What is the probability that the first head will appear on the even numbered tosses

Alternative approach - to see the first head on an even numbered toss we need to have:

1. Toss 1 is a tail (probability 1-R)

2. Renumbering Toss 2 as Toss 1, Toss 3 as Toss 2 etc. we are now want to see first head on an odd numbered toss

So

P(first head on even toss) = P(first head on odd toss) x (1-R)

but if we denote P(first head on even toss) by p then P(first head on odd toss) = 1-p, so

$p = (1-p)(1-R)$

$\Rightarrow p + p(1-R) = 1-R$

$\Rightarrow p = \frac{1-R}{2-R}$

Sanity checks: (i) p=0 when R=1 (ii) p approaches 1/2 as R approaches 0.

The probability that the first head will appear on the second toss is $(1 - R)R$.

The probability that the first head will appear on the fourth toss is $(1 - R)^3R$.

The probability that the first head will appear on the sixth toss is $(1 - R)^5R$.

In general, the probability that the first head will appear on the $2k$th toss is $(1 - R)^{2k - 1}R$.

Hence, the desired probability is $$p = \sum_{k = 1}^{\infty} (1 - R)^{2k - 1}R = (1 - R)R\sum_{k = 1}^{\infty} (1 - R)^{2k}$$ which is a geometric series with common ratio $(1 - R)^2$. If $R < 1$, we obtain \begin{align*} p & = (1 - R)R \cdot \frac{1}{1 - (1 - R)^2}\\ & = \frac{(1 - R)R}{1 - (1 - 2R + R^2)}\\ & = \frac{(1 - R)R}{2R - R^2}\\ & = \frac{(1 - R)R}{R(2 - R)}\\ & = \frac{1 - R}{2 - R} && \text{provided that $R \neq 0$} \end{align*} If $R = 1$, then heads will be obtained on the first toss, so $p = 0$. If $R = 0$, then heads will never be obtained, so again $p = 0$.