From: Christer Nilsson <janchrister.nilsson@...>
Date: 2006-01-08T03:56:06+09:00
Subject: Re: Dice Roller (#61)

Gavin Kistner wrote:
> On Jan 7, 2006, at 9:58 AM, Reinder Verlinde wrote:
>> ([1,21])d(16/[1,4])+3 =    (c/[x,y] = [c/y,c/x]  if x and y > 0)
>>
>> ([1,21])d([4,16])+3 =      ([a,b]d[c,d] = [a*c,b*d] if a,b,c, and d  
>> > 0)
> 
> I'm agree with you on that top line, but I'm not sure how you got to
> the second line from there.
> I think you meant:
> 
> ([1,21])d([1,4])+3 = ...
> 
> which results in:
> 
> [1,84]+3
> [4,87]

d(16/[1,4]) must be [16/4,16/1] => [4,16]

[1,21] d [4,16] must be [1*4, 16*21] => [4, 336]

I would like to introduce distribution. Using one normal dice we have an 
even distribution
  distr("d6") = [0,1,1,1,1,1,1]  Sum=6

Using two normal dices we have the following distribution
  distr("2d6")=[0,0,1,2,3,4,5,6,5,4,3,2,1]   Sum=36    (min:max) = 
(2:12)

P("2d6",12) = 1/36 = 2.8%


I have a question regarding the distribution for "(d2)d6".
In words, I'm first throwing a coin, to decide how many times I will 
throw a the dice.

[1,2] d [1,2,3,4,5,6]

My guess:

  distr("d6")  = [0,1,1,1,1,1,1]             Sum=6
  distr("2d6") = [0,0,1,2,3,4,5,6,5,4,3,2,1] Sum=36

Probability merge
distr("d6")     [0,6,6,6,6, 6, 6]             Sum=36
distr("2d6")    [0,0,1,2,3, 4, 5,6,5,4,3,2,1] Sum=36

distr("(d2)d6") [0,6,7,8,9,10,11,6,5,4,3,2,1] Sum=72

The probability of having one point is P("(d2)d6",1) = 6/72 = 8.3%

Can somebody agree or disagree on this?

Christer

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