Biased coin, probability of landing first heads on odd/even I am currently stuck with this question, as I prepare for my exam. I'm not sure what kind of answer it is looking for:
A biased coin is tossed until a head appears for the first time. Let $p$ be the probability of head, $0 < p < 1$.
a) What is the probability that the number of tosses required is odd?
b) How about even?
Here's what I have so far:
\begin{align*}
P(1) &= p\\
P(2) &= (1-p)p\\
P(3) &= (1-p)(1-p)p\\
P(4) &= (1-p)(1-p)(1-p)p\\
\end{align*}
and so on. I understand that it is multiplying by $(1-p)$ for each extra toss. How would I put this in terms of a probability?
 A: What are the probabilities that we have to keep in mind when solving this question? You have got the gist, namely that $(1-p)$ represents failure and $p$ a success.
Now, suppose there is success the first time itself. The probability of this happening is $p$.
Now, suppose there is success the third time . The probability of this happening is $p(1-p)^2$, because you have to fail twice before succeeding.
Now, suppose there is success the fifth time . The probability of this happening is $p(1-p)^4$, because you have to fail four times before succeeding.
Can  you see what the pattern is? The probability of then getting a head on an odd turn is $p(1 + (1-p)^2 + (1-p)^4 + \ldots)$.
$$
p(1 + (1-p)^2 + (1-p)^4 + \ldots) = p\displaystyle\sum_{i=0}^\infty ((1-p)^2)^i = \frac{p}{1-(1-p)^2} = \frac{1}{2-p}.
$$
Similarly, the probability that you will get heads on an even turn is $1-\frac{1}{2-p} = \frac{1-p}{2-p}$
Note: $0<p<1$, otherwise the geometric series convergence is questionable.
A: You have $P(N=n)=(1-p)^{n-1}p$. $\quad\color{seagreen}{\checkmark}$
And have noticed that $P(N=2k)=(1-p) P(N=2k-1)$ so $$\begin{align}\sum_{k=1}^\infty P(N=2k)~=~&(1-p)~\sum_{k=1}^\infty P(N=2k-1) \\[3ex] P(N\in2\Bbb N^+)~=~&(1-p)~P(N\in 2\Bbb N^+-1)\end{align}$$ So, since $P(N\in2\Bbb N^+)+P(N\in 2\Bbb N^+-1)=1$  therefore...
