Not a solution to the anthropic dice killer puzzle

I recently came up with what I thought was a solution to the dice killer puzzle. It turns out that I was wrong, but in the process of figuring this out I discovered a few subtleties in the puzzle that I had missed first time around.

First I’ll repost the puzzle here:

A mad killer has locked you in a room. You are trapped and alone, with only your knowledge of your situation to help you out.

One piece of information that you have is that you are aware of the maniacal schemes of your captor. His plans began by capturing one random person. He then rolled a pair of dice to determine their fate. If the dice landed snake eyes (both 1), then the captive would be killed. If not, then they would be let free.

But if they are let free, the killer will search for new victims, and this time bring back ten new people and lock them alone in rooms. He will then determine their fate just as before, with a pair of dice. Snake eyes means they die, otherwise they will be let free and he will search for new victims.

His murder spree will continue until the first time he rolls snake eyes. Then he will kill the group that he currently has imprisoned and retire from the serial-killer life.

Now. You become aware of a risky way out of the room you are locked in and to freedom. The chances of surviving this escape route are only 50%. Your choices are thus either (1) to traverse the escape route with a 50% chance of survival or (2) to just wait for the killer to roll his dice, and hope that it doesn’t land snake eyes.

What should you do?

As you’ll recall, there are two possible estimates of the probability of the dice landing snake eyes: 1/36 and 90%. Briefly, the arguments for each are…

Argument 1  The probability of the dice landing snake eyes is 1/36. If the dice land snake eyes, you die. So the probability that you die is 1/36.

Argument 2  The probability that you are in the last round is above 90%. Everybody in the last round dies. So the probability that you die is above 90%.

The puzzle is trying to explain what is wrong with the second argument, given its unintuitive consequences. So, here’s an attempt at a resolution!

Imagine that you find out that you’re in the fourth round with 999 other people. The probability that you’re interested in is the probability that the fourth round is the last round (which is equivalent to the fourth round being the round in which you get snake-eyes and thus die). To calculate this, we want to consider all possible worlds (i.e. all possible number of rounds that the game might go for) and calculate the probability weight for each.

In other words, we want to be able to calculate P(Game ends on the Nth round) for every N. We can calculate this a priori by just considering the conditions for the game ending on the Nth round. This happens if the dice roll something other than snake eyes N-1 times and then snake eyes once, on the final round. Thus the probability should be:

Screen Shot 2018-09-02 at 10.47.18 AM.png

Now, to calculate the probability that the game ends on the fourth round, we just plug in N = 4 and we’re done!

But hold on. There’s an obvious problem with this approach. If you know that you’ve been kidnapped on the fourth round, then you should have zero credence that the game ended on the third, second, or first rounds. But the probability calculation above gives a non-zero credence to each of these scenarios! What’s gone wrong?

Answer: While the probability above is the right prior probability for the game progressing to the Nth round, what we actually want is the posterior probability, conditioned on the information that you have about your own kidnapping.

In other words, we’re not interested in the prior probability P(Game ends on the Nth round). We’re interested in the conditional probability P(Game ends on the Nth round | I was kidnapped in the fourth round). To calculate this requires Bayes’ rule.

Screen Shot 2018-09-02 at 10.54.19 AM

The top term P(You are in the fourth round | N total rounds) is zero whenever N is less than four, which is a good sign. But what happens when N is ≥ 4? Does the probability grow with N or shrink?

Intuitively, we might think that if there are a very large number of rounds, then it is very unlikely that we are in the fourth one. Taking into account the 10x growth in number of people each round, it looks like for any N > 4, the theory that there are N rounds strongly predicts that you are in the Nth round. The larger N is, the more strongly it predicts that you are not in the fourth round. In other words, the update on your being in the fourth round strongly favors possible worlds in which the fourth round is the last one

But this is not the whole story! There’s another update to be considered. Remember that in this setup, you exist as a member of a boundless population and are at some point kidnapped. We can ask the question: How likely is it that you would have been kidnapped if there were N rounds?  Clearly, the more rounds there are before the game ends, the more people are kidnapped, and so the higher chance you have of being kidnapped in the first place! This means that we should expect it to be very likely that the fourth round is not the last round, because worlds in which the fourth round is not the last one contain many more people, thus making it more likely that you would have been kidnapped at all.

In other words, we can break our update into two components: (1) that you were kidnapped, and (2) that it was in the fourth round that you were kidnapped. The first of these updates strongly favors theories in which you are not in the last round. The second strongly favors theories in which you are in the last round. Perhaps, if we’re lucky, these two updates cancel out, leaving us with only the prior probability based on the objective chance of the dice rolling snake eyes (1/36)!

Recapping: If we know which round we are in, then when we update on this information, the probability that this round is the last one is just equal to the objective chance that the dice roll lands snake eyes (1/36). Since this should be true no matter what particular round we happen to be in, we should be able to preemptively update on being in the Nth round (for some N) and bring our credence to 1/36.

This is the line of thought that I had a couple of days ago, which I thought pointed the way to a solution to the anthropic dice killer puzzle. But unfortunately… I was wrong. It turns out that even when we consider both of these updates, we still end up with a probability > 90% of being in the last round.

Here’s an intuitive way to think about why this is the case.

In the solution I wrote up in my initial post on the anthropic dice killer thought experiment, I gave the following calculation:

Screen Shot 2018-09-02 at 12.54.23 PM.png

Basically, we look at the fraction of people that die if the game ends on the Nth round, calculate the probability of the game ending on the Nth round, and then average the fraction over all possible N. This gives us the average fraction of people that die in the last round.

We now know that this calculation was wrong. The place where I went wrong was in calculating the chance of getting snake eyes in the nth round. The probability I wrote was the prior probability, where what we want instead is the posterior probability after performing an anthropic update on the fact of your own kidnapping.

So maybe if we plug in the correct values for these probabilities, we’ll end up getting a saner answer!

Unfortunately, no. The fraction of people that die starts at 100% and then gradually decreases, converging at infinity to 90% (the limit of \frac{1000...}{1111...} is .9). This means that no matter what probabilities we plug in there, the average fraction of people will be greater than 90%. (If the possible values of a quantity are all greater than 90%, then the average value of this quantity cannot possibly be less than 90%.)

Screen Shot 2018-09-02 at 1.07.07 PM.png

This means that without even calculating the precise posterior probabilities, we can confidently say that the average probability of death must be greater than 90%. And therefore our proposed solution fails, and the mystery remains.

It’s worth noting that even if our calculation had come out with the conclusion that 1/36 was the actual average chance of death, we would still have a little explaining to do. Namely, it actually is the case that the average person does better by trying to escape (i.e. acting as if the probability of their death is greater than 90%) than by staying around (i.e. acting as if the probability of their death is 1/36).

This is something that we can say with really high confidence: accepting the apparent anthropic calculation of 90% leaves you better off on average than rejecting it. On its own, this is a very powerful argument for accepting 90% as the answer. The rational course of action should not be one that causes us to lose where winning is an option.

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