Solution:

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 $(2n+1)^{\text{st}}$ 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
$$\frac{p}{1 - p(1-p)} = \frac{1}{2}.$$
Cross-multiplying gives:
$$2p = 1 - p + p^{2}$$
$$0 = 1 - 3p + p^{2}$$
$$p^{2} - 3p + 1 = 0$$
Solving the quadratic equation:
$$p = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$
Since $p$ must be between $0$ and $1$, the solution is:
$$p = \frac{3 - \sqrt{5}}{2}$$