Maximizing the probability, such that the sum of the three numbers is at least one

Question

Source: AoPS

The sum of five real numbers is $8/5$. Let $P$ denote the probability that if you pick three different numbers randomly (from the five), then the sum of the three numbers is at least one.

Find the value of the five numbers, such that $P$ is maximum.

Note for equal numbers, $P=0$, so I just have no idea... Please help!

Answer 1

There are $10$ equiprobable triplets of the five numbers. Every number appears $6$ times in the triplets. Every pair of numbers appears $3$ times and the different triplets appear only one time.

If you concentrate $\frac 85$ in one number then the chance of getting a sum higher or equal than $1$ would be $\frac1{10}\times 6=0.6$. If you concentrate the total in $2$ different numbers then the same would be only $0.3$.

However if you divide $\frac85$ to three equal parts $(0.53)$ then the chance of getting a sum greater than $1$ would be will be $0.8$

Dividing the sum to four equal parts $(0.4)$ will result in $0.4$ .

I would divide $\frac85$ to $3$ equal parts.

Answer 2

I ran an optimization problem such that

if $x_1,x_2,x_3,x_4,x_5$ are the five real numbers then the constraints are

$x_1+x_2+x_3-y_1\ge0$

$x_1+x_2+x_4-y_2\ge0$

$x_1+x_2+x_5-y_3\ge0$

$x_2+x_3+x_4-y_4\ge0$

$x_2+x_3+x_5-y_5\ge0$

$x_2+x_4+x_5-y_6\ge0$

$x_1+x_3+x_4-y_7\ge0$

$x_1+x_3+x_5-y_8\ge0$

$x_1+x_4+x_5-y_9\ge0$

$x_3+x_4+x_5-y_{10}\ge0$

$x_1+x_2+x_3+x_4+x_5=\frac{8}{5}$

$x_i\gt0$

$y_i = 0$ or $1$

Objective function $P = \sum_{i=1}^{10}\frac{y_i}{10}$

When you ran this optimization problem Max of $P = 0.7$ for any three values of x being equal to $0.4$ and the other two equal to $0.2$

Solution $\boxed{x_i = 0.4, i = 1,2,3}$ and $\boxed{x_i = 0.2, i = 4,5}$

Answer 3

The optimization program obviously proves that $P=0.7$ is the best you can get, but that's a machine proof. Here is a more "traditional" proof. :)

Let the 5 numbers be $a \le b \le c \le d \le e$. If there is only one failing triplet (i.e., whose sum $<1$) then it has to be $\{a,b,c\}$, while if there is a second failure triplet then it has to be $\{a,b,d\}$, because any other triplet sum $\ge a+b+d \ge a+b+c$.

Assume for later contradiction that there are 0 to 2 failing triplets. Then in particular the following two triplets are both successful:

$$a+c+d \ge 1$$

$$a+b+e \ge 1$$

Summing these two implies: $2a + b + c+ d + e \ge 2$. Substituting $a+b+c+d+e = 1.6$, we have: $a \ge 2 - 1.6 = 0.4$, but since $a$ is the smallest number, the sum $a+b+c+d+e \ge 5a \ge 2$, which is a contradiction.

Therefore there are NOT o to 2 failing triplets, i.e. there are at least 3 failing triplets. Since there are constructive examples with exactly 3 failing triplets, those must be optimal.

Incidentally, the same argument also shows how to increase the sum $S$ (beyond $1.6$) s.t. one can have fewer failing triplets. Combining $a \ge 2 - S$ and $S \ge 5a$ implies: $S \ge 5 (2-S)$ or $6S \ge 10$ or $S \ge \frac{10}{6} = \frac{5}{3} = 1.666...$ Curiously, at this value of $S$, the optimal assignment is each number being $\frac{1}{3}$ and all 10 triplet sums $=1$, so $P=1$. In other words, the optimal probability $P$ jumps from $0.7$ to $1$ at $S=\frac{5}{3}$, and there is no value of $S$ for which the optimal $P$ is $0.8$ (two failing triplets) or $0.9$ (one failing triplet).