What is the best algebraic proof of the Monty Hall problem (e.g.: xor(a xor { C < B or B < C }) = 1)

Question

I want to look at the Monty Hall problem as a way of measuring the host's involvement in the game. Namely:

1. The game show host's knowledge and action results in a change of the outcome of the game
2. Since the host has a is nonzero effect on the game, then what is that value?

Example

Given 3 doors (a,b,c), solve for the most efficient & concise function of the game show host's involvement in the showm. m = A function that is the lesser of B or C

a xor m = 1

or

a xor { C < B or B < C } = 1

My goal is to quantify and isolate m

Question

• Is there any mathematical model, or proof that captures "Monty's knowledge" of the door's contents?

• Alternatively, is there a model proof that expresses the relative "side channel" of information that would cause a savvy person to switch?

I'm asking because I think that a difference in statistical behaviors are an indicator that some variables are not accounted for in the problem. (hence common sense often fails this problem)

Is there any algebraic proof, or other discipline of mathematics that would capture the data I'm seeking?

Edit

To make this more concrete, if a "Montecule" represents the if/then logic that the host goes through in order to play the game, then I want to isolate that on one side of an equals sign. Why? I think it can be a building block to Artificial Intelligence, and learning.

The components of a 'Montecule' consist of

• Private knowledge of which door contains the prize
• A shared values system (between host and contestant) of what is most important. (typically the car)
• Ranking of whats less important and exposing that (decision making)

It seems to me that if I can identify a 'Montecule', I can chain them together to form more complex learning and decision systems.

Answer 1

The Wikipedia page on the Monty Hall problems lists many different kinds of Monty's, from 'Angelic Monty' and 'Ignorant Monty' (Monty does not know where prize is) to 'Lazy Monty' (lazy Monty always opens the door closest to him ... I suppose you could also have a 'Health-Conscious Monty' who always opens the door furthest away), and I am sure you can think of some more.

It might also be interesting to play successive Monty Hall scenarios, where both Monty and the contestant take into account each other's previous 'moves' and thus Monty might switch from one strategy to a different one.

But all these different scenarios can still be analyzed using conditional probabilities (as the Wikipedia page does) ... you just need to make all the different conditions explicit. (and you're right, most treatments of the Monty Hall problem don't)

Answer 2

What is usually overlooked in solutions to the Monty Hall Problem - that is, the missing variable you refer to - is that you see which doors get opened. That needs to be a part of any model like what you want.

Let the random variable C represent the door where the car is, and the random variable M represent the door Monty opens. To simplify the model, I'm going to assume you always choose Door #1. Since your choice is independent of the car placement, we can do this without any loss of generality.

We can now write the event where the car is behind door #1 as "$C=1$", and the event where Monty opens Door #3 as "$M=3$". Note that $M=1$ or $M=C$ would end the game, so we needn't consider those events. If we see Monty open #3 (you can do this with #2 and get similar results), and want to know if switching is a good idea, we just need to evaluate the conditional probability $Pr(C=2|M=3)$. This is trivial to do using Bayes' Theorem:

$Pr(C=2|M=3) = Pr(M=3|C=2)*Pr(C=2)/Pr(M=3)$

The denominator of this expression can be found using the Law of Total Probability:

$Pr(M=3) = Pr(M=3|C=1)*Pr(C=1) + Pr(M=3|C=2)*Pr(C=2)$

Finally, we put this back in the first expression and simplify it by dividing out the terms for $Pr(C=1)$, $Pr(C=2)$, and $Pr(C=3)$ which are all equal to each other. There is a similar expression for your chances without switching that I'll just list, along with what you called the "player benefit" $B$:

$Pr(C=2|M=3) = Pr(M=3|C=2) / [Pr(M=3|C=1) + Pr(M=3|C=2)]$ $Pr(C=1|M=3) = Pr(M=3|C=1) / [Pr(M=3|C=1) + Pr(M=3|C=2)]$ $B = [Pr(M=3|C=2) - Pr(M=3|C=1)] / [Pr(M=3|C=1) + Pr(M=3|C=2)]$

I believe this is the model you seek. You just need to determine what the two probabilities are from your model of Monty Hall's behavior.

The usual assumptions, that everybody realizes they make, are that Monty will always open a door, will always open a door with a goat, and will never open the contestant's door. So $Pr(M=3|C=2)=1$. What the don't realize they need to assume, is that if the car is behind Door #1, that Monty chooses randomly between Door #2 and Door #3. So $Pr(M=3|C=1)=1/2$. This makes $Pr(C=2|M=3)=(1)/[(1/2)+(1)]=2/3$.

The people who think switching can't matter are actually using logic that is equivalent to this model. But since they don't consider how Monty could have opened Door #2, they think $Pr(M=3|C=1)=1$ and get the wrong answer. Those who argue that your initial chances with Door #1 can't change are logically wrong - this model shows it can change - but that happens only if $Pr(M=3|C=1)$ and $Pr(M=2|C=1)$ are not equal. So they happen to get the right answer.

Noting that $0<=Pr(M=3|C=1)<=1$, we can see that $B$ ranges from (1-1)/(1+1)=0 to (1-0)/(0+1)=1, but with an unbiased choice is is (1-1/2)/(1/2+1)=1/3.

You can implement other Monty Hall behaviors by changing how you assign the four probabilities.

Answer 3

I think we could argue informally that no algebraic proof exists, or no finite circuit exists, because the solution of the Monty-Hall Problem (MHP) involves measure theory, which includes measures of irrational sizes that cannot be simulated on a finite circuit. Any such circuit would need a comparison part comparing this measure to the probability 1/2, which is part of the solution to the MHP.

For instance, Monty's action can be described by the measure $\Pr(M=3\cup C=2 \cup P=2)$ [1] , describing the union of events where Monty has a choice, Monty (M) opening door labeled "3", the car (C) being behind door labeled "2", and the contestant/player (P) choosing door 2. Since this is a mathematical problem, this measure can take on any irrational number.

[1] The MHP solution, expressed in odds, is this measure divided by $\Pr(M=3\cup C=1 \cup P=2)$, where the result is 0, infinity or not a number, with the last one expressing that the problem is undefined, the first one that the contestant should switch doors, and the last one that the contestant should stay.