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luvgal
04-11-2005, 07:34 AM
i wonder would the probability of either switching or staying would be 1/2. if the host randomly picks the door between the two unchosen doors without knowing where the prize is. and if the host opens the unchosen door with the prize.. then the contestant lose.. but if the host opens the door with the goat then what is the probability of having the prize?
could anyone pls answer this question with proof..
thanx

elitegimp
04-11-2005, 11:41 AM
This is a simple counting proof -- here are all the possible outcomes of the game, all equally likely (important for figuring out the probability)

Call the door you choose A:
If the goat is in A, then it doesn't matter which door is chosen. You want to stay (if door B is opened, you win my staying... if door C is opened, you win by staying)
If the goat is in B, there is a 1/2 chance you lose automatically (door B opened), or a 1/2 chance you win if you switch (door C is opened)
If the goat is in C, there is a 1/2 chance you lose automatically (door C opened), or a 1/2 chance you win if you switch (door B is opened)

So you win by staying 2/6 times, you win by switching 2/6 times, and you lose because the door with the prize is opened 2/6 times.

Not surprising, because no information is gained by having a door opened so you still have a 1/3 chance of picking the door.

Note: if the door is opened and you don't lose automatically, there is now a 2/4 chance you win by staying and a 2/4 chance you win by switching... i.e. it's 50-50.

Siegmund
04-11-2005, 04:58 PM
[ QUOTE ]

if the host randomly picks the door between the two unchosen doors without knowing where the prize is. and if the host opens the unchosen door with the prize.. then the contestant lose.

[/ QUOTE ]

*Under these conditions*, 1/3 of the time you will lose automatically when the host exposes the prize, and the rest of the time you have the same chance whether you switch or not.

Note that those *aren't* the conditions that applied to the classical Monty Hall problem - in which the host DID know where the prize was, and ALWAYS exposed a goat. Under those conditions, your chance was indeed 1/3 if you stayed and 2/3 if you switched. If you play bridge, see also "Principle of Restricted Choice" for the same phenomenon at work.

TomBrooks
04-12-2005, 07:07 PM
[ QUOTE ]
...the classical Monty Hall problem - in which the host DID know where the prize was, and ALWAYS exposed a goat. Under those conditions, your chance was indeed 1/3 if you stayed and 2/3 if you switched.

[/ QUOTE ]
I tried this problem using this simulation (http://people.hofstra.edu/staff/steven_r_costenoble/MontyHall/MontyHallSim.html) and I won the big prize 3/4 (75%) of the time. I made 20 tries and won 15 times.

i wanna be me
04-13-2005, 12:57 AM
[ QUOTE ]
Note that those *aren't* the conditions that applied to the classical Monty Hall problem - in which the host DID know where the prize was, and ALWAYS exposed a goat. Under those conditions, your chance was indeed 1/3 if you stayed and 2/3 if you switched.

[/ QUOTE ]

1/3 for the initial choice because (before the goat door is open) it is 1/3 to have the prize - and the probability doesn't change given the goat door opening?

2/3 for switching because 1-1/3 = 2/3?

I just want to clarify this (cause my upper-year prof in econ said it would be 1/2 to switch and I told him that didn't make any sense).

elitegimp
04-13-2005, 02:13 AM
[ QUOTE ]

I just want to clarify this (cause my upper-year prof in econ said it would be 1/2 to switch and I told him that didn't make any sense).

[/ QUOTE ]

it does make sense -- econ professors can't do simple math (says a bitter math/econ dual major that suffered through a couple of econ classes junior year cuz I was "so close to fulfilling the requirements").

seriously, I had a professor who couldn't figure out why a finite difference wasn't _exactly_ equal to the derivative.

(but yes, your thinking is right -- 1/3 to pick right the first time, so 1/3 of the time you switch away from the winner and 2/3 you switch to the winner)

kevin017
04-13-2005, 02:32 AM
oh good lord, i have a friend who thinks he knows everything, trying to explain this concept to him took me forever. I even made cups with a ball under it, and did an experiment that ran all the possibilities (surprisingly easy), and he still didn't believe me, even when he came out a winner 2/3 of the time.

actually, here's what i did. say you pick door a, and its behind door a. then switching away makes you lose.
say you pick door a, and its behind door b, switching away makes you win.
say you pick door a, and its behind door c, switching away makes you win.

it takes all of one trial to see that its 1/3 loss, 2/3 win if you switch.