Expected utility is a weighted average
Multiply each payoff in an act's row by the probability of its state and add the products. An act that pays 2 with probability 1/3 and 1 with probability 2/3 scores 4/3.
When the states have probabilities, an act can be scored by the average of its payoffs weighted by those probabilities. That score is its expected utility, and the act with the largest score is the one the rule recommends. The probability at which two acts tie says how much the recommendation depends on your estimate.
Core ideas
The rules for ignorance used the table alone. Once the states have probabilities, the table can be summarised more finely.
Multiply each payoff in an act's row by the probability of its state and add the products. An act that pays 2 with probability 1/3 and 1 with probability 2/3 scores 4/3.
4/3 is not a number the act can pay. It is a number for comparing acts, and it has a meaning only as a comparison.
With two states, the probability at which two acts score equally divides the cases where one is recommended from those where the other is. Your estimate matters only if it is on the other side of that point.
The material
The umbrella with a probability of rain, then the tipping point.
Knowledge and Evidence introduced credences: degrees of belief that obey the laws of probability. Suppose your credence in rain is 1/3, and so 2/3 in no rain. Those two numbers are the new ingredient, and with them the umbrella table can be scored in a way the rules for ignorance could not.
The expected utility of act a is the sum, over states s, of the probability of the state times the payoff of a in it: EU(a) = Σ P(s)·u(a, s). The rule recommends an act with the largest expected utility.
For take, the weighted sum is 2·(1/3) + 1·(2/3) = 4/3. For leave it is (−3)·(1/3) + 3·(2/3) = 1. Take scores higher, so at a credence of 1/3 in rain the rule recommends the umbrella, which is the verdict maximin and minimax regret also gave.
The number 4/3 deserves a closer look, because the word “expected” misleads. No single outing with the umbrella pays 4/3; it pays 2 or 1. The figure is what the average would be if the same decision were faced many times with the same odds. For one decision it is a score, a number that ranks acts by weighing every state in proportion to how likely you think it is. That is the whole claim of the rule, and it is a substantial one.
With two states, let p be the probability of the first. Each act's score is a straight line in p, so the two lines cross at one value, and the recommendation changes only there.
For the umbrella, let p be the probability of rain. Take scores 2·p + 1·(1 − p) = 1 + p and leave scores (−3)·p + 3·(1 − p) = 3 − 6·p. They tie when 1 + p = 3 − 6·p, that is when 7·p = 2, so p = 2/7. Above 2/7 the umbrella is recommended and below it the act of leaving is. The lab prints that figure as the point where the expected utilities tie.
The tipping point is the most useful number in the lab, because it separates two questions. How likely is rain is a question about the world; whether the answer matters is a question about the table. If you are sure rain is more likely than 2/7, your exact estimate is irrelevant to what to do.
Win 1 or lose 1 with probability 1/2 each, against not betting at all. Betting scores 1·(1/2) + (−1)·(1/2) = 0 and not betting scores 0. The two tie, and the tipping point is p = 1/2. A bet is worth taking when your probability of winning is above that.
The lottery-ticket preset is the same comparison with the odds moved. A ticket costing 1 pays 500 if it wins, so buying scores 499 with probability 1/1000 and minus 1 otherwise, which comes to 499/1000 − 999/1000 = −1/2, against 0 for skipping. The lab recommends skipping and prints the tie at a probability of winning of 1/500, which is where the prize is exactly the odds against it. Only above that is buying recommended, and a lottery that sold tickets there would be paying out more than it took in.
One caution. The payoffs in the table are utilities, numbers that stand for how much you value each outcome, and for money amounts the two can come apart. The next lesson takes the difference up. Here the payoffs are taken as utilities from the start, so there is nothing to convert.
Interactive
the same payoffs, eight ways of choosing, each computed exactly
Method
Four steps. The last two apply only when there are two states.
One per state, none negative, summing to 1. A table whose probabilities do not sum to 1 describes some other problem.
Pair every payoff with the probability of its column and multiply. Keep the fractions; rounding early moves a tie.
The sum is the act's expected utility. The act with the largest sum is recommended, and a tie means the rule does not choose.
With two states write each score as a function of p, set them equal and solve. The answer tells you how wrong your estimate can be before the choice changes.
Take pays 2 in rain and 1 when dry; leave pays −3 and 3. The probability of rain is p.
The credence 1/3 is above 2/7, so the umbrella wins, but not by much: only 1/21 of probability separates them. A forecaster who revised rain from 1/3 to 1/5 would reverse the advice.
Given a table and a probability for each state, you should be able to compute each act's expected utility, name the act the rule recommends, and for two states solve for the probability at which the recommendation changes.
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