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Lets assume I have a distributions of outcomes $A=\{a_1, a_2, a_3, \ldots, a_n\}$. I also have a probability function $P$, with $\forall a\in A, P(a)\in[0,1]$ and $\sum_{a\in A}P(a) = 1$.

Now let assume I run my experiment $n$ times. I'm talking a REAL experiment where I don't know the full range of $A$ nor the details of $P$ ... those are what I am trying to reconstruct. I am trying to estimate the probability of $A$ containing a single element $\alpha$ such that $P(\alpha)=1$

If after $n$ runs I have at least $2$ different outcomes, I know that my experiment in not deterministic and that is good for me. But what if I do have $n$ time the same result ? What can I conclude ? What is the probability of my experiment being deterministic ? Can I achieve a lower bound on $P(a')$, the probability of the most probable event ($\forall a\in A, a\neq a' \implies P(a)\leq P(a')$)

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You can't ever prove the experiment is deterministic because there could always be some possible but improbable event. You could think of buying lottery tickets and the event being not winning the grand prize. After $1000$ tries you (very likely) have every try with the same result.

You can put a lower bound on the chance of the most probable event. If the chance of the most probable event is $p$ and you do $n$ trials, the chance you get the same result on every trial is $p^n$. For a $90\%$ confidence lower bound on $p$ you can say $p^n \gt 0.1, p \gt \sqrt [n]{0.1}$

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  • $\begingroup$ I know I cannot prove that the experiment is not deterministic, I just want an estimation of something that never happened happens. And If I have two events, one with $P(a)=90\%$ and the other with $P(b)=10\%$, the probability of having $n$ times the same is $0.9^n+0.1^n$, not $0.9^n$ ... so things are a bit more complicated then you say $\endgroup$
    – Amxx
    Commented Oct 19, 2017 at 14:24
  • $\begingroup$ The $0.9^n+0.1^n$ is strictly true, but $0.1^n$ is so much smaller that for any reasonable $n$ you can ignore it. $\endgroup$ Commented Oct 19, 2017 at 14:39
  • $\begingroup$ If I have $n$ events with proba $1/n$ I cannot ignore any of those. That is my issue here, I'd like to simplify the $\sum_{a\in A}P(a)^n$ by having an upper bound which removes some terms I don't want to carry ... but I don't find the way to do so without any assumptions on $A$ $\endgroup$
    – Amxx
    Commented Oct 19, 2017 at 14:42

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