I am waiting for you to Toss Toss

It only makes a difference who goes 1st is if Jess and I toss our 3rd head after the same number of turns. The probability of rolling one's 3rd head after k turns is $\binom{k-1}{2} 2^{-k}$ . Therefore the probability us rolling our 3rd head after the name number of turns is $\sum (\binom{k-1}{2} 2^{-k})^{2} = \frac{1}{4}\sum (\binom{k}{2} 2^{-k})^{2}$

I'm not sure of the best way to evaluate this sum. What I did was consider the function $\frac{1}{1-x}=\sum x^{k}$ , the differentiate twice and multiply by $x^{2}$ , then differentiate twice and multiply by $x^{2}$ . You end up with the probability being $\frac{345}{3^7} = 0.112$ .

Now the probability of me winning is the probability of my going 1st being the deciding factor + 1/2 * the probability of my going first not being the deciding factor. $0.112+\frac{1-0.112}{2} = 0.568$