Probability that landing the second heads requires at least $n$ tosses

Question

Calculate probability that landing the second heads requires at least $n$ tosses of a $p$-biased coin.

I'm having trouble calculating this possibility, since the variables are tripping me up.

Answer 1

You know that last event should be heads, besides you can order the possible sequences by the number of events needed to arrive to second heads.

So, if you have $2$ coin tosses then the only possible sequence is $HH$.

For $3$ coin tosses you can have $THH,HTH$ .

For $n$ coin tosses you can place a $H$ at the end and then choose where to put the first heads on the other $n-1$ positions, so there are $n-1$ possible sequences.

Because the coin is p-biased, you can write the probability of the required outcome as :

$$p^2+2(1-p)p^2+3(1-p)^2p^2+...+n(1-p)^{n-1}p^2$$

which can be expressed as :

$$\sum_{k=1}^{n} k(1-p)^{k-1}p^2 = \frac{p^2}{1-p} \sum_{k=1}^{n} k(1-p)^{k}$$

Using the result from here about closed form of the summation you obtain :

$$\frac{p^2}{1-p} \sum_{k=1}^{n} k(1-p)^{k}=\frac{p^2}{1-p}\frac{(1-p)(1-(1-p)^n) - np(1-p)^{n+1}}{p^2} = 1-(1-p)^n(1+np) $$

EDIT : I realized that I made all my reasoning about the event $tosses \leq n+1$ !

But since the right one is $tosses \geq n$ you can use substitution $n=k-2$ and property $ \mathbb{P}(tosses \geq k) = 1 - \mathbb{P}(tosses \leq k-1)$ to obtain :

$$\fbox {$\mathbb{P}(tosses \geq k) = (1-p)^{k-2}(1+(k-2)p)$} $$