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 Best Simple Explanation to the Monty Hall Problem
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.
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3 comments:
you said the 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.
but whay didn't you write it like this:
if Monty knows, 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 a goat. Switching good.
4. You picked goat B. Monty opens a goat. Switching good.
in this case ,the odds are still 50-50
the chance you pick the car on the first try is 1/3. the chance youre gonna pick a goat on your first try is 2/3. if you pick a goat the first time and switch, because the host knows where the prize is and HAS to open a goat door, you win. car on the first time 1/3, and goat and switch you win 2/3 of the time.
The chances of you guessing it right the first time are 1 in 3, no matter what happens afterwards. If there were 10 doors (one car, nine goats), you would have a 1 in 10 chance of getting it right even after Monty revealed 8 goats after your guess. Revealing goats has no bearing on your original odds. Think of it like this, you can't magically increase the chance that you guessed right from 1 in 3 ( or 1 in 10) to 1 in 2, no one is that good.
To confuse you more, if someone walked in the room and was told and to guess which door held the car (after one goat had been revealed) but NOT knowing which door was originally opened, his odds of guessing WOULD be 50-50, but that is because he has LESS information than you do.
Think of it like this, if you ask me if my wife is home now, I have a better chance of being right probably 90% or so, than you do, maybe 60% or so, even though neither one of us actually knows the answer to the binary choice (please no silly jokes about "oh I know where she is now ha ha"). I know when she usually works, when she shops and so on, you have to hypothesize based on usual working hours and so on for some hypothetical person. I know more than you in this case so my odds are statistically better.
That is how the Monty Problem works. You have more information to work with that someone who just randomly walked in, not knowing which was the original door chosen. That is why you have a 2/3 chance of being right based on switching doors, and he has only a 1/2 chance based on no information at all. There is NO contradiction or paradox here.
Hope this helps.
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