# Black-Scholes pricing of binary options

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)


0.390...

• You are missing the brackets around v*sqrt(T) – Phil-ZXX Jul 27 '18 at 21:02

$S_T$ 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.