How are stock options priced?

Question

I was reading the wikipedia article about the Black-Scholes model, and it says this:

"The key financial insight behind the equation is that one can perfectly hedge the option by buying and selling the underlying asset in just the right way and consequently "eliminate risk".[citation needed] This hedge, in turn, implies that there is only one right price for the option, as returned by the Black–Scholes formula (see the next section)."

Does this mean that an option is priced in such a way that the buyer and seller of the option should not turn a profit or a loss?

Also how would one hedge the option through buying and selling the underlying asset in order to eliminate risk?

Answer 1

Options pricing is related to game theory.

In sports, you like the Reds, I like the Greens. We wish to bet on a game. We can choose points to give the lower team's final score in order to make such a bet 50/50. Or knowing my Greens are far superior, I offer you odds. "If your Reds win, I will pay you 10 times your bet."

The B-S model does a good mathematical job of looking at historic volatility, and determining the odds of a price being exceeded in a given timeframe.

In Oct 2009 I wrote an article Will Gold Break $1250 by 2011? in which I described a strategy offering a 4 to one return if gold would increase 25% just over a year out. The options market literally said "give us $2400, and if gold closes over $1250 next January, we will give you $10,000." Similar betting was available on the downside. The market thought such a move over the 15 moths was not likely.

My friend Nasim Taleb will tell you that the market, any market, does not follow the bell curve the math produces. That six-sigma event should occur every million years, but in fact, occurs far more often. I have nothing against the model, so long as you understand what it is and what it isn't.