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 by Charlie)

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**.

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*Edited on ***January 5, 2013, 1:45 pm**