Lets See How Smart NTers are : The Monte Hall Problem

its switch its the first thing they talk to you about in any statistics class. stay with door one = 33.3% chance of being right switch to door two = 50% or 67%chance of being right when you picked door one you had a 1 in 3 chance, now its down to 2 choices and you have a higher chance of picking right if you pick theother door, i think. been a few years since ive had to use stat.
 
Originally Posted by K2theAblaM

I think you're all wrong. It's the same either way. You have a 50% chance of the car being behind door one or two.

BEFORE he revealed door three, you had a 33% chance. Now, it comes down to 50%, door 1 or 2. There is no point in switching, unless your gut tells you something.
WRONG!
 
its been a while since i delt with a stats question but bascially you have 33% chance that you will get the right door before the one door is opened
after the one door is open you have a 50 % chance that you will get the right door
there is a forumla for this i believe P(A n B), which will tell you the chances of you being right twice, the chance of you being right is P(A) times P(B).33*.50= there is a 16.5% chance that the next time a door is selected door number 1 will be the correct one, so it is better to switch, please let me know ifi am right, its been a while since i had a stats question lol
 
Dude, no. It's a 1/3 chance if you stay, 2/3 if you don't. It can't add up to less than 3/3, first of all, so those numbers should look wrong toyou even before you check them.
 
Originally Posted by Joseph Camel Jr

Dude, no. It's a 1/3 chance if you stay, 2/3 if you don't. It can't add up to less than 3/3, first of all, so those numbers should look wrong to you even before you check them.
yeah i think your right but if you round 16.5 to 17 then that equals 33+17+50=100%
happy.gif
 
smh @ myself that i had to think for that long about the problem to finally get it. i view myself as a smart dude too. this definitely took a hit on my swagfrom smartness level
 
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