The Encyclopedia of Economic Principals

Chapter 15

Choice Under Risk, Ambiguity, and Probability Weighting

Where expected utility breaks: certainty, losses, unknown odds and distorted weights.

Four of the chapter's worked examples, made interactive: the Allais menus, a loss-averse chooser facing a favorable bet, the Ellsberg urns under worst-case priors, and inverse-S probability weights. Change one value at a time and watch the figure, the numbers and the hand calculation respond.

Every example here is a constructed teaching example: it uses the hypothetical numbers of the chapter's worked examples, plus a few values added for comparison and labelled as such in each panel. Nothing here measures a real market, firm or household.

Demonstration 1 of 4

The Allais common consequence

Can one utility function make Maya choose A in menu 1 and D in menu 2?

Expected utility is linear in probabilities, so a consequence common to both options adds the same 0.89u(1) to each and cannot change the ranking. Removing it shifts both bars down by the same amount, as switching the menu shows.

Equation, written in LaTeX: u(1)>0.10u(5)+0.89u(1).

Equation, written in LaTeX: 0.11u(1)>0.10u(5).

Equation, written in LaTeX: 0.10u(5)>0.11u(1).

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Prizes are 0, 1 and 5 million with u(0) = 0. Utility is scaled so u(1) = 1, and the control sets u(5)/u(1). Menu 1 offers A (1 million for certain) or B; menu 2 removes the common 0.89 chance of 1 million, turning A into C and B into D. EU is expected utility.

Predict first. Can any utility ratio make A best in menu 1 and D best in menu 2?

Your prediction

Choose an example

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Figure: The Allais common consequence. Two bars for menu 1 at u(5)/u(1) = 1.2: A has expected utility 1.000 and B has 1.010.
Utility ratio u(5)/u(1): 1.2, Common chance of 1 million: 0.89 (menu 1: A or B)
Constructed example: the chapter's hypothetical menus (prizes 0, 1 and 5 million, probabilities 0.10, 0.89, 0.01 and 0.11, 0.10). The book gives no utility values; the ratios 1.0, 1.1, 1.2 and 1.5 are added.

Calculated values

u(5)/u(1)
1.2
EU(A)
1.000
EU(B)
1.010
0.11u(1)
0.110
0.10u(5)
0.120
Expected-utility choice
B

With u(0) = 0, u(1) = 1 and u(5) = 1.2: EU(A) = u(1) = 1.000 and EU(B) = 0.10 x 1.2 + 0.89 x 1 + 0.01 x 0 = 0.120 + 0.890 = 1.010. Either way the comparison reduces to 0.11u(1) = 0.110 against 0.10u(5) = 0.120, because the common 0.89 chance of 1 million cancels. B has higher expected utility in menu 1. In both menus expected utility picks the riskier option (B and D), so the observed pair A and D cannot occur.

Worked steps

  1. EU(A) = u(1) = 1.000
  2. EU(B) = 0.10 x 1.2 + 0.89 x 1 + 0.01 x 0 = 0.120 + 0.890 = 1.010
  3. Difference = 0.11u(1) - 0.10u(5) = 0.110 - 0.120 = -0.010
  4. B has higher expected utility in menu 1

Use the idea

When a choice flips after a shared outcome is added or removed, the flip points to special treatment of certainty, not to a particular utility curve.

Where the conclusion applies

Expected utility with u(0) = 0 and u(1) = 1. The book gives no utility values; the ratios u(5)/u(1) are constructed.

Check your understanding: At what u(5)/u(1) is expected utility indifferent between A and B?
0.11u(1) = 0.10u(5) gives u(5)/u(1) = 0.11 / 0.10 = 1.1.

Chapter 15 source: section "Allais paradox".

Demonstration 2 of 4

Loss aversion flips a favorable bet

How large must the gain be before a loss-averse chooser takes a 50-50 bet that risks 100?

Losses are scaled by lambda before weighting, so the kink at zero makes the loss side steeper than the gain side. A bet with positive expected value is rejected whenever the gain is below lambda times the loss.

Equation, written in LaTeX: V=0.5(150)+0.5[-2(100)]=75-100=-25.

Equation, written in LaTeX: V'=0.5(220)-0.5(200)=10,

Equation, written in LaTeX: 0.5g^*-0.5(2)(100)=0,

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The gamble pays the gain or loses 100, each with decision weight 0.5. Value is v(x) = x for gains and v(x) = -lambda(-x) for losses (alpha = 1). V is the prospect value and g* the gain at which V = 0.

Predict first. With lambda = 1.5, is the 150 or minus 100 gamble accepted?

Your prediction

Choose an example

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Figure: Loss aversion flips a favorable bet. Kinked value function with loss slope 2.0. The gain 150 has value 150 and the loss of 100 has value -200; the prospect value is -25.
Gain: 150, Loss aversion lambda: 2
Constructed example: the chapter's hypothetical gamble (gain 150 or 220, loss 100, lambda 2, threshold 200); lambda values 1, 1.5 and 2.5 are added for comparison.

Calculated values

Prospect value V
-25
Expected value
25
Threshold gain g*
200
Decision
reject

With alpha = 1 and weights 0.5, V = 0.5(150) + 0.5[-2.0(100)] = 75 - 100 = -25. V = -25 is below zero, so the chooser rejects despite expected value 25. The gain needed to accept is g* = 2.0 x 100 = 200, against 100 under expected value.

