Four fair six-sided dice, each numbered 1 to 6, are rolled once. What

Question

Four fair six-sided dice, each numbered 1 to 6, are rolled once. What is the probability that resulting four numbers are consecutive ?

A. 1/9
B. 1/18
C. 1/54
D. 1/108
E. 1/432

Fresh GMAT Club Tests' Question

sanjitscorps18 · Answer

Available patterns are 1,2,3,4 or 2,3,4,5 or 3,4,5,6

Now all these patterns can be arranged in 4! ways

Hence the probability = (4! + 4! + 4!)/6^4 = 1/18

IMHO Option B

gmatophobia · Answer

Method 1   - Using Principle of counting

Either of the three possible series of consecutive numbers is possible

1 2 3 4

2 3 4 5

3 4 5 6

Now, its not necessary that the numbers be in this order, the series can be in any order possible.

For example - With 1 2 3 4, there are 12 (i.e. 4!) possible combinations in which the numbers can be arranged

Hence total possible favorable outcomes = 4! * 3

Total outcomes =  6^4

Required Probability =  \frac{4! * 3 }{ 6^4 }  =  \frac{1}{18}

Method 2   - Using Probability

We can select the first number in 3/6 ways (either 1, 2, or 3)

The rest of the numbers (remaining three) can be selected in 1/6 ways (i.e. we select only consecutive numbers)

However there are 4! ways to arrange them -

(4 * 3 * 2)* \frac{1}{2} * \frac{1}{6} * \frac{1}{6 }* \frac{1}{6}

=  \frac{1}{18}

Option B

BrushMyQuant · Answer

Four fair six-sided dice, each numbered 1 to 6, are rolled once

As we are rolling 4 dice => Number of cases =  6^4

What is the probability that resulting four numbers are consecutive ?

Lets write down the possible cases 
(1,2,3,4), (2,3,4,5), (3,4,5,6)

Now, each of these cases can have combinations where the numbers  can come in any sequence

Ex: 1,2,3,4 can come in (1,3,2,4), (1,4,3,2), etc
So, we have 4 places and each number can take any of the 4 places
=> For 1st place we have 4 options (1,2,3,4)
=> For 2nd place we have 3 options (3 of the remaining numbers except the one which took place 1)
=> For 3rd place we have 2 options (Numbers except the ones which took place 1 and 2)
=> For 4th place we have only 1 option
=> Total number of cases = 4 *3 * 2 * 1 = 24

Each of the 3 cases will have 24 cases each 
=> Total Cases = 3 * 24 = 3 * 2 * 2 * 6 = 6 * 6 * 2

=>  P(Four numbers are consecutive)  =  \frac{6 * 6 * 2}{6^4}  =  \frac{1}{18}

So,  Answer will be B 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

Four fair six-sided dice, each numbered 1 to 6, are rolled once. What

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Four fair six-sided dice, each numbered 1 to 6, are rolled once. What

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards