Update
This post got a shout-out on an episode of the Modern Wisdom podcast, where Chris Williamson interviews Alex O’Connor (the ‘Cosmic Skeptic’). Very exciting! Here’s a link to where it gets brought up. I should also attribute the source of this puzzle: I first heard of it here, in Scott Aaronson’s online version of his quantum computing course (which I highly highly recommend, by the way). He attributes the puzzle to the philosopher John Leslie, though I haven’t been able to track down that source. (The addition of the 50%-escape-route is due to me.)
At the time I wrote this (about 3 years ago), I probably did think that the ‘anthropic’ calculation was the right one. Coming back to the puzzle after ~3 years, I am less sure. Like Alex, what I find most interesting about the puzzle is really that there are these two arguments that both feel fairly independently plausible, but end up at opposite conclusions. On the one hand: I’m in the room waiting to see how the roll of the dice comes out, and it seems crazy (if not blatantly contradictory) to say both that the dice are fair and that nonetheless my credence that they will land snake-eyes is > 90%. This is a very strong argument!
On the other hand: I am one of the captives, and I know that >90% of the captives end up dead (no matter how many rounds occur). So if I stay in the room I have a >90% chance of ending up dead. Stated with a more decision-theoretic flavor: Suppose the captives accept the first line of reasoning. Then they will all forsake the 50%-chance-of-survival route and stay in the room. But then >90% of them will die! On the other hand, if the captives follow the anthropic line of reasoning, then they all take the 50% route, and only 50% of them end up dying. Is it really rational for me to follow a strategy that I know does significantly worse in expectation?
Original post starts now
Today we discuss anthropic reasoning.
The Problem
Imagine the following scenario:
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 (ten times as many as he just let free).
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?
(Think about it before reading on)
A plausible-sounding answer
Your chance of dying if you stay and wait is just the chance that the dice lands snake eyes. The probability of snake eyes is just 1/36 (1/6 for each dice landing 1).
So your chance of death is only 1/36 (≈ 3%) if you wait, and it’s 50% if you try to run for it. Clearly, you are better off waiting!
But…
You guessed it, things aren’t that easy. You have extra information about your situation besides just how the dice works, and you should use it. In particular, the killing pattern of your captor turns out to be very useful information.
Ask the following question: Out of all of the people that have been captured or will be captured at some point by this madman, how many of them will end up dying? This is just the very last group, which, incidentally, is the largest group.
Consider: if the dice land snake eyes the first time they are rolled, then only one person is ever captured, and this person dies. So the fraction of those captured that die is 100%.
If they lands snake eyes the second time they are rolled, then 11 people total are captured, 10 of whom die. So the fraction of those captured that die is 10/11, or ≈ 91%.
If it’s the third time, then 111 people total are captured, 100 of whom die. Now the fraction is just over 90%.
In general, no matter how many times the dice rolls before landing snake eyes, it always ends up that over 90% of those captured end up being in the last round, and thus end up dying.
So! This looks like bad news for you… you’ve been captured, and over 90% of those that are captured always die. Thus, your chance of death is guaranteed to be greater than 90%.
The escape route with a 50% survival chance is looking nicer now, right?
Wtf is this kind of reasoning??
What we just did is called anthropic reasoning. Anthropic reasoning really just means updating on all of the information available to you, including indexical information (information about your existence, age, location, and so on). In this case, the initial argument neglected the very crucial information that you are one of the people that were captured by the killer. When updating on this information, we get an answer that is very very different from what we started with. And in this life-or-death scenario, this is an important difference!
You might still feel hesitant about the answer we got. After all, if you expect a 90% chance of death, this means that you expect a 90% chance for the dice to land snake eyes. But it’s not that you think the dice are biased or anything… Isn’t this just blatantly contradictory?
This is a convincing-sounding rebuttal, but it’s subtly wrong. The key point is that even though the dice are fair, there is a selection bias in the results you are seeing. This selection bias amounts to the fact that when the dice inevitably lands snake-eyes, there are more people around to see it. The fact that you are more likely than 1/36 to see snake-eyes is kind of like the fact that if you are given the ticket of a random concert-goer, you have a higher chance of ending seeing a really popular band than if you just looked at the current proportion of shows performed by really popular bands.
It’s kind of like the fact that in your life you will spend more time waiting in long lines than short lines, and that on average your friends have more friends than you. This all seems counterintuitive and wrong until you think closely about the selection biases involved.
