This problem is from Introduction to Probability by Blitzstein and Hwang (Chapter 2, exercise 40)
The ratings of Monty Hall's show have dropped slightly, and a panicking executive producer complains to Monty that the part of the show where he opens a door lacks suspense: Monty always opens a door with a goat. Monty replies that the reason is so that the game is never spoiled by him revealing the car, but he agrees to update the game as follows. Before each show, Monty secretly flips a coin with probability p of heads. If the coin lands Heads, Monty resolves to open a goat door (with equal probabilities if there is a choice). Otherwise, Monty resolves to open a random unopened door, with equal probabilities. The contestant knows p but does not know the outcome of the coin flip. When the show starts, the contestant chooses a door. Monty (who knows where the car is) then opens a door. If the car is revealed, the game is over; if a goat is revealed, the contestant is offered the option of switching. Now suppose it turns out that the contestant chooses door 1 and then Monty opens door 2, revealing a goat. What is the contestant's probability of success if he or she switches to door 3?
The issue is, almost every place I've seen this problem solved states:
- Flipping the coin is independent of position of the car or the choice of door that Monty opens
- Car’s position is independent from coin’s toss or Monty’s door opening
My intuition feels like there are errors within these assumptions and I would like somebody to check my work and tell me what (if any) that I am doing wrong. Thank you.
My work:
$W$ = Event that you win
$G_i$ = Goat behind the $i^{th}$ door
$M_i$ = Monty opened the $i^{th}$ door
$H$ = Heads
$T$ = Tails
$P(W|G_2,M_2)=P(W|G_2,M_2,H)*P(H|G_2,M_2)+P(W|G_2,M_2,T)*P(T|G_2,M_2)$
$P(W|G_2,M_2,H) = P(W|G_2,M_2,H,C_3)*P(C_3|G_2,M_2,H)$
$P(C_3|G_2,M_2,H) = \frac{P(G_2,M_2|C_3,H)*P(C_3|H)}{P(G_2,M_2|H)}= \frac{1*\frac{1}{3}}{\frac{1}{2}}= \frac{2}{3}$
$P(H|G_2,M_2) = \frac{P(G_2,M_2|H)*P(H)}{P(G_2,M_2)}=\frac{P(G_2,M_2|H)*P(H)}{P(G_2,M_2|H)*P(H)+P(G_2,M_2|T)*P(T)}= \frac{\frac{1}{2}*p}{\frac{1}{2}*p + \frac{1}{3}*(1-p)}$
$P(W|G_2,M_2,T) = P(W|G_2,M_2,T,C_3)*P(C_3|G_2,M_2,T)$
$P(C_3|G_2,M_2,T) = \frac{P(G_2,M_2|C_3,T)*P(C_3|T)}{P(G_2,M_2|T)}= \frac{\frac{1}{2}*\frac{1}{3}}{\frac{1}{3}}= \frac{1}{2}$
$P(T|G_2,M_2) = \frac{P(G_2,M_2|T)*P(T)}{P(G_2,M_2)}=\frac{P(G_2,M_2|T)*P(T)}{P(G_2,M_2|H)*P(H)+P(G_2,M_2|T)*P(T)}= \frac{\frac{1}{3}*(1-p)}{\frac{1}{2}*p + \frac{1}{3}*(1-p)}$