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On 9/02/2010 06:13 AM, Warp wrote:
> It might be more interesting and intuitive (especially when considering
> the original subject matter, ie. throwing dice in a tabletop game) if the
> probabilities were divided into ranges. For example, what is the
> probability of getting a value in the range 1-10, the range 11-20, the
> range 21-30 and so on.
>
> This distribution ought to be uneven.
>
Your rule 'folds' some of the values into others. Visualise this as a
10 x 10 table. Each cell has the starting probability of 1 / 100. The
axis where D1 = D2 (10 cells) remains unaffected. The cells from one
triangle off the axis get turned to 0 and their previous value added to
the cell reflected across the axis. So these all become probability 2 /
100.
There are 10 cases where D1 = D2, 45 where D1 < D2 and 45 where D1 > D2.
Lets say you fold the D1 < D2 cases into D1 > D2 by reversing the order
of the digits.
Now there are 10 cases (unchanged) where D1' = D2' each with probability
1 / 100 and 45 cases where D1' > D2' each with probability 2 / 100.
Considering the ranges:
- I'm going to use the range 00-99 because it makes things easier to set
out. Just add 1 to map it to 1-100.
- Each 'decade' has one of the D1' = D2' cases. The first decade (ie
00-09) has no case where D1' > D2'.
- Each subsequent decade has one more case where D1' > D2' than the
previous decade.
To set that out:
P(00-99) = (1 / 100 + (0 * 2 / 100)) = 1 / 100
P(10-19) = (1 / 100 + (1 * 2 / 100)) = 3 / 100
P(20-29) = (1 / 100 + (2 * 2 / 100)) = 5 / 100
...
P(n0-n9) = (1 / 100 + (n * 2 / 100)) = (1 + 2n) / 100
...
P(90-99) = (1 / 100 + (9 * 2 / 100)) = 19 / 100
Summing 1 + 3 + 5 ... + 19 = 100 just to confirm.
And graphing the cumulative values gives Andrel's chart.
Note that summing the sequence 1, 3, ... (1 + 2n) gives the value n^2.
Thus the curve is parabolic. Or at least the step up at the end of each
decade sits on a parabola. The intermediate steps track along flat or
step up at 45 degrees. to get to that point.
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