Anyway, I want to impress upon you that 90% really is the right answer, so I’ll throw some math at you. Let’s calculate in full detail what fraction of the group ends up surviving on average.
By the way, the discrepancy between the baseline chance of death (1/36) and the anthropic chance of death (90%) can be made as large as you like by manipulating the starting problem. Suppose that instead of 1/36, the chance of the group dying was 1/100, and instead of the group multiplying by 10 in size each round, it grew by a factor of 100. Then the baseline chance of death would be 1%, and the anthropic probability would be 99%.
We can find the general formula for any such scenario:
IF ANYBODY CAN SOLVE THIS, PLEASE TELL ME! I’ve been trying for too long now and would really like an analytic general solution. 🙂
There is a lot more to be said about this thought experiment, but I’ll leave it there for now. In the next post, I’ll present a slight variant on this thought experiment that appears to give us a way to get direct Bayesian evidence for different theories of consciousness! Stay tuned.
This just got a shout-out from Alex O’Connor on Chris Williamson’s podcast. The more you know…
Cool! I’ve added a short update mentioning where I first heard the puzzle.
After 10 rounds of this (which is more than likely to happen), more than the entire population of the world would be included – meaning you’d continually be recaptured and therefore your chance of dying would be 100%. Therefore your best chance of survival would be to teleport out and end the experiment with a 50% chance of success.
Is there not an error in your calculation?
The odds of the dice roll in each round is the same. Therefore, previous rolls do not impact the odds of future rolls.
Your calculation correctly shows the odds of getting snake eyes in the long run, but this doesn’t change your odds of the roll.
Even if every person on the planet was kidnapped in the final round, the odds of that final round are still 1/36.
Even in the final round you would be better off with the roll, because those previous rolls do not impact this roll. Thinking they do is the gambler’s fallacy.
I think the way I worded it (“chance of getting snake eyes in the nth round”) is a bit confusing: the probability I am looking at is really the probability of (1) getting to the nth round and (2) getting snake eyes in the nth round. To get to the nth round requires that you never got snake eyes in the previous rounds, which is where the factor of (35/36)^(n-1) comes from. We want both (1) and (2) because the goal of the calculation is to sum over all the different mutually exclusive possible outcomes.
I think it’s a false paradox.
The dice has no memory. It holds no information on previous rolls or the population held captive.
Every time, the odds are the same. Every time, regardless of whether you’re in the first or 50th iteration, you should go with the dice.
Taking the big picture view, ‘what percentage of people die’ makes no sense to the person in the situation.
Honestly, I don’t understand why it’s even supposed to be a puzzle. Statistical independence is stats 101.
Suppose that this killer announces this plan before starting it, and people people the announcement, and are strategising how to survive if they are captured.
And rather than a risky room, let’s say everyone has a risky teleportation device that takes them safely home, or kills them, with a 50/50 chance (so that people can plan around it).
Let’s imagine that every potential victim talks to each other and agrees with the same strategy of what to do if they are captured:
* If every potential victims deicdes that they’ll wait for the dice to be rolled (a ‘sit and wait’ strategy), then ~90% of the victims will die at the murderer’s hands, and 10% will released by the murderer.
* If every potential victim plans to press the buttom if captured, then 50% of the victims will die from teleportation accident, and 50% will escape through successful teleportation. (The dice-killer won’t get to kill anyone.)
What strategy should people agree upon?
I think one of the problems is that the expected value of the numbers of people that will be killed is infinite. Therefore this comparison of the 90% and 50% ratios is meaningless and they certainly don’t translate into survival probabilites for the indiviuals.
the probability that it terminates approaches 0 as n tends to infinity, so the probability that it goes on forever also tends to 0. So – even though it’s possible it does go to infinity, it’s not likely to contribute a lot to the probabilities.
I agree! He’s looking at it as a distribution and saying the probability of getting x in a row. However, like you said, the probability doesn’t change from 1/36 for each trial.
Your 90% calc is not wrong, however if you suggest it implys that one ought to take the escape route in order to maximize chance of survival, then I respectfully disagree. All this information is irrelevant in the situation that you are in ->facing only 2 options. You let him role the dice and walk away free
Isn’t it now just (35/36)^x * 1/2? which is even worse. no paradox here
This is not a paradox – the odds at a population level have no bearing on the odds at any particular ‘turn’.
You haven’t worked out the probability of getting killed to be 90%. You’ve worked out “the fraction of people that will die if you get snake eyes”. There’s a difference.
