How can I apply Monty Hall problem correctly?

Question

I previously wondered about the possibility of predicting events by exclusion and I was convened that this is a sort of 'Gambler's fallacy' my previous example was like this:

Early discovery of diseases help treating them early before complications occur. Suppose I estimate that a disease ,which is rare, occurs in 1% of patients coming to my department,. Among 1000 patients who came to the hospital, I excluded 800 of them ,by doing some cheap diagnostics,. Do the probability change among the rest 200 patients ?

Now I have a problem combining this with the Monty hall problem , So I repeat the question in a Monty hall style :

I have 10 doors,or patients,one of them has the rare disease(gold). Each door has either yes or no. you've select one door as it contains yes (so the probability that your door is correct is 1 in 10). Now Monty will show you that in the rest (9 doors), eight have 'no', will you change your choice? ,i.e. you will insist on your first choice, or move to the other door.

Answer 1

You pick the door, say $d_1$. The unconditional probability that it has gold is $P(A)=1/N$, where $A$ is an event that the gold is behind door $d_1$.

Next, you find out that doors $d_2,\dots,d_{N-1}$ have no gold, i.e. event $B$. That's new information, so you have to update your probabilities using Bayes theorem:

$$P(A|B) = \frac{P(B | A) \, P(A)}{P(B)}$$

Here, $P(B|A)=1/(N-1)$ is the probability that doors $d_1,\dots,d_{N-1}$ will be opened if the gold was behind the door $d_1$, or alternatively, the the door $d_N$ will not be opened given event $A$.

$$P(A|B) = \frac{\frac{1}{N-1} \, \frac{1}{N}}{P(B)}$$

Now, the to get $P(B)$ we need to sum all possibilities:

• if gold is behind $d_1$, then $P(B|A)P(A)=\frac{1}{N-1}\frac{1}{N}$
• if gold is behind one of $d_2,\dots,d_{N-1}$, then $P(B|d_1,\dots,d_{N-1})P(d_1,\dots,d_{N-1})=0$
• if gold is behind $d_N$, then $P(B|d_N)P(d_N)=\frac{1}{N}$

So, we get $P(B)=\frac{1}{N}\frac{N}{N-1}=\frac{1}{N-1}$

Finally, $$P(A|B)=\frac{1}{N}$$, i.e. the probability that gold is behind door $d_1$ given that it's not behind doors $d_2,\dots,d_{N-1}$ is $1/N$, which means that the probability that it's behind the door $d_N$ is $1-1/N=\frac{N-1}{N}$, so you have to switch.

The Monty Hall logic is stronger for more doors. Here's the intuition. Imagine an infinite or very large $N$. What is a chance that you'll pick gold by randomly picking a door? Basically, ZERO. It's somewhere else.

Now, Monty comes and shows every door where there is NO gold. So, you just open the remaining door, gold is almost surely there. The Monty Hall problem is most difficult to comprehend when $N=3$, once you increase $N$ it's more and more evident that you've got to switch the doors.

Answer 2

This sounds more like the old gag of sitting down with someone and saying to them with an incriminating look, "statistics show 1 out of 2 people is having an affair and it's not me".

In your context, say you estimate a disease prevalence of 1 in 1000 and you've ruled out the first 999 with your inexpensive test. That wouldn't mean the last person definitely has it unless you somehow knew at least one of those 1000 definitely had it.

In Monty Hall, there is a negative correlation in prize status between doors. In the patient setting, you would have to assume a negative correlation in disease status between patients. It's hard to imagine a setting where that would be the case.