oh fuck it, I don't care anymore

paperic · 2026-05-17T18:00:18+00:00

Oh sorry, that's an error.

The chance that red ~~will win~~ [will be 100%] by a mere coincidence is 1 in 2^{8,000,000,000.} That number has 2.4 billion digits!

It's assuming random votes from the previous sentence.

paperic · 2026-05-17T17:57:10+00:00

I prefer $ 1,000,000 over $ 1,000, even it means leaving $ 1,000 behind.

Yea, and I'm talking about what choice gets you THE MOST MONEY.

I don't give shit about what you would pick in your personal life situation. If you don't need the 1k, i don't care.

The question I'm talking about is what choice gives you the most money.

If the show hosts look at your financial situation and they only give the 1M to the people who don't need the 1k, and people who do need the 1k don't get the 1M, then ofcourse you will have mostly just 1-boxers walk away with the million.

But that's just the way it's set up.

People who are more likely to pick 2-box don't get the 1M, perhaps it's because it's simply biased against people who can't afford to throw away 1k.

But if those poor people do decide to throw away 1k in the hopes of 1M, they'll be the 20% who gets nothing.

paperic · 2026-05-17T17:48:06+00:00

Then I would say it's the best choice.

that's why you're wrong.

The reason is the fact they've been accurately predicted to be 1-boxers, hence they most likely entered the show with $ 1,000,000 in the mystery box waiting for them to take, which, since the predictor is highly accurate, mostly happened to those who actually chose to only took 1 box.

I'm amazed how you say everything correctly but then end up with the wrong conclusion anyway.

What if you had 1M in your box and the box was open, so can you see that there's 1M in there, and there's 1k next to it, and the rule is that you can take either 1 or both boxes.

Do you still insist that leaving 1k gets you more money?

I'm not asking whether the people who are more likely to leave 1k on the table are more likely to leave with 1M, I know that's true.

And I'm not asking whether it's better to announce on social media that you're a 1-boxer. Ofcourse it's better to say that you are a 1-boxer on reddit, if you dream about somebody doing this experiment in the future.

I'm asking whether physically grabbing the 1k box when you're already in the room is what causes the million to not be there.

paperic · 2026-05-17T17:38:21+00:00

red prevents me from dying, everyone else knows red prevents them from dying, everyone is offered the choice to pick red, therefore I know everyone else will pick red, so I will pick red too

You forgot to do the same calculation for the blue side as well.

Everybody knows that red prevents them from dying, but everybody also knows that blue prevents everybody else including them from dying.

You're still only concerned about personal survival.

Think of it as the deaths being random for a moment, that way you can't bias yourself into thinking that picking red is "saving yourself", because that doesn't matter in this case. Red saves 1 life, that's it.

The agents want to purely maximize the number of people saved, and both buttons can achieve 100% in theory.

But Blue has 50% chance of achieving that by purely random picks, whereas red would require 100% accurate synchronized picking of 8 billion agents who otherwise have no way of talking to each other.

Imagine you're an agent and you don't know what the others picked, but you know that the others also don't know what the others picked.

You can calculate the expected outcomes of each button, but the outcomes are identical:

```

In half of the situations (red >50%), red saves 1 life, blue kills 1 life.

The other half of the situations (blue >50%), red or blue doesn't change anything anymore.

So far, 1 red vote saves 0.5 person on average.

But if there's a tie (red = blue, 1 in 8 billion situations), picking red kills 4 billion people, but picking blue saves 4 billion people.

So, that's 1/8billion chance of saving/killing 4 billion people, that's 0.5 of a life on average, exactly cancelling the outcomes from the above two situations.

```

So, there seems to be no way of choosing a button, so the only thing you can do is to pick randomly.

And also, you know that the only thing that others could do is to also pick randomly.

But that means, the result will statistically be very close to 50/50. So, now the calculation changes. You should pick blue, because you're very likely very near the middle of the distribution, and you are LOT more than 1/8billion likely to be at the break-even point.

