The original question is quoted below.
The underlying stock price is now \$100, and tomorrow it will be either \$101 (with probability $p$) or \$99 (with probability $1-p$). A call option with value $c$ which expire tomorrow has exercise price \$100. Find the the value of $c$ under the Black-Scholes model. Ignore interest rate.
When I tackle this question, I first derive $p=\frac{1}{2}$. To use call option price formula, we need $S, E, r, T-t, \sigma$. From the question, it is clear that $$S=100, E=100, r=0, T-t=\frac{1}{365}$$ So we only need $\sigma$. Since $\sigma$ is measured by the standard deviation of the return $\frac{dS}{S}$, I proceed as follow: $$E(return)=(1/100)(0.5)+(-1/100)(0.5)=0$$ $$Var(return)=(1/100-0)^2(0.5)+(-1/100-0)^2(0.5)=0.0001$$ $$sd(return)=\sqrt{0.0001}=0.01$$ Apply the explicit price formula for call option under the Black-Scholes model, I found $$d_1=0.00026171, d_2=-0.00026171, N(d_1) = 0.500104407965456, N(d_2)=0.499895592034544$$ Thus, the desired price is $$SN(d_1)-Ee^{-r(T-t)}N(d_2)=0.0209$$ The procedure seems logical to me. However,since the profit from the call option is $$(1)(0.5)+(0)(0.5)=0.5$$ I expected the price to be close or equal to \$0.5. How come they differ so much?