Newcomb’s Problem

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Newcomb’s Problem

> Image Above is Generated Using Stable Diffusion

Background

Decision theory is a branch of mathematics (specifically probability). This branch analyses decision-making based on outcomes with known probabilities. The normative subsection of decision theory is concerned with identifying the most optimal decision that could be made by a fully rational agent. In this way, it is similar to the field of game theory. Indeed, one may advocate for identifying game theory as a subset of decision theory.


Decision theory has both academic and practical impetus. As to the former; one may be interested in considerations of hyper-rationality as an a-priori task. In other words, one may treat decision theory scenarios as novel ‘puzzles’ to be solved for their own sake. However, as to the latter, the field also has a large implication in highly practical areas. For example; decision theory can be applied in microeconomic or sociological analysis. Particularly if such analysis is conducted in an attempt to produce an optimised outcome decided by the actions of individual actors. More recently, decision theory has become critical to the fields of artificial intelligence, machine-learning, and computer science. In these latter fields, operating principles must be established by the programmer and thus the program must have a sophisticated understanding of the principles, of rationality, of basis for decisions, and of the processes for achieving optimal outcomes.

Outline

Newcomb’s problem is a complex sceneratio presented within the framework of normative decision theory. The problem was created by William Newcomb. However, it was first analysed by prominent philosopher Rober Nozick (2019)1 in 1969. Today it remains a much-debated topic of decision theory.


The scenario may be presented as such:


There exists a super-intelligent predictor (the controller), a player, and two boxes labelled A and B.


Box A is transparent and visibly contains $1,000

Box B is opaque. It is not clear what it contains.


The player is given two options:

  1. Take both boxes A ($1000) and B (?)

  2. Take only box B (?)


The player is only told three pieces of information:

  1. If the predictor has predicted that the player will take both boxes, then box B contains nothing

  2. If the predictor has predicted that the player will only take box B, then box B will contain $1,000,000

  3. The predictor has played this game 100 times so far. It has correctly predicted those player’s choices 100 times.


The player has not been told what the predictor has predicted. 

The player may not interact with either box before making their choice.

The player may not change their mind after asserting their choice.


Analysis

Below are the four possible outcomes based on the prediction and the players choice



Predicts A + B

Predicts B Alone

A + B

$1000

$1,001,000

B Alone

$0

$1,000,000


Prima facie, the optimal choice appears to be picking both boxes. If all outcomes are equally likely; the expected values would be as such:


E(x) of picking A + B will be $501,000 

[E(x) = x1P(x1) + x2P(x2) + … = $10000.5+ 1,001,0000.5]

E(x) of picking B alone will be $500,000

[E(x) = x1P(x1) + x2P(x2) + … = $00.5+ 1,000,0000.5]


E(A+B) > E(B), therefore B is the more optimal decision.


However; we have to remember that the controller is a reliable predictor (at least so far). If we are to accept the scenario in good faith (ie. not to say he might have gotten lucky), then it is rational to presume that his predictive power will continue (since we have no sufficient reason to conclude or infer otherwise). This makes the decision strikingly harder.


For the sake of simplified analysis, let us say the predictor actually predicted 99/100 of the last games (E(x) would break otherwise). Using this hypothetical past performance, it is rational to infer that the predictor would have a 99% chance of correctly identifying our choices:



Predicts A + B

Predicts B Alone

Chooses A + B

$1000

99%

$1,001,000

1%

Chooses B Alone

$0

1%

$1,000,000

99%


Under this knowledge, the E(x) change significantly:


E(x) of picking A + B will be $11,000 

[E(x) = x1P(x1) + x2P(x2) + … = $10000.99+ 1,001,0000.01]

E(x) of picking B alone will be $990,000

[E(x) = x1P(x1) + x2P(x2) + … = $00.01+ 1,000,0000.99]


E(B) > E(A+B), therefore A + B is the more optimal decision.


It should be noted that the E(B) relative to E(A+B) varies as the predictors' reliance decreases. The two E(x) functions can be generalised according to the following formulae:


E(A+B) = 1000p+1001000(1-p)

E(B) = 0(1-p)+1000000p


Where: ‘p’ is the predictor’s reliability (ie. the probability they will guess correctly).

Alternatively, the two functions can be graphed according the following formulae:


For E(A+B); f(x)= 1000p+1001000(1-p)

For E(B); f(x) = 0(1-p)+1000000p

Where x [0 , 1] and y [0 , 1001000]


According to the utility maximisation principle of decision theory, whereby one makes optimal decisions leveraging any information they have, it is always optimal to pursue the route of highest expected value (E(x)) and not the route of highest possible value. Therefore, the graph with highest functional value (on the y-axis in this case) and its corresponding decision represents the optimal choice. 


In this example, the two E(x) are equivalent at a reliability of 0.5005 (ie. predicting 5005 out of 10000 scenarios). At a higher reliability (ie. p > 0.5005) B alone is optimal. At a lower predictability (ie. p < 0.5005) A + B is optimal.


1Nozick, R 2019, Newcomb's Problem And Two Principles Of Choice, Wayback Machine, viewed 22 November 2022, <https://web.archive.org/web/20190331225650/http://faculty.arts.ubc.ca/rjohns/nozick_newcomb.pdf>

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