Finance-Theory · Unit 22 · Video 2 · Interactive Practice

Behavioral Finance: Loss Aversion and Ambiguity Aversion

IKey Formulas

Formula	Name	What you need
$\text{E}[X] = \sum_i p_i\, x_i$	Expected value	Each outcome and its probability
$X_{1+2}(s) = X_1(s) + X_2(s)$	Combined position	Both payoffs in the same outcome $s$
$X_{B+C}(s) > X_{A+D}(s)$ for every $s$	Strict dominance	An outcome-by-outcome comparison
$P(\text{win}) = \tfrac{1}{2}\cdot\tfrac{K}{100} + \tfrac{1}{2}\cdot\tfrac{100-K}{100} = \tfrac{1}{2}$	Coin-flip guarantee	Your own fair coin, any mix $K$

Key Insight: Facing gains, people avoid risk; facing losses, the same people seek it (the reflection effect). Each choice alone is defensible, but the popular pair A + D pays $10,000 less than B + C in every outcome.

IIVisualization 1 — Two Decisions, One Position

Taken together, the popular pair A + D pays exactly $10,000 less than B + C in every outcome.

Decision 1 (gains) Decision 2 (losses)

💡 In Tversky and Kahneman's study, when the combined options were shown side by side, everyone chose the dominant pair B + C.

IIIVisualization 2 — The Game-Runner's Mix vs Your Coin

Any fixed colour can be beaten by the game-runner's mix; your own fair coin pins $P(\text{win})$ at $\tfrac{1}{2}$ .

Red balls in urn B (K) 30 Your rule

💡 Even so, urn B sells at a large discount to urn A when both are auctioned: people shy away when they are uncertain about the risk itself, the same answer buyers gave toxic assets in late 2008.

IVQuiz Questions

Problem 1 · Expected Value of a Gain Gamble

Given: Choice A is a sure gain of $400. Choice B is a $\tfrac{2}{3}$ chance of winning $750 and a $\tfrac{1}{3}$ chance of winning nothing. Which choice has the higher expected value, and by how much?

✅ Correct! The gamble is worth $500 on average, $100 more than the sure $400: more risk for more expected return, so a risk-averse choice of A is a preference, not an error.

❌ Check the probability. The $750 arrives with probability

\tfrac{2}{3}

, not

\tfrac{1}{3}

; weighting it by

\tfrac{1}{3}

gives 250 and wrongly makes A look better.

❌ Not quite. $750 is the payoff if B wins, not B's expected value; weight it by its probability first.

❌ Not quite. Equal expected values would need B's expected payoff to be $400; compute

\tfrac{2}{3} \times 750

Show solution

In dollars:

\text{E}[A] = 400, \qquad \text{E}[B] = \tfrac{2}{3}\times 750 + \tfrac{1}{3}\times 0 = 500

\text{E}[B] - \text{E}[A] = 500 - 400 = 100

Choice B has the higher expected value, by $100. Taken alone, preferring the sure $400 is a legitimate risk-averse choice.

Problem 2 · The Loss Side

Given: Choice C is a sure loss of $300. Choice D is a $\tfrac{1}{3}$ chance of losing $900 and a $\tfrac{2}{3}$ chance of losing nothing. In the video's experiment, most people take the gamble on the loss side; suppose you choose D.

What is the expected value of D?

In Kahneman and Tversky's precise terms, the switch from avoiding risk over gains to seeking it over losses is called…

✅ Correct! C and D have the same expected value, so C is simply the less risky option; choosing D anyway is risk seeking over losses, the reflection effect.

❌ Check the probability. The $900 loss has probability

\tfrac{1}{3}

; the

\tfrac{2}{3}

belongs to losing nothing.

❌ Not quite. −$900 is the worst outcome, not the average; weight it by its probability.

❌ Not quite. Losing nothing is only the more likely outcome; the expected value also counts the

\tfrac{1}{3}

chance of losing $900.

❌ Close, but that is the other finding. The video uses "loss aversion" loosely for the pattern, but in Kahneman and Tversky's precise sense it means a loss hurts more than a gain of the same size pleases.

❌ Not quite. Herding is following what others do; here one person's attitude to risk flips between gains and losses.

❌ Not quite. Shying away from unknown odds is the urn B finding; every probability here is known.

❌ Check the expected value. Multiply each loss by its own probability and add.

