Probability in geometry: finding the probability of selecting a $5$-tuple such that a pentagon can be formed from its sides

Question

Consider all tuples of $5$ real numbers each less than $5$.
Find the probability of selecting a tuple such that a pentagon could be formed from its sides.

My attempt : I tried to visualize the problem for a triangle and could come up with a solution with graphs but I don't know how to approach this for the above one .

Part 2 : Solve the above problem if the sum of the elements of the tuple is 5.

Answer 1

Let $[5] = \{1,2,3,4,5\}$ and $X_i \sim \mathcal{U}([0,5)), i \in [5]$ be five iid random variables uniformly sampled over $[0,5)$. Let $Y_1 \ge Y_2, \cdots, Y_5$ be any reordering of them such that $Y_1$ is the largest number.

In order for $X_1,\ldots, X_5$ to form a pentagon, the necessary and sufficient condition is $$X_i < \sum\limits_{j=1,\ne i}^5 X_i\quad\text{ for }i \in [5]\quad\iff\quad Y_1 < Y_2 + Y_3 + Y_4 + Y_5\tag{*1}$$

Aside from events of probability zero, $Y_1 \in (0,5)$. When $Y_1 > 0$, define $$Z_2 = \frac{Y_2}{Y_1},\quad Z_3 = \frac{Y_3}{Y_1},\quad Z_4 = \frac{Y_4}{Y_1}, \quad Z_5 = \frac{Y_5}{Y_1}$$ Under the assumption $Y_1 > 0$, it is easy to see $Z_2,\ldots,Z_5 \sim \mathcal{U}([0,1])$. As a result, the probability we seek equals to

$$ \verb/Pr/[ 1 < Z_2 + Z_3 + Z_4 + Z_5] = 1 - \underbrace{\verb/Pr/[Z_2 + Z_3 + Z_4 + Z_5 \le 1 ]}_{\text{ volume of a right-angled 4-simplex of sides 1}} = 1 - \frac{1}{4!} = \frac{23}{24}$$

Update

For part 2, let $X = \sum_{i=1}^5 X_i$. Under the constraint $X = 5$, the condition on LHS of $(*1)$ becomes $$X_i < X - X_i,\;\forall i \in [5]\quad\iff\quad X_i < \frac{5}{2},\; \forall i \in [5]$$

Notice up to events of probability zero, the events $X_i \ge \frac{5}{2}$ for different $i$ are disjoint.
This means the conditional probability for forming a pentagon can be evaluated as

$$\begin{align}\verb/Pr/\left[ \bigcap_{i=1}^5 \left( X_i < \frac{5}{2} \right) : X = 5 \right] &= 1 - \verb/Pr/\left[ \bigcup_{i=1}^5 \left(X_i \ge \frac{5}{2}\right) : X = 5 \right]\\ &= 1 - \sum_{i=1}^5 \verb/Pr/\left[ X_i \ge \frac{5}{2} : X = 5\right]\\ &= 1 - 5 \verb/Pr/\left[ X_1 \ge \frac{5}{2} : X = 5 \right] \end{align} $$ Geometrically, the region determined by the relations $\frac{5}{2} \le X_1, 0 \le X_2, X_3, X_4, X_5$ and $X = 5$ is a 4-simplex whose sides is one half of the 4-simplex determined by the relations $0 \le X_1, X_2, X_3, X_4, X_5$ and $X = 5$. This implies $$\verb/Pr/\left[ X_1 \ge \frac{5}{2} : X = 5 \right] = \frac{1}{2^4}$$ and as a result,

$$\verb/Pr/\left[ \bigcap_{i=1}^5 \left( X_i < \frac{5}{2} \right) : X = 5 \right] = 1 - \frac{5}{16} = \frac{11}{16}$$

Answer 2

ad. 1

Hint 1

The pentagon can be formed, if each side is smaller than the sum of the rest, ie:

$$a_k < a_1+a_2+a_3+a_4+a_5 - a_k$$ for $k\in \{1,2,3,4,5\}$

This leads us to the inequality $$a_k < \frac{a_1+a_2+a_3+a_4+a_5}{2}$$

Hint 2

Assume, that the numbers are arranged in growing order, ie: $$0<a_1\leq a_2\leq a_3\leq a_4\leq a_5\leq 5$$ Now we have to check the situations, where $$a_5<a_1+a_2+a_3+a_4$$ Note, that if $a_1>\frac{5}{4}$, then $a_1+a_2+a_3+a_4>4a_1>5\geq a_5$. Of course $p_1=P\left(a_5<\frac{5}{4}\right)=\frac{3}{4}$

If not (ie. $a_1\leq \frac{5}{4}$), then let's take $a_2 > \frac{5-a_1}{3}$ (and $a_2\geq a_1$) with probability $p_2 = \frac{10+a_1}{3(5-a_1)}$ - our condition is satisfied.

If not ($a_2 \leq \frac{5-a_1}{3})$, let's take $a_3>\frac{5-a_1-a_2}{2}$ (and $a_3\geq a_2$) with probability $p_3 = \frac{5+a_1+a_2}{2(5-a_2)}$ - we can create the pentagon.

If not, let's take $a_4 > 5-a_1-a_2-a_3$ (and $a_4\geq a_3$) with probability $p_4 = \frac{a_1+a_2+a_3}{5-a_3}$ - again our condition is satisfied.

If not (ie. $5 > a_+a_2+a_3+a_4$) selection of $a_5$ will decide, if the pentagon can be constructed. Of course in that situation we can construct pentagon only if $a_5 < a_+a_2+a_3+a_4$ (where $a_5 \geq a_4$), so probability is $p_5=\frac{ a_+a_2+a_3}{5-a_4}$

If not, the pentagon can't be constructed.

Finally the probability can be computed by calculating the following integral:

$$P=p_1+q_1\left(\int\limits_0^\frac{5}{4}p_2+q_2\left(\int\limits_{a_1}^\frac{5-a_1}{3}p_3+q_3\left(\int\limits_{a_2}^\frac{5-a_1-a_2}{2}p_4+q_4\left(\int\limits_{a_3}^{5-a_1-a_2-a_3} p_5 da_4\right)da_3\right)da_2\right)da_1\right)$$ where $q_i=1-p_i$ are the probabilities of "if not's"

ad 2 Arrange the numbers in increasing order, ie: $$0<a_1\leq a_2\leq a_3\leq a_4\leq a_5$$ We have: $$a_1\in \left(0, \frac{5}{5}\right)$$ $$a_2\in \left(a_1, \frac{5-a_1}{4}\right)$$ $$a_3\in \left(a_2, \frac{5-a_1-a_2}{3}\right)$$ $$a_4\in \left(a_3, \frac{5-a_1-a_2-a_3}{2}\right)$$ $$a_5=5-a_1-a_2-a_3-a_4$$ Applying the following inequality to the biggest number $$a_k < \frac{a_1+a_2+a_3+a_4+a_5}{2}$$ we obtain $$\frac{5}{2}<a_1+a_2+a_3+a_4$$ So we have: $a_1 > \frac{\frac{5}{2}}{4}$ with probability $p_1=\frac{1-\frac{5}{8}}{1}=\frac{3}{8}$

If not then $a_2 > \frac{\frac{5}{2}-a_1}{3}$ with probability $p_2=\frac{\frac{5-a_1}{4}-\frac{\frac{5}{2}-a_1}{3}}{\frac{5-a_1}{4}-a_1}$

and so on...