This is an exercise from the probability book by Ross. This is not homework.

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Using conditional probability and the distribution of sum of two geometric random variables, the probability comes out to be $\frac{1}{(n-1)}$.

But I am not able to understand how can the same probability be deduced from just the hint without all the computations.

I am not sure but it seems to have to do with each of the previous $(n-1)$ tosses being equally probable of being the time of the first head. Is that correct?

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  • 1
    $\begingroup$ One head occurs among the first $n-1$ flips. On which flip did it occur? Does the coin favor showing heads at any particular time compared to any other time? $\endgroup$
    – whuber
    2 days ago
  • 1
    $\begingroup$ Please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ yesterday
  • $\begingroup$ @whuber As the coin is unbiased, all flips have the same probability of landing in heads. So no, the coin doesn't favor showing heads at any particular flip. $\endgroup$
    – Biggo
  • $\begingroup$ @whuber I am confused regarding the distinction between the event of getting a head in the i-th flip and the i-th flip being the time of first head. $\endgroup$
    – Biggo
  • $\begingroup$ Suppose the coin flips go T, H, T, H, .... You obtain a head on the i = 4th flip, but the first head was obtained on the i = 2nd flip. $\endgroup$
    – whuber


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