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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