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Multiply by $\left(\frac{2(7)}{3+7}\right)^{1/3} = 1.1187$

More generally, suppose that player $A$ rolls $n$ times and player $B$ rolls $m$ times (without loss of generality, we assume $m \geq n$). As others have already noted, the (unscaled) score of player $A$ is $$X \sim Beta(n, 1)$$ and the score of player $B$ is $$Y \sim Beta(m, 1)$$ with $X$ and $Y$ independent. Thus, the joint distribution of $X$ and $Y$ is $$f_{XY}(x, y) = nmx^{n-1}y^{m-1}, \ 0 < x, y < 1.$$

The goal is to find a constant $c$ such that

$$P(Y \geq cX) = \frac{1}{2}$$.

This probability can be found in terms of $c$, $n$ and $m$ as follows.

\begin{align*} P(Y \geq cX) &= \int_0^{1/c}\int_{cx}^1 nmx^{n-1}y^{m-1}dydx \\[1.5ex] &= \cdots \\[1.5ex] &= c^{-n}\left\{\frac{m}{n+m} \right\} \end{align*}

Setting this equal to $1/2$ and solving for $c$ yields

$$c = \left(\frac{2m}{n+m}\right)^{1/n}.$$

Multiply by $\left(\frac{2(7)}{3+7}\right)^{1/3} = 1.1187$

More generally, suppose that player $A$ rolls $n$ times and player $B$ rolls $m$ times (without loss of generality, we assume $m \geq n$). As others have already noted, the (unscaled) score of player $A$ is $$X \sim Beta(n, 1)$$ and the score of player $B$ is $$Y \sim Beta(m, 1)$$ with $X$ and $Y$ independent. Thus, the joint distribution of $X$ and $Y$ is $$f_{XY}(x, y) = nmx^{n-1}y^{m-1}, \ 0 < x, y < 1.$$

The goal is to find a constant $c$ such that

$$P(Y \geq cX) = \frac{1}{2}$$.

This probability can be found in terms of $c$, $n$ and $m$ as follows.

\begin{align*} P(Y \geq cX) &= \int_0^{1/c}\int_{cx}^1 nmx^{n-1}y^{m-1}dydx \\[1.5ex] &= \cdots \\[1.5ex] &= c^{-n}\left\{\frac{m}{n+m} \right\} \end{align*}

Setting this equal to $1/2$ and solving for $c$ yields

$$c = \left(\frac{2m}{n+m}\right)^{1/n}.$$

Edit

Upon re-reading this question, I noticed that the OP correctly determined that "the constant does not just depend on the ratio of throws". With this as a motivating point of interest, it is interesting to rewrite the answer as $$c = \left(\frac{2}{n/m + 1}\right)^{1/n}.$$ With this version, we see that the constant depends precisely on the ratio $n/m$, but is shrunk towards $1$ via the exponent $1/n$, in perfect agreement with OPs observations.

Multiply by $\left(\frac{2(7)}{3+7}\right)^{1/3} = 1.1187$

More generally, suppose that player $A$ rolls $n$ times and player $B$ rolls $m$ times (without loss of generality, we assume $m > n$). As others have already noted, the (unscaled) score of player $A$ is $$X \sim Beta(n, 1)$$ and the score of player $B$ is $$Y \sim Beta(m, 1)$$ with $X$ and $Y$ independent. Thus, the joint distribution of $X$ and $Y$ is $$f_{XY}(x, y) = nmx^{n-1}y^{m-1}, \ 0 < x, y < 1.$$

The goal is to find a constant $c$ such that

$$P(Y \geq cX) = \frac{1}{2}$$.

