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With 5 doors we get the following events.

A: Probability of choosing a door with a car behind it: 1/5

B|A: Likelihood of monty opened door B if door A contains a car: 1/4 ? Im not sure about this one

How can I proceed now?

With 5 doors he will open 3 doors. Door A is the door I choose, door B is the one monty left closed.

How can I obtain the probability that the car is actually behind door B? That is if i switched from door A from door B.

The result supposedly is 4/15 but i cannot achieve using bayes theorem.

Can someone guide me in the right direction? Thanks

Pedro Martins
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  • The result of what? What are you supposed to calculate? – Bram28 Jul 15 '20 at 20:23
  • I just edited my question @Bram28 – Pedro Martins Jul 15 '20 at 20:28
  • Isn't B the door that Monty opens? I think you want the probability of getting the car when swithcing to door C .... – Bram28 Jul 15 '20 at 20:33
  • With 5 doors he will open 3 doors. I said door B, but door B is an arbitrary name for the other door. Door A is the door I choose, door B is the one monty left closed. – Pedro Martins Jul 15 '20 at 20:35
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    Ah! Maybe put that in the question as well .. I thought Monty still just opened one door ... door B. In fact, isn't switching always supposed to be better? So why is the result less than $\frac{1}{2}$? That would suggest that switching to door B is worse than staying with door A. Hmmmm, so maybe the fact that you are having problems getting to the right solution is because you misunderstood the question? Maybe Monty is really opening just one door? – Bram28 Jul 15 '20 at 20:37
  • Im not sure. I found the 4/15 from here: https://math.stackexchange.com/questions/1032661/monty-hall-problem-with-five-doors – Pedro Martins Jul 15 '20 at 20:40
  • Aha! Yes, it's right there: Monty opens one of the other doors and then invites you to switch to one of the remainig closed doors. – Bram28 Jul 15 '20 at 20:41
  • Let us continue this discussion in chat. – Pedro Martins Jul 15 '20 at 20:42
  • This does not use Bayes' theorem explicitly, but you can see this directly using the symmetry: not switching yields you probability 1/5. If you switch, then you have equal probability for each of the other doors, which makes it $\frac{4}{5}\cdot \frac {1}{3}=\frac 4{15}$. – tomasz Jul 15 '20 at 20:44
  • You say the problem is from https://math.stackexchange.com/questions/1032661/monty-hall-problem-with-five-doors but the problem you describe is different. There monty shows you one door. You can stay, or you can switch. And if you switch there are three doors you can switch to. The prob is 4/15 for any of the other doors. The problem you gave Monty shows you three doors and leaves you one to switch to. The probability if you switch if 4/5. A completely different problem! And a completely different answer. – fleablood Jul 16 '20 at 01:04

1 Answers1

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You are not at all interested in the probability that Monty chooses a particular door.   You are entirely concerned about the probability that another door hides the prise when given that Monty always reveals one which does not.

So the events of interest are the car is behind the first chosen door (A) and the car is behind the door you would switch to (C).

Well, if the prise is behind the first door, switching surely loses, else if it is not behind the A-door, then after Monty reveals another door where it is not, the prize is equally likely to be behind each of the three remaining doors.

Thus we evaluate the marginal probability that switching wins as:

$$\begin{align}\mathsf P(C)&=\mathsf P(A)\,\mathsf P(C\mid A)+\mathsf P(A^\complement)\,\mathsf P(C\mid A^\complement)\\&=\tfrac 15\tfrac 03+\tfrac 45\tfrac 13\\&=\tfrac 4{15}\end{align}$$

Note that the probability that staying wins is $3/15$, so this game still makes switching the better option, though only slightly.

Graham Kemp
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