Dice numbering puzzle
James is having 2 dices.
Dice 1- is a normal dice numbered 1 to 6.
Dice 2- is also having 6 faces but with no numbers written on them .
He needs to mark numbers on that 2nd dice so that after throwing both dices simultaneously , all the numbers from 1 to 12 will have equal probability to occur.
How can he do it? and what will be the probability?
attach the cubes with a face on another some way. now it is a single cuboidal block . now throwing the block will reveal a number on top of marked dice.
label the unmarked dice with 12-number appeared on dice. repeat. the most distinct opposite faces ,and nearest faces of the dice follow the same rule.
We can mark all the sides of the second sides with 0,0,0,0,0,6,
We have to get the numbers from 1 to 12 to be generated as sum of the no.s appearing on the dice.
By doing this , we get that:
1 -> (1,0)
2 -> (2,0)
3 -> (3,0)
4 -> (4,0)
5 -> (5,0)
6 -> (6,0)
7 -> (1,6)
8 -> (2,6)
9 -> (3,6)
12 -> (6,6)
Every no. has an equal probability of appearance as sum of the no. on the dice.
Total combinations possible:
6C1 * 2C1 = 12
Hope it helps!!
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