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I have a coin. When throwing there is a $p$ percent chance to get head and $(1-p)$ for tail. I throw it once and get head. What is the expected number of throws to get and even number of heads and tails for the first time.

Let $T$ be that event. Then I have

$E[T] = (1-p) + p(2 + E[T])$. Solving this for $E[T]$ gives $E[T] = (1+p)/p$.

  • Do you want to have an equal amount of heads and tails or an even amount? For example would 2 heads and 4 tails be an accepted result? – Jakob W. Feb 20 '19 at 15:40
  • also, what is your actual question? – Jakob W. Feb 20 '19 at 15:43
  • If $p > \frac12$ there is a non-zero probability that you never get to an equal number of heads and tails. See https://math.stackexchange.com/questions/153123/hitting-probability-of-biased-random-walk-on-the-integer-line – David K Feb 20 '19 at 16:53

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