Alex's loaded dice

[bonanova]

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,

1. 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?
   
   
   
2. If so, what percentage of the original tetrahedron's volume
   would remain?
   

Drinks for a week to the bloke who has the answers.

[Guest]

nearly 11/27 of original dice???

[bonanova]

nearly 11/27 of original dice???

A bit on the low side.

You can't cut more than a certain amount if the new faces become triangles.

[Guest]

0.68 of original tetrahedron.

it is 1-(1/14)^3/2 + (2/14)^3/2 + (3/14)^3/2 +(4/14)^3/2)

[bonanova]

Alex slides a cold one down the bar to nobody.

Umm.. you're missing a paren, but I'll consider that a typo.

The intent is clear.

[Guest]

Say the area of any side id the Tetrahedron is x.

You make 4 cuts, each cut removes a surface area of a, b, c, and d respectively. And creates a new side with the surface are a, b, c, or d.

The new surface areas are now:

a

b

c

d

x-(b+c+d)

x-(a+c+d)

x-(a+b+d)

x-(a+b+c)

in that order, lets say a=1 then b =2, and c=3 and d =4 and x-(b+c+d) has to be 5... solving we get x= 14.

So original area: x = 14

Of the tetrahedron x*4 = 14*4 = 56

New surface area: 4x - 2(a+b+c+d) = 4*14 - 2(1+2+3+4) = 36

% = 36/56 = 64.3 %

Edited June 3, 2009 by adiace

[bonanova]

Say the area of any side id the Tetrahedron is x.

You make 4 cuts, each cut removes a surface area of a, b, c, and d respectively.

The surface areas are now:

a

b

c

d

x-(b+c+d)

x-(a+c+d)

x-(a+b+d)

x-(a+b+c)

in that order, lets say a=1, b=2, c=3, and d=4, then if x= 14 and your probabilities are in order of 1 to .

So original area: x = 14

Of the tetrahedron x*4 = 14*4 = 56

New surface area: 4x - 2(a+b+c+d) = 4*14 - 2(1+2+3+4) = 36

% = 36/56 = 64.3 %

Nice analysis.

But we need the percentage of volume that remains.

[Guest]

Oh opps...

Say the face surface area of any side id the Tetrahedron is x.

You make 4 cuts, this creates new sides with the face surface are a, b, c, or d.

The new face surface areas are now:

a

b

c

d

x-(b+c+d)

x-(a+c+d)

x-(a+b+d)

x-(a+b+c)

in that order, lets say a=1 then b =2, and c=3 and d =4 and x-(b+c+d) has to be 5... solving we get x= 14.

So original face area: x = 14

length of side from side from SA > s = 2(SA^1/2)/(3^1/4)

volume > [(2^1/2)/12]*(s^3) = [(2^1/2)/12]*[2/(3^1/4)]^3*SA^3/2

volume = constant * SA^3/2 = k*SA^3/2

Old Volume = V(x)

New Volume = V(x) - [V(a) + V(b) + V© + V(d)]

% of remaining volume is:

k*[14^(3/2)-(1^(3/2)+2^(3/2)+3^(3/2)+4^(3/2))]

----------------------------------------------------------- * 100

k*14^(3/2)

= 67.5 %

[bonanova]

Oh opps...

Say the face surface area of any side id the Tetrahedron is x.

You make 4 cuts, this creates new sides with the face surface are a, b, c, or d.

The new face surface areas are now:

a

b

c

d

x-(b+c+d)

x-(a+c+d)

x-(a+b+d)

x-(a+b+c)

in that order, lets say a=1 then b =2, and c=3 and d =4 and x-(b+c+d) has to be 5... solving we get x= 14.

So original face area: x = 14

length of side from side from SA > s = 2(SA^1/2)/(3^1/4)

volume > [(2^1/2)/12]*(s^3) = [(2^1/2)/12]*[2/(3^1/4)]^3*SA^3/2

volume = constant * SA^3/2 = k*SA^3/2

Old Volume = V(x)

New Volume = V(x) - [V(a) + V(b) + V© + V(d)]

% of remaining volume is:

k*[14^(3/2)-(1^(3/2)+2^(3/2)+3^(3/2)+4^(3/2))]

----------------------------------------------------------- * 100

k*14^(3/2)

= 67.5 %

Yeah, that's it.

[Guest]

I jotted this problem down and worked on it away from the computer - but forgot to write the last bit.

Now I've read someone's answer, which seems fine. Certainly, the solution for the bits to cut off agree with my results.

Anyway, 2 thoughts occur:

1)It's possible to construct such a solid using the net of a tetrahedron to start with (That's where I began my investigation - I'm a visual learner)

2)We can also calculate the lengths of the edges for this new solid with a bit of elementary maths.

Donjar