Theoretical probability of the sum of two dices

Question

you have two dices that have each have one side with one dot, two sides with two dots and three sides with three dots. you roll these dice and the possible sums are 2,3,4,5, and 6. What are the theoretical probabilities for each sum? Does The number of times you rolled effect your theoretical probability? Thanks :) 

Answer 1

You can express a given roll of two dice as (x,y), where x is the number that comes up on the first di and y is the number that comes up on the second di.

 

To get a sum of 2, you can roll:

(1,1)

 

To get a sum of 3, you can roll:

(1,2)

(2,1)

 

To get a sum of 4, you can roll:

(1,3)

(3,1)

(2,2)

 

To get a sum of 5, you can roll:

(2,3)

(3,2)

 

To get a sum of 6, you can roll:

(3,3)

 

 

 

Let's figure out the probabilities of each of those outcomes occurring.

 

 

Sum of 2:

(1,1)

 

There is one 1 on the first di and one 1 on the second. That means 1/6 of the first-di rolls will be 1 and 1/6 of the second-di rolls will be 1, for a total probability of:

 

(1,1) = (1/6)(1/6) = 1/36 probability for a sum of 2

 

 

Sum of 3:

(1,2)

(2,1)

 

There is one 1 on the first di and there are two 2's on the second. That means 1/6 of the first-di rolls will be 1 and 2/6 of the second-di rolls will be 2.

 

There are also two 2's on the first di and there is one 1 on the second, with probabilities of 2/6 and 1/6. (I'll skip this explanation from now on.)

 

(1,2) = (1/6)(2/6) = 2/36 = 1/18

(2,1) = (2/6)(1/6) = 2/36 = 1/18

 

Adding those probabilities together gives us the total probability:

 

1/18 + 1/18 = 2/18 = 1/9 probability for a sum of 3

 

 

Sum of 4:

 

(1,3) = (1/6)(3/6) = 3/36

(3,1) = (3/6)(1/6) = 3/36

(2,2) = (2/6)(2/6) = 4/36

 

3/36 + 3/36 + 4/36 = 10/36 = 5/18 probability for a sum of 4

 

 

Sum of 5:

 

(2,3) = (2/6)(3/6) = 6/36 = 1/6

(3,2) = (3/6)(2/6) = 6/36 = 1/6

 

1/6 + 1/6 = 2/6 = 1/3 probability for a sum of 5

 

 

Sum of 6:

 

(3,3) = (3/6)(3/6) = 9/36 = 1/4 probability for a sum of 6

 

 

 

Add those probabilities up to make sure we've got a total of 1:

 

1/36 + 1/9 + 5/18 + 1/3 + 1/4 = 1/36 + 4/36 + 10/36 + 12/36 + 9/36 = 36/36 = 1

 

 

 

As for the second part of the problem, that question isn't terribly clear. It might be asking: On a given one of your rolls, how likely is it that you got a given sum? In that case, the theoretical probability isn't affected by number of rolls. For a given roll, the probabilities remain the same.

 

Or, it might be asking: For a given set of n rolls, how likely is it that one of your rolls gave you the given sum? In that case, the theoretical probability for any given sum increases, since you have n chances to get it.