Using code for Newcomb's problem

Logic & Philosophy of Mathematics
21

Neither 2 nor 3 is assumed by anybody. The box is opened by everybody. There isn’t an option not to. The choice is to take the contents of the transparent box or not.

Our known physics does not support an infallible predictor. It’s pretty trivial to thwart the prediction if the goal is to do so instead of a goal of maximizing utility. So an infallible predictor indicates that known physics is wrong.

If 1 is not assumed, then you get into a case of trickery, where the box contents gets its contents teleported in or something, or it’s some kind of superposition that is entangled with the eventual choice, which again, violates know physics.

Not at all. I said it was my opinion, whereas the fallacy would require a conclusion of my opinion being correct, as you correctly point out with the Monty Hall thing. No, I’m simply stating that my opinion is that most people want to maximize their expected utility.
You stand out, advocating instead to ‘get more money than one random person making the opposite selection’. I’ve never met anybody else defending such a stance.

But it is a function of this knowledge.

A trivial truth table indicates otherwise:
Observed, Advice
Box w/ 1M 2B
Box empty 2B

If your friend ignores this knowledge, then his advice is functionally equivalent to that of a blind person with no knowledge.

Yes, which is why I said ‘might as well not have looked’.
He doesn’t use the knowledge, so there’s no point in acquiring it, even though it is asserted that he does, and therefore is not blind.

In particular St. Petersburg paradox

I actually didn’t get that one. The utility of the situation seems to be 2, and the problem includes a singularity, rendering the mathematics invalid. They claim that it has unbounded utility, and I don’t see that.
Maybe I need more time to find something I’m not seeing.

1000000, 1000, 99%

The probability that you get $1,000 given that you chose two-box is explicitly stated in Newcomb’s problem’s formulation.

You didn’t say ‘given that you chose two-box’. I put your quote back to show what you actually asked. There’s 0% of me getting 1000 because I don’t 2B in the original scenario.

That is correct. But you don’t seem to understand that a program can give you multiple results. It is up to you to decide what to do with that information.

OK, challenge accepted. You need at least a dollar to survive. There’s 1001 and 1000 in the boxes, 75% predictor accuracy.
Your opinion is that 1001>1000 and that happens 56% of the time, so you 1B. But that’s a 25% chance of death since the one box might be empty. Meanwhile, the 2B guy gets 1000 unconditionally, and a 25% chance of twice that.
Do you ignore this challenge?

I didn’t ask to put the $1 needed to survive clause in, but it hardly helps the 1B justification.

I did not address Simpson’s paradox because I did not find it relevant, and I did not agree with your assertions about it. I forget where you brought it up.

I cannot follow the code. It has zero comments, and I don’t know what the output is trying to convey. So far more context is needed for me to parse post 20.

Simpson’s paradox had and example of females being accepted at a lower rate at some university when in fact they’re accepted at higher rates than males in every department.

22

If we were only allowed to repeat the views of other people no progress would ever be made.

A trivial table hides Simpson’s paradox as well, which is a mathematical fact I already showed you in code.

That is precisely the point: the mathematics of expected value is wrong. The mathematics of expected value assumes an infinite number of trials, if in your opinion infinities are invalid, then the whole approach of expected value is invalid.

But we don’t need to use mathematics or infinities to see the problem, we can see it in code:

payoffs <- 2^(rgeom(1e6, 0.5) + 1)
mean(payoffs)
median(payoffs)

In this case the number of trials isn’t infinite, so the mean isn’t infinite, but it’s much bigger than the median.

If the mean is 23.5315, that doesn’t mean you should expect to obtain that value, the median is 2, which means 50% of trials result in $2. It’s obvious why.

That is the same reason why you shouldn’t expect to have the mean income of a country.

“Expected” value is a misnomer.

By you I obviously meant the plural you: a rational agent.

But fine, if you want to play that game, let’s say you are Bob, and the probability that you choose one-box is 100%. On the other hand the probability that Alice chooses two-box is 100%.

With that reframing, the questions are:

  1. What is the probability that Bob gets 1000000?
  2. What is the probability that Alice gets 1000?

The probability that Bob gets 1000000 is 99%, by complement the probability that Bob gets 0 is 1%. The probability that Alice gets 1000 is 99%, by complement the probability that Alice gets 1001000 is 1%.

