9. [6] Newton and Leibniz are playing a game with a coin that comes up heads with probability $p$. They take turns flipping the coin until one of them wins with Newton going first. Newton wins if he flips a heads and Leibniz wins if he flips a tails. Given that Newton and Leibniz each win the game half of the time, what is the probability $p$ ?
Answer: $\frac{3-\sqrt{5}}{2}$ The probability that Newton will win on the first flip is $p$. The probability that Newton will win on the third flip is $(1-p) p^{2}$, since the first flip must be tails, the second must be heads, and the third flip must be heads. By the same logic, the probability Newton will win on the $(2 n+1)^{\text {st }}$ flip is $(1-p)^{n}(p)^{n+1}$. Thus, we have an infinite geometric sequence $p+(1-p) p^{2}+(1-p)^{2} p^{3}+\ldots$ which equals $\frac{p}{1-p(1-p)}$. We are given that this sum must equal $\frac{1}{2}$, so $1-p+p^{2}=2 p$, so $p=\frac{3-\sqrt{5}}{2}$ (the other solution is greater than 1 ).