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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?

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    $\begingroup$ Hint: The advantage in the usual game derives from the fact that Monty knows the true answer and you can use that. Here, there is no such advantage. And, for clarity, I assume that the game ends (and you lose) if you accidentally expose the prize? $\endgroup$
    – lulu
    Feb 10 at 12:49
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    $\begingroup$ 50% each for door A and C because by chance he has eliminated wrong door B $\endgroup$ Feb 10 at 13:04
  • $\begingroup$ @lulu In this question I am only interested in the case when I don't expose the prize. The fact that we take advantage of Monty knowing the answer makes the last 2 probabilities in Event I: 0 and 1 respectively, (as in here ), so this should be different than Monty Hall, and my evaluation of probabilities in Event I are wrong? $\endgroup$
    – Notwen
    Feb 10 at 13:05
  • $\begingroup$ The fact that someone with less information has a different sense of the probability is irrelevant. That's always true. This variant is not the same as the standard one, because no information is involved. $\endgroup$
    – lulu
    Feb 10 at 13:11
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    $\begingroup$ Suppose, for simplicity, that you are always going to choose $A$ and open $B$, and that you will always switch (as you can rename the doors, this is really the general case). Then there are three equally likely cases (according to where the prize actually is). If it's behind $A$, you lose. If it is behind $B$, you lose, if it is behind $C$ you win. If you exclude the middle case, it's still equal probability for the other two. $\endgroup$
    – lulu
    Feb 10 at 13:14

3 Answers 3

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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.

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  • $\begingroup$ Thanks for the very great explanation! $\endgroup$
    – Notwen
    Feb 10 at 15:57
  • $\begingroup$ @Notwen You're quite welcome! :) $\endgroup$
    – Bram28
    Feb 10 at 15:57
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Since you emphatically stated not being interested in the case where the opened door shows a prize, we need a conditional probability for the condition that this did not happen. So your choice can be described as picking two distinct doors $x,y$ and then opening door $y$, finding nothing there. You might as well have chosen $y$ first, and then $x$ distinct from it (this is distinct from the original Monty Hall problem, where things would be very different if Monty were to open a door before you had made your choice). But the choice of $x$ produced no effect or information. So the problem becomes equivalent to "a randomly chosen door opens and no prize is behind it; which one of the two remaining doors should you choose?", to which the answer is obviously that (the initial probabilities for each door being equal) it does not matter.

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Let's look at it this way.

Without the loss of generality, we assume the winning door is $A$.

Strategy 1. Switch whenever possible

For Monty to switch and win (say event $S_1W$), he must first select one of the two losing doors ($B$ or $C$) and then follow it up with choosing the wrong door again. The probability of that happening is

$$P(S_1W) = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$$

Strategy 2. Stay with the original choice

For Monty to stick with his first choice and win ($S_2W$), he must choose the right door ($A$) first. That's it. Now matter which door he opens following it, he will win. So,

$$P(S_2W) = \frac{1}{3}$$

So, the probability of him winning is $1/3$, whether he makes the switch or not.

Hence, switching offers no advantage.

Also, in either case, the probability of Monty losing is also $1/3$. as is the probability of the game being cut short.

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