Probability question about rolling dice

Dice is to be rolled twice. What is the probability that the number that comes up on the first roll will be less than the number that comes up on the second roll?


I can't do probabilities...can someone show me a methodical approach to this question?

6 * 6 = 36 => possible outcomes(in pairs)
a = b , a > b , a < b
..6........ ... 30 ......
a < b => 15/36 = 5/12 or 41.67%

A normal dice has got 6 sides.
The number below 1 is 6, below 2 is 5 and below 3 is 4. meaning 1,2,3 are small and 4,5,6 are large.
There are altogether 36 variations. Taking away the doubles, ( 11, 22, 33, 44, 55 and 66)you are left with 30. Split this figure of 30 into two, then the probability would be 15:36. Answer 15 :36.in short 5:12

probability of any one combination appearing when we roll dice twice = 1/36
i.e., total of 36 combinations r possible............
combinations favor to your question r as below:
first roll second roll
1 2,3,4,5,6
2 3,4,5,6
3 4,5,6
4 5,6
5 6
total of 15 combinations. so the probability = 15/36

for 1 dice the P for any role is 1/6
for 2 throws the P for any combination , such as 2,1; 4,6 etc is [1/6][1/6] = 1/36

now lets find all that the first is less than the second

1,2 ; 1,3 ; 1,4 ; 1,5 ; 1,6
2,3 ; 2,4 ; 2,5 ; 2,6
3,4 ; 3,5 ; 3,6
4,5 ; 4,6
5,6

there are 15 combos that work thus ans = 15/36

1, 2
1, 3
1, 4
1, 5
1, 6
2, 3
2, 4
2, 5
2, 6
3, 4
3, 5
3, 6
4, 5
4, 6
5, 6

15 / 36
5 / 12