From: jg4o+@andrew.cmu.edu (James T Graves)
Newsgroups: rec.games.frp.dnd
Subject: Re: How many dice?
Date: 26 Jul 92 04:36:49 GMT
Organization: Sophomore, Math/Computer Science, Carnegie Mellon, Pittsburgh, PA

Excerpts from netnews.rec.games.frp.dnd: 25-Jul-92 Re: How many dice? by
mrw@ecl.psu.edu 
> > that you are assuming s/he meant using a standard binary number 
> > 
> >    000 = 1          (Where 0's are tails and 1's are heads)
> >    001 = 2
> >    010 = 3 ...
> >    111 = 8
> > 
> > In which case I agree, that if you were looking for d7 or d6 or something,
> > then you would not necessarially get it the first time (ie flipping d7 you 
> > get :  1... 1... 1!  Arrrrg! I rolled an eight!)  About the same as trying 
> > to get 1-7 probability out of a d8.
> > 
> > BUT....  if you flip as many dice as the number you are trying to get, and
> > add all the heads you get : 1...1...1...1...1...0...1...1!  I rolled
7! almo
> st
> > max damage!
> >    Then as your dXX (whatever you are rolling) gets bigger, you'll have
> > to flip a large # of coins, but it will be a _defined_ large number, as 
> > opposed to an *arbitrary* one.  :=)
> > 
> > - Mate
>  
> I admit that I am only a weenie engineer, and don't know all that much about
> statistics,but it seems to me that your method will not give you a flat
> probability distribution. You have a 1/8 chance of rolling an 8 on a d8, but
> you have only a (1/2)^8 = 1/256 chance of rolling an 8 in your coin-flipping
> scheme. Think of your coins as a set of d2's; then, like the 3-18 bell curve
> so familiar to all of us, your results will fall on a Gaussian curve. 

Sorry, looks like your weenie engineering probability knowledge led you
astray.  ;)

Using the method above, the coins are flipped in a defined order.  That
is to say, the masochist in question says, "Ok, I want to flip a d7, so
I think I'll just flip this penny three times and ignore any 8.  Heads
will represent one.  The first time will represent 2^0, the second 2^1,
the third 2^2."  

Since there is a 1/2 probability that the first coin toss comes up heads
(1), and the same probability that the second and third tosses *each
individually* come up heads, the total probability for all heads, or 000
(ehich would map to the die roll "1" of course), is 1/8.  However, the
probability of the first coin coming up *tails* is also 1/2, so 1/2 *
1/2 * 1/2 = a 1/8 probability that the number 001 (die roll 2) is
"rolled".  It's summer, and I don't feel like making a formal inductive
proof (okay people, quit thanking your favorite dieties and skip this
boring off-topic message already), but I suppose you can see the trend. 
Any given roll from {000, 001, 010, ... ,111} has a 1/8 probability of
occurring, since each *individual* coin toss has an equal 1/2 chance of
coming up heads or tails.

I think the misunderstanding comes from the knowledge that when a coin
is tossed a bunch of times, it is unlikely for all the tosses to come up
heads or tails.  However, pick any one _sequence_ of tosses, and it will
be equally unlikely.

Where the distribution becomes a curve is if you change the problem to:
"I'll flip a coin six times, and the number of heads plus one will give
me the number."  Then the probability distribution becomes: 

#   Prob.  Sequences
1 -  1/64  {000000}
2 -  6/64  {000001, 000010, 000100, 001000, 010000, 100000}
3 - 15/64  {000011, 000101, 001001, 010001, 100001, ... , 110000}
4 - 20/64  (these are left as an excercise for the reader.) ;)
5 - 15/64
6 -  6/64  {111110, 111101, ... , 011111 }
7 -  1/64  {111111}

But of course I'm pretty sure that's not what the original poster meant. :p


I wonder if diceless role playing was invented to eliminate messages
like these?

Jim


---------------------------------------------------
James T. Graves   
Carnegie Mellon University
internet: jg4o+@andrew.cmu.edu  
---------------------------------------------------
300,000 kilometers per second -- it's not just a good idea, it's the law.
Newsgroups: rec.games.frp.dnd