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Given an integer n, and an event n that happens with $P(\frac{1}{n})$, is the probability that e will happen in n trials bounded by any constant?

For example, if I had an n-sided fair die and a target value t, can I say with certainty that regardless of the value of n, the odds of rolling a t in n rolls are no worse than $\frac{1}{x}$ for some constant x?

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$$1-\left(\frac{n-1}n\right)^n\gt\lim\limits_{k\to\infty}1-\left(\frac{k-1}k\right)^k=1-\frac1{\mathrm e}=0.632$$

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    That is the limit but it is not a bound for any n because the function decreases to the limit. You need to add a little to it. Also this is the same limit I gave an hour earlier.2012-07-09
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    @MichaelChernick No idea what it is you are talking about: this is a bound for every $n$, nothing needs to be added and *an hour earlier* is in fact 13 minutes later. (For your interest, this is the last time I reply to this kind of raving comment from you. From now on, please pick your fights with someone else.)2012-07-09
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    I was not complaining about your answer. iwas somewhat miffed at the OP for checking your answer when I think mine came first and was better because I pointed out that to get an upper bound on the probability for any particular n requires adding something to the limit.2012-07-10
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    @Michael, did's answer came first. Sort by "oldest" and that will be obvious. I thought you'd have learned that people don't like it when you insinuate that they've copied your answer after our most recent spat over on CV. You may be well advised to "reign it in" in the future and keep such opinions to yourself. Similarly, getting "miffed" at the OP for what he/she chose to checkmark is your business but you'd be well served not to mention it as a comment under the accepted answer. It's not productive and it can actually be interpreted as quite rude.2012-07-10
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    @Macro I don't think I need to be told by you how I should behave. Often when answers come in at the same time it is reasonable to expect duplication as we were both writing at the same time. The post said 5 hours ago for one and 6 for the other and i got them confused. So i thought mine came in one hour before. His response to me was a little over the top. All I said was that my answer ponted out that the limit was not an upper bound and he just didn't get my point. I would of course apologize about the misunderstanding on the timing of the posts. I was not criticizing about repeating.2012-07-10
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    @Michael, re: *I don't think I need to be told by you how I should behave* - I'm not telling you how to behave, I'm suggesting a modification of your approach that would prevent these kinds of responses. There are *several* posters who've had arguments with you. This *very rarely* happens with the other members of the math/stat community here so it seems likely that its something you're doing that is eliciting these kind of responses. The behaviors I pointed out in my last posts are glaring examples. If you don't mind being perceived that way then, please, proceed. Like did, I'm done with it.2012-07-10
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    I may raise issues and be argumentative. But I don't mean to be offensive and I am sorry when I am misinterpreted that way. In this case I really did not mean to offend him. If arguments seem to come between me and other members of the community more often it happens with others it may be in the way I state thinks and argue but I don't think that is a bad thing and I think the debate can sometimes be worthwhile.2012-07-10
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    If I may explain, I accepted this answer because it provides the answer to the question I asked, and it also came first. I understand that, for any given n, the actual value is going to be above this constant by some amount, but for my purposes, knowing that such a constant limit exists is sufficient. I *will* take account of the epsilon as appropriate, and I appreciate it being pointed out.2012-07-10
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    @philosodad Thank you. Part of my problem was my incorrect assumption that my answer came first and by a large amount of time. I was wrong. So i shouldn't be very upset with you or with "did". Sorry for making such a big fuss! However "did" apparently did not appreciate that you were looking for an upper bound.2012-07-10
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If P(e)=1/n on any trial and trials are independent then the probability that e occurs at least once in n trials is 1- probability that it does occur in n trials =1 - [(n-1)/n]$^n$. This converges to 1-exp(-1) = [exp(1)-1]/exp(1). The convergence is from above but for large enough n take the limit plus a small ε for the bound.