Six Dollars, Whatever You Believe About the Odds

A share trades at 50 and by the end of the period it will be at 65 or at 40. You think the up state has a seventy percent chance. I think it has fifty. We are not going to agree, and it does not matter, because we have to agree on what the right to buy the share at 50 is worth.

It is worth 6. Not approximately, not under a convention, and not because we happen to share a taste for risk. The number contains no probability at all, and seeing why is the whole content of the question.

The discount rate that ruins it

The instinct trained by corporate finance is to average the payoff and discount it. At a seventy percent chance the option pays 15 or nothing, so the mean payoff is 10.50. The share itself is expected to go from 50 to $0.7 \times 65 + 0.3 \times 40 = 57.50$, a required return of 15 percent. Discount the option at the same 15 percent:

The expectation in (1) is right. The rate is wrong, and it is wrong at every belief you could hold: at a fifty percent chance the recipe returns 7.1429, at ninety percent 10.80, at ninety-nine percent 11.4672. Four different beliefs, four different wrong answers, and the option was 6 the whole time.

Build the payoff instead of forecasting it

Forget probabilities and ask a shopping question. How much share and how much cash reproduces the option in both states? The share count is the payoff spread divided by the price spread, and the cash is whatever makes the down state come out right:

The bill for those parts is $\tfrac{3}{5}\times 50 - 24 = 6$. If the option traded anywhere else, you would buy the cheap one, sell the dear one, and hold two positions that cancel in every state while pocketing the difference today. That is the entire argument, and it is finished.

There is no probability in the bill

Look at what (2) used: two future share prices, one strike and one rate. Write the general version, with the strike anywhere between the two outcomes, and your belief is not among the symbols:

The derivative in (3) is zero for a reason stronger than any numerical check could give: the variable is absent. A sweep over 999 values of the probability returns one single cost, and it is the same fraction each time. That is the point the question is really testing, because a sweep can only ever fail to find a dependence, while an absent symbol proves there is none.

Once the price cannot depend on beliefs, you may compute it under whichever beliefs are convenient. The convenient ones are the beliefs under which nothing earns a premium.

Risk neutrality as a change of bookkeeping

Find the probability that makes the share's own expected value equal its price compounded at the riskless rate, and use it on the option:

The same 6, from an unrelated calculation. Both routes agree across 540 parameter sets crossing spots, strikes, up and down moves and rates. Notice what $q$ is not. It is not a forecast, nobody holds it, and it has no claim to describe the world. It is the weight that turns a payoff into a price, and calling it a probability is a convenience of notation.

It is also where the no-arbitrage condition hides. Equation (4) gives a number in $(0,1)$ exactly when $S_d < (1+r)S < S_u$. If the riskless asset beat the share in both states, or lost to it in both, there would be a free lunch in the share alone and no consistent price for anything written on it.

The rate that does work, and why you cannot use it

The discounted-expectation approach is not wrong in form. It was wrong in the rate. Discount the option's mean payoff at the return the option itself must earn and it works perfectly:

And there is the circularity. The 75 percent in (5) was computed from the price, so it cannot be an input to the price. Change your belief and it moves: 25 percent at a fifty percent chance, 125 percent at ninety, 147.5 percent at ninety-nine, while the price sits at 6 through all of them. Your beliefs determine the return the option must earn; they do not touch what it costs.

The figure hands over a small bonus. The required return crosses zero at a belief of exactly $0.4$, which is the $q$ of equation (4). The artificial probability is the one belief under which this option would deserve no premium at all, which is the cleanest one-line definition of what risk-neutral pricing is doing.

Where the argument stops

Everything above rested on a portfolio existing. Two states and two instruments is exactly enough to solve two equations, and that balance is fragile in one direction only. Add a third possible ending price and there are three equations for two unknowns, no replicating portfolio in general, and the artificial probabilities are no longer unique. The price is then pinned only between bounds, and the bounds are wide.

Even in the two-state world the model-free bounds are worth carrying, because they are all you keep when replication fails: $\max\bigl(S - K/(1+r),\,0\bigr) \le c \le S$, which here reads $0 \le 6 \le 50$. Those hold from arbitrage alone with no model at all, and they are the last thing to go.

The other assumption is that the shopping list can actually be bought. Frictionless trading at one price, unlimited borrowing at the riskless rate, and no cost to holding the hedge. Put a spread on the share and the single price 6 spreads into an interval, wide enough on an illiquid name that the preference-free argument stops deciding anything. The theorem is exact; what it is exact about is a market that does not quite exist.

Sources and further reading

• The result that (3) and (4) illustrate — Fundamental theorem of asset pricing
• The weights of equation (4), and what they are not — Risk-neutral measure
• The one-period model extended to a tree — Binomial options pricing model
• The condition that fails when a third state is added — Complete market
• The rate the first section misused — Cost of capital

Every figure here is exact rational arithmetic rather than floating point: the replication was checked state by state, the independence in (3) by symbolic differentiation rather than by a sweep, the agreement between replication and equation (4) over 540 parameter sets, and the 9.1304 of equation (1) against the exact fraction it approximates.