## Flipkart Interview Question

**Country:**India

Can you explain question little bit more ? I could not get what does it mean "Sum of faces is evenly distributed between 1 to 12".

Well imagine case of 2 normal dice. The sums will be between 2-12 when you throw them. Also 2 will come less times ( 1 + 1) while 7 would come more times ( 4 + 3 , 3 + 4, 5 + 2, 2+5), therefore this is not even distribution.

You have to make sure that the probability of each number between 1-12 is equal. Having 3 faces 0 and 3 faces 6 does that. Calculate by yourself

what about writing, 12-17, so that possibility of coming 1-12 will be equal (always zero)

The probability generating function sum(pix^i) of the even distribution {1,2,...,12} is 1/12(x+x^2+...+x^12)=G(x)H(x) where G(x)=1/6(x+x^2+...+x^6), where G(x) is the probability generating function of the regular fair dice.

Solving for H(x), you get H(x)=1/2*(x^0+x^6)=1/6(3x^0+3x^6) so the unmarked dice has 3 0's and 3 6's.

if including 1 and 12 then three faces 0 and other three 6.

if 1 and 12 is not included then three faces 1 and other three 5.

actually, i'm not sure if your latter solution would work because

num - probabilty

#2 - 3/36

#3 - 3/36

#4 - 3/36

#5 - 3/36

#6 - 6/36

#7 - 6/36

#8 - 3/36

#9 - 3/36

#10 - 3/36

#11 - 3/36

six - regular dice rolls 5 + fixed dice rolls 1 (3/36) AND

regular dice rolls 1+ fixed dice rolls 5 (3/36)

seven- regular dice rolls 6 + fixed dice rolls 1 (3/36) AND

regular dice rolls 2 + fixed dice rolls 5 (3/36)

three 0s and three 6s.

- Adil March 06, 2012