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 Die Order (Posted on 2013-01-04)
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.

 See The Solution Submitted by Charlie No Rating

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 re(2): No Subject re1 | Comment 6 of 7 |
(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
 Posted by Ady TZIDON on 2013-01-05 13:42:20

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