Upper bound for a probability expression There are $M=\{1,\dots,m\}$ machines and $N=\{1,\dots,n\}$ are a set of $n$ numbers. Suppose $m<n$. On the $t^{th}$ trial, each machine $m$ picks a number from $N$ with probability $p_m(t,,n)$. If more than one machine picks the same number, those machines experience a collision.
All machines start off with independent uniform distributions at time $t=0$. A machines keeps the first number it got without a collision for the rest of the time with probability one. Those who still didn't have any trial without a collision, pick a number again with independent uniform distributions. (i.e. hold the number with the fist no collision event)
To make it clear: those who hold their number (after fist successful trial) still do collide with others, but they don't change the number since they had at least one no collision incident in the past for that number, others have to retrial. This goes on till everyone is collision free.   
What is the probability that there is at least one collision at time $t>0$?I am not sure if this probability can be found easily in closed form, but can we find an upper bound? And using that upper bound can we say that the sum probability (sum over time) of at least one collision goes to zero?
Edit:
Trying to prove: Suppose the event that at least one collision occurs at time $t$ is $C(t)$ with probability $P(C(t))$. Then I'm trying to say $\sum_tP(C(t))<∞$. Hoping to apply Borel-Cantelli lemma.
 A: Suppose $k$ machines do not have a fixed number yet. Then, there is some probability $0<p\leq 1$ that at least one machine will get a fixed number during the next round. The expected time needed for the first machine to find a fixed number is
$$\frac 1p=p+p(1-p)+p(1-p)^2+\dots$$
Because $p>0$, we know that $\frac 1p<\infty$. Thus, within a finite amount of time, there will be (at least) one machine less without a fixed number. For the new state, we can use the same argument. Because the number of machines $m$ is finite, we only need a finite amount of time $m$ times, which is finite. We know that $\frac 1p $ can't/won't go to infinity as the number of machines that is left decreases, because overall, there is only a finite amount of possible states the system can be in, and for every state we can calculate $p$ and then the the minimum of those values, which is a lower bound for all $p$'s .
The problem with this argument may be (I'm not sure) that it talks about the expected time for one machine to get a number, not about a worst case situation.
A: Here is a solution that works for $m <\infty$. Fix an arbitrary time and suppose $k<m$ machines have already settled (meaning they have already picked their choice in previous rounds). The probability that in the next round, a fixed unsettled machine becomes settled is:
$$\frac{n-k}{n}\cdot \left(1-\frac{1}{n}\right)^{m-k-1} \ge \frac{n-k}{n}\cdot \left(1-\frac{m-k-1}{n}\right)\ge \frac {m-k}{m}\cdot\left(1-\frac{m-1}{n}\right)\ge \frac{1}{m^2}$$
Let $U(t)$ denote the number of unsettled machines at the end of round $t$. Then by the above argument, in expectation, $\frac{1}{m^2}\cdot U(t)$ become settled in the next round. Repeating the argument, we get:
$$E[U(t)] \le E[U(0)]\cdot \left(1-\frac{1}{m^2}\right)^t = m \cdot \left(1-\frac{1}{m^2}\right)^t$$
Hence, by Markov's inequality we get
\begin{align}
Pr[U(t)\ge 1] &\le E[U(t)] \\
&\le  m \cdot \left(1-\frac{1}{m^2}\right)^t
\end{align}
Note that for a collision to happen, we need to be left with at least one unsettled machine from the previous round. Therefore, $Pr[C(t)] \le Pr[U(t-1)\ge 1]$.
This yields:
\begin{align}
\sum_t Pr[C(t)] &\le \sum_t Pr[U(t-1)\ge 1]\\
&\le \sum_t m \cdot \left(1-\frac{1}{m^2}\right)^{t-1}\\
&\le  m^3
\end{align}
