Finance-Theory · Unit 22 · Video 3 · Interactive Practice

Risk Versus Uncertainty: Knight, Ellsberg and the Gorilla

IKey Formulas

Idea	Statement	Example from the video
Risk (Knight)	Probabilities can be measured, by reasoning or from data	A fair die, $P(6) = \tfrac{1}{6}$ ; a mortality table
Uncertainty (Knight)	No probability can be attached	Urn B's undisclosed mix; early nanotechnology
Ellsberg choices	Red: prefer A $\Rightarrow q < \tfrac{1}{2}$ ; black: prefer A $\Rightarrow 1 - q < \tfrac{1}{2} \Rightarrow q > \tfrac{1}{2}$	$q = P(\text{red from Urn B})$ : no single $q$ fits both
Implied chances	$P(\text{red}) + P(\text{black}) < 1$	The choices cannot come from any single probability for Urn B

Key Insight: Measurable risk is priced down to a thin margin by competition (US life insurers earned a return on equity of only about 5.7% in 2025); Knight argued that great fortunes go to those who bear uncertainty, and the video adds that so few are willing to bear it that they face little competition.

IIRisk: The Individual Is Unknown, the Group Is Not

An insurer cannot say who will die this year, yet across many policyholders the yearly death rate is predictable.

Policyholders n 10

💡 The 1% yearly death rate is an illustrative assumption, not a figure from the video; any measured group rate gives the same narrowing, which is why the premium can be priced on the group's number.

IIIThe Silence of the Cemetery

Two sets of ventures can share the same average while one clusters and the other is a spike at zero plus one huge winner.

Ventures we hear from

IVThe Ellsberg Paradox: No Single q

Preferring Urn A for a red bet and again for a black bet demands $q < \tfrac{1}{2}$ and $q > \tfrac{1}{2}$ at once.

Red bet: choose Black bet: choose q = P(red from Urn B) 0.30

VQuiz Questions

Problem 1 · Risk or Uncertainty?

Which one of these is risk in Knight's sense, rather than uncertainty?

✅ Correct! A balanced wheel's odds follow by reasoning, like the video's fair die, so the randomness is measurable: risk.

❌ Not quite. This is the video's early-nanotechnology case: nobody can attach a probability to whether it is even a business. That is uncertainty.

❌ Not quite. An undisclosed mix is exactly Urn B's situation: no probability can be measured for the split, so this is uncertainty.

❌ Not quite. No data and no reasoning pin down the odds for an industry nobody understands; that is uncertainty, where the video says the large fortunes lie.

Show solution

Knight's test: can the probabilities be measured, either by reasoning (a fair die) or from data (a mortality table)?

A balanced roulette wheel's chance of red follows from counting its pockets, so it is risk. The new technology, the undisclosed split and the not-yet-understood industry all lack any measurable probability: they are uncertainty.

Problem 2 · Who Earns the Fortune?

As the video presents Knight's argument, who stands to make great fortunes, and why?

✅ Correct! Most people shrink from what they do not understand, so the few who take it on are paid accordingly.

❌ Not quite. Measurable risk is what competition prices down: the video's US life insurers earned a return on equity of only about 5.7% in 2025.

❌ Right group, wrong reason. The video's first qualification is that much of this uncertainty is not systematic risk, the only kind the CAPM pays a premium for.

❌ Not quite. The zero-profit condition holds only for risks that are well understood; the video's point is that it fails for uncertainty.

Show solution

Risk: once a risk can be forecast, anyone can price it, competitors do the same, and the margin is bid down (the life-insurance numbers).

Uncertainty: randomness nobody can parameterize. Because most people avoid it, those willing to bear it face little competition, so the reward is not competed away.

The CAPM offers no reason for the reward (much of the uncertainty is not systematic), and zero economic profit is an equilibrium only for well-understood risks.

Problem 3 · Red Over Black, and Red Over Yellow

Given: an urn holds 60 balls: 20 red, and 40 that are black or yellow in an unknown split. A ticket on a colour pays $120 if that colour is drawn. A person strictly prefers the red ticket to the black ticket, and also strictly prefers the red ticket to the yellow ticket. Suppose the person held a single probability $p$ for drawing black.