Show solution

In dollars:

\text{E}[D] = \tfrac{1}{3}\times(-900) + \tfrac{2}{3}\times 0 = -300 = \text{E}[C]

The expected values are equal, so the sure loss C is the less risky option. Taking D anyway means seeking risk over losses after avoiding it over gains: Kahneman and Tversky call that switch the reflection effect. Loss aversion, in their precise sense, is the separate finding that a loss hurts more than an equal gain pleases.

Problem 3 · Combine the Popular Choices

Given: Decision 1: A is a sure gain of $570; B is a $\tfrac{2}{3}$ chance of $900 and a $\tfrac{1}{3}$ chance of $0. Decision 2: C is a sure loss of $300; D is a $\tfrac{1}{3}$ chance of losing $900 and a $\tfrac{2}{3}$ chance of losing $0. One draw settles both gambles: in the $\tfrac{2}{3}$ outcome B pays $900 and D loses nothing; in the $\tfrac{1}{3}$ outcome B pays nothing and D loses $900.

What does the pair A + D pay in the 1/3 outcome?

How does B + C compare with A + D?

✅ Correct! B + C pays +$600 or −$300, A + D pays +$570 or −$330: the popular pair is strictly dominated, $30 worse whichever way the draw goes.

❌ Wrong outcome. +$570 is A + D in the

\tfrac{2}{3}

outcome, where D loses nothing; in the

\tfrac{1}{3}

outcome D loses $900.

❌ Wrong pair. −$300 is what B + C pays in the

\tfrac{1}{3}

outcome (B pays nothing, C loses $300).

❌ Not quite. −$900 is D alone; the sure $570 from A is still collected in this outcome.

❌ Not quite. Add the payoffs in each outcome: B + C gives +$600 and −$300, A + D gives +$570 and −$330.

❌ Not quite. The $30 gap appears in each outcome separately; it is not added across outcomes, since only one outcome happens.

❌ Not quite. In the

\tfrac{2}{3}

outcome B + C pays $600 against A + D's $570, so B + C is ahead there too: it wins in every outcome, not only on average.

❌ Check the 1/3 outcome. A pays its sure $570 and D loses $900 in that outcome.

❌ Compare outcome by outcome. Work out both pairs in the

\tfrac{2}{3}

outcome and in the

\tfrac{1}{3}

outcome.

Show solution

Add the two payoffs in the same outcome (in dollars):

Outcome	A + D	B + C	Gap
$\tfrac{2}{3}$ : B pays, D loses 0	570 + 0 = 570	900 − 300 = 600	30
$\tfrac{1}{3}$ : B pays 0, D loses 900	570 − 900 = −330	0 − 300 = −300	30

A + D pays −$330 in the $\tfrac{1}{3}$ outcome, and B + C pays $30 more in every outcome, so A + D is strictly dominated. As a check on the averages: $\tfrac{2}{3}\times 570 + \tfrac{1}{3}\times(-330) = 270$ and $\tfrac{2}{3}\times 600 + \tfrac{1}{3}\times(-300) = 300$ .

Problem 4 · A Biased Randomizer (Transfer)

Given: An urn holds 100 balls, red and/or black, in a mix the game-runner chooses before you name a colour. Naming the colour that is drawn wins $600. Instead of a fair coin, you roll a die: 1–2 means you name red, 3–6 means black. The game-runner knows this rule but not your roll. What is the lowest expected payoff the game-runner can force on you?

✅ Correct! Filling the urn with 100 red balls cuts your chance to

\tfrac{1}{3}

, an expected $200; only a fair coin makes the

K

terms cancel.

❌ Close, but… $300 is the fair-coin guarantee. With a biased die the

K

terms no longer cancel, so the game-runner can push the mix toward the colour you name less often.

❌ Not quite. $0 is the worst case for always naming one colour; the die still names red a third of the time, which wins against an all-red urn.

❌ That is the best case. $400 comes from an all-black urn, but the game-runner chooses the mix that hurts you most.

Show solution

With $K$ red balls out of 100:

P(\text{win}) = \tfrac{1}{3}\cdot\tfrac{K}{100} + \tfrac{2}{3}\cdot\tfrac{100-K}{100} = \frac{200 - K}{300}

This falls as $K$ rises, so the game-runner sets $K = 100$ : $P(\text{win}) = \tfrac{100}{300} = \tfrac{1}{3}$ , and the expected payoff is $\tfrac{1}{3}\times 600 = 200$ dollars, or $200. A fair coin would make the $K$ terms cancel, giving $\tfrac{1}{2}$ and $300 for every mix.

Solved: 0 / 4