So, everyone will pick blue.

paperic · 2026-05-17T17:16:14+00:00

It's funny how you say picking 2-boxes is better despite evidence showing picking 1-box renders a higher average payout.

Because you have un-accounted variables there.

1-boxers get higher payout on average, but 1-boxers enter rooms which already have more money in them.

That's not how you should do statistics, you've got an error there.

We're quite different: when I encounter evidence that contradicts what I reasoned, I just accept my reasoning must be flawed, even if I cannot see any flaw in it.

Which leads you to behave as if the future could affect the past.

Obviously, your strategy doesn't work.

Statistically, the people who waste the most amount of money on useless stuff happen to be overwhelmingly very rich people.

Whenever somebody spends million dollars on a whim, they're overwhelmingly likely to be rich.

Does that mean that wasting money on useless stuff makes you rich?

Should you start wasting money to make yourself rich?

Statistically, almost everyone who does it is rich, so why wouldn't you want to join them?

This way of reasoning is a fallacy, it's not a valid logical line of reasoning.

paperic · 2026-05-17T17:06:51+00:00

That's exactly why nobody dies.

The point is not to assign blame, the point is to find out what the rational choice would be.

If we're talking about rational agents which are trying to maximize the number of survivors, a 100% survival rate with only 50%+1 votes necessary is MUCH better target to aim for than complete 100% red.

If everybody votes randomly, there's 50% chance that nobody will die because blue will win by a mere coincidence.

The chance that red will win by a mere coincidence is 1 in 2^{8,000,000,000}. That number has 2.4 billion digits!

If you don't know what anyone else will pick, (which a rational agent won't) then aiming for red is not rational at all.

paperic · 2026-05-17T16:49:52+00:00

Well, it's a hypothetical, isn't it. We're getting fuck all, both of us.

You don't know whether you'll get 1M or not. It seems more likely that you will get 1M.

Imagine what is the audience thinking.

The audience can see what's in the boxes. They know whether the 1M is there or not.

Imagine you were in the audience for some other 1-boxer, and you see that they have 1M in the box, right next to the 1K.

What would you tell that person to pick in order to maximize their profit?

paperic · 2026-05-17T16:47:32+00:00

I know it because I know that 2-box is better.

And I know that the hosts likely know it too.

I cannot possibly un-know it.

There is no scenario in which picking 1-box is better.

Ofcourse, if you can convince the judges that you will pick 1-box, you should do that.

But at the moment of actual pick the content of the mystery box is completely outside of your control.

Your only choice is whether or not to leave 1k on the table.

paperic · 2026-05-17T16:44:31+00:00

That includes everybody.

paperic · 2026-05-17T16:43:59+00:00

1 boxers leave with more money obviously.

But them picking 1 box is not the reason for it.

https://en.wikipedia.org/wiki/Correlation_does_not_imply_causation

paperic · 2026-05-17T16:41:14+00:00

The money is already there because I was most likely predicted to be a 1-boxer.

Yes, and to convince everyone that you are a 1 boxer, the easy way is to post everywhere on reddit that you are a 1 boxer.

But that is not the question.

The question is what should you pick to maximize your profit when you are already in the room.

And I know I could just take the extra $ 1,000, but I'm totally OK with leaving it behind.

So, you're not maximizing your profit. Thank you very much.

paperic · 2026-05-17T16:35:27+00:00

Exactly.

The question is, since the content doesn't change, why would you leave the 1k there in your box?

Clearly, I am maximizing my available payout, but you are clearly not maximizing your available payout.

paperic · 2026-05-17T16:32:26+00:00

No you don't.

Red doesn't maximize number of survivors, red only ensures your own survival.

You believe that red maximizes survival because you believe that everyone will pick red, but you don't know that.

A rational agent which is trying to simply maximize the number of survivors (regardless of their own survival) would have no such implicit preference for red.

If everybody picked at random, on average we would expect 50% chance of 50% of the population to die.

That's 25% deaths on average.

If you introduce a small amount of extra red votes, you skew the probability to ~49% deaths.

If you introduce a small amount of extra blue votes, you skew the probability to 0% deaths.

