Warp wrote:

>   However, suppose that you use a variant of this: Throw two d10's and
> then the larger result is always used as the tens, and the other as the
> ones.
> 
>   What is the probability distribution now?

Pr(0) = 1/100
Pr(1) = 0/100
Pr(2) = 0/100
Pr(3) = 0/100
Pr(4) = 0/100
Pr(5) = 0/100
Pr(6) = 0/100
Pr(7) = 0/100
Pr(8) = 0/100
Pr(9) = 0/100
Pr(10) = 2/100
Pr(11) = 1/100
Pr(12) = 0/100
Pr(13) = 0/100
Pr(14) = 0/100
Pr(15) = 0/100
Pr(16) = 0/100
Pr(17) = 0/100
Pr(18) = 0/100
Pr(19) = 0/100
Pr(20) = 2/100
Pr(21) = 2/100
Pr(22) = 1/100
Pr(23) = 0/100
Pr(24) = 0/100
Pr(25) = 0/100
Pr(26) = 0/100
Pr(27) = 0/100
Pr(28) = 0/100
Pr(29) = 0/100
Pr(30) = 2/100
Pr(31) = 2/100
Pr(32) = 2/100
Pr(33) = 1/100
Pr(34) = 0/100
Pr(35) = 0/100
Pr(36) = 0/100
Pr(37) = 0/100
Pr(38) = 0/100
Pr(39) = 0/100
Pr(40) = 2/100
Pr(41) = 2/100
Pr(42) = 2/100
Pr(43) = 2/100
Pr(44) = 1/100
Pr(45) = 0/100
Pr(46) = 0/100
Pr(47) = 0/100
Pr(48) = 0/100
Pr(49) = 0/100
Pr(50) = 2/100
Pr(51) = 2/100
Pr(52) = 2/100
Pr(53) = 2/100
Pr(54) = 2/100
Pr(55) = 1/100
Pr(56) = 0/100
Pr(57) = 0/100
Pr(58) = 0/100
Pr(59) = 0/100
Pr(60) = 2/100
Pr(61) = 2/100
Pr(62) = 2/100
Pr(63) = 2/100
Pr(64) = 2/100
Pr(65) = 2/100
Pr(66) = 1/100
Pr(67) = 0/100
Pr(68) = 0/100
Pr(69) = 0/100
Pr(70) = 2/100
Pr(71) = 2/100
Pr(72) = 2/100
Pr(73) = 2/100
Pr(74) = 2/100
Pr(75) = 2/100
Pr(76) = 2/100
Pr(77) = 1/100
Pr(78) = 0/100
Pr(79) = 0/100
Pr(80) = 2/100
Pr(81) = 2/100
Pr(82) = 2/100
Pr(83) = 2/100
Pr(84) = 2/100
Pr(85) = 2/100
Pr(86) = 2/100
Pr(87) = 2/100
Pr(88) = 1/100
Pr(89) = 0/100
Pr(90) = 2/100
Pr(91) = 2/100
Pr(92) = 2/100
Pr(93) = 2/100
Pr(94) = 2/100
Pr(95) = 2/100
Pr(96) = 2/100
Pr(97) = 2/100
Pr(98) = 2/100
Pr(99) = 1/100

You can compute this quite easily in Haskell:

numbers = do
   a <- [0..9]
   b <- [0..9]
   let x = max a b
   let y = min a b
   return (10*x + y)

This tries every possible combination of D10 scores, and builds a list 
of the resulting score.

histogram =
   map (\xs -> (head xs, length xs)) $
   group $
   sort numbers

This sorts the above list, groups equal elements into little sublist, 
and then counts the length of all such sublists, producing a histogram 
chart for all scores with non-zero frequency.

With a little extra trickery and some formatting, you can create the 
above chart.

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

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