I am having trouble wrapping my head around the probability of success by the expected value in a series of Bernoulli trials.

The expected number of trials until success for a series of $\frac{1}{p}$ trials is $p$. The probability of getting a success by the $p$'th trial is $1-\:\left(1-\frac{1}{p}\right)^p$. Taking the limit to infinity this is about 63% $(1-\frac{1}{e})$.

So you're "expected" to hit a success before the expected number of trials required to get at least one success. As in, by the time you have completed the expected number of trials required for success you are more likely than not to have hit a success..

I just think, intuitively, this probability should be 0.5? Maybe I am missing something important in how I am thinking about the problem, I am just not sure what.


1 Answer 1


You're conflating the probability of success with the expected number of trials needed to observe that success.

While it is true that, when a single trial has probability of success $1/n$, then the expected number of trials needed to observe success is $n$, and that the probability that at most $n$ trials are needed is $1 - (1-1/n)^n$, ask yourself what it means as you "take the limit to infinity." If $n \to \infty$, then you are simultaneously changing the probability that any single trial will yield success. For instance, suppose a trial consists of flipping a fair coin; i.e., $n = 2$ and you are waiting for heads. Then the expected number of trials needed is $2$ and the probability that it takes $2$ or fewer trials to get heads is $1 - (1 - 1/2)^2 = 3/4$. That's all well and good. But now if you increase $n$, say $n = 20$, then what are you doing? In such a case, it would be like rolling a fair $20$-sided die until you roll the first $20$, and the probability that it takes you $20$ or fewer rolls is $$1 - (1 - 1/20)^{20} \approx 0.641514.$$ There is no intuitive reason why this value should tend to $0.5$ as the number of trials increases (and the probability of success decreases inversely).

This illustrates a second misconception you have, which is that the average or expected outcome is not necessarily the outcome for which the cumulative probability is $1/2$. In other words, the mean and the median are not in general the same.

  • $\begingroup$ Thank you for your answer, in writing a response to your answer I think I understand sort of the point I was missing. $\endgroup$ Commented Apr 16, 2022 at 17:54
  • $\begingroup$ @RandallPeckings The "counterintuitiveness" you're feeling is a result of something called memorylessness, which in turn arises from independence of events. Subsequent events aren't influenced by previous ones. The coin doesn't "remember" what it did before. Each trial is like the first trial. That's why it feels strange that, if you have not yet succeeded in, say, $m$ trials, that the average number of additional trials you need to succeed is the same as if you had not made any attempts in the first place. $\endgroup$
    – heropup
    Commented Apr 16, 2022 at 17:56
  • $\begingroup$ More than memoryless-ness I believe it is the strangeness of the statement that you're "expected" to hit a success by a value lower than the expected value of trials for success. $\endgroup$ Commented Apr 16, 2022 at 18:01

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