Monday, June 1, 2009

monty hall cond-prob-tree, simplified

http://bigblog.tanbin.com/2009/06/monty-hall-paradox.html has a tabular form of probability "tree" which I'm trying to simplify. After you pick a door and host reveals a door, let's paint the 3 doors

Amber on your pick
Black on the open
Cyan on the other

Now suppose immediately after (or before) your pick, a fair computer system (noisegen) randomly decides where to put the car. Therefore
p(A) := p(computer put car behind Amber) = 1/3
p(C) := p(computer put car behind Cyan ) = 1/3
p(B) := p(computer put car behind Black) = 1/3

Only after (not before) seeing the computer's decision, host chooses a door to open. My simplified probability Tree below shows switch means 66% win.

Note on the top branch -- if computer has put car behind the Black door then host has no choice but open the Cyan door. However, in this context we know B was opened, so this point is tricky to get right.

It's worthwhile to replace the host with another fair computer. This way, at each branching point we have a fair computer.

Amber door picked -33%--> Computer put car behind B -100%--> C opens  (A/B 0%) =>s/d switch
           
           
-33%--> Computer put car behind A -50%--> C opens (A 0%) => don't switch
-50%--> B opens (A 0%) => don't switch
           
           
-33%--> Computer put car behind C -100%--> B opens (A/C 0%) =>s/d switch

Now, what's prob(A has the car | B opens) ie p(chance of winning if we don't switch). Let's define
p(b) := p(B gets opened)
p(A|b) = p(b|A)p(A)/ { p(b|A)p(A)+p(b|C)p(C) } by Baye's thereom.
p(A|b) =50%33%/ { 50% 33% + 100% 33% } = 33%

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