A single die is rolled four times.
What is the probability that the numbers that come up will be in ascending order (not necessarily strictly ascending, but never a lower number after a higher)?
Intended to be solved without a computer.
(In reply to re: No Subject
re 1st question
For n- 1,2,3,4 it is a simple, albeit methodic counting, trying to discover a pattern.
Let us stay with a 6-sided dice:
1 throw- all 6 outcomes qualify, so a(1)=6. 2 throws- there are 6 doubles (all qualify), and half of the rest is strictly increasing so a(1)= 6+ 1/2*15= 21.
3 throws- there are:
a) 6 doubles (all qualify),
b)C(6,3)=20 strictly increasing triplets
c)for any 2 digits m & n( m<n) there are 2 "increasing" possibilities mmn or mnn; since there are C(6,2)=15 choices of 2 digits out of 6 , we get 2*15= 30 mixed triplets <br> (112,122,113,133,…566) - so a(3)= 6+ 20+30= 56.
You can figure out for 4 throws using similar reasoning mutatis mutandis to get a(4)=126.
Edited on January 5, 2013, 1:45 pm