Bayes' theorem - monty hall problem with five doors - incorrect final probabilities Consider the following revised five-door Monty Hall problem: Step 1, you choose two doors and your choice is: doors 1 and 2; Step 2, Monty Hall opens two doors (with no car behind them): doors 3 and 4; Step 3, Monty Hall asks you to decide which door among doors 1, 2, and 5 is your final choice. You will win the car if the car is behind the door you finally choose. Now apply Bayes’ theorem to calculate the probabilities that the car is behind doors 1, 2 and 5. (Note: you need to give detailed formulas and calculations step by step, rather than just present the final results).
I know what the final answer should be from intuition,
p(car behind door 1) = 1/5
p(car behind door 2) = 1/5 
p(car behind door 5) = 3/5
I tried to work through the question but the correct probabilities are not gained. This is my working so far:
p(car being behind door 1 | door 3 and 4 opened) = (2/3 * 1/5) / 2/3 
= 1/5 <-- correct 
p(car being behind door 1 | door 3 and 4 opened) = p (door 3 and 4 opened | car behind door 1) * p (car behind door 1) / p (Door 3 and 4 opened ) 
p(car being behind door 2 | door 3 and 4 opened) = (2/3 * 1/5) / 2/3
= 1/5 <-- correct
p(car being behind door 2 | door 3 and 4 opened) = p (door 3 and 4 opened | car behind door 2) * p (car behind door 2) / p (Door 3 and 4 opened ) 
p(car being behind door 5 | door 3 and 4 opened) = (1 * 1/5) / 2/3
= 3/10 <-- incorrect 
p(car being behind door5 | door 3 and 4 opened) = p (door 3 and 4 opened | car behind door 5) * p (car behind door 5) / p (Door 3 and 4 opened ) 
I think p(door3 and door4) != 2/3 - this is where I am possibly going wrong, my calculations say it should be 1/3 so that p(car being behind door 5 | door 3 and 4 opened) = 3/5, but that does not make sense. 
etc. 
bayes equation used: p(A|B) = p(B|A)*p(A)/p(B)
I am just not too sure. 
Please explain 
thanks 
 A: Short answer: You are right, $P(3\&4)$ should be $\frac{1}{3}$, since Monty can choose from 3 pairs of doors:3&4, 3&5, and 4&5. So, all your $\frac{2}{3}$'s should be $\frac{1}{3}$'s, and now it all works out.
Long and more detailed answer. Monty Hall problems are noriously confusing, so I really like to differentiate all of the following events:
Let $P1$ be the event that the prize is behind door 1 (same for 2,3,4,5)
Let $C1$ be the event that you choose door 1 
Let $E1$ be the event that Monty shows there is no prize behind door 1
For door 1, we now have:
$$P(P1|C1, C2, E3, E4)=\frac{P(E3,E4|P1,C1,C2)*P(P1|C1,C2)}{P(E3,E4|C3,C4)}$$
where:
$$P(E3,E4|P1,C1,C2)=\frac{1}{3}$$ (As stated in the short answer, given that the prize is behind door 1, Monty could have opened doors 3 and 4,  or doors 3 and 5, or doors 4 and 5. So, assuming these three possible events happen with equal likelihood (e.g. Monty has not preference to open up higher numbered doors if possible), then each of these three events has a probability of $\frac{1}{3}$)
$$P(P1|C1,C2)=\frac{1}{5}$$
(We assume that the prize was randomly put behind one of the doors, and that the picking of the doors is not going to change the location, e.g. The crew behind the scenes is not going to move the prize as a result of your picks!)
$$P(E3,E4|C1,C2)=\frac{1}{3}$$
(Without any knowledge of the prize, the three options (opening 3 and 4, or 3 and 5, or 4 and 5) are still equally likely, using the same assumptions as before)
If you don't immediately see this, you can also do:
$$P(E3,E4|C1,C2)=$$
$$P(E3,E4|P1,C1,C2)*P(P1|C1,C2)+P(E3,E4|P2,C1,C2)*P(P2|C1,C2)+P(E3,E4|P3,C1,C2)*P(P3|C1,C2)+P(E3,E4|P4,C1,C2)*P(P4|C1,C2)+P(E3,E4|P5,C1,C2)*P(P5|C1,C2)=$$
$$\frac{1}{3}*\frac{1}{5}+\frac{1}{3}*\frac{1}{5}+0*\frac{1}{5}+0*\frac{1}{5}+1*\frac{1}{5}=\frac{5}{3}*\frac{1}{5}=\frac{1}{3}$$
So, plug this in, and you get that the probability of the prize being behind door 1 after Monty has opened the doors is (still) $\frac{1}{5}$, which is what you got as well, even though you plugged in the wrong numbers into your formula.
Now, the formula for door 2 works similarly of course, but for door 5, you get $P(E3,E4|P5,C1,C2)=1$ (which is what you had as well)
