13.10.12

7-sided dice. With activities!

I spotted these dice in a hobby shop and couldn't resist. Seven sides!


These bring many questions to mind. First and foremost: are they fair? How can it be determined?

Trying to calculate by shape, angles or area would be insanely complicated. This could be resolved by experiment, though, by throwing them a large number of times... Activity 1: determine if the dice are fair.

Here's a thought experiment. These are pentagonal prisms, two pentagon faces with connecting rectangles. If the rectangles had very small thickness, the dice would be like coins and only land on one of the two pentagon faces. If the rectangles were very long, they'd be like pencils, and only roll one of the five rectangle faces. There must be a point somewhere in between where the probability of a rectangle or a pentagon are the same! Did the makers of these dice find that point?

Here's another thought: does the chance depend on the area of faces? Do equal areas = equal probabilities? Activity 2: devise a thought experiment/extreme case to disprove this.

While we're thinking of prisms with equal area faces... well that's interesting in itself. What would a square prism with equal area faces look like. A cube. What about a triangular prism? Say the triangle has edge length 1, what edge length would the rectangle need to have equal area? (The answer is interesting!) Activity 3: Investigate other equal area face prisms... any interesting patterns?

Notice that these dice have dot patterns split across the edges because an edge will be up. Clever design! The dots are colored to help distinguish them from the other "faces". Clever again! The pentagons have digits printed on them instead of dots patterns. Oh. Too bad. Was this a design choice to emphasize the "seven-ness" of the dice? I think dot patterns would look nice. Activity 4: Design dot patterns for all of the faces. (What dot patterns look good on pentagon faces? If not 6 and 7, then redesign all of the dot patterns so that each roll looks good).

2 comments:

  1. What a lovely argument! (There must be a point somewhere in between where the probabilities are equal.)

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  2. Thanks! I imagine another "proof by extreme cases" could be used to disprove that equal area of faces = equal probability.

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