Another perspective on the coin/die 50% problem.
Assume that you flip a coin, but this coin has an arrow printed on it (both sides), such that you can tell whether it landed heads or tails and which quadrant the arrow was pointed into. Now there are eight possible outcomes of each coin flip--heads-North, heads-East, tails-East, tails-South, etc.
Does having more possible outcomes change the probability that the coin will flip up M heads in N trials, where heads is a superposition of four of the possible outcomes in each trial? It shouldn't, even though we've added information to the trial that wasn't there before, and that's because we promptly forget that information.
Accordingly, I'll argue that a die should roll odds M times in N trials with exactly the same probability distribution that a coin will flip M heads in N trials.
Posted by Greg at June 9, 2005 12:38 PM
Yeah... so this was a rhetorical question, then? Because, duh. I had assumed that couldn't possibly be what you were asking.
I have a variant for you (which is related only in that it is also a dumb question that nontheless confuses its author): say I offer you a lottery where you roll a fair die and I give you $100 * the roll. And, I offer you another lottery where I give you $100 * (the roll + 1), except if you roll a 6 you only get $100. Which do you prefer?
A very little thought shows that you should have no preference, because the two lotteries are identical except for relabeling the sides of the die. But on the other hand, five times in six you'll get $100 more but one time in six you'll get $500 less! Do you really have no preference between that and nothing? What if it was more money, for example multiply the values by 100?
You're headed into human cognition theory as well as the decreasing value of money, there! That's non-mathematical territory, that is.
Actually, I just now found the standard deviation I was looking for: sqrt(n*p*(1-p)). Thus the variance of the number of times you roll a 1 on a die out of n trials is sqrt(5n/36).
One more step to go to achieve total RPG dice stat supremacy: What's the standard deviation of the sum of the results of M trials with N possible outcomes (i.e., σ(MdN))?
(The reason I was posing the question with the duh answer was that Elizabeth and I were arguing about it.)
Hmm...σ for a dN is N/sqrt(12); M independent trials, then, ought to produce σ of N sqrt(M/12). For 3d6, that's 3. For 10d10, that's about 9.
That was easier than I thought...
Well, say the standard deviation of 1 roll is x, then isn't it x sqrt( M ) from your earlier argument about adding the sums of squares? And, for 1 roll it's what, sqrt(sum_i=1..N (1/N) (i-(N+1)/2)^2)?
Is there some easy way to write equations in html? Really it should accept TeX. What I meant was \sqrt(\sum_{i=1 \hdots N} \frac{1}{N} (i-\frac{N+1}{2})^{2}).
For Md6, like we use in Champions, it simplifies greatly to sqrt(3M).
You've gotten to the answer, I think, but let me wander off into a slight more (IMO) direct way there.
Standard Deviation equals the square root of Variance.
Variance equals (the Expected Value of X^2) minus (the expected value of X)^2. That is, for a d6, it equals (1/6+4/6+9/6+16/6+25/6+36/6) - (3.5^2), or 35/12. Thus, Variance of a single die is easy to determine.
But here's the nice fact if you assume independance of each result... Just like Expected Value of a sum of independant trials is equal to the sum of the expected values, Variance of a sum of independant trials is equal to the sum of the Variances. Thus, the Variance of 5d6 is 5*Var(1d6).
And so, the standard deviation of MdN is equal to sqrt(M*Var(1dN)) or sqrt(M)*stddev(1dN).