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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.

 Submitted by Charlie No Rating Solution: (Hide) C(9,4) / 64 = 126 / 1296 = 7 / 72 as each choice of four integers from the range 1 through 9 corresponds with exactly one set of rolls that satisfy the desired condition. Consider the rolls outcomes labeled as a,b,c,d. The enhanced number sets as A,B,C,D, where A=a, B=b+1, C=c+2, D=d+3. The set A=3,B=5,C=6,D=8 corresponds to rolls of 3,4,4 and 5, and since there is inequality guaranteed by the choice of four out of nine, there is only one set, that can be placed in order, and will be in one-to-one correspondence with the original set. Alternative Method: As alluded to in Ady Tzidon's comment, separate track can be kept of sequences ending in 1, 2, ..., 6 and added up, starting with a sequence of 1, then 2, then 3 leading up to 4: 1+1+1+1+1+1 = 6 1+2+3+4+5+6 = 21 1+3+6+10+15+21 = 56 1+4+10+20+35+56 = 126 where each n'th term is the sum of the first n terms in the row before. Then, of course, the sum 126 is divided by 64.

 Subject Author Date sequel to my previous comment Ady TZIDON 2013-01-05 14:15:17 re(2): No Subject re1 Ady TZIDON 2013-01-05 13:42:20 re: No Subject Charlie 2013-01-05 08:39:24 No Subject Ady TZIDON 2013-01-04 18:50:51 LOL Hugo 2013-01-04 17:24:40 Answer Dej Mar 2013-01-04 16:03:51 Solution Jer 2013-01-04 14:02:54

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