Expected number of two successive tosses is the same?

An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is

1. $3$
2. $4$
3. $5$
4. $6$

My attempt :

I used a formula

The expected number of coin flips for getting $n$ consecutive heads is $(2^{n+1} - 2)$.

Similarly,

The expected number of coin flips for getting $n$ consecutive tails is $(2^{n+1} - 2)$.

That gives two ways to satisfy the criterion to stop, so stopping will be sooner, not later.

So,

The expected number of coin flips for getting $n$ consecutive same tosses is $\frac{(2^{n+1} - 2)}{2}$.

Hence, answer is $3$.

Can you explain in formal way please?

• The answer $3$ is correct : After the first toss, it takes $2$ tosses in average to copy the last result. – Peter Nov 21 '15 at 13:24

A simple recursion: Let $E$ be the answer you want and let $E_1$ be the expected number assuming you have tossed at least once (but have not yet won). then $$E=E_1+1$$
as tossing the coin once gets you to the state governed by $E_1$. But then tossing again either ends the game (if you get a match) or leaves you with expectation $E_1$ again. Thus $$E_1=\frac 12 1 \;+\; \frac 12(E_1+1)$$
this is easily solved to get $E_1=2$, whence $E=3$.