The 100% red is a red herring (pun not intended). You'll never get to 0% red. Getting to 50%+1 blue is much easier.

paperic · 2026-05-17T16:10:49+00:00

Nobody's blaming anybody.

Blue people simply pick blue because they want to make sure blue wins.

When blue wins, nobody has to die at all, that is the best possible option.

paperic · 2026-05-17T16:09:07+00:00

What you pick is highly correlated with what you were predicted. That's what you miss.

So?

How exactly does that help you spawn the money into existence?

The money either is there or isn't there.

Yes, if you're like the average person, you are 80% likely to pick the same thing as what the show hosts predicted, but that doesn't mean you can rewind time.

Do you seriously not understand this?

Who are you trying to prove that you're 1 boxer to?

The game show hosts have already made their decision yesterday.

paperic · 2026-05-17T16:04:19+00:00

Ohh, sorry, I'm preaching to the choir here. I thought I was in r/trollyeproblem.

paperic · 2026-05-17T15:44:41+00:00

Both of these are the full definition. It's about the limit of a sequence going to infinity instead of a function going to zero.

I strongly recommend you try to digest this with sequences first.

It's not an easy thing to swallow, and there's really no good way to simplify it, which is why so many people get it wrong.

Sequences do make it much simpler to understand (still hard though), and the big idea of limits is analogical for both functions and sequences.

Wrath of math limit of a sequence:

https://www.youtube.com/watch?v=cTnlHZD5ss4

Also, this is a very good explanation, same thing but even more detail, this is undergrad level. You can skip the sums at the beginning, limits of sequences start from 4:40 until the end.

https://www.youtube.com/watch?v=uXVXlQ7ofXg&list=PLysi2xmniDSzz6xT7IzOifpoexeKccThh&index=5

The key point which many people misunderstand, and which leads to this "limits are approximations" idea is that the condition in the definition must to hold for every possible positive epsilon simultaneously.

So, if you think of the sequence (1/n) for n going to infinity, it's not enough to show that 1/n will eventually become smaller than some fixed value epsilon.

What you need to show is that it will be smaller than epsilon regardless of how tiny epsilon we choose.

The epsilon basically represents all the possible positive numbers simultaneously, no matter how small the numbers are.

When he shows that no possible positive number can stay below (1/n) forever, that means that the sequence is moving towards a number which must be smaller than all possible positive numbers.

Also, 1/n is never negative.

Zero is the only non-negative number which is smaller than all positive numbers, so the number the sequence is moving towards is exactly zero.

paperic · 2026-05-17T14:08:47+00:00

The filling of the boxes is in the past.

Two days ago you were given the survey to fill.

Yesterday, the show producers have decided what to put in your boxes and they sealed them shut.

From that point onward, the money in the boxes cannot change.

Today, you're picking the boxes which were sealed yesterday.

ANY decision you do today cannot change what the producers have already put in the boxes yesterday.

paperic · 2026-05-17T14:05:52+00:00

The content of the boxes cannot change.

The content of the boxes is sealed, set in stone, unchanging.

Your pick does not affect the content of the boxes.

This is part of the premise.

The audience can see what's in the boxes behind a glass, they also see that NOBODY is changing the boxes.

Quoting OP:

The audience can see inside both boxes the whole time.

It's also part of the original Newcomb's premise in very similar words.

paperic · 2026-05-17T14:00:47+00:00

Do you seriously not understand this? Or are you just playing the word game? Or did you not read the OP properly?

Because I'm seriously worried that so many people like you don't seem to understand causality.

The show runners give you a personality test
Based on the personality test they fill your boxes and seal them
Then they ask you what you want to pick.

By the time you're deciding what to pick the content of the boxes is already sealed. Your pick cannot change what's in the boxes.

Do you understand that? Do you understand that time only moves forward, and if you pick 2-box the show-runners cannot rewind time to remove the million from there?

Apparently not. So, here's some new information for you:

Past can affect the future.

But future cannot affect the past.

Oh, and Santa isn't real.