Your 90% figure is staying so high because every round, you multiply the number of captives by 10. This doesn’t change _YOUR_ chances of survival. Your chances of survival aren’t based upon how many people are in the room with you.
This just feels like a murderous twist on the paradox of a coin flip, where people think, if they get multiple consecutive heads, that there’s some increased chance the next flip is a tails rather than a 50/50 split still.
Here are some thoughts about why the anthropic reasoning answer possibly doesn’t make sense if you reword the problem.
The only way you can get the calculation of a ~90% chance of dying is if you knew you would be in the game at some point before the game begins. I think the calculation relies on this assumption without stating it.
But this makes no sense as there can be no guarantee you will be part of the game as it is unbounded. The game could end in round one and in fact this is the only way you could be guaranteed to be in the game. Thus giving you a 90% chance of survival.
No matter what round you come in at there is a very high likelihood of being many more rounds as there is no dependence on the previous rolls. The population of people who die will almost certainly be massively greater than the number of people currently in the room.
I can’t see any paradox in the fact that you have a 90% chance of survival in any given round yet ~90% of people will die. Those people will be a great number of unlucky people but that bad luck will have massive consequences the further the game goes on.
Even if we accept this anthropogenic reasoning, one would need to know what the killer does if he gets to round 7 or so and captures the entirety of the human race. If he ends up capturing every human that exists and does not roll snake eyes, does he let them all go and then give up? If so, there is a strong change that he doesn’t actually kill anybody at all!
This is a good point, and I do think that the background assumption of an infinite (or at least potentially infinite) population is necessary for the argument to go through. I talked about this a bit in a follow-up post: https://risingentropy.com/infinities-in-the-anthropic-dice-killer-thought-experiment/
The problem with this reasoning here is that instead of “updating on all of the information available to you” it only looks at some of the information available while blatantly ignoring other information. In this case the information being ignored is that the event of any two people dying are not independent, and hence your chance of death is not equal to the fraction of all people who die. If the two people are in different groups then one person dying implies that the other lived, and if the two people are in the same group then they necessarily will have the same outcome. This dependence illuminates the (seemingly obvious, before we convoluted it) fact that the fraction of people who will die is not relevant to your odds of survival, only whether or not you happen to be in the final group. On average, the number of groups taken before snake eyes are rolled is 36 (expected value of the Geometric distribution), and so the average fraction of groups that will be killed is 1/36 or 2.78%. Thus, if the only information you have is that you are in one of these groups captured by the Dice Killer, you can reasonably conclude that you have a 35 in 36 chance of living, if you decide to stay.
If I were to set up the logic to argue against your “anthropic chance of death,” I would use the independence of the dice rolls between groups and induction to show that, at any point when you are captured, there is a 35 in 36 chance that you are released (and then somewhere down the line a much larger group that doesn’t include you is captured and killed), and only a 1 in 36 chance that you are in the final group.
But the probability for being in different groups isnt the same, its way more likely to be in the larger groups then to be the first guy for example
Absolutely! However, the odds the (large number of) people that were captured before you are independent of yours.
The series you have is a Lambert-type series. Any hope for a (nice) closed form solution should be squashed. If you write the series expansion for 1/(X^(n+1)-1) and swap the order of the series, you can show that that series is equal to \sum_{k=1}^{\infty} \frac{1}{X^k-pX}. This new series converges very quickly, so it is “better” than the series you have.
Thanks!
here’s my attempt at solving this paradox
https://www.youtube.com/watch?v=4YoUDuWe_CE
plz let me know what you think
I agree with many other comments that if you are the prisoner your chances of dying are low. Instead we can change our frame of reference and consider this problem from the outside, instead asking the question: “If we pick a prisoner who had at some point been captured, what is the probability that he died?”.
This question is still interesting as the first mistake is to assume that the answer is 1/36. However, the selection bias now applies, and the answer is 90%.
It comes down to two opposing ways of regarding the upcoming dice throw – undetermined or predetermined.
If the next dice throw is unknown but the outcome is determined, then the answer is indeed >90% that you are among the majority who are present at a snake eyes round.
For the latter, consider the scenario in which the dice has already been thrown until snake eyes was obtained. Also assume that the upper limit of throws was high enough to accommodate an enormous population. Suppose each dice thrown was recorded on video but is unknown to the killer or the potential victims. The killer proceeds as before, capturing victims. He then plays back a recording of each dice throw, releasing and capturing more prisoners until snake eyes turns up. In that scenario, any captured prisoner should assign a >90% likelihood that the dice throw about to be played back will come up snake eyes. However the killer will put that likelihood at 1/36.