The probability that Bob gets 990000 is zero.

That is not my challenge.

That “challenge” has absolutely nothing to do with what I said. And how am I supposed to address this “challenge”?

For the record, this is my challenge, which you completely ignored:

Nothing here assumes you need $1 dollar to survive, it’s a straight up expected value calculation, is it not?

They are not my assertions, they are the assertions of Edward H. Simpson. This is straight from his paper:

Male Female
Untreated Treated Untreated Treated
Alive 4/52 8/52 2/52 12/52
Dead 3/52 5/52 3/52 15/52

“This time we say that there is a positive association between treatment and survival both among males and among females; but if we combine the tables we again find that there is no association between treatment and survival in the combined population.”

If we focus on the left 2 \times 2 table (males), the survival of treated is 8/5, the survival of untreated is 4/3, so treated > untreated. If we focus on the right 2 \times 2 table (females), the survival of treated is 12/14, the survival of untreated is 2/3, once again treated > untreated.

But if we combine both it’s:

\frac{8+12}{5+15} = \frac{4+2}{3+3} \implies \frac{20}{20} = \frac{6}{6}

No association.

Do you not agree with mathematics?

What is there to follow?

counts <- array(
  data = c(4, 3, 8, 5, 2, 3, 12, 15),
  dim = c(2, 2, 2),
  dimnames = list(
    outcome = c('A', '!A'),
    treatment = c('B', '!B'),
    group = c('C', '!C')
  )
)

This is a 2 \times 2 \times 2 table, just like Simpson described:

CB C\bar{B} \bar{C}B \bar{C}\bar{B}
A a c e g
\bar{A} b d f h

We can easily obtain a 2 \times 2 slice of the C group:

> counts[,,'C']
       treatment
outcome B !B
     A  4  8
     !A 3  5

And a 2 \times 2 slice of the \bar{C} group:

> counts[,,'!C']
       treatment
outcome B !B
     A  2 12
     !A 3 15

Crucially, we can also marginalize over C:

> counts[,,'C'] + counts[,,'!C']
       treatment
outcome B !B
     A  6 20
     !A 6 20

The important metric is “association between treatment and survival”. Treatment is the second dimension (B), and survival is the ratio of alive to dead (A/\bar{A}). Therefore the metric is AB/\bar{A}B - A\bar{B}/\bar{A}\bar{B}.

In code:

assoc <- function(m) m['A','B'] / m['!A','B'] - m['A','!B'] / m['!A','!B']

We can now obtain the association for the two C groups:

> assoc(counts[,,'C'])
[1] -0.2666667
> assoc(counts[,,'!C'])
[1] -0.1333333

Both are negative.

But if we marginalize over C:

> assoc(counts[,,'C'] + counts[,,'!C'])
[1] 0

No association.

What exactly do you need explained?

I’ve simplified the code so hopefully you do understand.

So in every department males are accepted at a lower rate. According to your logic “the department doesn’t matter”, because every department slice gives the same result.

But when you do not slice by department, males are accepted at a higher rate.

Therefore your logic is wrong: slicing by department does matter.

23

I looked closer at St. Petersburg paradox. My mistake was presuming one lost money if tails comes up. No, it’s all gain.

But it isn’t. Mathematically, there is no bound to the initial bet that cannot be recouped given sufficient number of plays. Nobody would actually bet said arbitrarily high amount because the payoff isn’t enough to be worth the effort required. So the only paradox seems to arise from taking a pure mathematical situation and applying it to real life where payoff per hour matters, and diminishing marginal utility matters.

That means that expected utility calculations are valid for mathematical analysis, but not for real life where humans are not capable of flipping billions of coins for negligible payoff.

It’s one of the reasons I dumbed down the Newcomb thing to 1001/1000 in the two boxes. With those values, diminishing marginal utility is negligible.
The game is played but once, so payoff per minute is not an issue.

I’ll accept this, but I also don’t see why what Alice does or gets matters in the slightest to Bob. You put a lot of weight into this specific comparison, something I’ve not seen justified.

