Doors: A,B & C A = the prize is behind the chosen door A not A = the prize is behind one of the not chosen doors

Initially: P(A)=P(B)=P(C)=1/3

Pick up A; P(A)=1/3 P(notA)=P(B)+P(C)=1/3+1/3

Host opens door B (dud)

P(B)=0, hence P(notA)=2/3=P(B)+P(C)= 0 + 2/3

P(C) increases from 1/3 to 2/3 therefore is better to switch doors after the host has opened the first one.

Bayes' Rule notes:

Pick door A before the host opens a door: P(A) 1/3 (Assumption: the host will not open the door behing which the prize is)

P(Prize behind door C|Door B opened by the host)

P(Prize behind door C|Door B opened by the host)= P(door B opened by the host|prize behind door C)P(prize behind door C)/P(door B opened by the host) = 1(1/3)/1/2