Monty Hall problem

Michael Godin Arlotto (154aamga@pic.ucla.edu)
Tue, 14 Nov 95 15:36:28 -0800

Essay #1
Assuming the host is nuetral and that he is going to let us decide for
ourselves, wouldn't it be to our advantage to switch doors? I'm not sure if my
logic is right so if it isn't please let me know. Here it goes:
we have to strategies to choose from. We can either (1) switch all the
time, or (2) we can always stick with our first choice. So, the prob. for
winning is
P(winning)=P(car|goat on first choice)*P(goat on first choice)+
P(car|car on first choice)*P(car on first choice).
Now assuming (1),
P(car|goat on the first choice)=1
P(goat on the first choice)=2/3
P(car|car on the first choice)=0
P(car on the first choice)=1/3
so,
P(winning)=1*2/3+0*1/3=2/3.
Therefore, assuming (1) we have a 2/3 chance of winning if we switch.
Now assuming (2), our probabilities will be
P(car|goat on the first choice)=0
P(car|car on the first choice)=1
P(goat)=2/3, p(car)=1/3.
So
P(winning)=0*2/3+1*1/3= 1/3.
Therefore we have a 1/3 chance of winning with this strategy. So, we should
always switch. Right??
I'll be waiting for your response.

Thanks,
Mike

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