What Do Monty Hall and A Steak Have In Common? I Marinate Both

5 minute read


It took me half a PhD to finally understand the Monty Hall Problem, but I think I get it now. I’ve been tutoring a very bright student in probability (which is so ironic given I failed it way back when) and I’ve surprised myself with how effective I’ve been at solving little homework problems I would have previously not been able to solve. It isn’t like my PhD has been full of these sorts of problems and by virtue to their exposure I’ve gotten better at them. I’ve just been thinking about stats longer than I was thinking about stats in undergrad which sort of leads to this unconcious thinking about homework problems. More on that in the end.

Set Up

Ok, quickly though because I’m sure you have heard of the problem before:

  • Three Doors, 1 Car 2 Goats.

  • You pick a door

  • Monty opens a different door to show you a goat

  • Monty asks if you want to switch doors

  • What should you do?

I’ll tell you what you shouldn’t do: don’t label the doors before you choose them. If you do, that leads to some funky conditioning and argumentation which I think is what initially confused me. If you label them before choosing them, then you have to say stuff like “what is the probability Monty opens door B given I chose door A and A contains the goat?”.

Instead, label them after choosing. Let door $A$ be whatever door you choose, let $B$ be the door Monty opens to reveal the goat, and let $C$ be the remaining closed door. This set up maintains all the generality for the solution and will be easier to think about.

The question is now “how does the probability change if I were to select door C over door A after seeing door B”? In particular, let’s talk about relative changes rather than absolute because it will lead to some simplifiction. Let $A_w$ be the event that $A$ is the winning door, let $B_o$ be the event B is opened, and let $C_w$ be the event that $C$ is the winning door. We want to find $P(A_w \vert B_o)$ and $P(C_w \vert B_o)$.

But because we are interested in relative changes, we don’t need to find these probabilities completely. One thing you will see Bayesians write down a lot is

\[p(\theta \vert y ) \propto p(y \vert \theta) p(\theta)\]

This ignores the normalizing constant of $p(y)$. So I can just calculate

\[p(A_w \vert B_o ) \propto p(B_o \vert A_w) p(A_w)\]


\[p(C_w \vert B_o ) \propto p(B_o \vert C_w) p(C_w)\]

and then take their ratio.

The Math

A priori, all three doors could possibly have the car. This means $P(A_w) = P(C_w) = \frac{1}{3}$.

If $A$ contained the car, that means $B$ and $C$ would have goats. Monty can’t open your door and he can’t open the door with the car. So that means he can open one of two doors. So $P(B_o \vert A_w) = \frac{1}{2}$.

If $C$ contained the car, that means $A$ and $B$ would have goats. Monty can’t open your door and he can’t open the door with the car. So that means he can open one door (namely, $B$). So $P(B_o \vert C_w) = 1$.

This means

\[p(A_w \vert B_o ) \propto \dfrac{1}{2} \times \dfrac{1}{3}\] \[p(C_w \vert B_o ) \propto 1 \times \dfrac{1}{3}\] \[p(C_w \vert B_o ) = 2 p (A_w \vert B_o )\]

and now it becomes very clear that you double your chances of winning by switching doors. The actual probabilities are $1/3$ and $2/3$ but that doesn’t matter since we’re interested in relative probabilites.

There. The hardest undergrad probability probelm I ever encountered done in like 5 lines.

What Worked This Time?

I think what made the problem click so late in my life is not labelling the doors before choosing. When I did that, I would find myself getting lost in details and I’d ask “what is the probability door A has the car, given he opens door B? But I guess he doesn’t always open door B, sometimes he opens C? Wait, so then what if B has the car? UGH”.

The other thing that helped was not caring about $P(B)$. We could calculate it if we wanted, it involves the quantities we already computed in addition to $P(B_w \vert B_o)$ (which is 0 because Monty can never open the door with the car). Not caring about that left some mental space in my mind to think about what is going on.

Lessons Learned

This pattern of thinking, where I solve problems without actually thinking about them, has been happening a lot lately. I go on on walks (yay pandemic activities) and I come back with solutions to coding problems, or I take a shower and suddenly realize the solution to something that has been bugging me. I’d call it “passive problem solving” but I don’t think that is what it is. I think it is just giving myself time and room to think think about problems in different ways, even if I’m not aware I’m thinking about them. When I tell people I’m going to do this, I tell them I’m “going to let the problem marinate”.

And now that is what I am labelling the process. Problem solving through marination.