The folk down at Morty's couldn't remember the last time that
Alex had dropped by. Some had conjectured he had given up
puzzles. A few, less well informed, to be sure, guessed he
had given up drinking. Whichever was the case, or perhaps for
other reasons altogether, Alex had been conspicuous by his
absence for several months.
But last night he was there. And his barely concealed grin put
the boys on notice of a question that would put them once again
to the test.
Ya know how those basketball folk in the States draw numbered
ping pong balls to determine the teams' draft order? Well I say
that's a stone-age way of assigning variable probabilities. If they
had a brain in their collective head, they would design an unfair
die with faces numbered proportionately to the probability that
the number shows, when the die is rolled.
In fact, that's something that even you geniuses might be able
to do.
So here's the challenge. Suppose there are eight basketball
teams, numbered 1-8, and they numbered one ball with a "1",
two balls with a "2", and so on until they numbered eight balls
with an "8". The number eight team would have 8 times the
probability of winning as the number one team. And so on for
each of the other teams. You need to get the same result;
only do it by throwing an unfair 8-sided die.
I'll give you a regular tetrahedron - that's one of the five Platonic solids,
ya know, and tell you to slice off each of the four vertices, parallel to their
opposite faces, making them into new triangular faces. The four
original triangular faces would become hexagons. Now you have
an 8-sided die.
If you assume that the probability of a polyhedron
landing on one of its plane faces is proportional to the
area of that face, Can you construct in this manner an 8-sided die, with faces
numbered 1-8 with the property that the number on each face
is proportional to the probability of that face showing? If so, what percentage of the original tetrahedron's volume
would remain? Drinks for a week to the bloke who has the answers.