Odds of one person getting more odd numbers on a die than another

First player rolls a 6-sided die 100 times and the second player 101 times. What are the odds of the second player getting more odd numbers than the first one?

I tried solving this via listing the options which led to a nasty sum that I wasn't able to solve neither by hand nor by using wolfram alpha. Clearly there has to be a more clever solution using the properties of a binomial distribution, but so far I haven't come across anything useful.

Solution 1:

It's $\frac 12$.

Let $\psi$ be the probability that they are tied after $100$ throws (from each).

There are two ways player $2$ can win:

A. Player $2$ can be ahead after $100$ throws from each, probability $\frac 12\times (1-\psi)$.

B. They can be tied at $100$ and player $2$ might throw an odd number on the $101^{st}$, probability $\frac 12\times \psi$.

Thus the total probability that player $2$ wins is $$\frac 12\times (1-\psi)+\frac 12\times \psi=\boxed {\frac 12}$$