More Heads Than Tails

Use symmetry between heads and tails, and handle the tie case (5-5) separately.

$P(\text{more heads})=\frac{1-P(\text{tie})}{2}$.

1. 1. Symmetry

   By symmetry, $P(H>T)=P(T>H)$, and these two events plus the tie event $H=T=5$ partition the space.

2. 2. Tie probability

   P(exactly 5 heads) is $\left(\genfrac{}{}{0ex}{}{10}{5}\right)/2^{10}=252/1024$.

   $$P(H=5)=\frac{252}{1024}$$

3. 3. Solve

   So $P(H>T)=\frac{1-252/1024}{2}=\frac{772/1024}{2}=\frac{386}{1024}=\frac{193}{512}$.

   $$P=\frac{193}{512}$$

Answer: 193/512