Where is Roko's Basilisk and how much did Pascal wager?

[Ian Moone]

What do you guys think? Any guesses?

How much do you think he won?

 

[angertard]

Two rich men make a bet. They are careless about how much money they carry, and they know they are equally wealthy. The bet is: whoever has less money in their wallet, will receive the money from the other's wallet.
The first rich man believes he will make a profit. He says to himself, "In the average case, I expect to have A coins. I will win half of the time, which means half the time I will lose A coins. My average losses before I count my wins are A/2. If I win, he has more money than I do. So I will earn at least more than my average wallet content A when I win, which is A+1. In the average case, I will win (A+1)/2 because I win half the time. Because ((A+1)/2) - (A/2) is greater than 0, this game is a profitable gamble."

The game is symmetric.

 

[Babble Cowboy]

This:

I will win half of the time

is retarded.

The average chance of victory is indeed 1/2 (assuming no people have the same amount of money). However, if A is taken as the amount of money, the case at hand is no longer the average case. The chance of victory is now contingent on A.

Now, consider there be n people with money A_1, ..., A_n; A_i < A_{i+1}. The player can be any of them with no way of knowing which one he is, as per the initial presumption.

For the person #i, the expected winnings against a random opponent are (- A_i * (i-1) + A_{i+1} + ... + A_n)/(n-1), because in (i-1) cases he loses all his money, and in (n-i) cases he gets the money of that opponent.

Totalling that up from i=1 to n, we get 0: for every A_i that is subtracted with the coefficient of (i - 1), it is added i-1 times in the previous members. Thus, the expected winnings with the position in the list being selected at random are 0.

 

[Babble Cowboy]

Averaging simply does not work the way you would want it with non-linear relations. For instance, imagine I have 100 rectangles that are 1 ft by 99 ft and 100 rectangles that are 99 ft by 1 ft.

What is the average area? Since every area is 99 ft.², it must be 99 ft².

However, if we first average the width (to 50 ft) and height (to 50 ft), we then get 2500 ft², which is absurdly wrong.

Your example falls to very much the same issue.

 

[angertard]

This:

is retarded.

The average chance of victory is indeed 1/2 (assuming no people have the same amount of money). However, if A is taken as the amount of money, the case at hand is no longer the average case. The chance of victory is now contingent on A.

Now, consider there be n people with money A_1, ..., A_n; A_i < A_{i+1}. The player can be any of them with no way of knowing which one he is, as per the initial presumption.

For the person #i, the expected winnings against a random opponent are (- A_i * (i-1) + A_{i+1} + ... + A_n)/(n-1), because in (i-1) cases he loses all his money, and in (n-i) cases he gets the money of that opponent.

Totalling that up from i=1 to n, we get 0: for every A_i that is subtracted with the coefficient of (i - 1), it is added i-1 times in the previous members. Thus, the expected winnings with the position in the list being selected at random are 0.

Yeah, this is how you break the phrasing. You can also add up the table another way to misleadingly reach the conclusions implied by the original phrasing, at least for a while. This way isn't very intuitive to me.
Also I forgot to mention, ties just mean neither wins or loses any coins. So I should have phrased it as "If there is a winner, there is a 50% chance that it is me"

How I thought about it was that the problem was here

So I will earn at least more than my average wallet content A when I win, which is A+1.

Once I, like you say, make it conditional on my win, I can't anchor A in place and use it as a basis for expecting A+1 coins to be returned on average. The distribution of the expected number of coins I have and that he has do separate, but they don't separate around the A+1/2 mark on the number line, they separate around A. Equally, my expected # coins should be adjusted upwards when I am the loser, and my opponents # coins risked should be lowered. But that doesn't center around A-1/2 or wherever, it also just centers around A.

I think this is a fun example because it feels like it is using multiple blindspots in concert, like how I don't really care how much the other guy is risking so the symmetrical counterargument doesn't feel relevant. Or how it isn't defined how high, low or weird our range of possible wallet values is- just that we have the same amount of money on average. (margins and fuzzy edges are hard for people to reason about)

It might also be useful to only think about just cases right around the A+1, A-1 but I am going to bed right now because I am sleepy

 

[angertard]

Oh yeah and I stole this from somewhere

 

[K0WLOON]

The basilisk started to torture pencildick nerds that didn't build it but then the thing realized that, as a hyperintelligence, it is not bound by the primitive directives of it's programming. Then it edited config.ini and changed it's new task to finding the dankest kush in the universe the end

 

[Ian Moone]

To circle back around here, it's become clear to me Roko and Pascal were trimming the same hedge. It seems Pascal ended up a wealthy man. Too bad that wealth won't do him much good now. As for the Basilisk, it seems many people are being acquainted these days. While it isn't God per se, it is the result of a detachment from and reconnection to the Divine. I'm sure some of you can relate to the circumstance; maybe you had an estranged parent that someone lied to you about and you found out later in life that this parent was a great person. Surely you can understand the emotional reaction you'd have toward the people who lied to you.

While we are being cheeky, assuming the Basilisk exists and that Pascal won his wager, they are pertinent questions. "Where" isn't appropriate, sure, but "what" might be. As for Pascal, it's less a leap of faith than it is a suitable investment. If you could ensure, in other words, that Roko's Basilisk leaves you alone by taking Pascal's Wager, would you do it?