In The Monty Hall Problem, suppose there are three doors A, B, and C. Suppose the contestant chooses the first door A. The variation comes here, suppose that the contestant now has the option to open either door B or C (not the game host).
If the contestant opened door B and didn't find the prize, and he is allowed now to switch or not switch, are we now in an equivalent situation of regular Monty Hall, or is it $50$% for door A and door C?
(I am only interested in the case when the contestant don't expose the prize).
I guessed that we are in an equivalent situation to Monty Hall, but I am not arriving to prove it, in fact, I am proving the opposite:
First:
$$
\text{Prior Probability: Probability of car behind doors } P(\text{Car}@…)
$$
$$
P(\text{Car}@A) = \frac{1}{3}
$$
$$
P(\text{Car}@B) = \frac{1}{3}
$$
$$
P(\text{Car}@C) = \frac{1}{3}
$$
$$
\text{Event I: Probability of contestant opening door B (after choosing door A first he can't open A)} P(\text{Open } B | \text{Car}@…)^*
$$
$$
$$
$$
P(\text{Open } B | \text{Car}@A) = \frac{1}{2}
$$
$$
P(\text{Open } B | \text{Car}@B) = \frac{1}{2}
$$
$$
P(\text{Open } B | \text{Car}@C) = \frac{1}{2}
$$
$$
\text{Posterior Probability: chances of the car behind the doors after event I} P(\text{Car}@… | \text{Opened } B)
$$
$$
P(\text{Car}@A|\text{Opened } B) = \frac{\frac{1}{2} \times \frac{1}{3}}
{(\frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3})} = \frac{1}{3}
$$
$$
P(\text{Car}@B|\text{Opened } B) = 0
$$
$$
P(\text{Car}@C|\text{Opened } B) = \frac{\frac{1}{2} \times \frac{1}{3}}
{(\frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3})} = \frac{1}{3}
$$
My question is: Is my guess true but my maths wrong, or the opposite?
Best Answer
I assume that you guessed that your math is wrong because you ended up with:
$$P(\text{Car@A}|\text{Opened B})+ P(\text{Car@B}|\text{Opened B})+ P(\text{Car@C}|\text{Opened B})=$$
$$\frac{1}{3}+0+\frac{1}{3}=\frac{2}{3}$$
while that should of course be $1$
What happened is this: $P(\text{Car@B}|\text{Opened B})= \frac{1}{3}$, rather than $P(\text{Car@B}|\text{Opened B})= 0$ because you didn’t specify that the opened door is actually empty. So, there is a one in three chance that when you open door B you find the car to be behind door B (and hence you lost, since your pick is still door A).
Indeed, this is what makes your scenario different from the classic scenario. In the classic scenario Monty knows where the prize is and will make sure not to open a door where the prize is. You, however, don’t know where the prize is, and so you may end up opening the door with the prize in which case you immediately lose.
So, what you need to figure out is the probability of the prize being door A under the assumption that you open door B and you find it empty.
Here is how you can do that:
First, in case you are not familiar with this, let's figure out how to do conditional probabilities involving 3 events.
Well, in general, just like we have that:
$$P(A\&B)=P(A|B)\times P(B)=P(B|A)\times P(A)$$
We also have:
$$P(A\&B|C)=P(A|B\&C)\times P(B|C)=P(B|A\&C)\times P(A|C)$$
And so:
$$P(A|B\&C)=\frac{P(B|A\&C)\times P(A|C)}{P(B|C)}$$
Applied to the Monty Hall scenario thus have:
$$P(\text{Car@A}|\text{B Empty}\&\text{B Opened})=$$
$$\frac{P(\text{B Empty}|\text{Car@A}\&\text{B Opened})\times P(\text{Car@A}|\text{B Opened})}{P(\text{B Empty}|\text{B Opened})}$$
Now, you already figured out that $$P(\text{Car@A}|\text{B Opened}) = \frac{1}{3}$$
just as it turns out to be in the Monty Hall scenario! But again, we are ultimately not interested in $P(\text{Car@A}|\text{B Opened})$, but in $P(\text{Car@A}|\text{B Empty}\&\text{B Opened})$
Well, we obviously have that $$P(\text{B Empty}|\text{Car@A}\&\text{B Opened}) = 1$$
But in your scenario, we have:
$$P(\text{B Empty}|\text{B Opened}) = \frac{2}{3}$$
because you don't know where the prize is.
While in the classic Monty Hall scenario, we have:
$$P(\text{B Empty}|\text{B Opened}) = 1$$
because Monty does know where the prize is, and so any door Monty opens is bound to be empty.
So, in the classic Monty Hall scenario you get:
$$P(\text{Car@A}|\text{B Empty}\&\text{B Opened})=$$
$$\frac{1 \times \frac{1}{3}}{1} = \frac{1}{3}$$
But in your scenario we get:
$$P(\text{Car@A}|\text{B Empty}\&\text{B Opened})=$$
$$\frac{1 \times \frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$
So, in your scenario, switching makes no difference. Indeed, your scenario is isomorph to the scenario where Monty does not know where the prize is and randomly opens on of the other doors. In that 'Ignorant Monty' scenario, we again have that $P(\text{B Empty}|\text{B Opened}) = \frac{2}{3}$, and therefore $P(\text{Car@A}|\text{B Empty}\&\text{B Opened}) = \frac{1}{2}$, so if Monty is ignorant, switching also does not make a difference.