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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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2 Answers 2

up vote 6 down vote accepted

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

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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. –  Michael Chernick Jul 9 '12 at 18:07
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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.) –  Did Jul 9 '12 at 21:18
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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. –  Macro Jul 10 '12 at 17:47
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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. –  Macro Jul 10 '12 at 18:18
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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. –  philosodad Jul 10 '12 at 21:18

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.

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