Limit of lights in rooms

There are $k$ rooms with $n$ lights each. Each light is on with equal probability $p$ independently of other lights. As $k$ stays fixed and $n$ goes to infinity, what is the limit of the probability that Room 1 has the maximum number of lights on (possibly sharing this maximum with other rooms)?

By symmetry, this probability is clearly at least $1/k$ for any $n$. However, for fixed $n$ it is slightly larger than $1/k$ because the maximum can be equal for many rooms. Still I think the limit should be $1/k$.

Solution 1:

It's $1/k,$ because the probability of having a unique maximum tends to $1.$

For a fixed pair of rooms, with number of lights on $X_1$ and $X_2,$ let $p_n$ denote the probability $\mathbb P[X_1=X_2]$ of having an equal number of lights on. Note $X_1-X_2$ is a sum of $n$ i.i.d. $\{-1,0,1\}$-valued variables, where the $j$'th variable is $1$ if the $j$'th lightbulb in room $1$ is on, minus $1$ if the $j$'th lightbulb in room $2$ is on (so $0$ if they are both on or both off). By the central limit theorem, the probability of $X_1-X_2=0$ tends to zero as $n\to\infty:$ for any fixed $\epsilon>0,$ there is some $c$ such that $p_n\leq \mathbb P[-c\sqrt n\leq X_1-X_2\leq c\sqrt n]\leq\epsilon/2$ for sufficiently large $n.$

By a union bound, the probability of any pair of rooms having the same number of lights is at most $\binom k 2 p_n,$ which for any fixed $k$ will tend to zero as $n\to\infty.$