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Suppose you have an event that as a probability $p = 1/100$ of occurring. If I make $x=50$ independent attempts, I believe the probability $P$ of the event occurring at least once in those $50$ attempts is found by calculating the probability it will not happen $1-P$:

$$(1-P) = (1-p)^x = \left(1-\frac{1}{100}\right)^{50}$$ $$(1-P) = \left(1-\frac{1}{100}\right)^{50}$$ $$(1-P) = 60.5\%$$ $$P = 39.5\%$$

Now, let's assume the event did not occur in the $50$ attempts (a $60.5$% chance given our calculation above). What is the probability of the event occurring at least once in the next $50$ independent attempts? My intuition says that since the attempts are independent, the scenario "resets" and does not factor in my previous attempts, keeping my odds at the same $39.5$%. Am I correct in saying this?

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    $\begingroup$ If all the attempts are independent then you are correct. Incidentally, your $39.5\%$ is close to $1-\sqrt{e} \approx 0.3935$ and $1-(1-\frac1{1000})^{500}$ would be even closer $\endgroup$
    – Henry
    Commented Mar 24, 2021 at 23:26
  • $\begingroup$ Hi James, is my answer helpful or confusing $\endgroup$
    – Some Guy
    Commented Mar 24, 2021 at 23:31
  • $\begingroup$ @Henry You mean $2 -\sqrt{e}$? $1- \sqrt{e}$ is a negative number $\endgroup$
    – Some Guy
    Commented Mar 24, 2021 at 23:33
  • $\begingroup$ @Henry Actually, I calculated it to be $1-\sqrt{\frac1e} \approx .3935$ $\endgroup$
    – Some Guy
    Commented Mar 24, 2021 at 23:35
  • $\begingroup$ @SomeGuy I meant $1-e^{-1/2}$ but forgot the sign - thank you $\endgroup$
    – Henry
    Commented Mar 24, 2021 at 23:37

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Yes you are correct.

Many people may be confused and say the probability would change. This is called the gambler's fallacy. Here's an example. If people kept betting on red when they're gambling, but the spinner keeps showing up black, should they stop betting red and start changing to bet on black? No, each spin is independent, so the probability that the spinner will be red is the same as the probability that the spinner will be black, so there is no reason to switch colors. The reason the spinner kept coming up black is by pure chance, but each spin should not be affected by the spins before it. Likewise, the probability that the next $50$ events happen does not depend at all on the previous $50$.

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