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[Eeezee]

no text, thanks guys

 

[chuckywang]

Assuming Monty Hall knows what's behind the doors (which is a logical assumption), then you have a 33% chance of winning if you stay with your choice. You have a 67% chance of winning if you switch.

 

[Eeezee]

Originally posted by: chuckywang
Assuming Monty Hall knows what's behind the doors (which is a logical assumption), then you have a 33% chance of winning if you stay with your choice. You have a 67% chance of winning if you switch.

Okay, good, those probabilities add up to one (the other times that I've heard the problem, they claim you have a 50% chance if you switch and a 33% chance if you stay, which doesn't make any sense - the source I heard it from was probably just quoting the story wrong).

Now can you help me understand how you came to that solution?

 

[Farang]

Wikipedia has a very good explaination: http://en.wikipedia.org/wiki/Monty_Hall_problem

 

[chuckywang]

Originally posted by: Eeezee

Originally posted by: chuckywang
Assuming Monty Hall knows what's behind the doors (which is a logical assumption), then you have a 33% chance of winning if you stay with your choice. You have a 67% chance of winning if you switch.

Okay, good, those probabilities add up to one (the other times that I've heard the problem, they claim you have a 50% chance if you switch and a 33% chance if you stay, which doesn't make any sense).

Now can you help me understand how you came to that solution?

Without loss of generality, assume the first door contains the prize. Let's see the different scenarios:

1) You choose door 1. Monty opens one of door 2 or 3 (doesn't matter which, since they're all goats). If you switch, you lose.
2) You choose door 2. Monty opens door 3 (the goat). If you switch you win.
3) You choose door 3. Monty opens door 2 (the goat). If you switch you win.

In two of the three scenarios, you win by switching. In one of the three scenarios, you lose by switching. Hence switching wins 67% of the time.

We can mathematically formalize the solution, but this is just the "wordy" proof that people tell.

 

[Eeezee]

That makes sense to me, for some reason I was unable to write a good Monte Carlo that would prove it by using random chance without resorting to the use of multiple scenarios (ie what if the car is behind 1, what if it's behind 2, etc.). Thanks guys

What I wanted was (1/2)*(1/3) which is 1/6, so you have a 2/6 chance of winning if you don't switch, 4/6 chance of winning if you do

 

[Syringer]

One way that helped me understand the problem was to extrapolate it..say there were 50 doors. You chose one, so there's 49 left. Monty then opens up 48 other ones, all revealing goats--and gives you a chance to change. Looking at it this way, there's no doubt you'd change.