# A socialistic coin

Discrete Mathematics Level pending

Consider a hypothetical coin where the probability of getting the previous result, is half the probability of obtaining the previous result before it was tossed. For example, let $$p_{n}$$ be the probability of getting heads in the $$n^\text{th}$$ coin toss, and so $$(1-p_{n})$$ is the probability of getting tails in the $$n^\text{th}$$ coin toss. Then,

• If the result of the $$n^\text{th}$$ coin toss is heads then $$p_{n+1}=\dfrac{p_{n}}{2}$$.
• If the result of the $$n^\text{th}$$ coin toss is tails then $$(1-p_{n+1})=\dfrac{1-p_{n}}{2}$$ or $$p_{n+1}=\dfrac{1+p_{n}}{2}$$.

The coin is taken and tossed until two consecutive heads are obtained. If $$E(HH)$$ is the expected number of tosses to get two consecutive heads, find $$\left\lfloor 1000E(HH) \right\rfloor$$.

Details and Assumptions:

For the first toss, the probability of getting heads and the probability of getting tails are both 0.5.

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