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#Post#: 88--------------------------------------------------
Door game...
By: dinesh Date: November 15, 2012, 10:18 pm
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Suppose you're on a game show, and you're given the choice of
four doors:
Behind one door is a car; behind the others, goats.
You pick a door, say No. 1, and the host, who knows what's
behind the doors, opens another door, say No. 3, which has a
goat. [He doesn't open the door you chose.] He then says to you,
"Do you want to pick door No. 2 or No. 4 now ?" or "Would you
still keep your original choice? "
What would you do?
Which door will you pick?
Justify...
#Post#: 89--------------------------------------------------
Re: Door game...
By: kranthipls Date: November 15, 2012, 11:07 pm
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I will stick with the option one.
Justification: Once I fix in mind I will go blindly on that
decision
#Post#: 92--------------------------------------------------
Re: Door game...
By: dinesh Date: November 18, 2012, 12:03 am
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This question is a variation of the monte hall problem:
Suppose you're on a game show, and you're given the choice of
three doors: Behind one door is a car; behind the others, goats.
You pick a door, say No. 1, and the host, who knows what's
behind the doors, opens another door, say No. 3, which has a
goat. He then says to you, "Do you want to pick door No. 2?" Is
it to your advantage to switch your choice?
It is here again assumed that the contestant initially chooses
door 1.
Assuming the player picks door 1, the probability that the car
is behind door 2 and the host opens door 3 equals the
probability the car is behind door 2 times the probability the
host opens door 3 given the car is behind door 2 = 1/3 x 1 =
1/3.
The probability the car is behind door 1 and the host opens door
3 equals the probability the car is behind door 1 times the
probability the host opens door 3 given the car is behind door 1
= 1/3 x 1/2 = 1/6.
Given the player picked door 1, these are the only possibilities
in which the host opens door 3. Therefore, the probability the
host opens door 3 equals 1/3 + 1/6 = 1/2.
Finally, by the definition of conditional probability, the
conditional probability that the car is behind door 2 given that
the host opened door 3 equals the probability that the car is
behind door 2 and the host opened door 3 divided by the
probability the host opened door 3 = (1/3)/(1/2) = 2/3
Refer wikipedia for proof with Bayes formula and also visual
explanation :
HTML http://en.wikipedia.org/wiki/Monty_Hall_problem
Also
HTML http://montyhallproblem.com/
#Post#: 107--------------------------------------------------
Re: Door game...
By: kpr29 Date: November 18, 2012, 11:23 pm
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@Majeti: Too Hi-fi.............. can u summarize in 2
lines.............. What option will u stick to
it............... What ur probability says to stick
on............. Please emphasis this point............
kp.
#Post#: 132--------------------------------------------------
Re: Door game...
By: dinesh Date: November 19, 2012, 12:35 am
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By switching you have more chance of winning...
#Post#: 149--------------------------------------------------
Re: Door game...
By: kpr29 Date: November 19, 2012, 10:07 pm
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@Majeti: I also understood the same
thing..............Thanks......
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