A fair coin tossed repeatedly

Question

A fair coin tossed repeatedly and let $T$ be be the number of tosses till two consecutive tails are observed for the first time.

(a) Show that $E(T|\text{tail is observed in the first toss})=2+\frac{1}{2}E(T)$.

(b) Find similar formula for $E(T|\text{head is observed in the first toss})$.

(c) Compute $E(T)$.

What I can see easily is that

1. The value we get from (a)+The value we get from (b)=$E(T)$
2. We let $X_i=1$, if tail occurs, $0$ if head occurs. Then $T|N=n=X_1+X_2+\dots X_n$. But I can not find what $N$ follows!!

Need some hint here!

Answer 1

Let $\mathrm t_1, \mathrm t_2$ be the event of a tails in the first and second tosses, respectively, and likewise $\mathrm h_1, \mathrm h_2$ the event of heads in the first and second toss, respectively.

$(a)$ You begin by Law of Total Probability.

$$\mathsf E(T\mid \mathrm t_1) = \mathsf P(\mathrm t_2)\,\mathsf E(T\mid \mathrm t_1, \mathrm t_2) + \mathsf P(\mathrm h_2)\,\mathsf E(T\mid \mathrm t_1, \mathrm h_2)$$

Now you know the coin is fair, that if you have two tails in a row you stop, and if the last toss was a head you count the tries until then and start anew.

Carry on.

---

$(b)$ As above, so below.

$$\mathsf E(T\mid \mathrm h_1) = \mathsf P(\mathrm t_2)\,\mathsf E(T\mid \mathrm h_1, \mathrm t_2) + \mathsf P(\mathrm h_2)\,\mathsf E(T\mid \mathrm h_1, \mathrm h_2)$$

---

$(c)$ Total Probability again. $$\mathsf E(T)=\mathsf P(\mathrm t_1)\,\mathsf E(T\mid \mathrm t_1)+\mathsf P(\mathrm h_1)\,\mathsf E(T\mid \mathrm h_1)$$

Simplify the result to solve for $\mathsf E(T)$.