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I saw this solution someone gave to the Monty Hall Problem and was wondering if the notation was wrong:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car. Then if the contestant switches, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 2 3$.
If they don't, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 1 3$.

I know this is problematic and this is my attempt to fix it:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car, and $S$ the event that the contestant changes their mind after being shown another door. Then if the contestant switches, $P(W|S) = P(C \cap W|S) + P(C^c \cap W|S) = \frac 2 3$.
If they don't, $P(W|S^c) = P(C \cap W|S^c) + P(C^c \cap W|S^c) = \frac 1 3$.

Does this make sense? Is there a way to justify the solution I saw?

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  • $\begingroup$ It slightly depends on what you regard as the status of $S$ and $S^c$ as they are presumably not random events but instead strategic choices. I think both your formulations make sense. $\endgroup$
    – Henry
    Commented Oct 6 at 15:00
  • $\begingroup$ "I know this is problematic and this is my attempt to fix it" Why does it need fixing? It is just adding the conditions '$|S$' or '$|S^c$' which are already implied in the texts "if the contestant switches" and "If they don't". The additional notation can be both clarifying (more precise/detailed) or confusing (too verbose) and it is a matter of taste. $\endgroup$
    – Sextus Empiricus
    Commented Oct 6 at 15:48
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    $\begingroup$ IMHO, this reformulation runs into conceptual difficulties by attempting to use the mathematics of events to model making decisions as if they were inherently probabilistic and could be considered parts of the same sigma algebra used to model the outcomes of those decisions. This is how debates get started... . $\endgroup$
    – whuber ♦
    Commented Oct 6 at 16:47
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    $\begingroup$ Ah. I see the problem with my formulation now. Thanks. I guess my question is now about the probability function then. Do we have different probability functions under different decisions? Since the events are the same, so if the probabilities are different, then the difference must come from the probability function. Thank you. $\endgroup$
    – WinnieXi
    Commented Oct 7 at 7:55

2 Answers 2

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Perhaps this is not an answer to your specific question about notation, but is a simple way to show a frequentist approach to the Monty Hall (MH) problem.

One of three doors has a car behind it, and the other two doors each have a goat behind them.

Suppose the contestant intends to randomly select (rv X with observed value x) one of the three doors with equal probability. Let Sel(X) denote this door. Now MH always knows which door has the car behind it and he will always open one of the other doors (MH(X)) that has a goat behind it. This will leave the remaining door (Rem(X)) that is always offered to the contestant to switch to. Thus:

Pr(Sel(X) correct) = 1/3,

Pr(MH(X) correct) = 0, and

Pr(Sel(X) correct) + Pr(MH(X) correct) + Pr(Rem(X) correct) = 1, (since the car is behind one of the doors)

so Pr(Rem(X) correct) = 2/3

This is the structure of the game prior to the contestant initially selecting a door. Once a door is selected (Sel(x)), randomness is over, so now there is 33.3% confidence that Sel(x) has the car behind it and 66.7% confidence that the car is behind Rem(x). So switching to Rem(x) is good advice.

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  • $\begingroup$ Please, let's not rehash the MH problem: it has been discussed ad nauseam elsewhere. $\endgroup$
    – whuber ♦
    Commented Oct 8 at 13:27
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    $\begingroup$ O.k., fair enough. $\endgroup$
    – Graham Bornholt
    Commented Oct 8 at 17:52
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Is the notation problematic? Not really, but as pointed out in a previous comments, it is redundant with the text. However, there is a logic mistake in both attempts at the notation. The OP writes first:
if the contestant switches, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 2 3$.
But if the contestant switches, $P(C \cap W)=0$ and should not even be in the equation (if the contestant had picked the correct door to begin with, then switches, he/she can not win).
It gets rearranged as
$P(W|S) = P(C \cap W|S) + P(C^c \cap W|S) = \frac 2 3$.
Again, the term $P(C \cap W|S)$ does not belong here (and is equal to 0), as we are looking at the probability of winning by switching.

So, what may work better? Let's try the following, using the same terminology:
$P(W|S^c)=P(C)$
and
$P(W|S)=P(C^c)*P(C2|C^c)$
where $C2$= probability of picking the correct door the 2nd time.

Now, for the traditional 3 doors Monty Hall problem, we have $P(C)=\frac 1 3 $, and $P(C^c)*P(C2|C^c)=\frac {2}{3}*1=\frac 2 3$. as would be expected.
So then what is the point of $C2$. It helps because now you can use the same equations (and notation) to, for example, play with a 5 doors Monty Hall (where Monty only opens a single door before asking the contestant to switch or not). We would have
$P(W|S^c)=P(C)=\frac 1 5=\frac 3 {15}$ and
$P(W|S)=P(C^c)*P(C2|C^c)=\frac {4} {5}*\frac 1 3=\frac 4 {15}$ It still pays to switch, but not as much.

And you can play with any variant, where there are $n$ doors, and Monty opens $m < n-1$.

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