$33.3\bar{3}\%$ Of The Time, You Win Every Time

The average for all players must be same so lets calculate average for first player only. The probability that $r_{1}$ is greater than $r_{2}$ is $r_{1}$ because distributions are uniform. Similarly, probability that it is greater than $r_{3}$ is also $r_{1}$ . Hence the probability that it is greater than both of them is $r_{1}^{2}$ . Thus the density function in this case is:

$p(r)=r^{2}$

Hence the average value is given by:

$<r>=\frac{\int_{0}^{1}rp(r)}{\int_{0}^{1}p(r)}$

Note that in this case $\int_{0}^{1}p(r)\neq 1$

This gives us $\boxed{Average=0.75}$

By the Law of Symmetry, the average is when $r_1,r_2,r_3$ divide the range of numbers equally. This is when $r_1=0.25$ , $r_2=0.5$ , and $r_3=\boxed{0.75}$ . We can also see from this method that the average value of the loser is $0.25$ , and the average value of the remaining person is $0.5$ .

The average is the same for all players so it is equal to the average of the winning value for all games. We calculate the latter. We consider the unit cube $0\leq r_1\leq 1, 0\leq r_2\leq 1, 0\leq r_3\leq 1$ . Given $0\leq x\leq 1$ , the part of the unit cube where $x=\max(r_1,r_2,r_3)$ is a surface of area $3x^2$ . The average is then $\int_0^1(3x^2\cdot x)dx=\left[3\frac{x^4}{4}\right]_0^1=\frac{3}{4}$ .

The average value is the integral $\int_0^1 \int_0^1 \int_0^1 max(x,y,z) \, dx \, dy \, dz$ (divided by the volume of the unit cube, which is 1). By symmetry this is $3$ times the integral over the region where $z$ is the max. So we get $3 \int_0^1 \int_0^z \int_0^z z \, dx \, dy \, dz = 3 \int_0^1 z^3 \, dz = \fbox{3/4}$ .

The same proof shows that the average value for $n$ friends is $\frac{n}{n+1}$ .

Letting $r_0=0$ , we are looking for $E( \max \{ r_1,r_2,r_3\}) =E(M_0)$ where

$M_i= {\max}_j \{ r_i-r_j - \lfloor r_i-r_j \rfloor \}$

To complete the solution we need two observations:

First, $E(M_0)$ is unaffected by the value of $r_0$ . In particular, then, instead of $r_0=0$ we could assume that $r_0$ is chosen at random just like $r_1,r_2,r_3$ .

Second, $M_0+M_1+M_2+M_3=3$ .

Together, these imply that $E(M_0)=E(M_1)=E(M_2)=E(M_3)=\frac34$ .

I got the answer on the 3rd try, with no computation but pure logic, wow!