This probability can be found in terms of $c$, $n$ and $m$ as follows.

\begin{align*} P(Y \geq cX) &= \int_0^{1/c}\int_{cx}^1 nmx^{n-1}y^{m-1}dydx \\[1.5ex] &= \cdots \\[1.5ex] &= c^{-n}\left\{\frac{m}{n+m} \right\} \end{align*}

Setting this equal to $1/2$ and solving for $c$ yields

$$c = \left(\frac{2m}{n+m}\right)^{1/n}.$$

Multiply by $\left(\frac{2(7)}{3+7}\right)^{1/3} = 1.1187$

More generally, suppose that player $A$ rolls $n$ times and player $B$ rolls $m$ times (without loss of generality, we assume $m \geq n$). As others have already noted, the (unscaled) score of player $A$ is $$X \sim Beta(n, 1)$$ and the score of player $B$ is $$Y \sim Beta(m, 1)$$ with $X$ and $Y$ independent. Thus, the joint distribution of $X$ and $Y$ is $$f_{XY}(x, y) = nmx^{n-1}y^{m-1}, \ 0 < x, y < 1.$$

The goal is to find a constant $c$ such that

$$P(Y \geq cX) = \frac{1}{2}$$.

This probability can be found in terms of $c$, $n$ and $m$ as follows.

\begin{align*} P(Y \geq cX) &= \int_0^{1/c}\int_{cx}^1 nmx^{n-1}y^{m-1}dydx \\[1.5ex] &= \cdots \\[1.5ex] &= c^{-n}\left\{\frac{m}{n+m} \right\} \end{align*}

Setting this equal to $1/2$ and solving for $c$ yields

$$c = \left(\frac{2m}{n+m}\right)^{1/n}.$$

Multiply by $\left(\frac{2(7)}{3+7}\right)^{1/3} = 1.1187$

More generally, suppose that player $A$ rolls $n$ times and player $B$ rolls $m$ times. As others have already noted, the (unscaled) score of player $A$ is $$X \sim Beta(n, 1)$$ and the score of player $B$ is $$Y \sim Beta(m, 1)$$ with $X$ and $Y$ independent. Thus, the joint distribution of $X$ and $Y$ is $$f_{XY}(x, y) = nmx^{n-1}y^{m-1}, \ 0 < x, y < 1.$$

The goal is to find a constant $c$ such that

$$P(Y \leq cX) = \frac{1}{2}$$.

This probability can be found in terms of $c$, $n$ and $m$ as follows.

\begin{align*} P(Y \leq cX) &= \int_0^{1/c}\int_{cx}^1 nmx^{n-1}y^{m-1}dydx \\[1.5ex] &= \cdots \\[1.5ex] &= c^{-n}\left\{\frac{m}{n+m} \right\} \end{align*}

Setting this equal to $1/2$ and solving for $c$ yields

$$c = \left(\frac{2m}{n+m}\right)^{1/n}.$$

Multiply by $\left(\frac{2(7)}{3+7}\right)^{1/3} = 1.1187$

More generally, suppose that player $A$ rolls $n$ times and player $B$ rolls $m$ times (without loss of generality, we assume $m > n$). As others have already noted, the (unscaled) score of player $A$ is $$X \sim Beta(n, 1)$$ and the score of player $B$ is $$Y \sim Beta(m, 1)$$ with $X$ and $Y$ independent. Thus, the joint distribution of $X$ and $Y$ is $$f_{XY}(x, y) = nmx^{n-1}y^{m-1}, \ 0 < x, y < 1.$$

The goal is to find a constant $c$ such that

$$P(Y \geq cX) = \frac{1}{2}$$.

This probability can be found in terms of $c$, $n$ and $m$ as follows.

\begin{align*} P(Y \geq cX) &= \int_0^{1/c}\int_{cx}^1 nmx^{n-1}y^{m-1}dydx \\[1.5ex] &= \cdots \\[1.5ex] &= c^{-n}\left\{\frac{m}{n+m} \right\} \end{align*}

Setting this equal to $1/2$ and solving for $c$ yields

$$c = \left(\frac{2m}{n+m}\right)^{1/n}.$$