Probability of duplicates over time We have a set of $100$ unique types of items. Each item type has an infinite quantity of that item. Each day I get  $x$ number of random items from the set. None of those items will be the same as any others within the set of $x$. I accumulate these $x$ items day over day, but they always draw from the same pool of $100$ unique items.
What is the probability on any given day that I will get a duplicate of one of the items I’ve already gotten?
 A: This is something of a generalization of the birthday problem (q.v.).
The probability of no duplicate after $n$ days is
$$
\left(\frac{100}{100}\right)
\left(\frac{100-x}{100}\right)
\left(\frac{100-2x}{100}\right)
\cdots
\left(\frac{100-(n-1)x}{100}\right)
$$
so the probability of at least one duplicate within $n$ days is $1$ minus all that, or
$$
1-
\left(\frac{100}{100}\right)
\left(\frac{100-x}{100}\right)
\left(\frac{100-2x}{100}\right)
\cdots
\left(\frac{100-(n-1)x}{100}\right)
$$
For $x$ not too large, this probability reaches $1/2$ at about day $n \doteq 12/\sqrt{x}$.
The unconditional probability that you get, for the first time on day $n$, a duplicate of an item you already got, is
$$
\left(\frac{100}{100}\right)
\left(\frac{100-x}{100}\right)
\left(\frac{100-2x}{100}\right)
\cdots
\left(\frac{100-(n-2)x}{100}\right)
\left(\frac{(n-1)x}{100}\right)
$$
since you must first go $n-1$ days without getting a duplicate, and then duplicate one of the $(n-1)x$ existing items on the $n$th day.
ETA: Obviously, we assume here that the pigeonhole principle has not been invoked—i.e., that $(n-1)x \leq 100$.
A: Obviously, the probability of getting a duplicate on day one is 0. On day two, duplicity happens if we draw any of the $x$ items from the day before, so there is an $\frac{x}{100}$ probability of getting a duplicate. Assuming we make it to day three without having already gotten a duplicate, we have now drawn $2x$ unique items, so there is now a $\frac{2 x}{100}$ probability of getting a duplicate. In general, the probability of getting a duplicate on day $n$ (assuming we haven't already gotten duplicates on a previous day) is $\frac{(n-1)\cdot x}{100}$ for applicable $n$. Obviously after a certain number of days we will have to get something we've gotten before by pigeon-hole principle.
