You have been working on your plutonium-powered time machine for years, and now you finally succeeded!

At least, you thought you did...

After extensive testing with sending back and forth various objects, plant seeds and small animals, you decide it's time for the first human test.

All you have to do is let the time machine select a random date in time and transport you. Then, press the reverse button. That press will start a complex calculation to determine the exact amount of energy needed to make a similar jump in opposite direction of time. Once that is done, the leap will be initiated.

After selecting the random date option, you notice something strange. Sparks start shooting off the panel, and the time machine shakes heavily. Oh no... the heavy load caused a minor short-circuit. The heavy shaking causes you to hit your head and you pass out...

When you wake up everything seems normal. At least, as normal as a trip through spacetime can be. You are successfully transported to a different time, and all you have to do now is press reverse. You hold your finger over the button but pull it back the last second. In shock you read the travel log screen. You haven't just been sent to another time, you've been sent to FIFTY other times!

In other words, after you passed out, 50 parallel copies of you were created and scattered throughout spacetime. All in identical copies of the time machine.

You calculated before that this could happen and realise that pressing reverse is dangerous. It will start the return calculation and potentially send you back. However, each copy of the time machine processor is linked, such that if you and your copies press reverse more than once, the additional presses interfere with the calculation initiated by the first press. That will send you to some unknown place in spacetime from which you will never be able to return... But not pressing the button doesn't seem like a good option, as then you will stay where you are and never return home either. There is no way of communicating with your other selves.

When you take another look at the panel you notice that the battery of the time machine is starting to run out. You have to make a decision quick, or you will be too late to start the return cycle.

You weigh your options. If you don't press, you will know for sure that you will stay in an unknown time and never return home. However, if too many of you press the button, the return cycle is broken, and you will also end up lost in time and space.

You are desperate, but you know you have to keep your hopes up and rely on your awesome logic skills. Suddenly you realize something that may be helpful. Your other selves have the same logic skill as you. And they know you do too. And they know you know they know you do...

Is there a way you can make you way home? What is the best tactic to maximise your odds? And what chance of survival do you or your copies have?

To recap:
- 50 copies of you a trapped in a time machine
- All copies, including you, are equal, but independently assess the situation
- There is no way of communicating with your other selves
- There is a return button, but it only works if one person presses it. If multiple do, it sends all of them to a random place in time
- The return button only works for the person(s) that pressed it
- The return button can only be pressed once, otherwise the same will happen as if multiple people do: you overshoot your return trip
- You have perfect logic, and so do your copies
- No telepathy, or anything funky related to the fact that you are copies of yourself. The puzzle would work also for 50 random people with perfect logic.
- You have access to things you had in your pockets while travelling. Including your phone with calculator and other common apps you think are necessary. No internet though!

Inspired by a puzzle from Vsauce2 (Video includes answer to that question, so spoiler alert! https://www.youtube.com/watch?v=NkYCWqzBc7M )

  • 1
    $\begingroup$ If the "return button can only be pressed once" and the "return button only works for the person(s) that pressed it", how could more than one person return home? $\endgroup$
    – Jens
    Commented Sep 23, 2019 at 16:03
  • $\begingroup$ @Jens it's up to you to find out how many, if any can survive. Also, what the best tactic is and you personal chances of survival $\endgroup$
    – P1storius
    Commented Sep 23, 2019 at 16:20

5 Answers 5


You can get your chance of survival all the way up to

0.74%, or about 37% chance of the original scientist surviving.

For that, you need to find the best strategy. After you have come up with the strategy, it was shown that you can do it, which means you can count on the other "yous" in the other timelines to come up with it too. Having this important assumption out of the way, you

generate a random number from 1 to 50, and if it's a 50, you push the button. (The particulars of the randomisation method aren't important as long as you press the button with a 2% probability.)

This way, the probability of exactly one "you" pushing the button is maximised.

Since one is the maximum number of "yous" that can be saved (N.B. This is how I read the puzzle. I may be mistaken.), you happily accept whatever doom that awaits you if you get unlucky, and expect the other copies to do the same.

Since the situation is kind of an "altruistic prisoner's dilemma", every copy can easily reason that if they were to deviate from the strategy in any way, so would the other copies, resulting in a worse result for everybody, including themselves, so it is in every you's interest to stick to the strategy.

The probability of someone surviving with this strategy is

$(\frac{49}{50})^{49}\times\frac{1}{50}\times 50 \approx \mathbf{37\%}.$

The first term is the probability of getting 49 misses, and the second is the probability of getting a 50. Multiplying them together gives the probability of any given favourable outcome. The number of different favourable outcomes is 50 (it could be any "you" that rolls the sole 50), so we multiply the result by that, and get the final success rate as the result.

By symmetry, every copy is equally likely to get saved, so we get the odds of any given copy surviving

by dividing the overall success percentage by 50, for a total of approximately 0.74%.

The first term looks somehow familiar, and indeed, the more copies you fragment into, the closer the success probability is to

