POV-Ray : Newsgroups : povray.off-topic : Interesting dice problem : Interesting dice problem Server Time
5 Sep 2024 21:22:54 EDT (-0400)
  Interesting dice problem  
From: Warp
Date: 14 May 2009 11:32:45
Message: <4a0c399d@news.povray.org>
The rock-paper-scissors game between two people can substitute coin-tossing:
It's played until one of them wins, and that's equivalent to tossing a coin
to see who wins. (Let's forget about the slight non-randomness of human
behavior in this context. It's not the point here.)

  This method of deciding between two things at random between to people
(or two groups of people, eg. teams) is often used when there's no proper
equipment available (in this case a coin).

  Another way of putting it: Two people playing rock-paper-scissors once
(ie. until one of them wins) is equivalent to rolling a 1d2: If the first
person wins, it's like rolling a 1, and if the second person wins, it's
like rolling a 2.

  This gave me an idea for a small thinking problem:

- How should two people play rock-paper-scissors in such way that it's
completely equivalent to rolling a 1d6? Or is it theoretically even
possible?

- How about other die sizes, such as rolling a 1d10 or a 1d20? How about
more difficult sizes, such as 1d3 or 1d5? Is it possible to achieve a
perfect equivalent with rock-paper-scissors?

- How should they play to get the equivalent of throwing multiple dice?
For example, the equivalent of rolling 2d6? How about 3d6 or 4d6? How
about even more complicated situations, such as 2d20 or 3d20?

- Would the problem become easier if there were more than two people
available?

- Would the problem become easier if a more complicated variant of the
rock-paper-scissors game would be used (eg. one with five options, each
option beating two others and being beaten by the remaining two)?

  (If you don't understand dice notation, consult the
http://en.wikipedia.org/wiki/Dice_notation page.)

-- 
                                                          - Warp


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