On Sun, May 26, 2013 at 1:09 PM, Steven D'Aprano <steve+comp.lang.pyt...@pearwood.info> wrote: > You're right, all my dice are eight-sided and complex: > > 1+0i > 1+1i > 1-1i > -1+0i > -1+1i > -1-1i > > > :-)
Now THAT is a dice of win! >> Now, I have here with me >> a set used for maths drill (to be entirely accurate, what I have here is >> the company's stock of them, so there are multiples of each of these - >> anyone need to buy dice?) > > Are you serious? What's the cost, posted to Melbourne? $1 each, postage probably $5 for any number. Or there may even be option to pick up / hand deliver, depending on where in Melb you are. http://www.kepl.com.au/ - company's winding down, but we still have stock. > Oh, you mean ÷ (division sign)! Why didn't you say so? :-P I tend to stick to ASCII in these posts. :) > And another thing, shame on you, you mean × not x. It's easy to find too: > > py> from unicodedata import lookup > py> print(lookup("MULTIPLICATION SIGN")) > × I'm aware of that, but see above, I stick to ASCII where possible. The faces would be better represented with some of the other digits (the bolded ones, perhaps), but I used the ASCII digits. :) >> Plus, if you roll 2d6 (that is, two >> regular six-sided dice and add them up), 7 is statistically the most >> likely number to come up with. Therefore it IS random. > > Yes, but if you subtract them the most common is 0, if you multiply the > most common are 6 or 12, and if you divide the most common is 1. If you > decide on the operation randomly, using the +-×÷+ die above (ignoring > wildcards), the most common result is 6. The probability of getting a 7 > is just 1/15. > > from collections import Counter > from operator import add, sub, mul, truediv as div > ops = (add, sub, mul, div, add) > Counter(op(i, j) for op in ops for i in range(1, 7) for j in range(1, 7)) LOL! I never thought to go THAT far into the analysis..... Nice one! ChrisA -- http://mail.python.org/mailman/listinfo/python-list