By contrast, if each dice thrown is live and treated as a random/quantum event, the answer must be 1/36 for both the prisoners and the killer – regardless of how many people are in the room when snake eyes is obtained.
The crucial issue is whether it is fixed in advance that >90% of those captured will be present at a snake eyes round? This in turn relies on two assumptions: a) snake eyes will occur at some finite point in the future. b) the population is large enough to be captured for this round. Both assumptions depend on whether the upcoming dice throw is undetermined or predetermined.
If the future dice throw is a random quantum event or part of one, the chance of snake eyes can only be 1/36 for both the killer and the prisoners. In any quantum scenario, it is irrelevant what percentage have been captured just before snake eyes is obtained.
However, if a snake eyes round is a determined event (in the classic Newtonian sense), despite it being unknown how many rounds it took, then the anthropic sampling supersedes the odds of the dice. Here it is already fact that snake eyes was obtained in a finite number of throws. If you also know with that there are enough captured people to be present at all rounds, then it is indeed >90% likely that you are among the majority who are present at the snake eyes round, regardless of when it occurs.
You could imagine a scenario in which the dice has already been thrown until snake eyes was obtained. No-one knows what round snake eyes appeared. but it’s confirmed that the potential population to be captured is large enough to cover however many throws it took. Suppose all the rounds were recorded on video, to be played back chronologically by the killer instead of the dice being thrown live. Everything else is the same. Starting with one person, the killer captures ever increasing groups before each round until snake eyes turns up.
In the above scenario, being captured does automatically means a >90% likelihood that the current;y played-back throw will come up snake eyes. However, for the killer who kidnapped you it’s different. If the killer knows that there are enough people to be captured for many future rounds if required, the likelihood is approximately 1/36 for a snake eyes. Whereas, if it’s known that the population not yet captured is only a large enough for one more round, the odds for the killer would be slightly above 50/50 that snake eyes occurs on this round. For anyone captured it is still >90%.
I believe problem lies in the understanding of how you are added to the serial killers scenario.
If it is the case like you said that he chosen people completely randomly, then this would imply there is a chance that I am not added to the situation. This version of the scenario would lead to the intuitive answer of choosing to let the serial killer roll the dice, since there is a 5/6 chance that the scenario will play on.
However, if it was the case that I must be present in this serial killers killing spree, then it follows that I will always be more likely to be in the group that will be killed — 90% in the example. This is because in this scenario you can imagine that all dice must have already rolled in order to assign me to a group randomly, in which case you should chose to escape with a 50% chance, as there is a 90% chance that I am in the final group which will roll on snake eyes.
The crucial difference lies in whether I was added truly at random, in which case I cannot use the information that I am in the largest group, as there is a 5/6 chance that after the dice roll that I am not part of that group. In the second scenario as soon as I am playing the game, the game has sort of already played out, and I will know that there is a 90% chance that I am in the group where the dice will roll on snakes.
There really isn’t a paradox here at all. It isn’t inconsistent to say both that “if you find yourself picked then you have a 1/36 chance of dying” and “when the game is over 90% of the players will have died”.
The 90% is simply a consequence of the games structure, but any given turn has exactly a 1/36 chance of death.
If all the players were known up front (ie. a fixed pool of players) then there is a chance that no one would die. The 90% figure is only guaranteed with an infinite pool of players. In which case your chance of being picked to play in the first place is infinitesimal.
So, there are three independent parameters in the general problem: (0) the chance `p` that the individuals in a given round are let go (which is 35/36 in this particular case), (1) the population scaling factor `X` by which the number of victims in the current round is multiplied to select the victims of the next round (which is 10 in this case), and (2) the chance `r` that the individuals opting for the escape route are let go (1/2 in this case).
As shown by your formula, the anthropic probability of death `P`, depends on `p` and `X`.
However, note that the expected number of rounds `E` that the process is going to take is equal to `1/p`, which is given by the Geometric Distribution (https://en.wikipedia.org/wiki/Geometric_distribution) and depends only on `p`, not on `X`.
