Originally Posted by

**thepillow**
1. On any given roll the odds of getting a number greater than 4 is 2/6 = 1/3. (Only 5 and 6 are greater than 4, so this is 2 out of the 6 numbers.)

This means that the probability of getting a number that is 4 or less is 2/3. We'll use these numbers in a little bit.

We want to consider 2 cases:

**Case 1:** We get a number greater than 4 for the first time on the 3rd roll = Our first roll is less than or equal to 4 __and__ our second roll is less than or equal to 4 __and__ our third roll is greater than 4.

We can write this as: P(4 or less) x P(4 or less) x P(greater than 4) = (2/3) x (2/3) x (1/3)

**Case 2:** We get a number greater than 4 for the first time on the 3rd roll = Our first roll is less than or equal to 4 __and__ our second roll is less than or equal to 4 __and__ our third roll is less than or equal to 4 __and__ our fourth roll is greater than 4.

We can write this as: P(4 or less) x P(4 or less) x P(4 or less) x P(greater than 4) = (2/3) x (2/3) x (2/3) x (1/3)

We want to know about either case happening so we can add up these probabilities: (2/3)(2/3)(1/3) + (2/3)(2/3)(2/3)(1/3) = 4/27 + 8/81 = 12/81 + 8/81 = **20/81**

2. In order for there to be a range of 7 we need the largest number minus the smallest number to equal 7. From the numbers between 1 and 10, we can make 3 different pairs that satisfy this:

8 and 1; 9 and 2; 10 and 3

Let's take 8 and 1 as an example: along with our 8 and 1, we need 2 other discs out of the 6 numbers between 8 and 1 (we can't get a 9 or a 10 because that would change the range). So there are 6c2 = 15 different pairs we could choose for our other 2 discs.

This works out the exact same way for 9 and 2, as well as 10 and 3. So in each of the 3 cases there are 15 combinations for a total of 45.

The number of total groups of 4 that we can get from the 10 discs is 10c4 = 210.

So 45 of the 210 groups meet our requirements. 45/210 can be reduced to **3/14**.

Hope that helps! Let me know if I need to explain something better.