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Iosif Pinelis
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$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies \begin{equation*} M_n=EX_n\sim r_n \end{equation*}

\begin{equation*} M_n=EX_n\sim r_n \end{equation*}

(which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, $N_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13},
\begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\ll r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies \begin{equation*} M_n=EX_n\sim r_n \end{equation*} (which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, $N_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13},
\begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\ll r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies

\begin{equation*} M_n=EX_n\sim r_n \end{equation*}

(which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, $N_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13},
\begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\ll r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

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Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229

$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies \begin{equation*} M_n=EX_n\sim r_n \end{equation*} (which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, $N_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13}, 
\begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\le r_n^2. \tag{15}\label{15} \end{equation*}\begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\ll r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies \begin{equation*} M_n=EX_n\sim r_n \end{equation*} (which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, $N_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13}, \begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\le r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies \begin{equation*} M_n=EX_n\sim r_n \end{equation*} (which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, $N_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13}, 
\begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\ll r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

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Iosif Pinelis
  • 127.8k
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  • 107
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$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies \begin{equation*} M_n=EX_n\sim r_n \end{equation*} (which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, $N_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}(1-1/n)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*}\begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13}, \begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\le r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies \begin{equation*} M_n=EX_n\sim r_n \end{equation*} (which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}(1-1/n)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13}, \begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\le r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

$\newcommand{\Si}{\Sigma}$Let $X_n$ be the maximum number of prank cigarettes any pack receives, so that $M_n=EX_n$. Note that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. So, by Theorem 1 or formula (6) of Raab and Steger, \begin{equation*} X_n\sim r_n:=\frac{\ln n}{\ln\ln n} \tag{1}\label{1} \end{equation*} in probability (as $n\to\infty$).

We will also prove

Proposition 1: \begin{equation*} EX_n^2\ll r_n^2. \end{equation*}

(As usual, we write $A\ll B$ to mean $A=O(B)$.)

By Proposition 1 and the de la Vallée-Poussin theorem, $X_n/r_n$ is uniformly integrable, which implies \begin{equation*} M_n=EX_n\sim r_n \end{equation*} (which agrees with Qiaochu Yuan's heuristics/conjecture).

Proof of Proposition 1: Let $[n]:=\{1,\dots,n\}$. Note that \begin{equation*} EX_n^2\ll\sum_{m\in[n]}mP(X_n\ge m) =\Si_1+\Si_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} \Si_1:=\sum_{1\le m\le 4r_n}mP(X_n\ge m),\quad \Si_2:=\sum_{4r_n<m\le n}mP(X_n\ge m). \tag{4}\label{4} \end{equation*} It is easy to bound $\Si_1$: \begin{equation} \Si_1\le\sum_{1\le m\le 4r_n}m\ll r_n^2. \tag{5}\label{5} \end{equation}

Let us now bound $\Si_2$. Recall that $X_n=\max(N_1,\dots,N_n)$, where $(N_1,\dots,N_n)$ has the multinomial distribution with parameters $n;\frac1n,\dots,\frac1n$. Note that for each $m\in[n]$ \begin{equation*} P(X_n\ge m)\le\sum_{i\in[n]}P(N_i\ge m) =nP(N_1\ge m). \tag{7}\label{7} \end{equation*} Next, $N_1$ has the binomial distribution with parameters $n$ and $\frac1n$, and hence \begin{equation*} \begin{aligned} P(N_1\ge m)&=\sum_{k=m}^n\binom nk\frac1{n^k}\Big(1-\frac1n\Big)^{n-k} \\ &\le\sum_{k=m}^n\binom nk\frac1{n^k}\le\sum_{k=m}^n\frac1{k!}\ll\frac1{m!}. \end{aligned} \tag{11}\label{11} \end{equation*} Further, for $m>4r_n$, eventually (that is, for all large enough $n$), \begin{equation*} m!\ge(m/e)^m=\exp(m\ln(m/e)) \\ \ge\exp\Big(4\frac{\ln n}{\ln\ln n}\,\ln\frac{\ln n}{\ln\ln n}\Big)\ge n^3. \tag{13}\label{13} \end{equation*} So, by \eqref{4}, \eqref{7}, \eqref{11}, and \eqref{13}, \begin{equation*} \Si_2\le\sum_{4r_n<m\le n}mnP(N_1\ge m) \\ \ll\sum_{4r_n<m\le n}mn\frac1{n^3}\le1\le r_n^2. \tag{15}\label{15} \end{equation*} Now Proposition 1 follows immediately from \eqref{3}, \eqref{5}, and \eqref{15}. $\quad\Box$


Quite similarly one can show that \begin{equation*} EX_n^p\sim r_n^p \end{equation*} for each real $p>0$.

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