Monte Carlo Option Pricing

Risk-neutral pricing says a European option is worth its discounted expected payoff under $Q$: $V_0 = e^{-rT} E^Q[h(S_T)]$. Under $Q$ the stock is lognormal, so one draw of $Z \sim N(0,1)$ gives a terminal price

Monte Carlo pricing draws $n$ independent values of $Z$, computes the payoff on each, and averages:

The law of large numbers makes $\hat V_n$ converge to $V_0$. Take $S_0 = K = 100$, $r = 5\%$, $\sigma = 20\%$, $T = 1$, a call whose Black-Scholes price is 10.45. The draw $Z = 1$ gives $S_T = 100e^{0.23} = 125.86$ and a discounted payoff of $25.86\,e^{-0.05} = 24.60$. The draw $Z = -1$ gives $S_T = 84.37$, and the call pays nothing. Those two paths average to 12.30, a poor estimate, but a million paths land within a few cents of 10.45.

A European payoff only needs $S_T$, so each path is one normal draw. A path-dependent payoff (Asian, barrier) needs the whole path, built step by step with the same formula over each $\Delta t$.