Hardest Math question ever?!?!? ?!
Question: Hardest Math question ever!?!!?!!? !?
Hardest Math question ever!?!!?!!?
Create a pair of 6-faced dice so that the probability of rolling the sums 2 - 12 is equally likely!. You may use numbers over the regular 1-6 and 0 but no negatives!.
Normal dice have a probability like this:
2 - 1:1 (1)
3 - 2:1, 1:2 (2)
4 - 1:3, 3:1, 2:2 (3)
5 - 1:4, 4:1, 2:3, 3:2 (4)
6 - 1:5, 5:1, 2:4, 4:2, 3:3 (5)
7 - 1:6, 6:1, 2:5, 5:2, 3:4, 4:3
8 - 2:6, 6:2, 3:5, 5:3, 4:4 (5)
9 - 3:6, 6:3, 4:5, 5:4 (4)
10 - 4:6, 6:4, 5:5 (3)
11 - 5:6, 6:5 (2)
12 - 6:6 (1)
* 2 months ago
Additional Details
2 months ago
apparently there's more than one answer =)
2 months ago
Since we're allowed to use zero,
there doesn't need to be 3 occurences for each number!.
There can be only one occurence or maybe 2!. =)
My teacher says it's possible!.!.!.Www@Enter-QA@Com
Create a pair of 6-faced dice so that the probability of rolling the sums 2 - 12 is equally likely!. You may use numbers over the regular 1-6 and 0 but no negatives!.
Normal dice have a probability like this:
2 - 1:1 (1)
3 - 2:1, 1:2 (2)
4 - 1:3, 3:1, 2:2 (3)
5 - 1:4, 4:1, 2:3, 3:2 (4)
6 - 1:5, 5:1, 2:4, 4:2, 3:3 (5)
7 - 1:6, 6:1, 2:5, 5:2, 3:4, 4:3
8 - 2:6, 6:2, 3:5, 5:3, 4:4 (5)
9 - 3:6, 6:3, 4:5, 5:4 (4)
10 - 4:6, 6:4, 5:5 (3)
11 - 5:6, 6:5 (2)
12 - 6:6 (1)
* 2 months ago
Additional Details
2 months ago
apparently there's more than one answer =)
2 months ago
Since we're allowed to use zero,
there doesn't need to be 3 occurences for each number!.
There can be only one occurence or maybe 2!. =)
My teacher says it's possible!.!.!.Www@Enter-QA@Com
Answers:
Hm!.!.!.Wrong section =)Www@Enter-QA@Com
wrong section and use a caunculater get a brainWww@Enter-QA@Com