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How do you calculate the expected value of the number of failures until the first success is reached where the probability will change after a failure. Let $p$ be the probability of success. Let $X$ be a random variable denoting the number of trials until success. Therefore, $E[X] = \sum_{k=0}^{\infty} (1-\frac{m-1-k}{m})^{k}(\frac{m-1-k}{m})$. Notice that subtracting $k$ will modify the probability after each failure.

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    $\begingroup$ You need to state clearly exactly how you're supposing $\ p\ $ to change after each failure. From the formula you give I'm guessing that $\ p\ $ starts off at $\ \frac{m-1}{m}\ $ and decreases by $\ \frac{1}{m}\ $ after each failure. If that's correct, then $\ p\ $ becomes negative after $\ m\ $ successive failures, which is meaningless for a probability, and the probability of such a sequence of failures is $\ \frac{\left(m-1\right)!}{m^{m-1}}>0\ $. $\endgroup$ – lonza leggiera Apr 23 at 7:37
  • $\begingroup$ Also, the general term $\ \left(\frac{k+1}{m}\right)^k\left(1-\frac{k+1}{m}\right)\ $ of your series diverges to $\ -\infty $ as $\ k\rightarrow\infty\ $, and hence so does the series itself. $\endgroup$ – lonza leggiera Apr 23 at 7:37

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