When you are given the opportunity to use the escape route, you know that there might have been some number of rounds before the one for which you are selected. You just don’t know how many. In this case, the expected number of rounds `E` that “your” process will take isn’t just `1/p`. It’s given by the following conditional probability: the expected number of rounds *given* that there has already been a number `n` of rounds before which let the victims go. This is in fact `n + 1/p`. In other words, it doesn’t matter how many rounds there have been before the one in which you participate: the expected number of rounds that it will take for your process to finish is always `1/p` *counting from yours*. So, you should not take the escape route chance with these parameters and should let the dice roll instead.
Another way to see it is to realize that even if at the end of the process, approximately 90% of the victims will die, it doesn’t mean that when you’re picked, you have 90% of chance to be in the last round. The percentage of people that will end up dying converges to 90.304% as the rounds go. So, even though the number of people dying changes dramatically, the percentage of people dying does not change much from one round to another and it doesn’t affect the probability of you being in the last round.
By increasing the number of people involved each round, it gives a counter-intuitive result where the number of people who die in the killer’s game is always more than 90% of the kidnapped victims. However that still doesn’t change the independence of an individual round.
Here’s another similar paradox:
The local casino has a coin-flip game where you have exactly a 50% chance of winning or losing if you play. You can bet any amount of money – if you win you double your bet, if you lose you lose your bet. I have a foolproof gambling strategy for this game – we simply double our bet every round if we lose.
Say we start with a $100 bet. If we win, we’ve made $100. If we lose we are down $100, but we simply play again, betting $200 this time. If we win that, we’ve made $100 (100 lost on the 1st round, 200 gained on the 2nd round). If we lose that one, we just play again, betting $400. Again, if we win we’re up $100 (lost $100 and $200 on rounds 1 and 2, but gained $400 on round 3). And again, if we lose, we just bet again at $800 dollars. We just keep playing until we win and then restart. You can see that this strategy will give us infinite money.
Except it clearly won’t work. Think about why, and then think about how the escalating stakes is also part of this dice killer paradox.
To me this seems like a confusion between probability and a statistic. When the killer finally lands on snake eyes, the amount of people he kills compared to the amount of people he’s kidnapped is around 90% which is a statistic of how many people have died, but in each round the chance of landing snake eyes is still 1/36. On top of that, the chances of the killer landing snake eyes slightly decreases after each round, because it depends on him not hitting snake eyes the previous round. (1/36 in the first round, 35/36 x 1/36 in the second round, (35/36)^2 x 1/36 in the third round, and on and on). In any case, no matter the round you wake up in, you still have a 1/36 (~2.78%) chance that the killer lands on snake eyes, compared to your 50% chance of escape. If your luck isn’t there and you end up dying, you just end up in the “90%” of people that died because of the exponentially increasing kidnapping count, but every individual in each round still had a 35/36 chance of making it out alive.
I solved it with a simulation. I posted my solution here. https://medium.com/@elec.mike/solving-the-anthropic-killer-paradox-5ed99b655550
The short answer is that it matters what the killer does when he runs out people in the population, because that adds a massive correction factor to the probabilities, bringing them back in line with what you expect. The more people in the population, the rarer the case that he runs out of people to kill, but the larger the impact on the probabilities when he does run out of people. With an infinite population, it’s infinitely rare but has an infinite impact when it does happen.
Another way to look at it: even if there’s a massive population (trillions of people), if somehow you find yourself chosen by the killer, you can reason that it’s very likely that he’s almost run out of people to choose, because that’s when the most people are chosen, so you need to take into account what he will do when he runs out of people. If he will just kill everyone he has on hand, it’s likely you’ll die and this “corrects” the probability of just the dice. If he will just let everyone go and stop killing, it’s likely you’ll be let go this “corrects” anthropic figures to be in line with the dice.
This doesn’t change as the population is made larger. So it should follow that if there’s an infinite number of people in the population, you should still reason that if you’re likely picked when the killer is about to run out of people to pick. After all, of all the infinite people he *could* have picked, there’s exactly 0% chance that he will pick *you* in any *finite* round, so you’re likely picked in an infinitely-later round.
The bias is that you know 100% that you will be captured and face the dice roll. for it to be unbiased you would have to give everyone in the world a number, and thats the order that you get kidnapped in (in a world with an absurdly large but finite population). so then there is a chance that it ends before you are kidnapped. by guaranteeing that you get kidnapped you’re essentially saying: I already know how many rounds this thing goes, and I’m just gonna use a random number generator to give you the order in which you’re kidnapped, which obviously gives a 90% chance you’ll be in the last group. To get the true 1/36 odds there would have to be a chance that you were never kidnapped in the first place.