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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