How to equalize the chance of throwing the highest dice? (Riddle) I just invented the following riddle, doing statistics work. (I actually need the answer!)
Riddle:

Imagine a dice game with the aim of throwing the highest dice.
  The dice are special and have infinite sides with numbers ranging from 0 to 1! (uniform, no bias)
There are 2 players: Player-A has 3 dice to throw, player-B has 7
  dice. This means player-B has a chance of 7/10 of winning, which is
  to throw the highest number of all 10.
Now, to bring fairness to the situation, the players agree to multiply each number 
  thrown by player-A by a certain constant. What is the value of this
  constant, so that each player has a 50% chance of winning?

Can you find a general formula to determine this constant, based on the amounts of dice the 2 players have?
(And in case this is a known problem: Do you know how this is called?)

Considerations/ Spoiler:
The adjustment-constant does not just depend on the ratio of throws (3:7 in this case); instead, the absolute number is important. For example, if the players had 300 and 700 throws, then this constant would be much closer to 1. 
My intuition: I think a good estimate is to assume a homogeneous distribution of the throws: For example the 3 throws are at decimals 0.25, 0.5 and 0.75! Now the highest number would be 0.75! Do the same with player-B and you get the ratios of the expected highest numbers (-> the adjustment-constant). Unfortunately that's just my intuition and I am not sure if this is correct.

EDIT:
I am thankful for all the answers but surprised that nobody used an approach similar to my described one. For completeness, here I explain where I was wrong:
I assumed the expected maximum of throws would be 1-1/(n+1), which is correct, as simulated by the following script:
import numpy as np import matplotlib.pyplot as plt

x,y,y2 = [],[],[] for n in range(1,21):
    x.append(n)
    y2.append(1-1/(n+1))
    temp = []
    for _ in range(10000):
        sample = np.random.random_sample(n,)
        temp.append(max(sample))
    y.append(np.mean(temp))


plt.scatter(x,y) 
plt.plot(x,y2) 
plt.title("Mean max = 1/(n+1)")     
plt.xlabel("Number of throws") 
plt.ylabel("Mean max of throws") 
plt.show()


Which means, if we used a constant c to multiply each of the n throws of player A, the expected maximum would be equal to the m throws of player B, if we use this formula for c:
 (or)

But this is wrong, because the riddle does not try to equalize the mean of the maxima. Instead it wants to equalize the rank-sum of the 2 players distributions of maxima. (if we ranked each maximum throughout both distributions)
Here, just for illustrative purposes, I show how my formula is unable to accurately fit the median of maxima:

 A: I'd like to try to put pieces of comments and answers together into a simulation, and into a plan for an analytic solution.
As @whuber says in his Comment, the maximum $X_1$ of three independent
standard uniform random variables has $X_1 \sim \mathsf{Beta}(3,1)$ and the maximum $X_2$ of seven independent
standard uniform random variables has $X_2 \sim \mathsf{Beta}(7,1).$ This is easy to prove analytically.
Then, as implied by @MikeP's Answer, $X_1^{3/7} \sim \mathsf{Beta}(7,1).$ This is also easy to prove analytically. Thus $X_2$ and $X_1^{3/7}$ have the same distribution.
Below are simulations in R of the distributions of $X_1, X_2,$ and $X_1^{3/7},$ each based on samples of size $100\,000.$
Histogams show the simulation results along with the
density functions of $\mathsf{Beta}(3,1)$ [red curve] and 
$\mathsf{Beta}(7,1)$ [blue], as appropriate.
set.seed(1120)
x1 = replicate(10^5, max(runif(3)))
mean(x1)
[1] 0.7488232     # aprx E(X1) = 3/4
par(mfrow=c(1,3))
hist(x1, prob=T, col="skyblue2")
 curve(dbeta(x,3,1), add=T, col="red", n=10001)

x2 = replicate(10^5, max(runif(7)))
mean(x2)
[1] 0.8746943     # aprx E(X2) = 7/8
hist(x2, prob=T, col="skyblue2")
 curve(dbeta(x,7,1), add=T, col="blue", n=10001)

mean(x1^(3/7))
[1] 0.8743326     # aprx 7/8
hist(x1^(3/7), prob=T, col="skyblue2")
 curve(dbeta(x,7,1), add=T, col="blue", n = 10001)
par(mfrow=c(1,1))


A: I did not solve the problem analytically but I performed a simulation with 100 different $a/b$ ratios varying from 0.01 to 1. $a$ is the number of dice of player A and $b$ is the number of dice of player $b$. For each ratio I simulated 1000 games and computed the multiplicative constant.
This what I got: 
For the dice I assumed a uniform distribution between 0 and 1.

If we take the same ratio the expected value for the multiplicative constant is the same. I tested with a ratio of $0.5$ timing $a$ and $b$ up to a factor of 2000. Here the results as scatter plot and density distribution
:

A: 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
$$c = \left(\frac{2m}{n+m}\right)^{1/n}.$$
A: I don't believe that a linear scaling factor will equalize the odds, or at least I cannot determine one.  However, there is a power factor that can.
If you raise player-A's score to the $\frac{3}{7}$ you should have a fair game.  Obviously, since scores are between 0 and 1, raising it to a power of between 0 and 1 (not inclusive) will actually increase it.
Why?
The way I figure it, the probability of a score not exceeding $S$, is equal to $1 - S^n$.
If we set $1 - S_1^{n_1} = 1 - S_2^{n_2}$
$$1-S_1^{n_1}=1-S_2^{n_2}$$
$$S_1^{n_1}=S_2^{n_2}$$
$$n_1 log(S_1) = n_2 log(S_2)$$
$$log(S_1) = \frac{n_2}{n_1} log(S_2)$$
$$S_1 = S_2^{\frac{n_2}{n_1}}$$
