Is this informal solution to the Monthy Hall problem wrong? And why?

Question

I recently opened my high school book about probability and statistics and I found an informal solution to the Monthy Hall problem. I translate it here:

"Let's suppose that the first door chosen by the player is wrong and hides a goat. The host doesn't have any choice and opens the door with the other goat. In this case if the player changes door, he wins. If, instead, the first door chosen by the player is the right one, and the player changes door he loses.

So we can conclude that if the player changes door he wins if and only if his first choice was the wrong one, and this event has probability 2/3."

This informal solution seems very convincing to me, but in my (flawed?) understanding I could apply it even if the presenter didn't know what there is behind the doors (while if I try to give a formal solution to the problem, at least using probability samples, this information changes the result).

Here the same solution modified:

"Let's suppose that the first door chosen by the player is wrong and hides a goat. The host opens the door with the other goat by pure chance. In this case if the player changes door, he wins. If, instead, the first door chosen by the player is the right one, and the player changes door he loses.

So we can conclude that if the player changes door he wins if and only if his first choice was the wrong one, and this event has probability 2/3."

My intuition here tells me that the fact that the host chooses a goat by chance doesn't change my initial probability of choosing a goat, that is still 2/3.

So there are two possibilities:

1) I can't see why the informal solution the book gave can't be applied anymore if the presenter doesn't know what there is behind the doors. In this case I ask: Why can't I apply that solution anymore?

2) The informal solution the book gave is wrong. In this case I ask: Why is it wrong?

I don't know much about probability theory: just the definition of a probability space and what conditional probability is.

Answer 1

In the actual Monty Hall problem, before we open any doors, a player can win by switching if and only if his initial choice was wrong, and this has probability $\frac 23$. He can win by not switching if and only if his original choice was right; this has probability $\frac13$. So it is clearly better to switch.

In the modified version where the host doesn't know which door is which, a player can win by switching if and only if neither the door he originally chose nor the door the host opens contains the prize. This has probability $\frac 23\times \frac 12=\frac13$. He can win by not switching if and only if his original choice was right; this still has probability $\frac13$. So both choices have an equal chance.

Answer 2

You picked a door, and Monty opened another to reveal a goat, and now you need to evaluate the probability that the car is behind the remaining door (so you win if you switch).

There will certainly be a car behind the remaining door if you picked a goat given that Monty did so too. This is true whether Monty cheated or was fair.

However, what is the probability that you picked a goat when given that Monty picked a goat?

• When Monty Cheats he always picks a goat.   If you picked a goat, he doesn't have any choice but to pick the other; if you picked a car he could pick either goat.

$$\mathsf P_{\small \!C}(Y\mid M)=\dfrac{\mathsf P_{\small \!C}(Y\cap M)}{\mathsf P_{\small \!C}(Y\cap M)+\mathsf P_{\small \!C}(Y^\complement\cap M)}=\dfrac{\tfrac 23\tfrac 11}{\tfrac 23\tfrac 11+\tfrac 13\tfrac 22}=\dfrac 23$$

• When Monty is fair, there was a chance he could've picked the car.   If you picked a goat, one from the two doors he could've picked had a goat; if you picked a car he could only pick either goat.

$$\mathsf P_{\small \!F}(Y\mid M)=\dfrac{\mathsf P_{\small \!F}(Y\cap M)}{\mathsf P_{\small \!F}(Y\cap M)+\mathsf P_{\small \!F}(Y^\complement\cap M)}=\dfrac {\tfrac 23\tfrac 12}{\tfrac 23\tfrac 12+\tfrac 13\tfrac 22}=\dfrac 12$$