Alternative, more intuitive, formulation of Monty Hall problem

Here it is:

Three tennis players. Two are equally-matched amateurs; the third is a pro who will beat either of the amateurs, always.

You blindly guess that Player A is the pro; the other two then play.

Player B beats Player C. Do you want to stick with Player A in a Player A vs. Player B match-up, or do you want to switch? And what’s the probability that Player A will beat Player B in this match-up?

And here’s the background.

It started when Josh Miller proposed this alternative formulation of the Monty Hall problem:

Three boxers. Two are equally matched; the other will beat either them, always.

You blindly guess that Boxer 1 is the best; the other two fight.

Boxer 2 beats Boxer 3. Do you want to stick with Boxer 1 in a Boxer 1 vs. Boxer 2 match-up, or do you want to switch?

I liked the formulation in terms of boxers (of course, and see data-based followup here), but Josh’s particular framing above bothered me.

My first thought was confusion about how this relates to the Monty Hall problem. In that problem, Monty opens a door, he doesn’t compare two doors (in his case, comparing 2 boxers). There’s no “Monty” in the boxers problem.

Then Josh explained:

When Monty chooses between the items you can think of it as a “fight.” The car will run over the goat, and Monty reveals the goat. With two goats, they are evenly matched, so the unlucky one gets gored and is revealed.

And I pieced it together. But I was still bothered:

Now I see it. The math is the same (although I think it’s a bit ambiguous in your example). Pr(boxer B beats boxer C) = 1 if B is better than C, or 1/2 if B is equal in ability to C. Similarly, Pr(Monty doesn’t rule out door B) = 1 if B has the car and C has the goat, or 1/2 if B and C both have goats.

It took me awhile to understand this because I had to process what information is given in “Boxer 2 beats Boxer 3.” My first inclination is that if 2 beats 3, then 2 is better than 3, but your model is that there are only two possible bouts: good vs. bad (with deterministic outcome) or bad vs. bad (with purely random outcome).

My guess is that the intuition on the boxers problem is apparently so clear to people because they’re misunderstanding the outcome, “Boxer 2 beats Boxer 3.” My guess is that they think “Boxer 2 beats Boxer 3” implies that boxer 2 is better than boxer 3. (Aside: I prefer calling them A, B, C so we don’t have to say things like “2 > 3”.)

To put it another way, yes, in your form of the problem, people easily pick the correct “door.” But my guess is that they will get the probability of the next bout wrong. What is Pr(B>A), given the information supplied to us so far? Intuitively from your description, Pr(B>A) is something close to 1. But the answer you want to get is 2/3.

My problem with the boxers framing is that the information “B beats C” feels so strong that it overwhelms everything else. Maybe also the issue is that our intuition is that boxers are in a continuous range, which is different than car >> goat.

I then suggested switching to tennis players, framing as “two amateurs who are evenly matched.” The point is that boxing evokes this image of a knockout, so once you hear that B beat C, you think of B as the powerhouse. With tennis, it seems more clear somehow that you can win and just be evenly matched.

Josh and I went back and forth on this for awhile and we came up with the tennis version given above. I still think the formulation of “You blindly guess that Player 1 is the pro” is a bit awkward, but maybe something like that is needed to draw the connection to the Monty Hall problem.

Ummm, here’s an alternative:

You’re betting on a tennis tournament involving three players. Two are equally-matched amateurs; the third is a pro who will beat either of the amateurs, always.

You have no idea who is the pro, and you randomly place your bet on Player A.

The first match is B vs. C. Player B wins.

Players A and B then compete. Do you want to keep your bet on Player A, or do you want to switch? And what’s the probability that Player A will beat Player B in this match-up?

This seems cleaner to me, but maybe it’s too far away from the Monty Hall problem. Remember, the point here is not to create a new probability problem; it’s to demystify Monty Hall. Which means that the problem formulation, the correct solution, and the isomorphism to Monty Hall should be as transparent as possible.

P.S. Josh noted that the story was also discussed by Alex Tabarrok, and a similar form of the problem was studied by Bruce Burns and Marieke Wieth in 2004.