Via @sigmaleph on Tumblr:

You are playing the Monty Hall problem.

However, you secretly know one of the goats is the former pet of an eccentric billionaire who lost it and is willing to pay an enormous amount for its return, way more than the car is worth. You really want that goat. The host is unaware of this. After you pick your door, as is traditional, the host opens one door, which he knows doesn’t have the car. He reveals a hoad, which you can tell is the ordinary goat and not the secretly valuable one. The host offers to let you switch doors. Should you?

The Monty Hall problem has become a commonplace in maths outreach – I remember my Differential Equations lecturer quizzing us about it in his customary mid-lecture break in the late 90s, I’ve used it in workshops and I’ve done a MathsJam talk about variants of it. It crops up on reddit from time to time when angry teenagers who’ve misunderstood the setup feel the need to insist that they’re right and the well-established, proof-based maths community consensus is wrong.

I hadn’t seen this variant before. I like it. And I’ll explain how I solve it beneath the line. As with anything involving a sports car, you should expect spoilers.

Let’s look at the usual Monty Hall setup.

  • With probability $\frac{1}{3}$, you pick the car. Monty then has an equal chance of revealing the ordinary goat or Billy Gates’ Gruff.
  • With probability $\frac{1}{3}$, you pick the ordinary goat. Monty then reveals the special goat.
  • With probability $\frac{1}{3}$, you pick the special goat. Monty then reveals the ordinary goat.

So the probabilities are:

  • Car behind your door, ordinary goat revealed: $\frac{1}{6}$
  • Car behind your door, special goat revealed: $\frac{1}{6}$
  • Ordinary goat behind your door, special goat revealed: $\frac{1}{3}$
  • Special goat behind your door, ordinary goat revealed: $\frac{1}{3}$

If you were interested in the car, sticking would only win a third of the time, while switching would win two times in three. Take that, angry teenage redditors.

However, we explicitly don’t want the car – we want the goat that hasn’t been revealed. Intuitively (and in this case, the intuition is correct), if we don’t win the car, we win the other goat, so we do the thing that’s least likely to win us the car – to maximise our chances of winning the goat GOAT, we should stick.

It holds up if you do the sums: looking back to the probabilities, the probability of Monty revealing the ordinary goat was $\frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ (which makes sense, there’s no reason for one goat to be more likely than the other.)

The probability of the special goat being behind your chosen door given that Monty has revealed the other is, according to Bayes’ rule, $\frac{1}{3} \div \frac{1}{2} = \frac{2}{3}$. So the sums agree that you should stick with your original choice.

Neat puzzle! Do you have a better explanation?