A probability problem by Shiv Kumar

First, it's obvious that P(sum=25) = P(sum=10). Proof: Consider 25(sum of dices) as 35-10, 6(number of dice) as 7-1, 5 as 7-2, 4 as 7-3, 3 as 7-4, 2 as 7-5 and 1 as 7-6. To achieve sum 25, we can minus 10 from 35, such as 25=6+6+5+5+3 is as same as 35-10=(7-1)+(7-1)+(7-2)+(7-2)+(7-4). Actually if we choose other such as 25=30-5, we can calculate the answer, but it's a little bit more complex.

Hence, we are finding the probability of sum is 10 by 5 dices. It is equal to (Ways of sum=10 by 5 fair dices) /6^5. The required ways is same as separate 10 balls into 5 regions. But each regions must at least has one ball, because a dice cannot show number 0.Hence it's as same as separate 10-5=5 balls into 5 regions, but a region can have no ball. To create 5 regions, we need four stick.

Example : |••|•||•• means 5=0+2+1+0+2( | represent stick, • represent ball) , •|•|•||•• means 5=1+1+1+0+2, ••••||||• means 5=4+0+0+0+1.

Hence, it's obvious that the ways is (5+4)!÷5!÷4! = 9C4. Ans: 9C4/6^5=7/432

I am having Trouble calculating the number of possibilities to get with five dice, without keeping track of repetitions ( e.g. 1,5,5,5,5 = 5,1,5,5,5 is the same as I understand it) In My drawings I get up to 4 dice and get a total of 126 different combinations. But is there a pattern to get up to 5 Dice ? I assume one could visualize it as a Pyramid pointing up at the Length 6 triangle. but if for example I take the last triangle with the 6666 and I put a one in front for the 5th dice, then it would be the same as the bottom triangle 1666 plus a 6 so there are some that cancel. How to calculate this ?

Count all the ways that sum to 25.

zero 'six': 5,5,5,5,5 = 5C5 = 1

one 'six': 6,5,5,5,4 = 5C1 * 4 = 20

(5,5,5,4), (5,5,4,5), (5,4,5,5), (4,5,5,5)

two 'six': 6,6,5,5,3 = 5C2 * 6 = 10*6 = 60

(5,5,3), (5,3,5), (3,5,5), (4,4,5), (4,5,4), (5,4,4)

three 'six': 6,6,6,5,2 = 5C3 * 4 = 10*4 = 40

(5,2), (4,3), (3,4), (2,5)

four 'six': 6,6,6,6,1 = 5C4 * 1 = 5

(1)

P(X=25) = (1 + 20 + 60 + 40 + 5) / 6^5 = 126 / 6^5 = 7 / 432

We have to write 25 as a combination of 6a+5b+4c+3d+2e+1f, where a,b,c,d,e,f are non negative integers 25 = 4(6) + 1, there are 5 arangements, 25 = 3(6) + 5 + 2, there are 5!/3! = 20 arangements, 25 = 3(6) + 4 + 3, there are 5!/3! = 20 arangements, 25 = 2(6) + 2(5) + 3, there are 5!/(2! .2!) = 30 arangements, 25 = 2(6) + (5) + 2(4), there are 5!/(2! .2!) = 30 arangements, 25 = 6 + 3(5) + 4, there are 5!/3! = 20 arangements, 25 = 5(5) , there are 1 arangement, Total there are 126 arangements. The probability 126/6^5 = 7/432