Preferring the red ticket to the yellow ticket is equivalent to

Which values of $p$ fit both preferences?

✅ Correct! Red over black needs

p < \tfrac{1}{3}

and red over yellow needs

p > \tfrac{1}{3}

: no single probability for black produces both, the same contradiction as the video's two urns.

❌ That is the other preference.

p < \tfrac{1}{3}

is what red over black requires. Yellow's chance is

\tfrac{2}{3} - p

; compare that with red's

\tfrac{1}{3}

❌ Not quite. At

p = \tfrac{1}{3}

yellow also has chance

\tfrac{1}{3}

, so the person would be indifferent, not strictly prefer red.

❌ Not quite. At most 40 of the 60 balls are black, so

p \le \tfrac{2}{3}

in every possible urn;

p > \tfrac{2}{3}

cannot happen.

❌ Not quite. At

p = \tfrac{1}{3}

all three colours have chance

\tfrac{1}{3}

, so both strict preferences fail.

❌ Not quite. These values satisfy red over yellow, but red over black needs

p < \tfrac{1}{3}

❌ Not quite. These values satisfy red over black, but then yellow's chance

\tfrac{2}{3} - p

exceeds

\tfrac{1}{3}

, so the person should prefer yellow to red.

❌ Not quite. Compare red's known chance

\tfrac{1}{3}

with yellow's chance

\tfrac{2}{3} - p

❌ Not quite. Put the two conditions on

p

side by side.

Show solution

All three tickets pay the same prize, so the person prefers whichever wins with the higher probability.

The chances are $P(\text{red}) = \tfrac{20}{60} = \tfrac{1}{3}$ , $P(\text{black}) = p$ and $P(\text{yellow}) = \tfrac{2}{3} - p$ .

\text{red over black} \iff \tfrac{1}{3} > p

\text{red over yellow} \iff \tfrac{1}{3} > \tfrac{2}{3} - p \iff p > \tfrac{1}{3}

The two conditions contradict each other, so no single probability for black explains both choices. Both times the person picked the ticket whose odds are known: the line Knight drew between risk and uncertainty.

Problem 4 · Implied Chances That Fall Short

Given: the same urn (60 balls: 20 red; 40 black or yellow, split unknown). Someone will pay at most $40 for a ticket paying $120 if red is drawn, at most $24 for a ticket paying $120 if black is drawn, and at most $24 for a ticket paying $120 if yellow is drawn. Treating each price divided by the prize as the probability it implies (an assumption: the person values a ticket at its expected payoff), find the sum of the implied probabilities of red, black and yellow.

✅ Correct! The implied chances add to less than one, the signature the video describes: the prices cannot come from any single probability for the unknown split.

❌ Not quite. Consistent beliefs would sum to

1

, but compute these prices' implied chances: they fall short.

❌ Not quite.

\tfrac{1}{5} + \tfrac{1}{5} = \tfrac{2}{5}

covers black and yellow only; add red's implied

\tfrac{1}{3}

❌ Not quite.

\tfrac{2}{3}

is the known chance of black or yellow, not the sum of the implied probabilities.

Show solution

Implied probabilities (price ÷ prize):

P(\text{red}) = \tfrac{40}{120} = \tfrac{1}{3}, \qquad P(\text{black}) = P(\text{yellow}) = \tfrac{24}{120} = \tfrac{1}{5}

\tfrac{1}{3} + \tfrac{1}{5} + \tfrac{1}{5} = \tfrac{5}{15} + \tfrac{3}{15} + \tfrac{3}{15} = \tfrac{11}{15} < 1

Red's price matches its known chance. The shortfall sits entirely in the colours whose split is unknown: black or yellow certainly has chance $\tfrac{2}{3}$ , yet the two tickets imply only $\tfrac{2}{5}$ . That is the pattern the video calls aversion to ambiguity.

Solved: 0 / 4