Is Three Times Seven The Optimum?

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wheelsYou will remember that dice are made of six square sides, carrying one to six pips on each sides. The pips (or points, or eyes) are arranged such that pips on opposing sides add up to seven: one opposes six, two opposes five, and three opposes four. Each pair adds up to seven.

Why is that, I wonder?

What stops me from manufacturing regular dice with sides adding up to six (one, opposing five, and two, opposing four) and nine (three, opposing six)? Or four (one, opposing three), six (two, opposing four), and eleven (five, opposing six)?

I can only think of one reason: maybe, designing dice such that the sum of pips on opposing sides equals seven, is not at all about the sum being seven. Maybe, the argument is that all pairs of opposing sides add up to the same number results in fairly balanced dice (the weight across all axis levels, so that the centre of gravity is in the geometrical centre of the body)?

I tried a few alternative designs, and have not yet found a design that produces equal sums of pips on all pairs of opposing sides, other than seven.

Can you?

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3 thoughts on “Is Three Times Seven The Optimum?

  1. Danke fuer die Blumen. Ich arbeite noch an den Banner-bildern, die gefallen uns am besten.

  2. Und ich hab immer gedacht, Mathematiker beschäftigen sich mit dem, was die Welt im Innersten zusammenhält! Deine neue Seite ist übrigens schöner als die alte, zumindest auf den ersten schnellen Blick

  3. Roy commented by snail-mail:
    Incidentally, if dice are to have opposing pairs of sides adding up to the same number, that number HAD to be 7. Each die has 1,2,3,4,5,6 which totals 21, and there are only three pairs of opposing sides, so each pair has to sum 21/3=7.

    Thanks, Roy!

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