*When 4 dice are rolled, what is the probability that the 4th die has a number not obtained earlier?*

Consider the following cases:

(1) The first three die all have the same number. This can occur in 6 ways; as the fourth die can take on 6 values there are a total of 36 ways this can happen.

Of the 36 ways, 30 have the fourth die with a different value than the first three. (e.g. if the first three dice showed ones, the fourth could show any of 2,3,4,5,6 , etc...)

(2) The first three die are all different. This can occur in 6x5x4=120 different ways. The fourth die can take on 6 values, so there are a total of 720 ways the first three dice can show different numbers.

Of these, there are 360 ways for the fourth to be different from the first three. (Note that this agrees with the number of ways all 4 dice can show different numbers = 6x5x4x3.)

(3) The final case is when exactly two of the first three dice have the same number. There are 6 choices for the number to be duplicated; the duplicate die has no choice and the third die has 5 choices so this can occur in 30 ways. But there are three configurations with two of the die being the same: xxy,xyx,yxx. So there are 90 ways to have two of the first three dice showing the same number. The fourth die has 6 choices for a total of 540.

Of these 540: We have rolled two numbers so the fourth die can only take on 4 values. Thus there are 6x1x5x3x4=360 ways for the fourth die to differ. (6 choices for the duplicate number, the duplicate is fixed, the third die has 5 choices, there are three ways to get 2 out of 3, and the fourth die has 4 choices)

(4) Note that the total:36+720+540=1296 agrees with the total number of throws possible with 4 die; namely `6^4=1296` .

(5) The number of ways to have the fourth die differ from the previous three is 30+360+360=750.

**So the probability is **`750/1296~~58%`

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