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Here is a question from my probability textbook:

Three different persons have each to name an integer not greater than $n$. Find the chance that the integers named will be such that every two are together greater than the third.

Here's what I did. After computing the cases up to $n = 7$ (which I'm not typing out here due to being too lazy), I was able to observe that we have the recursion$$p_1 = 1, \quad p_n = {{(n-1)^3 p_{n-1} + {{3n(n-1)}\over2} + 1}\over{n^3}}$$However, I don't know how to solve it. Can anyone help me?

Edit: I bountied the question. I'd like to see a complete self-contained solution solving the recurrence I give without reference to external sources such as OEIS, Wikipedia, etc.

Emperor Concerto
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  • Calling the chosen integers $a,b$ and $c$, the only way to fail to be a triangle is if $a\ge b+c$, or $b\ge a+c$, or $c\ge a+b$. These three events have the same probability, are disjoint, so you can find $P(a\ge b+c)$ and then multiply by $3$. $P(a\ge b+c)$ can be found with a summation. That is, if $a=2$, there is $1$ choice for $(b,c)$, if $a=3$, there are three choices for $(b,c)$, namely $(1,1), (2,1)$ and $(1,2)$. Sum over all possible values of $a$ of the number of choices for $(b,c)$. – Mike Earnest Aug 16 '21 at 19:43
  • Good job finding the recurrence (+1)! I find it easier to think of the recurrence as counting the number $S(n)$ of triples $(a,b,c)\in{1,2,\ldots,n}^3$ satisfying the constraints. Hint: Look at the difference $\Delta(n)=S(n)-B(n)$ between $S(n)$ and the number of negative cases $B(n)=n^3-S(n)$ for small values of $n$. You will see how $\Delta(n)$ behaves. Of course, Mike's way may be easier. But if follow up on my hint, you will know what the recurrence must give, making it easier to prove it by induction on $n$. – Jyrki Lahtonen Aug 16 '21 at 19:54
  • Only posting that as a comment/hint, because that kind of experimentation is a useful strategy when solving a new (contest) problem. – Jyrki Lahtonen Aug 16 '21 at 20:01
  • Ok. Luckily that does not change the validity of my advice. In a non-contest you may not need to carry out all that experimentation, but you may still see something you missed earlier :-) – Jyrki Lahtonen Aug 16 '21 at 20:08
  • Just to be clear: Does the sample space include $0$ through $n$, or just $1$ through $n$? – Brian Tung Aug 16 '21 at 20:17
  • Hi @BrianTung, it just includes $1$ through $n$. – Emperor Concerto Aug 16 '21 at 20:19

2 Answers2

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Letting $w_n$ be the number of permitted configurations we get the recursion

$$ \begin{align} w_n &= 1 + 3(n-1) + 3 \binom{n-1}{2} + w_{n-1}\\ \tag1 &=w_{n-1} + 1 + \frac{3}{2} n (n-1) \end{align} $$

Explanation: in the first line of $(1)$, each term in the RHS corresponds to the configurations that have $3,2,1,0$ values equal to $n$. The initial condition is $w_1=1$, or also $w_0=0$.

Noting that $p_n = w_n/n^3$, this concides with your recursion.

To solve this, one can postulate $w_n = a_1 n + a_2 n^2 + a_3 n^3$, replace on $(1)$ and solve for $a_i$. (This also can be attacked via generating functions, see eg). Or, noticing that $w_n - w_{n-1}$ is the discrete analog of the derivative, we can integrate the other side:

$$w_n - w_{n-1} = g_n \implies \sum_{k=1}^n g_k + w_0 = w_n $$

Then $$\begin{align} w_n &= \sum_{k=1}^n \left[1 + \frac{3}{2} k (k-1) \right] \\ &= \sum_{k=1}^n 1 + 3 \sum_{k=2}^n \binom{k}{2} \\ &= n + 3 \binom{n+1}{3} {\hskip 1cm} \\ &= \frac12 n^3+\frac12 n \end{align} $$

where we've used the Hockey-stick identity.

See also OIES A006003 where many alternative interpretations and results about this sequence are given.

The desired probability is then

$$p_n = \frac{w_n}{n^3}=\frac12 + \frac{1}{2n^2}$$

leonbloy
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Calling the chosen integers $a,b$ and $c$, the only way to fail to be a triangle is if $a≥b+c$, or $b≥a+c$, or $c≥a+b$. These three events have the same probability and are disjoint, so you can find $P(a≥b+c)$ and then multiply by $3$.

$P(a≥b+c)$ can be found by counting favorable cases with a summation. For each possible value of $a$, we must find the number of ways to choose positive integers $1\le b,c\le n$ for which $b+c\le a$. This is the same as counting triples of positive integers $(b,c,d)$ for which $b+c+d=a+1$. Using stars and bars, the number of solutions is $\binom{a}2$, so the total number of $(a,b,c)$ is $$ \sum_{a=2}^n\binom{a}2=\binom{n+1}3 $$ which is a special case of the hockey-stick identity. Finally, $$ P(\text{triangle ineq. holds})={1-3\cdot P(a\ge b+c)}=1-3\cdot \frac{\binom{n+1}3}{n^3}=\frac12+\frac{1}{2n^2} $$ In this limit, this probability is $1/2$. In fact, if you let $a,b,c$ be continuously chosen uniformly over the range $[0,1]$, then the probability that none of the three is larger than the sum of the other two is exactly $1/2$. The same proof works as well, using an integral instead of a sum.

Mike Earnest
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