With that reframing, the questions are:

  1. What is the probability that Bob gets 1000000?
  2. What is the probability that Alice gets 1000?

Given your reframing, the answer to both is the (unstated) accuracy rate of the predictor. I presume you mean 99% if not otherwise stated.

The probability that Bob gets 990000 is zero.

Agree. I’ve never said it was nonzero. That value represents his expected utility, not his expected outcome, which is probably 1M but maybe zilch.

By utilizing the reasoning you’ve been pushing for this entire thread. Challenge apparently declined. You decline to say what choice should be made, and if 1B is the wrong choice, and your program correctly says 1B will probably yield more money than Alice (specifically just Alice) gets, then why is the program a useful tool? Why can’t the program simply output what should be chosen (1B or 2B)?

For the record, this is my challenge, which you completely ignored:

That which you labeled as a challenge was one concerning a player who needs at least $1 to survive. And yet even in the original scenario, he makes a choice that comes with a 1% chance of death as opposed to certain survival.
But OK, a different challenge, one I apparently missed.

Sorry, but what am I doing this to? The Newcomb thing doesn’t come with a ticket price at all. It’s free money.

According to your expected utility I should buy the ticket no matter what

With numbers that large, diminishing marginal utility plays a far more significant role. Unless I want to buy a country (mine is certainly up for sale), I as a normal human would find little more utility in a trillion than a million.

I would have to sell some assets and ask for a loan. 99 times out of 100 I would end up with nothing except debt.

From this I presume a million dollar lottery ticket with 1% chance of winning a trillion. I agree, I would not play that.

In what example of mine (except the need $1 to survive one, which was your idea) is diminishing marginal utility relevant?

Simpson’s paradox:

I agree with all the above. I just don’t see the relevance to the Newcomb thing or any of the variants we’ve explored. Are you trying to conclude that mathematics is faulty? It isn’t. There were no mistakes above.

Just a question. Why are all the figures expressed as /52? Yes, there are 52 subjects (20 male, 32 female), but wording it like that makes it sound like 4 of 52 untreated males survived, 3 of those 52 untreated males died, and the other 45 untreated males met some mystery fate. I know what it means, but it’s still a funny way to express it.

What is there to follow?

I don’t know R, and I don’t know what is being attempted with this uncommented code. A little context would help. OK, you’re apparently doing something with the Simpson’s data.
That at least helps.

So in every department males are accepted at a lower rate. According to your logic “the department doesn’t matter”, because every department slice gives the same result.

That’s right. One doesn’t have to know which department is being referenced to know that the department accepts female applicants at a higher rate than the acceptance rate of male ones.
The department matters for other things, like say which departments have far higher overall acceptance rates, and which departments have far higher percentages of male or female applicants. Your suggestion that “the department doesn’t matter” is ambiguous since it omits which attribute of the department is being disussed.

But when you do not slice by department, males are accepted at a higher rate.

That’s also correct. Hence the gist of the Simpson’s thing.

Therefore your logic is wrong: slicing by department does matter.

My logic doesn’t deny that slicing by department matters. It just doesn’t matter for the particular question I indicate just above.

24

Because nobody would be dumb enough to blindly trust the expected value mathematical formula.

But it’s exactly the same situation with Newcomb’s problem: you are not going to be playing billions of trials, just one.

In one round Bob gets 1001, Alice gets 1000. That’s it.

So now you moved the goalpost from expected value to diminishing marginal utility. No one is doing that for Newcomb’s problem. Either way you would need to provide an actual formula to calculate that, and the crossover point is not going to be 51%.

It may not matter to Bob, but it matters to a third party analyzing who is more rational.

What is your question? Are you asking me if 100% > 99%? The answer is yes.

That has absolutely nothing to do with my challenge.

Fine:

cat(sprintf('Choose %s\n', ifelse(p^2>=0.5,'1B','2B')))

I don’t understand what’s unclear to you. In Simpson’s paradox this is a slice of males counts[,,'C'], in Newcomb’s problem this is a slice of two-boxers whose prediction was incorrect money_c2[!pred_c2_p2].

