The queue at the tap

Four neighbours queue at one tap. The tap fills a ten-litre bucket in two minutes. Three neighbours carry one bucket each; the fourth has three buckets and stands at the front.

$4$ neighbours, $1$ tap, one bucket of $10$ L takes $2$ min. Three have $1$ bucket; one has $3$ buckets and stands first.

$4$ people, a bucket $=2^{\prime }$; order $A=(3,1,1,1)$ and order $B=(1,1,1,3)$ buckets.

If the one with three buckets goes first, he fills for six minutes. The others wait six, eight and ten minutes. Added up, that is twenty-four minutes of waiting, six minutes each on average.

Biggest job first: he fills for $3\times 2=6$ min, and the others wait $6$, $8$ and $10$ min. Total wait: $0+6+8+10=24$ min; average: $24÷4=6$ min.

Order $A$: waits $(0,6,8,10)$, total $W_A=24$, mean ${\bar{w}}_A=\frac{24}{4}=6$.

But he smiled and said, ‘You go first.’ Now the others wait nothing, two and four minutes, and he waits six. Added up, the waiting is twelve minutes, three minutes each on average.

He says ‘you go first’. Now the waits are $0$, $2$, $4$ and $6$ min. Total: $0+2+4+6=12$ min; average: $12÷4=3$ min.

Order $B$: waits $(0,2,4,6)$, total $W_B=12$, mean ${\bar{w}}_B=\frac{12}{4}=3$.

Either way everyone was done after twelve minutes. The queue did not get faster, only fairer: the average wait fell from six minutes to three, just by putting the short jobs first.

Either way the end is at $6\times 2=12$ min. The average wait fell from $6$ to $3$ min, just by putting the short jobs first.

End for both: $6\times 2=12^{\prime }$. But $W_B=\frac{W_A}{2}$.