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Suppose, for example, that on Day $n$, $n$ agents have each bet $n$ times in succession. That regime fits the form above with $f(n) = n$. Will that do? No. An argument I got from Ewain Gwynne shows that, again, $m=\frac{1}{4}$ does best. Indeed, by Hoeffding and the union bound, we have that with probability at least $1-2ne^{-2\epsilon^2n}$

That is, the lower bound for $S^m_n$ is higher than the higher bound for $S^ {m'}_n$. And with probability at least $1-4ne^{-2\epsilon^2n}$, both $S^m_n$ and $S^{m'}_n$ lie within their bounds simultaneously, in which case $S^m_n > S^{m'}_n$. Finally, noting that for any $\epsilon$, $1-4ne^{-2\epsilon^2n} \to 1$ as $n \to \infty$ and $\frac{1}{4}$ maximizes $(1-m)(1+2m)$ over $m \in [0,1]$, we have the result.

Suppose, for example, that on Day $n$, $n$ agents have each bet $n$ times in succession. That regime fits the form above with $f(n) = n$. Will that do? No. An argument I got from Ewain Gwynne shows that, again, $m=\frac{1}{4}$ does best. Indeed, by applying Hoeffding's Inequality to $K$, we have that with probability at least $1-2e^{-2\epsilon^2n}$:

\begin{equation} (1-m)^{n(\frac{1}{2} + \epsilon)}(1+2m)^{n(\frac{1}{2} - \epsilon)} \leq B^m_{n,i} \leq (1-m)^{n(\frac{1}{2} - \epsilon)} (1+2m)^{n(\frac{1}{2} + \epsilon)} \end{equation}

for any $\epsilon$, $n$, and each $i=1,\ldots,n$. By the union bound, we have that with probability at least $1-2ne^{-2\epsilon^2n}$, all the $B^m_{n,i}$ lie within these bounds simultaneously. So with probability at least $1-2ne^{-2\epsilon^2n}$:

That is, the lower bound for $S^m_n$ is higher than the higher bound for $S^ {m'}_n$. So with probability at least $1-2ne^{-2\epsilon^2n}$, we have $S^m_n > S^{m'}_n$. Finally, noting that for any $\epsilon$, $1-2ne^{-2\epsilon^2n} \to 1$ as $n \to \infty$ and $\frac{1}{4}$ maximizes $(1-m)(1+2m)$ over $m \in [0,1]$, we have the result.

Question. Let $S^m_n$ be the sum of row $n$. I'm interested in comparing $S^m_n$ for different values of $m$ as $n \to \infty$. Say that $m$ does better than $m'$ if $\mathbb{P}(S^m_n > S^{m'}_n) \to 1$ as $n \to \infty$. Say that $m$ does best if $m$ does better than $m'$ for all $m' \neq m$. Does there exist $f$ such that some $m$ other than $\frac{1}{4}$ or $1$ does best?

Next, suppose that on Day $n$, one agent has bet $n$ times. That regime may be represented by a sequence of random variables $Y^m_n$, where $Y^m_i$ is the agent's bankroll after betting $i$ times, so is $(1+2m)^K(1-m)^{i-K}$, where $K$ ~ Bin($i,\frac{1}{2}$). Say, similar to above, that $m$ does better than $m'$ if $\mathbb{P}(Y^m_n > Y^{m'}_n) \to 1$, as $n \to \infty$. By the Kelly Criterion, $m=\frac{1}{4}$ does best.

  Here's the idea behind the Kelly Criterion. The expected growth rate of betting $m$ is $(1-m)^{\frac{1}{2}}(1+2m)^{\frac{1}{2}}$. Each time the agent wins, she multiplies her bankroll by $(1+2m)$ and each time she loses she multiplies it by $(1-m)$. The agent's actual growth rate on Day $n$ is the $n$-th root of her bankroll on Day $n$. So, by the Law of Large Numbers, for large $n$, her actual growth rate tends to be close to her expected growth rate, and the Kelly proportion $\frac{1}{4}$ maximizes that quantity.

Summarizing so far: Betting proportion 1, which maximizes expected value, does best when many agents each bet once. Betting proportion $\frac{1}{4}$, which maximizes expected growth rate, does best when one agent bets many times. Both results are applications of LLN. I hope to find a regime in which some proportion other than $\frac{1}{4}$ or 1 does best. My idea was to look at an $n$ by $f(n)$ structure (on Day $n$, $f(n)$ have each bet $n$ times), for some $f$, in the hope that a 'weighted average' of the many-agents-bet-once and one-agent-bets-many-times regimes will lead to some proportion between $\frac{1}{4}$ and 1 doing best.

Suppose, for example, that on Day $n$, $n$ agents have each bet $n$ times in succession. Will that do? No. An argument I got from Ewain Gwynne shows that, again, $m=\frac{1}{4}$ does best. Indeed, by Hoeffding and the union bound, we have that with probability at least $1-2ne^{-2\epsilon^2n}$

Question. Let $S^m_n$ be the sum of row $n$. I'm interested in comparing $S^m_n$ for different values of $m$ as $n \to \infty$. Say that $m$ does better than $m'$ if $\mathbb{P}(S^m_n > S^{m'}_n) \to 1$ as $n \to \infty$. Say that $m$ does best if $m$ does better than $m'$ for all $m' \neq m$. Does there exist $f$ such that some $m$ other than $\frac{1}{4}$ or 1 does best?

Next, suppose that on Day $n$, one agent has bet $n$ times. That regime fits the form above, with $f(n) = 1$ for all $n$. By the Kelly Criterion, $m=\frac{1}{4}$ does best. Here's the idea behind the Kelly Criterion. The expected growth rate of betting $m$ is $(1-m)^{\frac{1}{2}}(1+2m)^{\frac{1}{2}}$. Each time the agent wins, she multiplies her bankroll by $(1+2m)$ and each time she loses she multiplies it by $(1-m)$. The agent's actual growth rate on Day $n$ is the $n$-th root of her bankroll on Day $n$. So, by the Law of Large Numbers, for large $n$, her actual growth rate tends to be close to her expected growth rate, and the Kelly proportion $\frac{1}{4}$ maximizes that quantity.

Summarizing so far: Betting proportion 1, which maximizes expected bankroll, does best when many agents each bet once. Betting proportion $\frac{1}{4}$, which maximizes expected growth rate, does best when one agent bets many times. Both results are applications of LLN. I hope to find a regime in which some proportion other than $\frac{1}{4}$ or 1 does best. My idea was to look at an $n$ by $f(n)$ structure (on Day $n$, $f(n)$ have each bet $n$ times), for some $f$, in the hope that a 'weighted average' of the many-agents-bet-once and one-agent-bets-many-times regimes will lead to some proportion between $\frac{1}{4}$ and 1 doing best.

Suppose, for example, that on Day $n$, $n$ agents have each bet $n$ times in succession. That regime fits the form above with $f(n) = n$. Will that do? No. An argument I got from Ewain Gwynne shows that, again, $m=\frac{1}{4}$ does best. Indeed, by Hoeffding and the union bound, we have that with probability at least $1-2ne^{-2\epsilon^2n}$