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If there's an event that has $N$ chance to occur, what is the probability that over the span of $M$ attempts it will happen exactly M*N times?

Example: $0.1$ chance to get $A$, $0.9$ to get $B$. What is the probability that over $10$ repeats it will be one $A$ and nine $B$'s? And over $100$ repeats to be $10:90$? Intuition says that it should be (1-chance)^times but then the chance is the lower the more repeats there are. However, in real life, the more repeats, the more the ratio approaches probability ratio.

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  • $\begingroup$ If $N=.1$, what does $N/M$ mean? That is meant to be an integer, no? $\endgroup$ – lulu Apr 16 '18 at 15:54
  • $\begingroup$ Oh, sorry. I meant M*N $\endgroup$ – Stepan Filonov Apr 16 '18 at 15:58
  • $\begingroup$ Well, but this is just a binomial distribution. The probability that it happens exactly $k$ times in $M$ trials is $\binom Mk\times N^{k}\times (1-N)^{M-k}$. $\endgroup$ – lulu Apr 16 '18 at 16:02
  • $\begingroup$ Thanks a lot, I wasn't familiar with that topic at all. $\endgroup$ – Stepan Filonov Apr 16 '18 at 16:03
  • $\begingroup$ You need $M \times N$ to be an integer to have any chance of hitting it exactly $\endgroup$ – Henry Apr 16 '18 at 16:17
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The chance the event happens is $\theta$. Say you repeat the experiment $N$ times. You want to know the chance to find the event happening $\theta M$ times. This can be calculated with binomial probability function. Unfortunately, in this case $\theta M$ is not nescesseraly an integer and the binomial distribution is not well defined. If we dont care (too much) though we find $$ P(X=\theta M) = {M\choose\theta M } \theta^{M\theta}(1-\theta)^{M(1-\theta)}$$ Say $\theta = 0.1$ and we calculate the probability for $M=10,100,100$, we find as probabilities $0.387$, $0.132$ and $0.042$. This might seem strange, because you expect is to be close to this ratio as $M$ grows large. However with $M$ growing larger the probablity of $\textit{exactly}$ hitting this ratio becomes less likely. If you would look at the probability of finding the average ratio with a margin of, for example, 1% you would find the probability of being within this margin increases as $M$ grows (just like you would expect!)

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  • $\begingroup$ Thanks, that became very clear to me now. $\endgroup$ – Stepan Filonov Apr 16 '18 at 17:10

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