Monty Hall problem

Michael Godin Arlotto (154aamga@pic.ucla.edu)
Thu, 16 Nov 95 09:10:40 -0800


This is the note I wrote a few days ago. For some reason you
are unable to read my message any longer.

Assuming the host is nuetral, you can follow one of two strategies.
The contestant can either (1) always switch or (2) always keep his
original choice. The probability equation would look like this:
P(car)=P(car|1st choice is a goat)*p(1st choice is a goat)+
P(car|1st choice is a car)*P(1st choice is a car).
Following the 1st stragegy, the probabilities would be as follows:
p(1st choice is a car) = 1/3
p(1st choice is a goat) = 2/3
p(car|1st is a goat)=1 since always switching
p(car|1st is a car)=0 for the same reason
therefore,
P(winning car) = 1*2/3 +0*1/3 = 2/3.
Following the 2nd strategy, the probabilities would be:
P(car|1st choice is a car) = 1 since always keeping your first choice
P(car|1st choice is a goat) = 0 for the same reason
therefore,
P(winning car)=0*2/3 + 1*1/3=1/3.
Since 2/3>1/3, you would want to stick with the first strategy
of always switching. Tell me what you think.

--Mike



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