$\frac{1}{e} \approx 0.367879$

  • $\begingroup$ "The particulars of the randomisation method aren't important as long as you press the button with a 2% probability." Well, if you all generate it from a PRNG, and you all have the same seed, then that doesn't work. $\endgroup$ Commented Sep 24, 2019 at 18:07
  • 1
    $\begingroup$ @Acccumulation as I mentioned in another comment, you can use some environmental cue to desync the RNG. For example, select a visible object at random, turn a sufficiently complex description of it (e.g. a description of what a cloud looks like or the shape created by a collection of visible stars) into a hash (say, by adding together the A1Z26 values of its description and taking the result mod 50), and generate that many throwaway random numbers before generating the one you use. $\endgroup$
    – hdsdv
    Commented Sep 24, 2019 at 19:05
  • $\begingroup$ This is exactly the answer I was aiming at. Great job!. Bonus points for incorporation e in your answer :). The first time I saw the answer for a similar problem it sounded very counterintuitive, because why not press anyway? Otherwise you will face certain doom?! But you address the point well in the paragraph about the altruistic prisoner's dilemma. $\endgroup$
    – P1storius
    Commented Sep 25, 2019 at 8:31
  • $\begingroup$ While I do agree with your solution, there still is a problem. It implies total cooperation between clones having opposing goals. Suppose you are one clone and you did the randomization and it says not to press. At that moment, you have the choice to let it be or, well, press the button anyway. There is in theory a 37% chance that nobody pressed the button. That is a big incentive to try your luck and press the button. You really need to care for your other selves enough to accept to sacrify your life for one of them. It is a kind of prisonner's dilemma I guess. $\endgroup$
    – Florian F
    Commented Sep 25, 2021 at 8:54
  • $\begingroup$ Oops, in fact you shortly mentionned the prisonner's dilemma. $\endgroup$
    – Florian F
    Commented Sep 25, 2021 at 8:58

My updated answer then is:

You have a 0% chance of returning home

My logic behind this:

If you don't push the button you will not return home that is for certain. If you were to somehow decide not to push the button, so would every other version of you and nothing would change. Therefore for any chance you need to push the button. But yet again by doing this the other versions you will have the same logic so they will push the button. There is no scenario where you will return home.

  • 1
    $\begingroup$ I think the question asks for exactly one person to press it at any time, not all persons to press it at the same time. $\endgroup$ Commented Sep 23, 2019 at 13:05
  • 1
    $\begingroup$ @JaapScherphuis I was wondering about this but the line "The return button only works for the person(s) that pressed it" made me think more than one person can push it as long as its at the same time $\endgroup$
    – Karm
    Commented Sep 23, 2019 at 13:26
  • $\begingroup$ @Karm if multiple people press it at the same time, the same happens as when multiple people press it at different times. They will not make it back $\endgroup$
    – P1storius
    Commented Sep 23, 2019 at 13:48
  • $\begingroup$ @Karm there is a flaw in your reasoning. Let's say, for example, that I rot13(syvc n pbva naq bayl chfu gur ohggba vs vg pbzrf hc urnqf.) Assuming all the copies behave identically, we will have arrived at a state where not all of the copies end up pressing the button. $\endgroup$
    – hdsdv
    Commented Sep 23, 2019 at 21:59
  • 1
    $\begingroup$ @Karm no, because there's no reason to think that all copies will land in identical, or even similar environments. Including net gravitational force, atmospheric conditions, what surface they're standing on, et cetera. And even so, all I have to do is pick some type of object that I can see (stars, e.g., although each of us could conceivably pick whatever is handy) and count them. Since we are all in different places and times, we will spend more or less time counting and no longer be in sync. As for your second question - see my response to William below. $\endgroup$
    – hdsdv
    Commented Sep 24, 2019 at 8:16


you were sent to different times and you can see the year in which each clone was transported to,

then you could:

create an order based on the year you traveled to. The clone transported to the earliest year would press the button first, followed a minute later by the second earliest year, etc.

  • 2
    $\begingroup$ How would the clones know who was in the earliest year? They can't communicate. $\endgroup$
    – Jens
    Commented Sep 23, 2019 at 23:15
  • $\begingroup$ And one clone could decide to use the latest year. $\endgroup$
    – Florian F
    Commented Sep 25, 2021 at 8:30

Here is my answer:

It's a solid Shmaybe.

Wait wait wait!! Don't downvote yet!! The explanation is waaaaaaay more fun:

Consider the following: If there was a single strategy which could be shown to have the highest probability of getting a single guy home, and all time-travelers independently decide that they will follow that strategy even if it means that they will not get home, then there is a non-zero likelihood of someone getting home. Presumably, that is the rational thing to do, as they give the greatest chance of anyone getting home, and that one person, considering that they are smart enough to build a time machine, might be able to rescue all the clones.

So what strategy can one employ? Simple logic shows you that you must introduce probability. If you didn't, all other clones, being equally rational, would arrive at the same decision, so either everyone or no one would press the button.

The classic solution is to :

Have each of the 50 clones give themselves a one-in-fifty chance of pressing the button. The chance that exactly one of them presses the button is roughly ${{49} \over {50}}^{49},$ or $0.37160171437$%. As the total number of clones increases, the chance of one getting home gets closer and closer to $1/e$. However, this still leaves a $1-({49 \over 50} )^{50} = 1 - (49 / 50) ^ 50 = 0.63583031991$% chance of not a single person having pressed to button. Unfortunately, that presents us with the same problem. Considering that it is likely that no one has pushed it, the logical thing to do is to "cheat" and push it, but then every one would do that, and then the chances of anyone getting home are 0.

  • 1
    $\begingroup$ The logical thing is not to cheat, since doing so guarantees failure. The logical thing is to play the odds, which maximizes your own odds (since all the other copies will behave in the same way as you do). There are epistemological/metaphysical questions about causality, but certainly if you choose to cheat, you are guaranteeing your failure. While choosing not to cheat creates a nonzero probability of success. $\endgroup$
    – hdsdv
    Commented Sep 24, 2019 at 4:51

I'm borrowing from @Bass's answer. However this solution focuses on getting all mes back to my time. I'm not sure I'd wanna see myself and 49 at that, but here goes:

1. Divide the remaining time into 50 slots
2. Since the OP gives the opportunity of all the necessary apps on phone, I'm assuming there is a Random Number generator app.
3. I use the app to generate a random number between 1-50.
4. Press the button when the generated number matches the currenttime-interval-number.

So, I'm hopeless with statistical probability and have no idea about the exact numbers.


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