I'm trying understand something basic about Black-Scholes pricing of binary options. In my example above, the current price is over the strike price. The volatility is extreme but I'm still having trouble understanding why the price of the binary option (which I'm interpreting as the probability of expiring in the money) would be below 50 (50% odds). Assuming a random walk from the current price, isn't it more likely that it would expire above the strike?

Black-Scholes gives an implied price of ~ 0.390. Which I interpret as a 39% chance of expirying in the money? Why wouldn't' it be more then 50%?

S = 110 #current_price
K = 100 #ATM strike
v = 1.20 #annualized volatility
r = 0.00 #interest rate
T =  0.44 #days remaining (annualized)

d2 = (log(S/K) + (r - 0.5 * v**2) * T) / (v*sqrt(T))
print exp(-r * T) * norm.cdf(d2)


  • $\begingroup$ You are missing the brackets around v*sqrt(T) $\endgroup$
    – Phil-ZXX
    Jul 27, 2018 at 21:02

1 Answer 1


$S_T$ is log-normal distributed and therefore skewed. In particular $E[S_T]=S=110$ (no drift), but $Q(S_T>S)<Q(S_T<S)$. For example if S=K=100 you don't get a value of 0.5 as you might expect, but a lower value since you have norm.cdf($-0.5\sigma\sqrt{T}$)$<$ norm.cdf($0$)=0.5.

enter image description here

  • $\begingroup$ Thanks Andrew, just want to make sure I understand E[S(T)]=S=110. E is expected value which is also the mean? So while the expected value is 110, more the probability distribution at T falls below the K of 100? $\endgroup$
    – Snapula
    Jul 30, 2018 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.