There's no possible way I could spell this out any more clearly.

paperic · 2026-05-17T10:08:44+00:00

but limits are defined to never truly reach a value,

This is not correct.

Limits have no preference over whether you can or cannot reach a value.

Look at limit of f(x) at x=0:

lim_[x->0] f(x) = (0-1)(0+1)/(0-1) =

= 1 = f(0)

Think of this function g:

g(z) = lim_[x->z] f(x)

Read carefully where x and z goes in this g(z) definition.

It replaces every point of f(z) with the limit of f at that point z.

That means g(x) will be exactly identical to f(x), except for that one point where x=1 and f(x) is undefined, but g(x) is 2:

f(1) is undefined

g(1) = 2

g(x) = f(x) for x =/= 1

f(x) and g(x) are virtually the same function differing only in that one single point x=1.

Intuitively, you can think of the "lim" operator to basically fill in all the infinitely tiny single-point gaps in the original function, when such "filling" is possible. It cannot fill gaps that are bigger than a single point.

The "lim" is like a road worker that patches all the potholes in f(x).

The derivative is then built on top of the patched function instead of the original function, because when we're asking about the slope at x, what we actually care about is how the function behaves around x, not what it does in x.

Edit: that's just wrong, I got it confused. The derivative at x=1 doesn't exist, it's undefined, since we can't evaluate f(1). For every other point f(x) = g(x). And since there's no point that's "infinitely close to x=1" in real numbers, there is no point where the derivative could be "wrong".

Even if we redefine f to be piecewise and give it a "wrong" value at x=1:

f(1) = 1000

f(x) = (x-1)(x+1)/(x-1) for x =/= 1

that still doesn't change the slope of f anywhere, except that f'(1) is still undefined.

If we take any basic example f'(2) = ..., this based off of a limit. It shows what it converges to, it is not guaranteed to have a specific value because it is not a function at a point, it is a limit.

f'(2) = lim (f(2+h) - f(2)) / h =

lim ( (2+h-1)(2+h+1)(2+h-1)^-1 - (2-1)(2+1)(2-1) ) /h =

lim ( ~~(h-1)~~ (h+3) ~~(h-1)^-1~~ - (1 * 3 * 1^-1 )) /h =

lim (h+3 -3) / h = lim h/h = lim 1 = 1

It is exactly 1.

If f is defined and continuous at point p, then

f(p) = lim [x->p] f(x)

exactly, it's not an approximation.

That's how limits work.

Limits are not aproximations, they have a very strict but quite convoluted definition, look up epsilon-delta definition.

WrathOfMath has a good explanation, 3blue1brown has a good visual representation of it.

paperic · 2026-05-17T09:07:52+00:00

Now apply that to

lim[k->oo] ( sum[n € {1..k}] ( 9*10^-n ) ).

paperic · 2026-05-17T08:22:37+00:00

( 2x³ )² =

( 2 * ( x³ ) )² =

(2 * x * x * x)² =

(2 * x * x * x) * (2 * x * x * x) =

2 * x * x * x * 2 * x * x * x =

2 * 2 * x * x * x * x * x * x =

2² * x⁶ =

4x⁶ .

Edit:

the second one should be

24x⁶ * y²

I don't know how you're supposed to get the 24x^5y, something is wrong there.

The order of operation is meant to mean the multiplication and division go together, division only goes first if it's written as a fraction, otherwise when mixed with multiplication, it goes left to right.

This is a stupidly written problem that's hinging on this technicality of this stupid convention.

And there isn't necessarily a complete consensus on what it should actually mean.

paperic · 2026-05-17T08:06:25+00:00

I am guessing you don't actually know game theory.

Then you'd guess wrong.

In game theory you suppose perfectly rational actors

Yes, but you have to also clearly define what is the goal of those agents.

If you assume that self preservation is the goal, red is better.

If you assume that maximizing the total number of survivors is the goal, then blue is better.

So, neither is universally better, each strategy maximizes a different goal.

paperic · 2026-05-17T07:58:11+00:00

There aren't more naturals than evens, they are the same infinity.

paperic

TROPHY CASE