Book problem: If the amount of change in a pocket is assumed to be uniformly distributed from $0$ to $99$ cents, how many people must be in a room until it is at least a $50\%$ chance that two will have the same amount of change?
Book answer: $P(10)\{\text{no duplicate}\} = 0.5031$
My solution: I interpret the question as asking for the chance that 'at least' two people will have the same amount of change.
I flip the problem to find when no one will have the same amount of change. If there were $k$ people in the room, $k < 100$, then the number of ways that they would each have different amounts is the same as the number of permutations of $k$ from the 100 options, so $\frac{100!}{(100-k)!}$.
The number of total scenarios for the $k$ people is $100^k$.
Hence the probability of no duplicates in the room is the quotient
$$ \frac{100!}{100^k(100-k)!}. $$
So for at least two people to have the same amount of change in their pockets, the opposite of no duplicates, would be
$$ 1 - \frac{100!}{100^k(100-k)!} $$
and we want this to be at least $1/2$.
Question: Is this not $k = 13$ which seems to contradict the book answer of $10$?
Book: "Methods of Mathematics" by Hamming