Unit 4, lesson 30

Expected Value and Decisions

How expected value supports decisions, with gross and net payoff, fair price, break-even probability and expected cost, and why the decision criterion and model risk must be stated.

  • Video 30
  • Expected Value

A ticket can have a positive expected net payoff and still lose most of the time, and two choices can share the same mean while carrying very different risks. Expected value becomes a decision tool only after the quantity being averaged, the costs, the probabilities and the rule for choosing are written down. This lesson separates gross and net payoffs, finds fair prices and break-even probabilities, compares expected costs, and shows where an expected-value comparison stops being the whole answer.

Key terms

Expected payoff

Definition In a decision problem, the payoff of an action is a random variable GG that records, in fixed units, what the action returns in each outcome of a stated probability model. Its expected payoff is E[G]E[G], assumed finite. If the action also costs a fixed amount cc in every outcome, then GG is called the gross payoff, the net payoff is N=G−cN = G - c, and

E[N]=E[G]−c.E[N] = E[G] - c.

A realization is the value that actually occurs on one trial.

In plain words Write the final result of every outcome, fee included, before averaging. A gross payoff with a positive mean does not yet say whether the action leaves you ahead on average. A single realization can differ from the expected payoff, which need not even be a possible result.

Example A constructed training ticket pays 2020 tokens with probability 0.10.1 and nothing with probability 0.90.9, and it costs 22 tokens. The expected gross payoff is 20⋅0.1=220 \cdot 0.1 = 2 tokens. The net payoff is 1818 tokens with probability 0.10.1 and −2-2 tokens with probability 0.90.9, so

E[N]=18⋅0.1−2⋅0.9=0,E[N] = 18 \cdot 0.1 - 2 \cdot 0.9 = 0,

which agrees with E[G]−c=2−2E[G] - c = 2 - 2. Each realization is a gain of 1818 tokens or a loss of 22, never 00.

Fair price

Definition For a gross payoff GG with finite expected value, the fair price in the expected-value sense is the fixed fee that makes the expected net payoff zero. Since E[G−c]=E[G]−cE[G - c] = E[G] - c, it is

c∗=E[G].c^{*} = E[G].

A fee below c∗c^{*} gives a positive expected net payoff, and a fee above it gives a negative one. A game whose fee equals c∗c^{*} is called fair in this sense.

In plain words The fair price is what the payoff is worth on average under the model. The word fair is narrow here. It does not mean that winning and losing are equally likely, that the outcomes are equally attractive, or that a participant gets the fee back.

Example For the training ticket, c∗=E[G]=2c^{*} = E[G] = 2 tokens, so the 22-token fee is fair in the expected-value sense. A buyer still loses the fee with probability 0.90.9. For a general fee cc the expected net payoff is 2−c2 - c, positive when c<2c < 2 and negative when c>2c > 2.

Break-even probability

Definition Suppose the expected net payoff of an action, or the difference between the expected costs of two actions, depends on an uncertain probability pp in the model. A break-even probability is a value of pp in [0,1][0, 1] at which this quantity equals 00. For a ticket that pays a>0a > 0 with probability pp and nothing otherwise, and has a fixed fee c≥0c \ge 0, the expected net payoff is ap−cap - c. It is 00 at

p∗=ca,p^{*} = \frac{c}{a},

positive for p>p∗p > p^{*} and negative for p<p∗p < p^{*}. This is a probability only when c≤ac \le a. If c>ac > a, every outcome of the ticket is a loss.

In plain words Instead of trusting one uncertain number, find the threshold at which the expected net payoff changes sign and the conclusion flips. The question then becomes whether the true probability lies above or below that threshold, which is often easier to judge than its exact value.

Example For the training ticket with prize 2020 and fee 22, the expected net payoff is 20p−220p - 2, so the break-even probability is p∗=2/20=0.1p^{*} = 2/20 = 0.1. At p=0.06p = 0.06 the expected net payoff is −0.8-0.8 tokens, and at p=0.14p = 0.14 it is 0.80.8 tokens. The two assumptions differ by only eight percentage points, yet they lead to opposite conclusions.

Expected cost

Definition If an action has a random cost CC with finite expected value, its expected cost is E[C]E[C]. Under the expected-cost criterion, an action with the smallest expected cost is preferred. The cost CC has to include every charge the decision is meant to cover, not only the usual or advertised one.

In plain words Weight each possible cost by its probability, including rare but expensive outcomes. A lower expected cost does not mean a lower cost on every occasion.

Example Method A always costs 33 tokens. Method B costs 11 token, or 1616 tokens when rework is needed, and rework happens with probability 0.10.1. The cost CBC_B of method B then has expected value

E[CB]=1⋅0.9+16⋅0.1=2.5 tokens,\begin{aligned} E[C_B] &= 1 \cdot 0.9 + 16 \cdot 0.1 \\[0.4em] &= 2.5 \text{ tokens}, \end{aligned}

which is below the 33 tokens of method A, although on a rework job B costs 1616. With a rework probability pp, the expected cost of B is (1−p)+16p=1+15p(1 - p) + 16p = 1 + 15p. It is lower than 33 exactly when p<2/15≈0.133p < 2/15 \approx 0.133, so 2/152/15 is the break-even rework probability.

