pop quiz question

hey dudes my sister brought home a sheet of interesting maths puzzles she'd been given at school becasue its maths weeks here in ireland.

anyway one of them has me a bit confused!!

you have an "n" digit number
each digit can assume only one of 3 values 1,2,3,

so in this n digit number what percentage of the time do all 3 values (1,2,3) appear at least once!

best i can see is 2/9 of the time, regardless of the length of the number!! but i'm thinking this may be wrong, because if the number grows drastically to say a billion digits, and each digit can only be one two or three, then surely most of the time in this string of a billion digits 1,2,3 will appear and so the percentage should rise as the value n increases??
actually i'm 100% sure i'm wrong with my 2/9ths, i'll work on it a bit more and post again!!

i wont go through my workings just yet, i'd be interested to hear what others have to say

Hi,

Nice to have maths weeks.

Total combination of number with "n" digit for 1,2,3 = 3^n

P(3 values appear at least once)
= (P(Total) - P(No "1") - P(No "2") - P(No "3"))/P(Total)

P(No "1") = P(No "2") = P(No "3")

Combination with "1" would be just Combination of 2 and 3
XXX, 3C3
XXY, 3C2
XYY, 3C1

Combination above = Sum of nCr, for r =1 to n
As Sum of nCr, for r = 0 to n => 2^n,
Sum of nCr, for r =1 to n => (2^n)-1

P(No "1") = (2^n)-1

P(3 values appear at least once)
= (P(Total) - P(No "1") - P(No "2") - P(No "3"))/P(Total)
= (3^n - 3((2^n)-1))/3^n

I check it for n = 3 to 14, and it corrects.
Hope it is the answer

hey soft
yeah your exactly right. i've come up with the same answer but a different way!!

the set of all combinations has 3^n members

the set containing only 1's and 2's (including all 1's and all 2's)= 2^n members
the set containing only 1's and 3's (including all 1's and all 3's)= 2^n members
the set containing only 2's and 3's (including all 2's and all 3'2)= 2^n members

the 3 sets above have 3*2^n members but 3 members are repeated
eg 11111.......1 appears in the first and second set
22222......2 appears in the first and third set
33333......3 appears in the second and third set

to the total unique members in theese 3 sets is (3*(2^n))-3

so that the set containing members with all 3 numbers appearing is
(3^n)-(3*(2^n)-3) = (3^n-3((2^n)-1)
and expressed as a percentage of all combinations= (3^n-3((2^n)-1)/(3^n)

:)