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An experiment is repeated, and the first success occurs on the 8th attempt. What is the success probability for which this is most likely to happen?

So we want to find $p$ between $0$ and $1$ which maximizes $(1-p)^{7}p$. To do this we could take the derivative and find all the critical points:

$$(1-p)^7 - 7(1-p)^6p = 0$$

But I don't know how to get the roots of this equation by hand. What can I do instead to solve this problem?

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    $\begingroup$ That is only the correct likelihood if it was determined before the experiment that there would be $8$ attempts. This is known as the "stopping condition." If the stopping condition is that you continue until you get a success, then you need a different likelihood. $\endgroup$ – Sean Lake Oct 10 '16 at 4:53
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    $\begingroup$ Try factoring out $(1-p)^6$ !!! $\endgroup$ – Ted Shifrin Oct 10 '16 at 4:53
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    $\begingroup$ @SeanLake The question implies that we continue until the first success. $\endgroup$ – Parcly Taxel Oct 10 '16 at 5:07
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    $\begingroup$ @SeanLake It is a geometric distribution, not a binomial distribution $\endgroup$ – GoodDeeds Oct 10 '16 at 5:10
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    $\begingroup$ It doesn't matter how the experiment was supposed to be run. The likelihood is simply $\mathcal{P}(\text{observation}|\text{parameter})$, and our observation is the result of $8$ first trials, not "the number of first success was $8$ and something else happened after that" nor "there was one success in first $8$ trials (but the order is unspecified)". For our observation, the likelihood is $(1-p)^7 p$. It doesn't even matter if we intended to do $100$ trials but stopped after the $8$th trial because of a fire alarm. $\endgroup$ – JiK Oct 10 '16 at 7:53
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Go ahead and solve for where the derivative is zero: $$(1-p)^7-7(1-p)^6p=0$$ $$(1-p)^6(1-p-7p)=0$$ $$(1-p)^6(1-8p)=0$$ $$1-p=0\text{ or }1-8p=0$$ $$p=1\text{ or }p=\frac18$$ We reject $p=1$ because then the experiment would succeed on the first try. $p=0$ can also be rejected for obvious reasons, so $p=\frac18$.

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  • $\begingroup$ Faster, I think, to rule out $p=0$ and $p=1$ rather than to apply the second derivative test. $\endgroup$ – user14972 Oct 10 '16 at 7:38
  • $\begingroup$ @Hurkyl that has been taken into account. $\ddot\smile$ $\endgroup$ – Parcly Taxel Oct 10 '16 at 7:44
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    $\begingroup$ Where does your $p=0$ come from? You reject it obviously, but there are only two extremal points, right? $\endgroup$ – example Oct 10 '16 at 14:06
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    $\begingroup$ @example because $0$ is an endpoint which you would check normally $\endgroup$ – b_pcakes Oct 10 '16 at 19:26
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$$(1-p)^7-7(1-p)^6p=(1-p)^6(1-p-7p)=(1-p)^6(1-8p)$$ Thus, for extremum, $$(1-p)^6(1-8p)=0$$ $$p=1 \text{ or }p=\frac18$$

The boundary values are $p=0$ and $p=1$, both which constitute zero probability. Moreover, by the second derivative test, it can be seen that $p=\frac18$ is a point of maximum.

Hence, $$p=\frac18$$

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  • $\begingroup$ (but the second derivative test is superfluous here -- the maximum must be either at $p=0$, $p=1/8$, or $p=1$, so as soon as you rule out $p=0$ and $p=1$, you're done) $\endgroup$ – user14972 Oct 10 '16 at 7:38
  • $\begingroup$ @Hurkyl I agree, that was my intention of using "Moreover,..", to show that it is another way to get the same thing. $\endgroup$ – GoodDeeds Oct 10 '16 at 7:39

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