If the trolley is moving at light speed by the time it hits the station, it is impossible for anyone to get on or off because—from the trolley’s perspective—no time passes between stops. Ergo, the number of passengers on it must be the same every stop.

If the initial number of passengers is odd or a non-zero integer, this inability to board/unboard would contradict the rules.

Thus, in order to satisfy all the conditions, the initial number of people on the trolley must be 0. As an even number it will be subject to halving, but 0/2=0, so the rules are satisfied.

Hence, pulling the lever is the optimal solution as 0 people will die. QED.

Motherfucker here making a math problem for infinite series from a philosophical question. Is nothing sacred /s

After an infinite number of loops?

After an infinite number of loops I'd want to be killed.

There are two super interesting problems in here.

One is: would you bet human lives on a conjecture being true? The collatz conjecture does hold for every number we have tried, but there have been conjectures that were disproven with a very large counterexample. You could kill countless humans if wrong, so even if you think the chance of a counterexample is low, is it low enough to outweigh that potentially very hight value counterexample?

The second one is: Let's say the collatz conjecture holds, and the number of passsengers just loops 4 -> 2 -> 1 -> 4 -> 2 -> 1 eventually. What is the 'final' number, when the trolley is done with the infinite loops? It can't be 1, because that is always followed by a 4. And it can't be 4 because it's always followed by 2 and so on. But it has to be one of those, because any other number is not possible. It reminds me of the Vsauce Video Supertasks, which comes to the conclusion that we can't know the answer to these type of questions.

So in conclusion, flipping the switch will either give you an arbitrarily large number of deaths, or an unknown number of deaths. Fun!

The conjecture has been checked by computer for all starting values up to 2^71 ≈ 2.36×10^21.

So you're probably good to go.

If the number of passengers is unknown, we can't guarantee it is strictly positive or even an integer.

Sure, accidentally killing lim {-2,-1,-2,-1...} people is good compared to lim {4,2,1,...} or 5, but only if it's an integer that's a 50/50 on being negative. If it's not an integer, which is infinitely more likely if we truly have nothing to go off and have to assume it was randomly chosen, then reality might break upon reaching the station.

This one is easy.

Either you kill 5 people, or you pull the lever and kill none. Because an infinite loop never ends.

All the rest is just fluff with no bearing on the question.

@mech

In the amount of time it takes to describe the situation, I could go free the people tied to the tracks.

It only accelerates to light speed, therefore it will need infinite time to complete the loops. Thus the risk is not the killing but getting stuck.

If the conjecture holds, naturally there is a small cycle so people can get on and off and use the train as a form of teleporting to the future.

If there are different loops, then still people can take turns.

Even if there are values that diverge, if it can be shown that at least one event of division occurs with a certain average frequency in the infinite divergence, then at any such point all previous guests can exit and the train can be ridden for one such span.

Only if there are no cases of division and endless steps of 3n+1 in the limit, would people be trapped on the train at no subjective time passing, and in essence time travel into the infinitely far future where they are killed.

Passengers: 4 > 2 > 1 > 4

So that's better, I guess?

I’d be too confused to make a decision

how fast is it going if i dont pull it? what kind of smear are we looking at?

2n+1 is not in the Collatz conjecture.

Mathematics is not ready for such carelessness.

And I did a dumb. Withdrawn.

No matter what wouldn't this grow to infinite passengers? Is that supposed to be the point?

Because any even number is going to halve itself down to ~~1, which is odd~~ an odd number, then double plus one will always make another odd number so it would grow to infinity.

Edit: Misread the problem, read replies for explanations

I am too tired to think too hard about this, so, SURE!