Consider exercise 5.5 from Shreve volume 1:
For part (I), I understand how you can use reflection to show that $P(M_n^*\geq m, M_n=b)=P(M_n=2m-b)$. However, it seems to me that this latter probability is just a binomial, and hence:
$$P(M_n=2m-b)={n\choose 2m-b}(1/2)^n$$
This is not equal to the equation given; for example $n=6,m=2,b=0$ is a counterexample. What am I missing?