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I have a bag that contains coins, these coins could be biased coins, and each coin has a certain pre-determined probability of head/tail (independently of the other coins). This pre-determined probability is derived from a uniform distribution over [0,1].

I draw a coin at random and flip it until I get a tail, then I threw it away and withdraw another coin and flip it again and again until I get a tail, and so on.

Then the expected distance between consecutive tails is given by: $$\int_{0}^{1}\frac{1}{1-p}*1*dp$$ where p is probability of getting a head, 1 is the density of uniform [0,1], $\frac{1}{1-p}$ is expected value of geometric with prob. success p.

I didn't understand how we characterized the above from the expected distance between consecutive tails to be in the form shown above. Does anyone know the logic behind coming up with the integral above? Thank you.

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  • $\begingroup$ I think your first sentence is putting everybody off here. What do you mean by "..bag of coins that are uniformly distributed over $[0,1]$..." $\endgroup$ – jay-sun Apr 12 '13 at 0:21
  • $\begingroup$ @jay-sun I have a bag that contains coins, these coins could be biased coins, and each coin has a certain pre-determined probability of head/tail (independently of the other coins). This pre-determined probability is derived from a uniform distribution over [0,1]. I hope it is more clear now. $\endgroup$ – Kabamaro Apr 12 '13 at 0:28
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Hint: Compute the expected number of flips needed to get a tail using a single biased coin whose probability of producing a head is $p$.

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