text: # The coupon collector's most collected coupon
  
  Suppose a coupon collector is collecting a set of $$n$$ coupons that he receives one-by-one uniformly at random.
  
  If the collector stops exactly when the collection is complete, we know the expected number of coupons in his collection is $$n*H[n]$$. What is the expected number of copies, $$M$$, of his most collected coupon?
  
  When $$n = 2$$, then $$M = 2$$ because after the first coupon he will collect the other coupon with probability p ~ Geom(1/2). So he will collect the same coupon one additional time on average, and that coupon is certain to be his most collected coupon.
  
  I don't know the exact value for any $$n$$ larger than 2, but found some approximate values by simulation:
  
  n M
  3 2.8415
  4 3.4992
  5 4.0259
  6 4.4633
  7 4.8377
  8 5.1649
  9 5.4560
  
  EDIT 1:
  
  For small n enumerating the small possibilities converges faster than simulation, and from this approach I hypothesize that $$M[3] = (15 + 6 \sqrt{5})/10$$ but don't have any real argument to support that claim.
  
  EDIT 2:
  
  With some more thought, I find that M has an explicit sum formula. The probability the collection terminates with a given distribution of coupons $$v = \{c_1, ..., c_{n-1}\}$$ other than the final collected coupon is just the multinomial coefficient $$(c_1; ...; c_{n-1})$$ over $$n^\text{total # of coupons in the collection}$$. This is an infinite sum over n-1 variables. Here is Mathematica code the computes the sum for terms where the collection has no more than k copies of any coupon:
  
  M[n_, k_] := Sum[Max[v]*(Multinomial @@ v)/n^Total[v], {v, Tuples[Range[1, k], n-1]}]
  
  
  Mathematica can't actually evaluate this though for $$k \rightarrow \infty$$ though.
  
  • You might be interested in Exercise 2.20 of Boucheron, Lugosi, and Massart's ""Concentration Inequalities: A Nonasymptotic Theory of Independence", which asks to prove an upper bound of $$e\log n \cdot ce^{W((1-c)/(ce))}$$for this expectation, when collecting $m = cn\log n$ coupons. ($W$ being the Lambert function.) May 21 at 6:52
  
  ## 1 Answer
  
  Value of $$M_3$$. I will denote $$M$$ as $$M_n$$ to emphasize the dependence on $$n$$. Then by using A230137, we get
  
  \begin{align*} M_3 &= \sum_{n=2}^{\infty} \frac{\sum_{k=1}^{n-1} \binom{n}{k} \max\{k,n-k\}}{3^n} \\ &= \sum_{n=1}^{\infty} \frac{\left( 4^n + \binom{2n}{n} \right)n - 2(2n)}{3^{2n}} + \sum_{n=1}^{\infty} \frac{\left( 4^n + \binom{2n}{n} \right)(2n+1) - 2(2n+1)}{3^{2n+1}} \\ &= 3\left(\frac{1}{2} + \frac{1}{\sqrt{5}}\right) \\ &\approx 2.84164, \end{align*}
  
  which is the same as what OP conjectured.
  
  Asymptotic Formula of $$M_n$$. A heuristic seems suggesting that
  
  $$M_n \sim e \log n \qquad \text{as} \quad n \to \infty,$$
  
  The figure below compares simulated values of $$M_n$$ and the values of the asymptotic formula.
  
  Heuristic argument. First, we consider the Poissonized version of the problem:
  
  1. Let $$N^{(1)}, \ldots, N^{(n)}$$ be independent Poisson processes with rate $$\frac{1}{n}$$. Then the arrivals in each $$N^{(i)}$$ model the times when the collector receives coupons of type $$i$$.
  
  2. Let $$T$$ be the first time a complete set of coupons is collected. Then we know that $$T$$ has the same distribution as the sum of $$n$$ independent exponential RVs with respective rates $$1$$, $$\frac{n-1}{n}$$, $$\ldots$$, $$\frac{1}{n}$$. (This is an example of the exponential race problem.) For large $$n$$, this is asymptotically normal with mean $$\sum_{i=1}^{n}\frac{n}{i} \sim n \log n$$ and variance $$\sum_{i=1}^{n}\frac{n^2}{i^2} \sim \zeta(2)n^2$$. This tells that $$T \sim n \log n$$ in probability, and so, it is not unreasonable to expect that
  
  $$M_n = \mathbf{E}\Bigl[ \max_{1\leq i\leq n} N^{(i)}_T \Bigr] \sim \mathbf{E}\Bigl[ \max_{1\leq i\leq n} N^{(i)}_{n\log n} \Bigr]. \tag{1}$$
  
  holds as well.
  
  3. Note that each $$N^{(i)}_{n\log n}$$ has a Poisson distribution with rate $$\log n$$. Now, we fix $$\theta > 1$$ and choose an integer sequence $$(a_n)$$ such that $$a_n > \log n$$ and $$a_n/\log n \to \theta$$ as $$n \to \infty$$. Then
  
  \begin{align*} \mathbf{P}\Bigl( N^{(i)}_{n\log n} > a_n \Bigr) &\asymp \frac{(\log n)^{a_n}}{a_n!}e^{-\log n} \\ &\asymp \frac{(\log n)^{a_n}}{(a_n)^{a_n} e^{-a_n+\log n} \sqrt{a_n}} \\ &\asymp \frac{1}{n^{\psi(\theta) + 1 + o(1)}}, \qquad \psi(\theta) = \theta \log \theta - \theta, \end{align*}
  
  where $$f(n) \asymp g(n)$$ means that $$f(n)/g(n)$$ is bounded away from both $$0$$ and $$\infty$$ as $$n\to\infty$$. This shows that
  
  $$\mathbf{P}\Bigl( \max_{1\leq i\leq n} N^{(i)}_{n\log n} \leq a_n \Bigr) = \biggl[ 1 - \frac{1}{n^{\psi(\theta) + 1 + o(1)}} \biggr]^n \xrightarrow[n\to\infty]{} \begin{cases} 0, & \psi(\theta) < 0, \\[0.5em] 1, & \psi(\theta) > 0. \end{cases}$$
  
  So, by noting that $$\psi(e) = 0$$, we get
  
  $$\max_{1\leq i\leq n} N^{(i)}_{n\log n} \sim e \log n \qquad \text{in probability}. \tag{2}$$
  
  This suggests that $$M_n \sim e \log n$$ as well.
  
  4. I believe that this heuristic argument can be turned into an actual proof. To this, we need the following:
  
  1. We have to justify the relation $$\text{(1)}$$ in an appropriate sense. We may perhaps study the inequality
  
  $$\max_{1\leq i\leq n} N^{(i)}_{(1-\varepsilon)n\log n} \leq \max_{1\leq i\leq n} N^{(i)}_T \leq \max_{1\leq i\leq n} N^{(i)}_{(1+\varepsilon)n\log n} \qquad \text{w.h.p.}$$
  
  and show that the probability of the exceptional event is small enough to bound the difference of both sides of $$\text{(1)}$$ by a vanishing qantity.
  
  2. Note that $$\text{(2)}$$ alone is not enough to show this. So, we need to elaborate the argument in step 3 so that it can actually prove the relation $$\mathbf{E}\bigl[ \max_{1\leq i\leq n} N^{(i)}_{n\log n} \bigr] \sim e \log n$$.