In both cases the sliced or stratified table gives you one result. In both cases the unsliced or marginalized table gives you a different result.

Do I really need to provide you with a table side-by-side so you can see it’s exactly the same issue?

Because Simpson wants the sum of all the terms to be 1 in order make comparisons with other problems from this paper and other papers. It doesn’t matter for the main calculation.

Just like the prediction in Newcbom’s matters to know how much money people got.

No it isn’t. We know what we care about is the association, which has a clear formula. For department A the ratio for males is 512 / 313 and the ratio for females is 89 / 19.

We know exactly what is being lost in a ratio: the total amount.

Which is exactly what is being lost in Newcomb’s problem.

In Newcomb’s problem slicing by prediction matters precisely for the very same reason.

25

Any expected human utility calculation should take diminishing marginal utility into account. At 1001 and 1000, it is completely negligible, and no, that’s not it, kind of like Suny saying he expects 1000 more if he two boxes, ‘and that’s it’. You’re both ignoring the rest of the story.

[quote]No one is doing that for Newcomb’s problem. Either way you would need to provide an actual formula to calculate that, and the crossover point is not going to be 51%.
[/quote]
An actual formula is different for each person. For 1M/1K, the crossover point would be higher than 51% for most people.
To quote the St Petersburg site:
“A common utility model, suggested by Daniel Bernoulli, is the U(w) = ln(w) (known as log utility). It is a function of the gambler’s total wealth w, and the concept of diminishing marginal utility of money is built into it. The expected utility hypothesis posits that a utility function exists that provides a good criterion for real people’s behavior; i.e. a function that returns a positive or negative value indicating if the wager is a good gamble. For each possible event, the change in utility ln(wealth after the event) − ln(wealth before the event) will be weighted by the probability of that event occurring.”

it matters to a third party analyzing who is more rational.

Ah, again just like Suny, assigning anybody who takes your stance to be the rational one, rather than addressing the arguments showing the alternative to be better.

Much better. This makes your stance clear, and doesn’t just deflect it away with ‘the tool can be used (or not) in the decision process’.
This code shows that the answer is in no way a function of the money in each of the two boxes. It says that with 1000 in the transparent box, 10 in the opaque box, and 99% prediction record, you should choose 2B.

Sure you want to go with that, or should the code maybe be slightly more complex?

We know what we care about is the association, which has a clear formula. For department A the ratio for males is 512 / 313 and the ratio for females is 89 / 19.

We know exactly what is being lost in a ratio: the total amount.

You put emphasis on ‘association’, but not sure what the word means if it doesn’t mean ‘ratio’ or ‘total amount’. The example in the wiki site showed a false concern about the acceptance rate of males vs females (which is a ratio). I don’t know where ‘association’ fits into that.
That aside, the data on a per dept basis showed no such bias in any dept.

In Newcomb’s problem slicing by prediction matters precisely for the very same reason.

What is ‘slicing by prediction’? Does this mean that perhaps the prediction percentage is different for predicted 1B than it is for predicted 2B? I mean, that’s a real issue. Suppose 99% of all players choose to 1B, and the predictor predicts 1B all the time. It now has a reputation of 99% prediction accuracy, and yet the few that choose 2B all go home with more.

26

No, it’s the exact opposite. Suny begins with his conclusion, I begin with an analysis of actual data.

The question “who is more rational?” in the case of Sunny is a tautology: “assuming two-boxing is more rational, two-boxing is more rational”.

I do not make any assumption.

Not true. If the money in the mystery box is $1, then the probability doesn’t matter, and two-box is always better.

Yes I’m sure. 99% of the time I’ll get $10 rather than $0.

There’s an entire field of economics devoted precisely to the fact that the expected value is misleading: ergodicity economics. I can give you the canonical example where the average of the population is one, but the actual outcome you get is completely different.

Association could mean different things depending on the problem, but in this case it’s very obvious we are talking about:

\frac{a/b}{c/d}

So it’s a ratio of ratios: (512/313)/(89/19).

The acceptance rate is already a ratio. 2/4 and 20/40 is the same rate, but it hides the total number.

But it showed the opposite result when ignoring the department.

