On 8-2-2010 20:08, Orchid XP v8 wrote:
> Orchid XP v8 wrote:
> 
>> You can compute this quite easily in Haskell:
> 
> ...and yet, an exhaustive analysis of the data quickly demonstrates that 
> Andrel was correct in the first place:
> 
> Pr(X > Y) = 2/100
> Pr(X = Y) = 1/100
> Pr(X < Y) = 0/100
> 

exhaustive analysis? What exhaustive analysis?

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>>> You can compute this quite easily in Haskell:
>>
>> ...and yet, an exhaustive analysis of the data quickly demonstrates 
>> that Andrel was correct in the first place:
>>
>> Pr(X > Y) = 2/100
>> Pr(X = Y) = 1/100
>> Pr(X < Y) = 0/100
>>
> 
> exhaustive analysis? What exhaustive analysis?

Reading the entire dataset by hand - the brute-force approach. ;-)

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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On 8-2-2010 21:47, Orchid XP v8 wrote:
>>>> You can compute this quite easily in Haskell:
>>>
>>> ...and yet, an exhaustive analysis of the data quickly demonstrates 
>>> that Andrel was correct in the first place:
>>>
>>> Pr(X > Y) = 2/100
>>> Pr(X = Y) = 1/100
>>> Pr(X < Y) = 0/100
>>>
>>
>> exhaustive analysis? What exhaustive analysis?
> 
> Reading the entire dataset by hand - the brute-force approach. ;-)

Ok, I'll make a note of that. If ever I have the pleasure to interview 
you for a job, I will pull that from my pocket.

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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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> 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.

Correction:

That last term should be (2n - 1) rather than (1 + 2n).

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>> any multiple of 11 has a probability of 1/100.
>> if the numbers are not the same the one where the 10 has a higher value
>> than the 1 has a probability of 2/100. the rest has probability 0
>
>  But getting values eg. in the range 10-20 is a lot less probable than
> getting values in the range 80-90. This would suggest that the probability
> distribution is not very even.

Well yes, because as andrel said, numbers where the unit is higher than the 
ten have a probability of 0.

So in 10-19 the only numbers you can get are 10 (P=2/100) and 11 (P=1/100) - 
overall P=3/100

whereas in 80-89 you can get 80-88, total P=9/100.

So you're 3 times more likely to get a number in the range 80-89 as you are 
10-19.

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