Problem Solving

Jacwis wrote:Hi

I need help with the following question.

Four people each will roll a fair die once. What is the probability that at least two will roll the same number?

A) 50%
B) 65%
C) 80%
D) 72%
E) 40%

I wanted to understand the problem by adding up all the cases

All possible rolls:
Since there are 6 number options for each roll, we get:
6*6*6*6 = 1296

Case 1: Exactly 2 rolls are the same (such as 1123)
From 4 rolls, the number of ways to choose a pair to yield the same number = 4C2 = (4*3)/(2*1) = 6.
Number options for this pair = 6. (Any of the 6 numbers on the die)
Number options for the third roll = 5. (Any of the 5 remaining numbers on the die)
Number options for the last roll = 4. (Any of the 4 remaining numbers on the die).
To combine these options, we multiply:
6*6*5*4 = 720

Case 2: Two pairs (such as 1122)
Number of ways to divide 4 rolls into 2 pairs = 3.
(First way: 1st and 2nd rolls are the same; 3rd and 4th rolls are the same but different from the 1st and 2nd.)
(Second way: 1st and 3rd rolls are the same; 2nd and 4th rolls are the same but different from the 1st and 3rd.)
(Third way: 1st and 4th rolls are the same; 2nd and 3rd rolls are the same but different from the 1st and 4th.)
Number options for first pair = 6. (Any of the 6 numbers on the die)
Number options for the second pair = 5. (Any of the 5 remaining numbers on the die)
To combine these options, we multiply:
3*6*5 = 90

Case 3: Exactly 3 numbers are the same
From 4 rolls, the number of ways to choose three to yield the same number = 4C3 = (4*3*2)/(3*2*1) = 4.
Number options for these 3 rolls = 6. (Any of the 6 numbers on the die)
Number options for the fourth roll = 5. (Any of the 5 remaining numbers on the die)
To combine these options, we multiply:
4*6*5 = 120

Case 4: All 4 numbers are the same
Number options = 6. (Any of the 6 numbers on the die)

Case 1 + Case 2 + Case 3 + Case 4 = 720 + 90 + 120 + 6 = 936
Thus:
P(at least 2 people roll the same number) = 936/1296 ≈ 0.72 = 72%

The correct answer is D.

MUCH easier to solve as follows:

P(at least 2 numbers are the same) = 1 - P(all 4 numbers are different)

The first person can roll any of the 6 numbers.
P(2nd number is different from the 1st) = 5/6
P(3rd number is different from the first 2 numbers) = 4/6
P(4th number is different from the first 3 numbers) = 3/6
To combine these probabilities, we multiply:
5/6 * 4/6 * 3/6 = 5/6 * 2/3 * 1/2 = 5/18

Thus:
P(at least 2 numbers are the same) = 1 - 5/18 = 13/18 ≈ 0.72 = 72%