Finance-Theory · Unit 22 · Video 4 · Interactive Practice

The Dutch Book Theorem: Inconsistent Beliefs Create Arbitrage

IKey Formulas

Formula	Name	What it says
$P(A) + P(\text{not } A) = 1$	Coherence	The only defence against a Dutch book
$\text{E}[B] = P(A)\,w - \big(1 - P(A)\big)\,\ell = 0$	A bet that looks fair	Win $w$ if $A$ , lose $\ell$ if not: fair at odds $\ell : w = P(A) : \big(1 - P(A)\big)$
$\text{price} = 100 \cdot P(E)$	Ticket view	Your value for a ticket paying $100 if $E$ happens
$\text{Profit} = 100\,\big\lvert P(A) + P(\text{not } A) - 1 \big\rvert$	Dutch book profit	Per $100 pair of tickets, earned in every state

Key Insight: Each bet is judged on its own belief and looks fair on its own; the incoherence only shows when the bets are combined, and then the other side wins in every state.

IIVisualization 1 — The Ticket View

Priced at your beliefs, the two $100 tickets cost exactly $100 together only when your probabilities sum to one.

IIIVisualization 2 — Scaling the Two Bets

With the video's beliefs $P(A) = \tfrac{1}{2}$ and $P(\text{not } A) = \tfrac{3}{4}$ , which stakes on $B_1$ and $B_2$ win in both states?

IVQuiz Questions

Problem 1 · Pricing the Pair of Tickets

Given: You believe $P(A) = \tfrac{4}{5}$ and $P(\text{not } A) = \tfrac{1}{3}$ . Each ticket pays $120 if its event happens, and you will trade either ticket at the price your belief makes fair. Find the riskless profit the other side makes from one pair of tickets.

✅ Correct! You pay $136 for a pair that returns $120 whatever happens, so the other side keeps $16 in every state.

❌ That is the price of the pair. You pay $136 for both tickets, but exactly one of them pays $120, so the other side's profit is the difference.

❌ The beliefs are added, not subtracted. Both tickets are sold to you, so their prices add; the profit is how far the sum overshoots one,

\tfrac{4}{5} + \tfrac{1}{3} - 1 = \tfrac{2}{15}

, times the $120 ticket.

❌ Check the ticket size. The overshoot

\tfrac{2}{15}

is right, but these tickets pay $120, not $100.

Show solution

Your price for each ticket is the payout times your probability:

120 \cdot \tfrac{4}{5} = 96, \qquad 120 \cdot \tfrac{1}{3} = 40

So you pay $96 + $40 = $136 for the pair. Exactly one of $A$ and not $A$ happens, so exactly one ticket pays, and the other side hands back $120 in either state:

136 - 120 = 16 = 120\Big(\tfrac{4}{5} + \tfrac{1}{3} - 1\Big) = 120 \cdot \tfrac{2}{15}

The profit is $16, the same in every state.

Problem 2 · Which Side of the Trade?

Given: You believe $P(A) = \tfrac{2}{3}$ and $P(\text{not } A) = \tfrac{1}{2}$ , and you will buy or sell, at your own prices, tickets that pay $60 if their event happens. Which trade gives the other side a riskless profit?

✅ Correct! Your prices sum to $70 for a pair worth $60 in every state, so the other side sells to you and keeps $10 whatever happens.

❌ That is the wrong direction. Buying both for $70 and collecting $60 loses $10 in every state. Buying both works only when your probabilities sum to less than one.

❌ One ticket alone carries risk. Selling only the ticket on

A

for $40 gains $40 if

A

fails but loses $20 if

A

happens; the riskless profit needs both tickets.

❌ Fair one at a time is the trap. Each ticket does look fair on its own belief, but together they cost $70 for a sure $60 payout.

Show solution

Your prices: ticket on $A$ is $60 \cdot \tfrac{2}{3} = 40$ , ticket on not $A$ is $60 \cdot \tfrac{1}{2} = 30$ , so the pair costs $70.

