Conditional expectation for coin toss

Question

A fair coin is tossed repeatedly and let T be the number of tosses before two consecutive tails occur for the first time.

Show that E(T | the first toss resulted in tail) = 2 + ½E(T)

Well T will have geometric distribution but some how I can't find the answer. Please help.

Answer 1

Half the time, the second toss is tails, so it takes $2$ tosses. The other half the time it's heads, so we've done $2$ tosses and we're back where we started. $$E(T|\text{first toss tails})=\frac12\cdot2+\frac12(2+E(T))$$

Answer 2

Given that the first toss is T, we have

• TT with probability $\frac{1}{2}$

• THTT that is the first goal (4 tosses) if the second is H, that is $T+2$ with probability $\frac{1}{2}$

Concluding, we can rewrtite our contitional random variable in the following way

$$ (T|\text{First toss is tail}) = \begin{cases} 2, & \frac{1}{2} \\ T+2, & \frac{1}{2} \end{cases}$$

With expectation as showed before and according to your statement