Hilbert’s Hotel and Gödel’s Freezer

A friend of mine, Flavien Hirsch, invented a fascinating paradox in probability, called Hilbert’s Hotel and Gödel’s Freezer. Technically speaking it is the same thing as the Sleeping Beauty problem, which I’ve written about here, but I find it much nicer because it’s more symmetrical, you actually care about the answer, and intuition leads you astray. Without further ado, here it is:

Gödel has a Freezer which contains 4 people in a state of suspended animation, with you among them. He picks 2 of them randomly, with uniform probability, to send to Hilbert’s Hotel, a nice place which is never full. The other 2 are doomed to stay in the Freezer forever. Being a bit mad and a bit evil, Gödel decides to do the transfer in a capricious way. He sends them via two tram trips, each with 3 people: one of the chosen ones, who stays at Hilbert’s Hotel at the end of the trip, and the doomed ones, who are taken back to the Freezer. Not being too evil, Gödel erases the memory of the doomed ones after each trip so that they won’t be tortured by their dream of freedom. You wake up in the tram, are told about the whole setup, and ask yourself: what’s the probability that I’m going to Hilbert’s Hotel?

There are two answers that seem obviously correct, but conflict with each other:1

  • $1/2$ – As 2 out of 4 people are going to Hilbert’s Hotel, and by assumption they were picked randomly and uniformly, the probability of each person must be $1/2$. Since everybody will wake up with certainty and ask themselves the question, you don’t learn anything by waking up, so the probability stays $1/2$.
  • $1/3$ – Out of the 3 people in the tram only one is going to Hilbert’s Hotel. Since there’s nothing distinguishing anyone there, they must all have the same probability of being the chosen one, which must therefore be $1/3$.

Now suppose there are $n$ people instead of 4, with still only 2 doomed ones. The probability $1/2$ becomes $(n-2)/n$, while $1/3$ stays $1/3$. The probability that you will go to Hilbert’s Hotel is therefore rather high. But you wake up in the tram – which will happen in any case – and now suddenly you should be sad that your probability is only $1/3$? How can that possibly be true?

The difficulty here is that we’re trying to work with an informal, intuitive notion of probability. That’s not good enough. If we want to have actual answers we need to figure out what probability means. We should first notice that the paradox is about subjective probability. The probability that Gödel used to pick the chosen ones can be taken to be objective, but that’s in the past, now the objective probability that you’ll go to Hilbert’s Hotel is either zero or one, you just don’t know what it is. So even if we had an amazing and universally agreed upon definition of objective probability that wouldn’t help us at all.

What definition of subjective probability do we have then? I’m happy with the decision-theoretical one: you say that an event $E$ happens with subjective probability $p$ if $p$ is the fair price you assign to a bet that pays 1€ if $E$ happens, and nothing otherwise. Fair here means that you don’t expect to win or lose money from this bet, so you’d be fine buying or selling the bet for this this price. For example, if $E$ happens with objective probability $q$, and you pay $p$ for the bet, then with probability $q$ you win $(1-p)$€, and with probability $1-q$ you win $-p\,$€. The expected value is $q(1-p)-(1-q)p=q-p$, so you will expect to win money if $p < q$, and you will expect to lose money if $p>q$, so the fair price is $p=q$.

You wake up in the tram then, and consider, what is the fair price of a bet that pays 1€ if I end up at Hilbert’s Hotel? There are two cases to consider: if you are a chosen one, you’ll pay $p$€ and win 1€, so your net gain is $(1-p)$€. But if you’re a doomed one, you’ll pay $p$€ and win nothing, with a net gain of $-p$€. In this trip. But if you are a doomed one you will certainly do another trip (or will have already done it). So you’ll do the same bet twice, paying in total $-2p$€. The bet just became less attractive, and it stands to reason that you’ll be willing to pay less for it! The expectation value is $(1-p)/2 -2p/2 = (1-3p)/2$, which is zero for $p=1/3$.

But wait. Is it fair to make the doomed one pay twice for the bet? Well, the bet is imaginary in any case, but the doomed one is who wakes up twice and asks themselves the question twice. Paying for the bet twice is just a formalisation of being wrong twice. Intuitively, you might not want to guess you’re a doomed one because even though the probability is $1/2$, when you’re wrong you’ll be wrong twice.

Maybe we shouldn’t count the mistake of the doomed one twice, because their memory will be erased between the trips? Wiseman suggested this would be natural in my original post about the Sleeping Beauty. I find it rather artificial. If we erase their memory after the second trip as well (as the statement of the paradox says), should we then never count their mistake? And thus assign subjective probability 1 to going to Hilbert’s Hotel? Or in a completely different situation, should I assign probability 1 to winning a game of Russian roulette, as my memory will be erased if I lose? I think this is really changing the question from being wrong about your prediction than knowing that you were wrong.

It does raise an interesting question, though, of what does it take to make the subjective probability of going to Hilbert’s Hotel match the objective probability $1/2$. One could get rid of the multiple awakenings, by saying that Gödel will only let the passengers of the tram wake up in the first trip. But this is not enough, because now the probability that you will wake up if given that you are a chosen one is $1/2$, while the probability that you will wake up given that you are a doomed one is still 1, so the expectation value is $(1-p)/4 – p/2 = (1-3p)/4$, so the fair price is still $1/3$.

It does work if Gödel decides to wake up each of the four people exactly once, say all passengers of the tram wake up once, in the first trip, and the remaining chosen one wakes up alone in the second trip (and somehow doesn’t notice that the other two passengers are still asleep). Now everybody will wake up once, and only once, with probability 1, and the expectation value is $(1-p)/2 – p/2 = 1/2 – p$, and finally we have $p=1/2$. But the problem clearly changed.

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1 Response to Hilbert’s Hotel and Gödel’s Freezer

  1. ARaybold says:

    Ever since I failed the Monty Hall problem, I have been suspicious of arguments that include the phrase “you don’t learn anything from…”

    In this case, I reckoned that the n-3 people not on the trolley are guaranteed a place in the hotel, so we always have three people in hopes of one place, and if you are not that person, then you are somewhere along in a series of n-2 disappointments.

    I think there is something Monty-Hall-esqe about this view, in that the people not on the trolley are analogous to the box that Monty Hall cannot open.

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