That’s why in Simpson’s paper the same table was used for two scenarios. In one conclusion the overall association seemed to be wrong — like in the UC Berkeley example. But that’s why Simpson provided another example with cards where the overall association is the correct conclusion.

The point is not that one conclusion is inherently better than the other, the point is that they are different.

Slice OB TB Winner
C 1000000 1001000 TB
!C 0 1000 TB
Marginal 1000000 1000 OB

Simpson’s paradox says when slicing by group you get one result (TB), and not slicing by group you get another (OB).

I don’t understand why the similarity is not obvious to you.

In Simpson’s paradox when you look at two ratios: 512/313 and 89/19, the later one is bigger, but the structure of the calculation hides the fact that the amount of people in the former is much bigger. And the same is happening in Newcomb’s problem when you compare 1000000 with 1001000: that is hiding the fact that the amount of people in the former group is much bigger.

I explained that in my essay: money_c1[pred_c1_p1] that is a slice by prediction.

27

The code you showed me does not reflect that.

Seems like a subset of the actual data to me.

So you should pick that rather than the option that gets you 1000 99% of the time, rather than 1010? Your code does not assert your comment above, It just says you should 1B in this case, based purely on prediction rate and ignoring the rest of the ‘actual data’.

Expected value is misleading. 10 (maybe 0) being worse than 1000 (maybe 1010) is a myth put out by those that pay attention to ‘expected utility’ theory.

Ah. Gotcha. Thx, except I don’t like the marginal row since it ignores much of the data, and that ignored data matters. Still OB in that case, but for different numbers. The table you show also discards the predictor percentage, which is also relevant, especially considering the fact that your code pays attention to no other bit of data.

The ‘marginal’ row gets super misleading in the 1001/1000/75% case, and you’re never really addressed that case except to double down without justification. There’s negligible diminishing marginal utility with those numbers, so appealing to DMU is fallacious twice over since taking it into account makes your position even worse.

28

Because the code I showed you wasn’t meant for that case.

But the subset is representative of the infinite data. Statistics works because of the nature of probability.

Do you not believe in statistics?

OK, so you changed the scenario.

Any programmer worth his salt would tell you that it’s impossible to write a program that considers 100% of the cases. My program was for Newcomb’s problem. There’s no program that can deal with an infinite number of alternative problems you can conceive.

If you want my program to deal with a certain number of n possible scenarios, you have to list them all.

Marginalizing takes into consideration all the data, by definition.

No it doesn’t. 1000000 and 1000 are there because of the percentage. If the percentage was 1%, then than row would be 0 and 1001000.

But I don’t think you understand Simpson’s paradox at all.

Even if we used your expected value formula, the table would be:

Slice OB TB Winner
C 1000000 1001000 TB
!C 0 1000 TB
Marginal 990000 11000 OB

The conclusion of Simpson’s paradox is the same: slicing gives you one result, not slicing gives you another.

But the most interesting corner case is when the accuracy is 100%, in that case your expected value and my most-likely heuristic return the same:

Slice OB TB Winner
C 1000000 1001000 TB
!C 0 1000 TB
Marginal 1000000 1000 OB

You agree that when p=1 a one-boxer gets 1000000 and a two-boxer gets 1000, correct?

But what does slicing by prediction means in this case? It means supposing Omega got the prediction wrong for a one-boxer — which is impossible — the result would be 1001000. This makes no sense.

But according to you, the two-boxer mindset is valid when p=0.9999999999, just not when p=1. That is totally irrational to me.

No it is not.

Slice OB TB Winner
C 1001 2001 TB
!C 0 1000 TB
Marginal 1001 1000 OB

The conclusion is the same: slicing by prediction gives you one conclusion, not slicing gives you a different one.

That’s Simpson’s paradox.

29

Just Newcomb scenarios, but with variable amounts in each of the two boxes, plus variable prediction accuracy. Ideally we’d also list the player’s net worth. For additional purposes, we’ll limit each variable to the limit of precision of whatever your program uses.

My program was for Newcomb’s problem.

Because the code I showed you wasn’t meant for that case.

