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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?

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

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