Decision criterion

Definition A decision criterion is an explicit rule that uses the outcome distributions of the available actions to choose among them. Maximizing expected net payoff and minimizing expected cost are decision criteria. So are maximizing the worst possible outcome, minimizing the probability of exceeding a budget, and minimizing expected cost among the actions whose probability of exceeding that budget is at most a chosen level. Different criteria can lead to different choices among the same actions, even when every probability is known exactly.

In plain words Expected value answers one question, which action does best on average. Before using it, say that this is the question being asked, and list the constraints it does not capture, such as a loss that cannot be afforded.

Example Suppose a single job must not cost more than 66 tokens. Method B has the lower expected cost, 2.52.5 against 33, but it exceeds the budget with probability 0.10.1, while method A never does. A criterion that rules out any chance of exceeding the budget picks A. Equal means need not carry equal risk either. Not buying the training ticket gives a net payoff of 00 for certain, the same expected value as buying it, while buying means losing 22 tokens with probability 0.90.9 or gaining 1818 with probability 0.10.1.

Model risk

Definition Model risk is the possibility that a decision is misguided because the probability model behind it does not describe the situation well enough. The probabilities may be wrong, a cost may be missing, or an outcome may be left out of the table. An expected value computed from a model describes that model, and a sensitivity check recomputes the decision under other plausible assumptions to see whether it changes.

In plain words Correct arithmetic on the wrong table can give a confident wrong answer. Ask which assumption the conclusion rests on and how far that assumption can move before the decision flips.

Example The comparison of methods A and B used a rework probability of 0.10.1. If the jobs at hand actually need rework with probability 0.20.2, the expected cost of B is 1+15⋅0.2=41 + 15 \cdot 0.2 = 4 tokens, more than the 33 of method A, and the preferred method changes. Since the break-even value is 2/15≈0.1332/15 \approx 0.133, the decision rests on which side of that threshold the true rework probability lies, so a rework rate measured on a different kind of job deserves caution. In the same way, if the training ticket also carried a handling charge left out of its table, the computed expected net payoff of 00 would be too high.

Check yourself

  1. A constructed ticket pays 1212 tokens with probability 0.250.25 and nothing otherwise, and its fee is 22 tokens. Find the expected gross payoff, the expected net payoff, the fair fee and the probability of a negative net payoff. Does the positive expected net payoff mean that a gain is more likely than a loss?
Show the answer

The expected gross payoff is 12⋅0.25=312 \cdot 0.25 = 3 tokens. The net payoff is 1010 tokens with probability 0.250.25 and −2-2 tokens with probability 0.750.75, so the expected net payoff is 10⋅0.25−2⋅0.75=2.5−1.5=110 \cdot 0.25 - 2 \cdot 0.75 = 2.5 - 1.5 = 1 token, which matches 3−23 - 2. The fair fee equals the expected gross payoff, 33 tokens.

A negative net payoff occurs whenever there is no prize, which has probability 0.750.75. So a loss is three times as likely as a gain. The mean is still positive because the gain of 1010 tokens is five times the loss of 22 tokens, and that factor of five outweighs the factor of three in likelihood.

  1. Method A costs 44 tokens every time. Method B costs 22 tokens, or 1212 tokens when rework is needed, and rework happens with probability pp. For which pp does B have the smaller expected cost? With p=0.15p = 0.15, compare the expected costs and find the probability that B costs more than A on a single job.
Show the answer

The expected cost of B is 2(1−p)+12p=2+10p2(1 - p) + 12p = 2 + 10p, which is less than 44 exactly when p<0.2p < 0.2. So the break-even rework probability is 0.20.2.

With p=0.15p = 0.15 the expected cost of B is 2+1.5=3.52 + 1.5 = 3.5 tokens, lower than the 44 tokens of A. Still, B costs 1212 tokens, more than A, whenever rework is needed, and that happens with probability 0.150.15.

  1. A buyer purchases 1010 of the original training tickets, each paying 2020 tokens with probability 0.10.1 and costing 22 tokens, and the ten draws are independent. Each ticket is fair in the expected-value sense. Is the buyer guaranteed to break even? Find the probabilities of ending behind, exactly even and ahead.
Show the answer

No. The fees total 2020 tokens and every win returns 2020 tokens, so the buyer ends behind with no wins, exactly even with one win and ahead with two or more.

By independence the probability of no win is 0.910≈0.3490.9^{10} \approx 0.349. Exactly one win can happen in ten disjoint ways, one for each ticket, and each way has probability 0.1⋅0.990.1 \cdot 0.9^{9}, so the probability of exactly one win is 10⋅0.1⋅0.99≈0.38710 \cdot 0.1 \cdot 0.9^{9} \approx 0.387. The probability of ending ahead is about 1−0.349−0.387=0.2641 - 0.349 - 0.387 = 0.264. A zero expected net payoff per ticket describes the model average, not a promise about any finite number of tickets.

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