What is the probability that the first head will appear on the even numbered tosses
Question
$\text{Consider a coin with probability R to be heads. What is the probability}$ $\text{that the first head will appear on the even numbered tosses?}$
My Approach
let the required Probability$=P$.
Hence we can write our eauation as-:
$$P=(1-R) \times R+(1-R) \times (1-R) \times (1-R) \times P $$
$$P(1-(1-R) \times (1-R) \times (1-R))=(1-R) \times R $$
$$P=\frac{(1-R) \times R }{(1-(1-R) \times (1-R) \times (1-R)}$$
Am i correct?
Answer is given as-:
$$P=\frac{(1 - R)}{(2 - R)}$$
Solution 1:
Alternative approach - to see the first head on an even numbered toss we need to have:
Toss 1 is a tail (probability 1-R)
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.
Solution 2:
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 $p = 0$.
Solution 3:
Your equation for $P$ isn't quite right. Let's write $T_k$ for the outcome of the $k$-th toss: $T_k \in \{H,T\}$, and $P(T_1 = H) = R$ and $P(T_1 = T) = 1-R$. We then need to carefully split cases: $$ P(\text{first heads on even}) = P(\text{first heads on even} \mid T_1 = H)P(T_1 = H) \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = T)P(T_1 = T, T_2 = T) \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = H)P(T_1 = T, T_2 = H) \\ = P(\text{first heads on even} \mid T_1 = T, T_2 = T) \cdot (1-R)^2 \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = H) \cdot R(1-R). $$ Now observe that $$ P(\text{first heads on even} \mid T_1 = T, T_2 = T) = P(\text{first heads on even}) $$ (by the Markov property, if you like), and $$ P(\text{first heads on even} \mid T_1 = T, T_2 = H) = 1 $$ since this is a 'success'. Hence, writing $p = P(\text{first heads on even})$, we have $$ p = p(1-R)^2 + R(1-R). $$
This is the same as you had except that you had a factor $(1-R)^3$. I've put in all the details so you can see exactly why it's a power of 2, not a power of 3.
Intuitively, I think you've got the idea, except that you were thinking that if we don't get $TH$, then we need to go $TTT$ to not fail -- which is correct -- but this first $T$ it taken into account with the probability $p$, so we could write down straight away $$ p = R(1-R) + p(1-R)^2; $$ the danger of this, as you have found out, however, is that if there's a slight mistake, then it's not at all clear where it came from.
Hopefully this helps! :)