Thursday, July 4, 2013

Math Puzzle

I spent way too much time today trying to figure out this math puzzle. The results he wants are equivalent to rolling 1d4 twice and keeping the lowest result:

He wants to use d2s exclusively, and you can reproduce a d4 with two 1d2s easily enough. If you're actually sticking to coins, for instance, you could say "flip a shiny penny, a shiny nickel, a dull penny and a dull nickel. Face-up coins get their full value, tails-up coins get zero value, take the lower value of shiny+shiny and dull+dull." That would give you 6 cents 1/16th of the time, 5 cents 3/16ths of the time, 1 cent 5/16ths of the time, and 0 cents 7/16ths of the time.

The trick is that he wants all d2s to have uniform face values, and because d2s only have two faces, all of those other tricks he mentions are going to come out the same as just adding the dice together. At least, everything else I've tried so far has come out that way, and I'm pretty sure you can't get the 1/3/5/7 ratio just by straight summing the roll.

Okay, [(3d2 - 2)*1d2]/2, round up, seems to work. It's not the friendliest looking thing, but I don't think the solutions can get much better.

Anyways, I was looking at that chart I drew of the 1/3/5/7 ratio, and I thought, hey, that kind of looks like my bullseyes. But, instead of summing two dice like the 2d6 rolls I graphed before, it chooses between two independent rolls! That's more flexible, mechanically; and it's nice to see that the stripes don't thin towards the rim. I guess I'm now hoping to find a 1dX roll that thins evenly and infinitely towards the center. So I can roll two of them for every roll, and allocate them between disconnected aspects of the action.