sebmathguy

There's actually no need to settle for finite truncations of a decision agent. The unlosing decision function (on lotteries) can be defined in first-order logic, and your proof that there are finite approximations of a decision function is sufficient to use the compactness theorem to produce a full model.

If p + q = 1, then p(A or B) = 1. The equivalence statement about A and B that we're updating can be stated as (A or B) iff (A and B). Since probability mass is conserved, it has to go somewhere, and everything but A and B have probability 0, it has to go to the only remaining proposition, which is g(p, q), resulting in g(p, q) = 1. Stating this as p+q was an attempt to find something from which to further generalize.

My first reaction to the second question is to consider the case in which p + q = 1. Then, the answer is clearly that g(p, q) = p + q. I suspect that this is incomplete, and that further relevant information needs to be specified for the answer to be well-defined.

import java.util.*;

public class absentMindedDriver {

static Random generator = new Random(); static int trials = 100000; static double[] utilities;

static int utility(double x) { if(generator.nextDouble() < x) { //the driver guesses return 0; } if(generator.nextDouble() < x) { //the driver guesses return 4; } return 1; //the driver missed both exits }

static double averageUtility(double x) { int sum = 0; for(int i = 0; i < trials; i++) { sum += utility(x); //iteratively generates the sum to ... (read more)

If anybody wants to try simulating this problem, here's some java code:

`import java.util.*;

public class absentMindedDriver {

static Random generator = new Random(); static int trials = 100000; static double[] utilities;

static int utility(double x) { if(generator.nextDouble() < x) { //the driver guesses return 0; } if(generator.nextDouble() < x) { //the driver guesses return 4; } return 1; //the driver missed both exits }

static double averageUtility(double x) { int sum = 0; for(int i = 0; i < tria... (read more)