the meaning of with probability at least 1-\delta In the theoretical analysis of some algorithm for stochastic optimization, we often need to prove that something like
$$error\leq\epsilon,~~~~(under~some~conditions)$$
holds with probability at least $1-\delta$. But I found that many works use the specific number like $\delta = 1/4$ in their result even if the more general conclusion for arbitrary $\delta\in(0, 1/2]$ has been made. I was wondering that is there any particular meaning assigned to this specific number.
 A: As long as $1-\delta>1/2$ can be achieved, then repeatedly apply the algorithm, and by the law of large numbers, we can make the probability of error $\le \epsilon$ as large as possible, so it's not very important what $\delta$ is as long as $\delta<1/2$. (Although sometimes, it's interesting to optimize $\delta$ for specific $\epsilon$.)
This is really universal in TCS. For example in the definition of BPP, the number of $1/3$ and $2/3$ can really be replaced by $\epsilon$ and $1-\epsilon$ for any $\epsilon<1/2$.
A: There is no significance to $1/4$, except that it is a constant smaller than $1/2$, By the so-called "median trick," in many problems you can reduce the error probability to arbitrarily small $\delta>0$, at the cost of a factor $O(\log(1/\delta))$ in resources/time/sample complexity. [Note that this factor is not always optimal, but often is.]
From this (related) question:

For randomized algorithms $\mathcal{A}$ taking real values, the "median trick" is a simple way to reduce the probability of failure to any threshold $\delta > 0$, at the cost of only a multiplicative $t=O(\log\frac{1}{\delta})$ overhead.
Namely, if the $\mathcal{A}$'s output falls into a "good range" $I=[a,b]$ with probability (at least) $3/4$, then running independent copies $\mathcal{A}_1,\dots,\mathcal{A}_t$ and taking the median of their outputs $a_1,\dots,a_t$ will result in a value falling in $I$ with probability at least $1-\delta$ by Chernoff/Hoeffding bounds.

The choice of $1/4$ (instead of say $1/3$ or $1/1000$) then only affects the hidden constant inside the $O(\cdot)$.