If you only handle one specific case, then why not just print ‘OB’ instead of the needless conditional? Then it’s not the program telling you which to choose, it’s just you telling the program what to say, very much like Suny’s ‘rational’ friend that sees the contents.

I agree with that. You didn’t show a table for 1001/1000/75
OK, you do, but only with that ‘marginal’ line that doesn’t in any way represent the difference between 75% and 99% predictions.
Your table with expected utility still labels the line ‘Marginal’ instead of EU. That’s misleading.
Is the below wrong?

Slice OB TB Winner
C 1001 2001 TB
!C 0 1000 TB
EU 751 1250 TB

I never said that, and you know it. If p was that, 1B is the answer for 1001/1000. But I asked about p=.75, and you avoid any analysis besides doubling down on the one most probable data point against one preferred rival.

30

@noAxioms @felipec I just saw the Veritasium video on this. I’m a 1Ber myself.

I’ve been trying to rationalize this and what I came up with was tweaking the situation to the following:

Imagine that instead of a prediction, Omega has built a quantum entangled device that reads your mind and upon your decision whether you will 1B or 2B the opaque box either stays empty or is filled with a measly 1,000,000 USD (terrible exchange rate currently). The device is highly accurate, it predicted the right choice in the past 67 cases but it is not 100% accurate.

Would a 2Ber still choose 2B in that setup?

Or in other words, why does the passage of time make this a different choice?

Oh, and great discussion! :slight_smile:

EDIT: the Veritasium video on this subject: https://www.youtube.com/watch?v=Ol18JoeXlVI&t=1254s

31

Does it read your mind before you are in the room or not?

Also, there is the Newcomb topic, here might not be the best place.

32

Funny coming from the group telling you that not taking all of the money on the table gets you more money.

There is a probability approach telling you to two-box and I already showed that so this is simply false.

Good thing we are not comparing apples to oranges here. So yes, 1,001,000 is greater than 1,000,000 therefore you should two-box. Being against that is denying basic arithmetic.

Yeah, it’s not like the prediction being made way before your choice and the box already being filled accordingly are a major part of the problem.

Which is again ridiculous.

Let’s say you have a lottery where you pay $1 to enter and you have a 40% chance of winning $1,000,001. Your analysis would say the probability of ending up with a million is 40% and the probability of ending up with -$1 is 60%, therefore you shouldn’t play this lottery. This is ridiculous. Not taking the sums into account, as long as one is simply higher than the other, is stupid.

We can even do it the other way around. It’s not a lottery now, it’s another person with $1,000,000 and for them, it’s 60% chance of winning $1 dollar. Your analysis says that they should play the game because they have more chance of winning.

33

There is no money on the table.

I already addressed this example. You are deliberately using $1 to hide the cost. Make it $100,000 and now it’s obvious why I wouldn’t play it (not unless I had $100,000 to throw around).

That topic is just you going on in circles repeating the same circular reasoning: “two-box is more rational because two-box is more rational”.

34

Easy:

r <- sum(outer(c(0,1) * b, c(0,1) * b + a, '>') * outer(c(1-p,p), c(p,1-p))) >= 0.5
cat(sprintf('Choose %s\n', ifelse(r,'1B','2B')))

Which simplifies to:

r <- (b>a)*p^2 >= 0.5

a is the money in the transparent box, b is the money in the mystery box.

Are you going to get more money if the prediction is 99% accurate?

You don’t understand what the table is showing. The whole table is expected value. 1001 is the expected value given C. The marginal row is the expected value marginalizing over C.

You are conveniently avoiding the point I’m making. I’m explaining to you how this maps to Simpson’s paradox and why the two-box approach of slicing by prediction makes no sense, especially when p=1.

But by all means completely ignore my point.

35

Exactly. I thought of a similar scenario where Omega stores “prediction was accurate” in a device, and 99.99% of the time it would show that, the rest of the time “prediction was inaccurate”. The contents of the box are filled after your choice.

The usual retort of two-boxers that “you didn’t cause the outcome” disappears. This makes it possible to simulate Newcomb’s problem: the statistics are exactly the same.

A second retort is “the prediction was made before my choice” (which is nonsense since that’s how all predictions are made) can be addressed with the device coming up with “prediction was accurate” after you made the choice. I can easily write a program that does that. This once again doesn’t change the statistics.

