Let $h:[U]\to[m]$ be a uniform hash function, and $x_1,\dots,x_n\in[U]$ be the elements currently stored in the hash table, and we want to look up the query $q\in[U]$. Finally let $X_i$ be the indicator variable for the event that $x_i$ collides with $q$:
$$X_i = \begin{cases}1 && h(x_i)=h(q)\\0 &&h(x_i)\ne h(q)\end{cases}
\Rightarrow P(X_i=1) = \frac{1}{m}$$
Then the time $T$ to loop up $q$ is equal to the number of elements that collide with it:
$$T=\sum_{i=1}^n X_i$$
By linearity of expectation we have:
$$\mathbb{E} [T] = \sum_i\mathbb{E} X_i = \frac{n}{m}=\alpha$$
And finally by Markov's inequality:
$$ P(T>t) \le \frac{\alpha}{t}$$
Now if we set $t:={\alpha}/{\delta}$, we will have:
$$P(\text{look-up time}\ge \frac{\alpha}{\delta}) \le \delta$$
Think of $\delta$ as the failure rate. So for $\delta=0.1$, the query time is bounded by $10\alpha$ with probability $0.9$.
If you strengthen the assumption about $h$ to a $k$-wise independent hash, the failure rate rouhgly drops to $P(\text{fail})\approx\delta^k$ and so you can have a much better complexity/failure-rate trade-off.
Now, a more conservative analysis is, what is the length of the longest chain in your table? In this case, we can define indicator variables $X_{i,j}$ for $i,j\in[n]$, that indicate the event that $x_i$ and $x_j$ collide. Clearly, $P(X_{i,j})=1/m$. Then the expected total number of collisions $S$ is:
$$\mathbb{E} S = \sum_{i,j\le n} \mathbb{E} X_{i,j} = \binom{n}{2}\frac{1}{m}$$
If the maximum chain length is $\ell$, there is $\binom{\ell}{2}$ pairs of elements in that chain that collide with each other, implying $S\ge \binom{\ell}{2}$. We have:
$$P(\text{max chain length}\ge \ell) \le P\left(S\ge \binom{\ell}{2}\right) \le\frac{ \mathbb{E} S} { \binom{\ell}{2}}\approx (n^2/m\ell^2) $$
If you set $\ell:=n^2/m\delta$ the probablistic bound is:
$$P(\text{max chain length}\ge \ell) \le \delta \qquad \text{ for } \ell = \frac{n}{\delta \sqrt{m}}$$
So if $m\approx n$, your maximum length grows like $\ell\approx \sqrt{n}/\delta$ with $1-\delta$ probability. This is because the simple uniform hash assumption is really weak. The more independence you assume, the better the guarantee. For a universally uniform (meaning completely iid) hash function, you will get a logarithmic maximum chain $log(n)$ for $m\approx n$.