Worked steps

  1. V = 0.5(150) + 0.5[-2.0(100)] = 75 - 100 = -25
  2. Expected value = 0.5(150) - 0.5(100) = 25
  3. Threshold: 0.5g* - 0.5(2.0)(100) = 0 gives g* = 200
  4. V = -25 is below zero, so the chooser rejects despite expected value 25

Use the idea

Before calling a refused favorable bet irrational, compute the gain-to-loss ratio and compare it with plausible loss-aversion coefficients.

Where the conclusion applies

Linear value on each side (alpha = 1), equal decision weights of 0.5 and outcomes coded relative to current wealth. Curvature or unequal weights change the threshold.

Check your understanding: What gain makes the gamble acceptable with lambda = 2.5?
0.5g* = 0.5(2.5)(100) gives g* = 250; any gain above 250 is accepted.

Chapter 15 source: section "Prospect theory".

Demonstration 3 of 4

Worst-case priors in the Ellsberg urn

Why does a maxmin chooser prefer the known urn for both colors?

Under maxmin expected utility the red bet uses the lowest red share and the black bet the lowest black share. A symmetric prior set pushes both below 50, so K wins for both colors. Revealing the composition collapses the set to one probability.

Equation, written in LaTeX: 0.5(100)=50.

Equation, written in LaTeX: 0.2(100)=20.

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Urn K holds 50 red and 50 black balls; urn A holds 100 balls in unknown proportions. A correct color pays 100. The prior set [low, 1 - low] lists every plausible red share in A, and the maxmin chooser values each bet at its worst case.

Predict first. If the prior set shrinks to [0.35, 0.65], does Maya still prefer K for both colors?

Your prediction

Choose an example

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Figure: Worst-case priors in the Ellsberg urn. Three bars: known urn 50, ambiguous red 20, ambiguous black 20, with the prior set [0.20, 0.80].
Lowest plausible red share: 0.2, Composition of urn A: Unknown
Constructed example: the chapter's hypothetical urns (prize 100, prior set [0.2, 0.8], revealed 65 red); lowest red shares 0.35 and 0.5 are added for comparison.

Calculated values

Known urn
50
Red bet on A
20
Black bet on A
20
Maxmin gap per color
30
Red choice
K
Black choice
K

With the prior set [0.20, 0.80]: Known urn K: 0.5 x 100 = 50; Ambiguous red: worst case 0.20(100) = 20; Ambiguous black: worst case 1 - 0.80 = 0.20, so 0.20(100) = 20. Both ambiguous bets are worth 20, so Maya prefers K for both colors, a maxmin gap of 30 each. No single prior can explain betting on K for red and for black.

Worked steps

  1. Known urn K: 0.5 x 100 = 50
  2. Ambiguous red: worst case 0.20(100) = 20
  3. Ambiguous black: worst case 1 - 0.80 = 0.20, so 0.20(100) = 20
  4. Red: bet on K; black: bet on K

Use the idea

When a client avoids an option with unknown odds, ask what range of probabilities they find plausible; the worst case of that range is what they are pricing.

Where the conclusion applies

Linear utility, a symmetric prior set and maxmin evaluation. Smooth ambiguity models would give values between the worst case and the average. The control for the prior set has no effect once the composition is revealed.

Check your understanding: With the prior set [0.5, 0.5], what is the maxmin gap?
Both bets are worth 0.5(100) = 50, the same as K, so the gap is 0 and Maya is indifferent.

Chapter 15 source: section "Ellsberg paradox".

Demonstration 4 of 4

Overweighting rare, underweighting likely

How can the same chooser buy a long shot and refuse a likely gain?

An inverse-S weighting function lies above the 45 degree line at small probabilities and below it at large ones. The same chooser therefore pays for long shots and gives up likely gains for a sure amount, while knowing the probabilities exactly.

Equation, written in LaTeX: 0.01(1{,}000)=10,

Equation, written in LaTeX: 0.05(1{,}000)=50,

Equation, written in LaTeX: 0.80(100)=80,

Equation, written in LaTeX: 0.65(100)=65,

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p is the objective probability of the prize and w(p) the decision weight. Value is linear in gains, so the gamble is worth w(p) times the prize. The rare case is 1 percent of 1,000 against 30 for certain; the likely case is 80 percent of 100 against 70 for certain.

Predict first. At w(0.01) = 0.03, does the rare prize still beat 30 for certain?

Your prediction

Choose an example

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Figure: Overweighting rare, underweighting likely. Left: the 45 degree line with the weight 0.050 at p = 0.01. Right: bars for the gamble worth 50 and the sure amount 30.
Comparison: Rare: 1% of 1,000 vs 30, Decision weight: Book (0.05 or 0.65)
Constructed example: the chapter's hypothetical comparisons (1 percent of 1,000 against 30, weight 0.05; 80 percent of 100 against 70, weight 0.65); midpoint weights 0.03 and 0.725 are added for comparison.

Calculated values

Decision weight
0.050
Weighted value
50
Sure amount
30
Choice
gamble

With book weight w(0.01) = 0.050, the gamble is worth 0.050 x 1,000 = 50 against 30 for certain, so the gamble wins. The probability itself is unchanged; only the weight it receives differs.

Worked steps

  1. Weight (book weight): w(0.01) = 0.050
  2. Weighted value = 0.050 x 1,000 = 50
  3. Compare with 30 for certain: the gamble wins

Use the idea

When a product sells rare large payouts or sure discounts against likely gains, compare the linear value with a weighted value before reading the choice as a belief error.

Where the conclusion applies

Linear value for gains and known probabilities. The weights are constructed, and a weight at one probability says nothing about weights elsewhere.

Check your understanding: What w(0.80) makes the chooser indifferent against 70 for certain?
w(100) = 70, so w = 0.70.

Chapter 15 source: section "Inverse-S probability weighting".