This is probably a pretty really easy equation to derive, but I’m a math idiot. I want to know what the premium tells us about what the market views as the probability of a the stock price in the future. Example: Let’s say stock is at 20 dollars today: The call has a strike price of 30 The call’s premium is 1 dollar 1 year to expiration. According to those numbers, what are the odds that the stock price will be in the money 1 year from today? Sorry if this is poorly worded.

vols?

Anyone who wants to explain this can make any assumptions they want about the Greeks. It might be easier to do using a put because then you don’t have to explain a theoretically infinite loss as a person writing a call would.

Here’s a free online calculator that will show you the probability of closing ITM: http://www.hoadley.net/options/optiongraphs.aspx?

Hi dude. Assuming you are using a lognormal return distribution, you need to know two out of three of the following: vols, dividend yield, 1 year interest rate. Or if you assume zero dividends, you need to know one of the other two. Step 1 is to solve for the implied volatility using the Black Scholes formula and the inputs that you mentioned. There isn’t a closed form way to do this. You need to guess a starting vol, then move the vol up or down until you get the price that matches your question. http://en.wikipedia.org/wiki/Black–Scholes The probability of the call being in the money is Phi(d2), which is one of the intermediate calculations from the Black Scholes formula.

My understanding is that there is a rule of thumb that the delta of an option can be interpreted (roughly) as the chance that the option will expire in the money. Don’t know if that’s close enough for you. Clearly the delta is in the BS equation that tells you the price (and therefore the time value, indirectly), but there is more to it than that. My sense is that the delta of the option tells you the chance that the option will be in the money *tomorrow* (or the next instant), given where it is now. The approximation probably gets messed up with long-dated options because most assets have an expected appreciation between now and expiration. For example, if the option is ATM right now, it’s delta would be 0.50, which makes sense because a day from now, the underlying might be higher or lower with roughly equal probability, but if it’s an ATM LEAP, then the chance of the underlying being in the money is at expiry (let’s say a year from now) is considerably higher than 50%, because we know that the stock is ATM today and likely to be at a higher price in a year, barring any systemic catastrophes or corporate screw-ups. If the option is deep ITM, then its delta is almost 1, and so is the probability of being ITM tomorrow. If it is deep OTM, then its delta is almost 0, as is the probability of being OTM tomorrow. I don’t really deal with options much on a daily basis, and it’s been a while since I’ve done L2, but I do find this calculation handy.

LPoulin’s website doesn’t work for me. but morningstar has a “chance of breakeven” column for real-life options. http://quote.morningstar.com/Option/Options.aspx?ticker=CSCO

strictly speaking, the probability of the call expiring in the money at maturity is N(d2) from the black-scholes formula, as Ohai said. even more strictly speaking, this is the risk-neutral probability. how do you want to use it, is the key question. if you intend to use it somehow in the same sense as we use everyday probabilities ("the market thinks the likelihood of the option finishing in the money is X%), i wouldnt touch it. it could be very misleading, it’s probably only useful to call this thing “probability” if you are writing an academic paper. again, technically, to get to the “real-world” probability from the black-scholes formula, the simplest adjustment is to substitute the risk-free rate with the rate of return for the stock (perhaps from a CAPM). this won’t give you the correct option price of course, but the N(d2) that you read from the formula will, in theory, show the “real-world” probability as opposed to the “risk-neutral” one. The real world probability is always higher. this approach above is all too academic, and requires extra effort. perhaps that’s one of the reasons to just use the delta, N(d1) as a proxy, as Bchad said. N(d1) is always higher than N(d2), so in a sense the delta N(d1) is a closer estimate for the real-world probability than the actual risk-neutral probability N(d2). finally, if the options market is rich you could probably extract the delta from option prices without relying on greeks or the black-scholes formula. say you have two calls with strikes K1, K2 fairly close to each other: call(S,K1) and call(S,K2). you can use the equality K1/K2*call(S,K2)=call(S*K1/K2,K1). then delta=[call(S,K1)-K1/K2*call(S,K2)]/[S-S\*K1/K2]. This would only work well if K1 and K2 are strikes fairly close to each other. The advantage of the approach, however, is that it’s “model-independent” so you dont need black-scholes or any other assumptions about the stock price movement.

First of all you need to get a correct vol model, realized volatility is biggest sucker in this game, so don’t trust that! And as people have noted, trader’s back of envelope calculation for your question is delta, but that’s like when you are seeing 100 contracts in one minutes! And I don’t think delta or N(d1) would be such helpful for such longer maturity (1 year), because your delta itself depends on time to maturity and can increase/decrease depending on it’s moneyness and the rate of increase/decrease on how much “in the money”/“out of money” the option is. Basically, if you are trading for short term, near maturity options, then it makes sense, gives you basic idea about how much in the water you are. But again, a swing in market will scrap al your calculations, you can never make a deterministic guess with options, mostly you can enhance your risk/return metric or trade the sentiment or maybe hedge!