# Probability of Average

### Problem

### Solution 1

To have a whole average, the largest and the smallest numbers of the three drawn need to have the same parity. If they do, the third number is determined uniquely.

There are $\displaystyle {10\choose 2}=45$ pairs of even numbers in the set $1,2,\ldots,20$ and as many pairs of odd numbers. This means there are exactly $90$ ways to choose three numbers out of twenty so that one of them is the average of the other two. Altogether, there are $\displaystyle {20\choose 3}$ ways to choose three numbers, all equiprobable, giving the sought probability as

$\displaystyle P=\frac{\displaystyle {10\choose 2}+{10\choose 2}}{\displaystyle {20\choose 3}}=\frac{90}{1140}=\frac{3}{38}.$

### Solution 2

For each ball $i=2,\ldots,10,$ the number of pairs which has $i$ as average is $i-1,$ which total $\displaystyle \frac{9\cdot 10}{2}=45.$ The same for $i=11,\ldots,19,$ due to symmetry: $i\mapsto 20-i.$ Hence the total number of choices is $45\cdot 2=90,$ out of $\displaystyle {20\choose 3}.$

### Solution 3

Let the smallest number chosen be $k$. To satisfy the average condition, the other two balls need to be numbered $k+d$ and $k+2d$ for some positive integer $d$. Moreover, $k+2d\leq 20$. Thus, if $k$ is odd, $d\in \{1,2,...,10-k/2\}$; if $k$ is even, $d\in \{1,2,...,10-(k+1)/2\}$. Thus, the required probability is

$\displaystyle \begin{align} P&=\frac{1}{C(20,3)}\left[\sum_{k\in \{1,3,5,...,19\}}\left(10-\frac{k+1}{2}\right)+ \sum_{k\in\{2,4,6,...,20\}}\left(10-\frac{k}{2}\right)\right] \\ &=\frac{2}{C(20,3)}\sum_{m=1}^{10}(10-m) \\ &=\frac{2\cdot 6 \cdot 45}{20\cdot 19\cdot 18}=\frac{3}{38}\sim 0.079. \end{align}$

### Solution 4

Number of overlapping triples $(i-2k, i-k, i)$

for $k=1, \ldots,$ $9$

for $k=1,$ $i \in [3,20],$ so the number is $18$

for $k=2,$ $i \in [5,20],$ so the number is $16$

... and so on

for $k=9,$ $i \in [19,20],$ so the number is $2.$

So number of triples = $\displaystyle 2\cdot\frac{9\cdot 10}{2} =90$ while total possible is $\displaystyle {20\choose 3}.$

### Acknowledgment

This is a slight modification of problem 320 from J. G. McLoughlin et all, Jim Totten's Problems of the Week, World Scientific, 2013.

Solution 2 is by Hélvio Vairinhos; Solution 3 is by Amit Itagi; Solution 4 is by Joshua Miller. Vrund Shah came up independently with Solution 1.

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