. You picked car. Monty opens a goat. Switching bad.
2. You picked goat A. Monty opens a goat. Switching good.
3. You picked goat B. Monty opens a goat. Switching good.
Two good switch, one bad switch. 2/3 chance it's worth switching.
If Monty doesn't know, the possibilities are:
1. You picked car. Monty opens Goat A. Switching bad.
2. You picked car. Monty opens Goat B. Switching bad.
3. You picked goat A. Monty opens Goat B. Switching good.
4. You picked goat A. Monty opens car. No switch option.
5. You picked goat B. Monty opes Goat A. Switching good.
6. You picked goat B. Monty opens car. No switch option.
So in this case there is 2/6 chance switching is bad, 2/6 chance switching is good, and 2/6 chance you don't have the option to switch. If you *do* have the option, that leaves you 2/4 and 2/4 good or bad. (Fractions not minimised to avoid confusing people.)
Posted by: RavenBlack | April 15, 2008 at 05:56 PM
Tuesday, April 22, 2008
The Monty Hall problem is always good for an argument. For me, though, the basic version is pretty straightforward. Well, after I found the right way to visualize it anyway. (Basically if he knows where the prize is and will always open one of his two doors without the prize, you're effectively being given the opportunity to trade your one door for his two.) However, the question of whether the assumption that Monty knows where the prize is affects the benefit of switching. It does and it's not intuitive--at least to me. The below comment does a good job of succinctly explaining.
Posted by Gordon Haff at 9:27 AM