How does having more items than slots guarantee that two of them share a slot?

It is a certainty, not a probability. If more items are placed into fewer slots, at least one slot must hold two items, because if every slot held at most one there would not be enough slots to hold them all. The conclusion follows from counting alone; it does not depend on how the items are distributed, which is why it is a guarantee rather than a likelihood.

The smallest case shows the whole argument. Thirteen people, twelve months. Suppose no two share a month. Then January holds at most one person, February at most one, and so on through December. Adding those twelve caps gives at most twelve people in the room. But the room holds thirteen, so the supposition is false: some month holds at least two people. Nothing was assumed about which months are likely, or about the people. The same line works with any labels: days of the year, hash buckets, possible friend counts. It is not an averaging argument either. It does not say that on average a slot has $n/k$ items. It says that the fullest slot has at least $n/k$ rounded up, which is a statement about the worst slot.

The ceiling is the arithmetic. If $n=25$ and $k=12$, then $25/12$ is a little over 2, so the ceiling is 3. Suppose no month held three people. Then each month held at most two, so twelve months could hold at most 24 people. But there are 25, so some month holds at least three. That is stronger than two sharing: it names how many must pile up in one slot. The same calculation with 1000 people and 365 days gives $\lceil 1000/365 \rceil = 3$, so three people share a day of the year. With 1,000,000 people and 1,000,000 days, the ceiling is 1, which is true and useless: it says only that some day has at least one person.

Notice what the ceiling does not say. It does not say which slot is full, or that every slot is used, or that the items are evenly spread. It says at least one slot reaches the ceiling. That is enough to rule out the possibility that every slot holds at most $\lceil n/k \rceil - 1$. The proof is always the same: assume every slot holds at most one less than the ceiling, multiply by the number of slots, and contradict $n$.

The natural guess is that this is probabilistic, that sharing becomes likely as the number of items approaches the number of slots, and that the guarantee is a statement about chance. It is not. The counting argument above never mentions chance. It rules out the alternative: no sharing. Because it needs no assumption about how items are placed, it can prove impossibility. That is why the same principle settles claims that otherwise invite endless argument, such as no compression scheme shrinking every input: there are fewer short files than long ones, so two inputs must map to the same short file. A probabilistic statement would require a model: each item independently chooses a slot uniformly, or birthdays are evenly spread across months. The pigeonhole argument requires no model. It works if every item chooses the same slot, if the slots have different sizes, if the labels are chosen adversarially. That is why it is a certainty. The price of needing no model is that the conclusion is weak: it tells you a collision exists, not how many collisions there are, not which items collide, and not how likely a collision is in a random placement. Probability enters only when you ask a different question: what is the chance that no two share? For birthdays, that chance falls below one half at twenty three people, since 1.18 times the square root of 365 is about 22.5. Certainty needs 366 people, one more than the number of days. The guarantee and the threshold are on different scales.

The guarantee is a floor on the worst case, not a forecast of the typical case. Put one million items into one million slots. The principle says some slot holds at least one item, which is true and useless. The typical case, assuming items fall independently and uniformly, has about 368,000 empty slots (one million divided by $e$, approximately 367,879), about half a million colliding pairs (one million squared divided by two million), and a fullest slot holding several items. Those numbers need the uniformity assumption. The certainty does not, which is why it can prove that two items must share, and why it cannot tell you how often that will happen or how bad it will be. For a hash table with a thousand buckets and a thousand items, the guarantee says only that at least one bucket has an item. In fact collisions are everywhere: the expected number of empty buckets is $1000/e$, about 368, and the expected number of colliding pairs is $1000^2/(2\cdot 1000)=500$. That is why hash tables are built to handle collisions with chains or probing rather than to avoid them. If you need to know whether a collision is likely, you need the square root rule and a distribution. If you need to know that a collision exists, the pigeonhole count is enough.