An 80% Chance of $20 Does Not Make the Option Worth $16

A stock trades at 100 today. In a year it will be worth 130 or 70, and you are told the real chances: 130 happens with probability 0.8, 70 with probability 0.2. Interest rates are zero. What is the right to buy the stock at 110 in a year worth today?

The payoff is easy. At 130 the right to buy at 110 is worth $130 - 110 = 20$, and at 70 it is worthless, because nobody exercises a right to pay 110 for something trading at 70. So it pays 20 with probability 0.8 and 0 otherwise, an expected payoff of 16. Rates are zero, so there is nothing to discount. The answer is 16.

The answer is 10. And the calculation above contains no arithmetic error at all, which is what makes it worth taking apart slowly.

Build the thing out of stock and a loan

Start somewhere that involves no probability whatsoever. Ask what portfolio of stock and cash pays exactly 20 in the up state and exactly 0 in the down state. Let it hold $\Delta$ shares and borrow $B$ in cash. The two conditions are

Subtract one from the other and the cash cancels, which fixes the share count immediately:

Substituting back gives $B = 70/3$. So the portfolio is a third of a share, funded by borrowing $70/3$. Check it against both futures. If the stock rises, $130/3 - 70/3 = 20$. If it falls, $70/3 - 70/3 = 0$. Those are the payoffs of the option, in every state of the world, with no exceptions and no residual risk.

Two things that pay the same amount in every possible future have to cost the same amount today. Otherwise you buy the cheap one, sell the dear one, pocket the difference and walk away with a position that nets exactly zero at the horizon whatever happens. The portfolio costs

Ten. Notice what did not appear anywhere in that derivation: the numbers 0.8 and 0.2. They were never used, because the portfolio matches the option state by state, and a match in every state does not care how likely the states are.

The same 10, dressed as an expectation

There is a second route, and it is the one usually taught first. Look for the pair of chances under which today's price is the fair average of next year's. Write it as an equation in $p$:

This is linear in $p$ and has the unique solution $p = 1/2$. Price the option as an expectation under those chances instead of the real ones:

With $R = 1$ that reduces to $p = (S - d)/(u - d)$, and here $(100 - 70)/(130 - 70) = 1/2$. It is worth being explicit about how much of a coincidence that clean one half is. It comes out only because rates are zero and 100 sits exactly halfway between 130 and 70. Shift the spot to 110 with the same two outcomes and $p = 40/60 = 2/3$. Nobody should walk away from this problem believing the balancing chances are always even.

What is actually wrong with 16

The failure is not the multiplication. $0.8 \times 20 = 16$ is exact, and it is the correct expected payoff of the option under the real probabilities. What is missing is a discount rate, and the reason it is missing is that the problem never gave you one.

Look at the stock under the real chances. Its expected price in a year is

against a spot of 100. Nobody is paying 100 today for something expected to be worth 118 in a year unless they are being compensated for bearing risk, so the stock carries a risk premium of about 18 percent. A real-world expectation of a risky payoff has to be discounted at a rate that reflects that risk, and 118 discounted properly comes back to 100.

The option is a geared claim on the same stock. Equation (2) says it behaves like a third of a share bought with borrowed money, so its own risk premium is larger than the stock's, and it is nowhere in the data you were given. Taking the expected payoff of 16 and discounting it at the risk-free rate of zero is exactly the step that has no justification. It prices a risky claim as if it were safe.

Replication sidesteps the whole question by never forming a real-world expectation in the first place. Sweep the real up-probability across every value in $(0,1)$ and the naive expectation of the payoff wanders over almost the entire interval from about 0.02 to 19.98, while the replicating portfolio and its cost of 10 do not move at all. That invariance is the lesson, and the 16 is just the value the naive route happens to produce at 0.8.

Where the argument stops working

Two states and two traded assets, the stock and cash, make the system in equation (1) a square system with an invertible matrix. That is what completeness means here, and it is the reason the price is a single number rather than a range.

Add a third possible outcome, say 130, 100 or 70, and equation (1) becomes three conditions on two unknowns. In general there is no solution, no replicating portfolio, and no unique price. What survives is a band: the set of prices that admit no arbitrage, whose endpoints come from the cheapest portfolio that dominates the option and the dearest one it dominates. Pinning a number inside that band requires a genuine assumption about preferences or about the dynamics, which is a different kind of statement from anything used above.

The other quiet assumption is that you can trade both assets freely in both directions at the quoted prices. Short selling, borrowing and lending at the same rate, no fees. Relax those and equation (3) becomes a bid and an ask rather than a price. Nothing in the reasoning breaks; the conclusion just softens from a point to an interval.

Sources and further reading

• The model this problem is one step of — Binomial options pricing model
• The measure behind equations (4) and (5) — Risk-neutral measure
• Why matching payoffs force matching prices — Rational pricing
• The completeness argument in the last section — Complete market

Every figure above was verified in exact rational arithmetic rather than floating point, which matters here because 0.8 is not exact in binary. The price 10 was confirmed three ways: by the risk-neutral expectation, by solving the two-state linear system for the share count and the cash position independently, and by checking that every other quote tested (0, 9, 11, 16 and 20) admits a position that gains in both states. Sweeping the real probability over every value $k/1000$ for $k = 1, \dots, 999$ left the replication price at exactly 10 throughout, and a 20000-draw simulation of the hedged position produced a profit and loss set containing nothing but zero.