Die shaving (not a male-grooming thread)

[Changabooniggiwan]

A futuristic casino has invested in a new set of craps dice:

Each die is a cube of flawless diamond (with perfectly uniform density). The numerals on the dice are etched in such a way that they have no effect whatsoever on the outcomes of rolling the dice, and we will assume, for this puzzle, that the frequencies of the outcomes are entirely a product of surface area (i.e. faces with a higher surface area are more likely to land 'down'). The dice are manufactured to very high specifications - all angles are exactly 90 degrees, and all faces are planes. The only slight variable being side lengths, which are produced to a tolerance of +/-1%. Since each side of the die are a mere 15mm (+/- tolerance), the tolerance is virtually undetectable by humans, but any variation beyond 1% will be automatically detected by the casino's security systems.

You are somehow able to pinch one of these dice, and wish to gimmick it by shaving planes off the sides such that the die favours a higher mean outcome. The die you have stolen is the maximum size of a die within tolerance (i.e. 101% side lengths).

What are the dimensions of the die when you finish tampering with it, and what is its mean outcome?

[ralphmerridew]

My answer: Dimensions could be anything within specs; mean outcome is exactly 3.5. If all angles are exactly 90, then areas of opposite faces will be equal. Unless the numbering is changed, opposite faces sum to 7. Thf, P(1) == P(6), P(2) == P(5), P(3) == P(4).

Mean == P(1)*1 + P(2)*2 + P(3)*3 + P(4)*4 + P(5)*5 + P(6)*6
Mean == P(1)*(1+6) + P(2)*(2+5) + P(3)*(3+4)
Mean == (P(1) + P(2) + P(3))*7
Mean == (1/2)*7 == 3.5

[Changabooniggiwan*]

Dang, I was hoping somebody would at least frustrate themselves first. Well done, Mr Merridew. Let's extend this problem then, into it's more complex cousin, and say:

What if we allow angles to be included in the +/-1% tolerance, but still insist on plane faces?

[Zag]

What, exactly, are you shaving it with?

[Changabooniggiwan*]

A quantum-multiphasic laser, naturally.

[Tony Gardner]

I figured it couldn't be that straightforward - but apparently it was

Quick stab at the extended version:

The largest increase of the average outcome will be reached by maximizing the 6 area and simultaneously minimizing the 1 area. No additional tampering allowed, since that would create curved surfaces (I think).
Maximizing the 6 area means not reducing the entire die's size, so the 6 ara remains 101x101 (rescaling the 15 mm to 100 for clarity). Minimizing the 1 area means this becomes 99x99. Easy goniometry shows that the other four surfaces are trapezoids with height 101, shorter side 99, and longer side 101 - and hence of area 100x101. Dimensions of the new die become 99x99 for the 1 area, 101x101 for the 6 area, and all the other sides are sqrt(1^2+1^2+101^2). Divide by (100/15) to arrive at the new dimensions for the non-rescaled die
The mean outcome becomes (1*99^2+3.5*4*100*101+6*101^2)/(99^2+4*100*101+101^2)=3.5166

Typing this, I realize this probably isn't correct because it does increase the mean outcome if you reduce all the 'other' sides to 99% of the original. Doing this means that the dimensions become 101 for all four sides of the 6 area, and 99 for all other.
The surface area of a single trapezoid becomes100xsqrt(9800). This changes the mean outcome to 3.5168.

Shoot away...

[Tony Gardner]

Actually, I think my entire above post really doesn't add anything because it discusses the case of sides being within 1% tolerance, not angles

[Zag]

[Changabooniggiwan*]

[Changabooniggwan*]

Oooh, spanner in the works:

Just for kicks, I calculated for a 3% tolerance, and got 3.48105 - lower than with a 1% tolerance. I guess the reduced areas of the 2, 3, 4, 5 sides overtakes the gain from reducing the area of the 1 side. Hmm.

I'll check that again

[Changabooniggiwan*]

Oops. My mistake. It does get higher.

[fadeblue]

What happens if you also increase the 5 face to a trapezoid with legs of 101? That brings the mean up, but it also turns the 3 and 4 faces into kites, which will bring the mean down. Anyone feel like calculating that?