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

cto multiply each of thenthrows of player A, the expected maximum would be equal to themthrows of player B, if we use this formula for c:But this is wrong, because the riddle does not try to equalize the

meanof 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:

**Answer**

### 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}.

**Attribution***Source : Link , Question Author : KaPy3141 , Answer Author : knrumsey*