The best answer to "how would you price an option without Black-Scholes", and a favourite because it exposes whether you understand replication.
One-step setup
Stock at 100. In one period it goes to 120 or 80. Interest rate zero. Price a call struck at 100.
The call pays 20 in the up state, 0 in the down state.
Replicate it
Find a portfolio of stock and cash that pays the same in both states. Hold d shares and B cash:
- Up: 120d + B = 20
- Down: 80d + B = 0
Subtracting: 40d = 20, so d = 0.5, and B = -40.
So the option is exactly equivalent to holding half a share and borrowing 40. That portfolio costs 0.5 times 100 minus 40 = 10.
The call is worth 10. If it traded anywhere else, you would arbitrage the difference.
The point
Notice what never appeared: the probability of an up move. It is irrelevant. Two traders who violently disagree about whether the stock will rise must still agree on the option's price, because the replication argument does not care.
This is the single most important idea in derivatives pricing, and stating it explicitly is what a good answer does.
Risk-neutral probabilities
Rewrite the answer as a weighted average of payoffs:
10 = q(20) + (1-q)(0), so q = 0.5
That q is the risk-neutral probability - the weight under which discounted prices are martingales. It is not anyone's belief about the world; it is the pricing weight implied by no-arbitrage.
In general q = (1 + r - d) / (u - d) for up factor u and down factor d.
Extending
Add steps and the tree branches. Each node is priced by the same one-step argument working backwards from expiry. As the number of steps grows and each becomes small, the model converges to Black-Scholes.
The binomial tree also handles American options naturally: at each node, compare the continuation value with the immediate exercise value and take the larger. Black-Scholes cannot do this, which is why trees survive in practice.