I was watching this video on option greeks
and the guy said (at 34 min): If an option has a delta of 34, it has a 34% probability of expiring in the money? Is it possible to understand it intuitively without getting into the math of Black Scholes formula?
Just for clarification, delta and probability of expiring in the money are not the same thing. What the guy meant was that delta is usually a close enough approximation to the probability.
One way to think about it is to look at the probabilities and deltas of In the Money, Out of the Money, and At the Money options.
In these cases, the delta and probabilities are about the same. In fact if you look at an options chain with delta and probabilities, you can see that they are all about the same. In other words, there is a linear relationship between delta and probability.
Here are a couple links to other answers around the web:
Hope this answer helps!
The delta is the ratio between a price change in the underlying and price change in the derivative. So a delta of .5 means that if the stock price goes up $1, the option will go up $.50. The reason that the delta and the probability of being in the money are (roughly) equal is that an increase in stock price is useful to a holder of the option only if the option ends up being in the money. So if the stock goes up $1, and there's a 50% chance of the option ending up in the money, then the expected value of that stock price increase is $.50.
BTW, joebloggs's characterization of an ATM options having a 50% chance of going up or down is often a reasonable approximation, but is not always true.
What the other answers seem to miss is that there isn't just a vague or qualitative relation between delta and the probability of expiring in the money. Though the two are not equal in general, there is a precise limiting case in which they converge (and so in many practical cases they may be very close). This is the case where the Black-Scholes assumptions hold and the expected relative changes in the underlying price by expiration are small enough to be linearized (e.g., the option is close to expiration). Then the lognormal distribution converges to a plain normal one.
We can then state Acccumulation's observation more precisely, say for a call. If the current underlying price moves $1 higher while the other parameters (including volatility) are unchanged, then the (normal) probability distribution of price at expiration shifts, with its mean, additively upward by $1. The current option value is the expectation of its value at expiration. The in-the-money values increase by $1 while the out-of-the-money values remain unchanged (worthless). By linearity of expectation, the change in option value (delta) therefore equals the in-the-money probability.
The two will diverge for longer-dated options on volatile underlyings, or when Black-Scholes assumptions are strongly violated.
For OTM stock calls (strike $5, stock $1), its a cheap but low probability bet. Option price is not sensitive to stock price as the latter increases a bit. It will be like 0.1 or 0.2x delta.
However, when stock rises reasonably close to strike like $3/share, people think its more likely and start buying option. Delta goes to maybe 0.3-0.35x with some concern stock may fall back.
As stock price goes up and crosses $5 strike, more people get confident and buy more options as much as the rise in stock. Delta goes to 0.7-0.8x. Nearing to the expiry of the option, for each $1 rise in stock value, the option also rise by close to the same. Delta is 0.9-1x.
