Suppose 4 dice are rolled, what is the probability that 1 number appears at least twice?

Question

Savannah Anderson · Accepted Answer

The probability is #13/18 #

Let's number the dice with 1,2,3, and 4. We first count the number of ways a roll of the four dice does not have a number that appears at least twice. Whatever is on top of the first die, there are 5 ways to have a different number on die 2.

Then, assuming that we have one of those 5 outcomes, there are 4 ways to have a number on die 3 that is not the same as on dice 1 and 2. So, 20 ways for dice 1, 2, and 3 to have all different values.

Assuming we have one of these 20 outcomes, there are 3 ways for die 4 to have a different number than dice 1, 2, or 3. So, 60 ways altogether.

So, the probability of NOT having two numbers the same is #60/6^3 = 60/216#, as there are #6^3# different outcomes for rolling three six-sided dice.

The probability of the opposite, i.e. having at least two, equals 1 minus the above probability, so it is #1 - 60/216# = #(216-60)/216 = 156/216#=#13/18#.

Kennedy Adams · Answer

undefined To find the probability that at least one number appears at least twice when rolling four dice:

1. First, calculate the total number of outcomes when rolling four dice. Since each die has 6 sides, the total number of outcomes is 6^4.

2. Next, calculate the number of outcomes where at least one number appears at least twice. To do this, calculate the number of outcomes where all four dice show the same number (6 outcomes), the number of outcomes where three dice show the same number and the fourth die shows a different number (6 choices for the repeated number and 5 choices for the different number, times 4 ways to arrange which die shows the different number), and the number of outcomes where two pairs of dice show the same number (6 choices for the first pair, times 5 choices for the second pair, times 3 ways to arrange which pair shows which number).

3. Add up the number of outcomes from step 2.

4. Divide the number of outcomes where at least one number appears at least twice by the total number of outcomes from step 1 to find the probability.

Calculations:

1. Total number of outcomes = 6^4 = 1296.

2. Number of outcomes where at least one number appears at least twice:
   - 6 outcomes with all four dice showing the same number.
   - 6 * 5 * 4 = 120 outcomes with three dice showing the same number and the fourth die showing a different number.
   - 6 * 5 * 3 = 90 outcomes with two pairs of dice showing the same number.

3. Total number of outcomes where at least one number appears at least twice = 6 + 120 + 90 = 216.

4. Probability = Number of outcomes where at least one number appears at least twice / Total number of outcomes
             = 216 / 1296
             = 1/6.

So, the probability that at least one number appears at least twice when rolling four dice is 1/6.