#### Posers and Puzzles

1.  talzamir
Art, not a Toil
28 Jun '12 10:21
Rolling a standard six-sided die until a goal number > 1 is reached. How many ways are there do it? And is it so that there are as many ways involving an even number of rolls as an odd one, no matter what the target number is?

For example;
to reach 2; 1 way with an even # of rolls (1, 1) and 1 way with an odd # of rolls (2)
to reach 3; 2 ways each; (1, 2; 2, 1) ; (3; 1, 1, 1)
to reach 4; 4 ways each; (1, 3; 3, 1; 2, 2; 1, 1, 1, 1) ; (1, 1, 2; 1, 2, 1; 2, 1, 1; 4)
2.  forkedknight
Defend the Universe
29 Jun '12 16:54 / 2 edits
It doesn't work for 7, there is one additional way to make seven with an even number of rolls (32) than there are with an odd number of rolls (31)

I think this would be true if you took any n-sided die and tried to roll n+1.

Take a coin (1,2) for example, and try to roll a 3.
odd: (111)
even: (12, 21)

As soon as you can't roll your goal number in a single roll, there's no way to offset rolling all 1's
3.  talzamir
Art, not a Toil
01 Jul '12 11:43
So it would appear. That also shows that the # of ways to reach the target number n is not 2^(n-2) ways with an odd number of rolls + 2^(n-2) with an even number of rolls for 2^(n-1) total. But if not that, then what?

It would seem to me that

odd(n) = 1, 1, 2, 4, 8, 16 for n = 1..6
odd(n) = even(n-1) + ... + even(n-6) for n > 6

even(n) = 0, 1, 2, 4, 8, 16 for n = 1..6
even(n) = odd(n-1) + ... + odd(n-6) for n > 6

if so, then

odd(6k+1) = even(6k+1) + 1 for all k = 0,1,2,3,...
and odd(6k+2) = even(6k+2) - 1 for all k = 1, 2, 3, ...
odd(n) = even(n) for all other values of n
total(n) = odd(n) + even(n) = 1, 2, 4, 8, 16, 32, 63, 125, 248, 492, 976, ...

so it starts like a geometric sequence, but falls short of that after n = 6.
4. 03 Jul '12 07:28 / 5 edits
If you pre-seed 6 initial null elements with 0,0,0,0,0,1 then you can generate the number of ways of rolling each number as the sum of the previous 6 entries.

It looks to me that the positions of odd elements in this series are predictable, elements (0,1) are odd then elements (7,8), then elements (14,15).

These number are achievable with an odd number of total combinations of rolls, therefore the ways of throwing them with odd rolls and even rolls cannot be equal
5. 04 Jul '12 00:55 / 1 edit
Originally posted by iamatiger
If you pre-seed 6 initial null elements with 0,0,0,0,0,1 then you can generate the number of ways of rolling each number as the sum of the previous 6 entries.

It looks to me that the positions of odd elements in this series are predictable, elements (0,1) are odd then elements (7,8), then elements (14,15).

These number are achievable with an odd numb ...[text shortened]... ions of rolls, therefore the ways of throwing them with odd rolls and even rolls cannot be equal
similarly, we can denote odd even pairs by (x,y) where x is the even entry and y is the odd entry, there are an even number of ways of throwing zero, and an odd number of ways of throwing one, and each later even entry is the sum of the six previous odd entries and vice versa, so:

{0,0,0,0,0,(1,0)}(0,1),(1,1),(2,2),(4,4),(8,8),(16,16),(32,31),(62,63),(124,124),(246,246),(488,488),(968,968),(1920,1920),(3809,3808),(7554,7555),(14985,14985)

It is clear that all totals which are multiples of 7 will have one more even way of throwing the number than odd ways, and all totals which are one more than a multiple of 7 will have one more odd way than even ways.