The truth is two-boxers don’t have a leg to stand on.

36

There is $1,000 you’re leaving on the table and I am taking so I am going with more money, basic arithmetic.

It’s not obvious.

Repeating it won’t make it true.

37

You get $1,000, I get $1,000,000. You don’t get more money. Period.

That’s literally the only thing you do.

38

It’s kind of amazing that both of you argue your positions the same way, and make fun of each other for doing it. Look at just one aspect, and declare ‘case closed’, ignoring the full story.
And yes, repetition doesn’t make it true. I don’t see counters to my examples where the strategy fails.

You are completely misrepresenting the position of felipec. His argument is based on one arbitrary person (Bob) to compare himself to. So not 40%, but rather 28%. He would decline to risk the dollar just because probably having a dollar more than specifically Bob is far more important than hitting any jackpot.

Thank you. It precisely states your position when it has been foggy in some places. Remind me not to hire you as a money manager.

I don’t care. It’s not about expected outcome. It’s about expected utility, especially with taking DMU into account.

You don’t address what everybody else points out: taking a big risk of losing it all just to ‘probably’ get an outcome that is larger by 1. It’s a bad gamble: close to 48/52 odds that Bob will get 1000 (or even 2000) more than you. But hey, expected outcome (52/48) is you’ll beat Bob by that $1, and apparently it’s all about gloating to Bob and not about utility at all.

I’m talking about expected utility, not expected value, something which hasn’t been yet defined. The table shows expected outcome.

But I was illustrating the 75% case, which especially shows weakness in your strategy.

This makes no sense at all. It sounds like you expect ‘entanglement’ can be used to deliberately cause some outcome. It can’t. That would be FTL message sending, which is forbidden.

It probably matters only to a 2-boxer.

And yes, such posts probably belong in the other thread. We’re using this one to make fun of felipec’s program/strategy.

Newcomb's Paradox
39

Once again that couldn’t be further from the truth. I’m not starting with an assumption, I’m starting from FACTS.

I don’t look at one aspect, my approach takes everything into consideration.

Then why are you doing it right here?:

That is you literally repeating “it’s about expected utility” with zero justification, just ipse dixit.

Wrong. I would bet the $1, precisely because losing it doesn’t matter at all. The $1 is just a red herring you and Suny are adding to the problem just to delude yourselves into thinking that you are actually doing a rational calculation. The calculation with $1 or $0 is virtually the same.

Finance has absolutely nothing to do with a thought experiment that is never repeated.

If you think if a hammer is good in one case, then it’s good in every single case, you are the one with the bad heuristics.

But you don’t live in alternate universes, you live in one universe with one result.

If the bet was your life for option a) with 48% or option b) with 52%, you would take option b) every time. Because when it matters, every rational agent will pick the most likely outcome.

The only cases were you would pick option a) is when losing doesn’t matter.

And that’s why you completely avoided my example, because you know it proves you wrong.

I addressed every single one of your examples, why aren’t you addressing mine? That’s intellectually dishonest.

It doesn’t matter, you can pick whatever formula you want. When p=0.9999 Simpson’s paradox would still be present.

You are utterly confused.

For some reason you decided to conflate two arguments 1) (1/2)^{1/2} is the correct cross-over point, and 2) Simpson’s paradox explains why the two-box approach is wrong.

You can criticize my strategy for argument 1) all you want, that has absolutely nothing to do with argument 2), which you have not addressed at all.

I advice you to focus on one argument, because you are doing a poor job at both, completely avoiding both of my challenges.

40

That’s how you solve problems, you ignore noise. I do see the correlation but it doesn’t imply whatever one-boxers want it to imply.

What do you mean?

No, I don’t think so. Look at the justification before:

So, if you pick someone playing the lottery, you’ll have a 40% chance of the person having the million and a 60% chance of losing $1. If you pick someone not playing the lottery, you’ll have a 100% chance of that person having gained nothing, so $0.

We only look at the pair where playing the lottery gives you more money, which happens 40% of the time. So, you multiply 0.4 * 1 and you get 40%.