# Hard Probability Question

#### MathStudent1999

##### Junior Member
4. Markum rolled a 6-sided number cube with the numbers from 1-6 on the faces cube. He kept track of each number he rolled and stopped as soon as he rolled any of the numbers four times. He stopped after the 18th roll and the sum of the numbers was 65. What is the sum of the numbers he could have rolled 4 times?

I tried testing each number one by one, but it didn't work because I did not factor in there was 18 rolls. Now looking at this question I have absolutely no idea of what to do next. I am completely stumped.

Can someone give me some hints on this question? (Not the answer)

#### soroban

##### Elite Member
Hello, MathStudent1999!

This problem has four solutions.

4. Markum rolled a 6-sided number cube with the numbers from 1-6 on the cube's faces.
He kept track of each number he rolled and stopped as soon as he rolled any of the numbers four times.
He stopped after the 18th roll and the sum of the numbers was 65.
What is the sum of the numbers he could have rolled 4 times?

Suppose he rolled 18 times and got each number three times.

Then (in some order) he got: .{1,1,1, 2,2,2, 3,3,3, 4,4,4, 5,5,5, 6,6,6}.
. . The sum of these numbers is 63.

To get a sum of 65, replace a number n with n+2.

There are four solutions:

. . {1,1, 2,2,2, 3,3,3,3, 4,4,4, 5,5,5, 6,6,6} . . Sum = 12

. . {1,1,1, 2,2, 3,3,3, 4,4,4,4, 5,5,5, 6,6,6} . . Sum = 16

. . {1,1,1, 2,2,2, 3,3, 4,4,4, 5,5,5,5, 6,6,6} . . Sum = 20

. . {1,1,1, 2,2,2, 3,3,3, 4,4, 5,5,5, 6,6,6,6,} . . Sum = 24

#### MathStudent1999

##### Junior Member
Hello, MathStudent1999!

This problem has four solutions.

Suppose he rolled 18 times and got each number three times.

Then (in some order) he got: .{1,1,1, 2,2,2, 3,3,3, 4,4,4, 5,5,5, 6,6,6}.
. . The sum of these numbers is 63.

To get a sum of 65, replace a number n with n+2.

There are four solutions:

. . {1,1, 2,2,2, 3,3,3,3, 4,4,4, 5,5,5, 6,6,6} . . Sum = 12

. . {1,1,1, 2,2, 3,3,3, 4,4,4,4, 5,5,5, 6,6,6} . . Sum = 16

. . {1,1,1, 2,2,2, 3,3, 4,4,4, 5,5,5,5, 6,6,6} . . Sum = 20

. . {1,1,1, 2,2,2, 3,3,3, 4,4, 5,5,5, 6,6,6,6,} . . Sum = 24
Thanks for the explanation! Based in what you said, I agree there is 4 solutions. The four numbers would be 3, 4, 5 and 6. Adding them together I got 18.