Since $\tfrac{2}{3} + \tfrac{1}{2} = \tfrac{7}{6} > 1$ , the pair is overpriced: the other side sells you both. Exactly one ticket pays $60, so the other side's profit is

70 - 60 = 10 = 60\Big(\tfrac{2}{3} + \tfrac{1}{2} - 1\Big)

in every state. Selling only one ticket is a risky bet, and buying both loses $10 for sure.

Problem 3 · Building the Book from Two Bets

Given: You believe $P(A) = \tfrac{2}{3}$ and $P(\text{not } A) = \tfrac{3}{5}$ , so you accept two bets. $B_1$ : you receive $1 if $A$ occurs and pay $2 if it does not. $B_2$ : you receive $2 if $A$ does not occur and pay $3 if it does. I take the other side of 30 copies of $B_1$ . Find how many copies of $B_2$ I take so my profit is the same in both states, and that profit.

How many copies of $B_2$ ?

My profit in each state?

✅ Correct! With 30 copies of

B_1

and 18 of

B_2

I make $24 whether or not

A

occurs.

❌ That only breaks even if $A$ occurs. With 10 copies, if

A

occurs I pay $30 on

B_1

and collect $30 on

B_2

: zero, while I make $40 if

A

fails.

❌ Equal copies are not equal profits. With 30 copies of each I make $60 if

A

occurs but $0 if it does not: no loss, but no riskless profit either.

❌ The ratio is upside down. With 50 copies of

B_2

I lose $40 if

A

does not occur. Set the two state profits equal and solve.

❌ Check the copies. Write my profit in each state as a function of the copies of

B_2

and set the two equal.

❌ That is the number of copies of $B_1$ . Compute the profit in a state: if

A

occurs I pay $30 on

B_1

and collect

3 \times 18

B_2

❌ That is only what $B_2$ pays me if $A$ occurs. Subtract the $30 I pay you on

B_1

in that state.

❌ That is the profit with 10 copies, and only if $A$ fails. With the equalizing number of copies the two states give the same, smaller profit.

❌ Check the profit. Add up my two positions in one state of the world.

Show solution

Both bets look fair to you: $\text{E}[B_1] = \tfrac{2}{3}(1) - \tfrac{1}{3}(2) = 0$ and $\text{E}[B_2] = \tfrac{3}{5}(2) - \tfrac{2}{5}(3) = 0$ .

I take the other side of 30 copies of $B_1$ and $k$ copies of $B_2$ . My profit in each state:

A \text{ occurs: } -30 + 3k, \qquad A \text{ does not occur: } 2(30) - 2k = 60 - 2k

Setting them equal: $-30 + 3k = 60 - 2k$ , so $5k = 90$ and $k = 18$ .

-30 + 3(18) = 24, \qquad 60 - 2(18) = 24

I make $24 in every state. Check with the ticket view: $\tfrac{2}{3} + \tfrac{3}{5} - 1 = \tfrac{4}{15}$ , and this book is a pair of $90 tickets, $90 \cdot \tfrac{4}{15} = 24$ .

Problem 4 · Which Beliefs Are Safe?

Given: Four investors each state a pair $\big(P(A),\ P(\text{not } A)\big)$ . Which investor cannot be Dutch-booked on this event?

✅ Correct!

\tfrac{1}{3} + \tfrac{2}{3} = 1

: coherent. Coherence does not fix the numbers, it only requires the sum to be one.

❌ That sum is $\tfrac{16}{15}$ . It overshoots one, so the other side sells this investor both tickets.

❌ That sum is $\tfrac{14}{15}$ , below one. A shortfall is exploitable too: the other side buys both tickets cheaply and collects the full payout in either state.

❌ That sum is $\tfrac{7}{6}$ . It overshoots one, so selling both tickets is a free lunch.

Show solution

Add each pair:

\tfrac{2}{3} + \tfrac{2}{5} = \tfrac{16}{15}, \quad \tfrac{1}{3} + \tfrac{3}{5} = \tfrac{14}{15}, \quad \tfrac{1}{3} + \tfrac{2}{3} = 1, \quad \tfrac{2}{3} + \tfrac{1}{2} = \tfrac{7}{6}

A sum above one is booked by selling both tickets; a sum below one by buying both. The profit vanishes only when the sum is exactly one, so only $\big(\tfrac{1}{3}, \tfrac{2}{3}\big)$ is safe.

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