Dice problems aren’t too common on the GMAT. As for the exact probability of getting a dice problem is something only privy to those over at GMAC.

While it is a good idea to know your dice basics, doing so will only help you in the case of an easy dice problem. And one thing is for sure – if you are looking to break a 700 on the GMAT the probability of an easy dice problem (if you end up getting a dice problem) is very, very low.

Below are three excruciatingly difficult dice problems. They will pop up only if you are heading towards a 51Q. For those still brushing up on their math fundamental don’t waste time on such challenge questions.

## Practice Questions

1. What is the probability of rolling the same number exactly three times with five six-sided dice?

- 1/5
- 5/18
- 35/216
- 125/648
- 225/1296

2. What is the probability of rolling three six-sided dice, and getting a different number on each die?

- 1/12
- 1/3
- 4/9
- 5/9
- 7/18

3. A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on the dice in his left hand is greater than the sum of the dice in his right?

- 7/108
- 5/54
- 1/9
- 2/17
- 1/4

## Explanations:

1. Rolling any number on a dice three times in a row is equal to the number of throws , where 3 represents the number of throws and 6 is the number of different ways to get three of the same number (e.g. 1, 1, 1 2, 2, 2 3, 3, 3…).

Next we have to use the combinations formula to determine how many ways three out of five can be the same. . We want to multiply this number to

We are not finished yet – there is one little twist to the problem. The question says “exactly three times.” Meaning we have to discount all the instances in which we roll the same number four times and those instance in which we roll the same number five times.

One way to do so is by multiplying by the probability that the fourth and fifth dice will NOT land on the same number as the three dice. Because there are five other possibilities on dice that will not compromise “exactly three numbers” we can multiply by

Finally this gives us

2. For the first die, we can roll any one of six numbers. For the second die, we can roll any number save for the number we rolled on the first die, giving us 5 possibilities. For the third die, we can roll four different numbers (we can’t roll the number we rolled on the first or second die.

6 x 5 x 4 = 120 possibilities out of 216 total possibilities. (For total possibilities we get 6 x 6 x 6 = 216).

120/216 = 5/9

3. Perhaps the hardest of the bunch, this question requires that we find a solution that doesn’t take much longer than 2 minutes. A quick way – or at least relatively quick way – is to determine the number of instances in which our roll of one die will yield more than two die. Intuitively, this is a good place to start because the number of instances in which rolling a great number with one die than with two is much smaller than vice versa.

The only way to roll higher on one die is if the magicians rolls between 2 and 5, inclusive, with two dice. Were he to roll a six with two dice than there is no way he could eclipse that number by rolling one die.

Below is the probability of rolling a certain number with two dice.

‘2’ – 1/36

‘3’ – 2/36

‘4’ – 3/36

‘5’- 4/36

Now the only numbers a magician can roll with the one die and win is between 3 and 6, inclusive. The chances of rolling any are always 1/6.

Next, we have to combine the probability distribution relating to two dice with that relating to the one.

The chances of a magician rolling any given number with one die are 1/6. So let’s start with the lowest number he can roll: a ‘3.’ To win with this roll, he will have to roll a ‘2’ with two dice, the odds of which are 1/36. So .

So the chances of him rolling a ‘3’ with one die and winning are 1/216.

Let’s repeat this logic for the next roll, ‘4.’ Chances of rolling are 1/6. Only way a ‘4’ wins is if he rolls a ‘2’ or a ‘3’ with two dice.

Odds of rolling a ‘2’ with two dice + odds of rolling a ‘3’ with two dice = . Combine this with the odds of rolling a ‘4’ ( which is 1/6): .

If he rolls a ‘5’ with one die, he can win if he rolls a ‘4’ with two dice, the probability is 3/36. He can also win if he rolls a ‘2’ or a ‘3’ with two dice, the number outcomes we just found: 3/36. So we add

Next, if he rolls a ‘6’ on one die, he can beat ‘2’ through ‘5’ with the two dice. Number of ways to roll a ‘5’ = 4/36. Combining this with the odds of rolling a ‘6’ on one die with the odds or rolling ‘2’, ‘3’, ‘4’, or ‘5’ with the two dice we get: 10/216.

You may be wondering why I left the denominator as 216. Well, this allows us to add up all the instances he can possibly win:

(Answer B)

Hi,

I do not get the first part of the question #3.

Why are the possibilities of rolling the dice for each number, this way :

Below is the probability of rolling a certain number with two dice.

‘2’ – 1/36

‘3’ – 2/36

‘4’ – 3/36

‘5’- 4/36

If I have 2 dices the result of each one would not be just 1/6 for each, as one does not depend from the other one?

Thanks,

Lucy

Hi Lucy 🙂

Happy to help! To determine the probability of rolling a certain number with two dice, we have to think about the possible combinations that lead to these values. You’re correct that the number rolled on each die is independent of the other. So, to find the probability that x is rolled on one die and y on the other, we multiply the probabilities of the individual rolls together (e.g. P(x,y) = P(x)*P(y) = 1/6*1/6 = 1/32). With that in mind, there is a 1/36 chance that a given combination x,y is rolled.

To determine the probabilities of rolling a certain sum, we need to figure out how many possible ways we can come up with that sum. Let’s look at the situations you’ve listed to see what I mean 🙂

2: there is only 1 way to roll 2 (D1 = 1, D2 = 1). For that reason, the probability of rolling a 2 is 1/6*1/6 = 1/32.

3: there are 2 ways to roll 3 (D1 = 1, D2 = 2 OR D1 = 2, D2 = 1). So, the probability of rolling 3 is the sum of the probability of these two situations:

P(1,2) + P(2,1) = 1/6*1/6 + 1/6*1/6 = 2/36

4: there are 3 ways to roll 4 (D1 = 1, D2 = 3; D1 = 3, D2 = 1; D1 = 2, D2 = 2). So the probability of rolling 4 is P(1,3) + P(3,1) + P(2,2) = 1/6*1/6 + 1/6*1/6 + 1/6*1/6 = 3/36

5: there are 4 ways to roll 5 (D1 = 1, D2 = 4; D1 = 4, D2 = 1; D1 = 3, D2 = 2; D1 = 2, D2 = 3). So the probability of rolling 5 is P(1,4) + P(4,1) + P(3,2) + P(2,3) = 1/6*1/6 + 1/6*1/6 + 1/6*1/6 + 1/6*1/6 = 4/36

Does that make sense, Lucy?

As a side note, from this analysis, we can see that there is always an even number of ways to roll an odd sum and an odd number of ways to roll an even sum.

I hope this clears up your doubts! Happy studying 🙂

SOLUTION FOR Qn.1)

I will take first no. as 1 1 1 X X.

in place of X i can place any 5 no.s except 1.

Total no. of ways for arranging this 5 digit no.

(6*1*5*5*5!)/(3!*2!).

6 is because of that in place of 1 i can place 2,3,4,5,6.

Total 6 numbers.

3! is because of that 1 is repeated 3 times.

2! is because of that x may repeat for 2 times.

The total no. of outcomes=6^5.

SO,

Ans=(6*1*5*5*5!)/(3!*2!*6^5).

Can you solve this sum and explain me how it is done !

Company employs 8 proff on their staff.Their respective probability of remaining in employment for 10 yrs are 0.2 ,0.3,0.4,0.5,0.6,0.7,0.8,0.9.The probability that after 10 yrs at least 6 of them still work in the company is?

is the answer 8.35400(6)%?

It’s good to point out that this question is beyond the scope of the GMAT due to the amount of calculations. With that in mind, the correct answer is 0.19. If you’re interested how this question can be solved, I recommend checking out the thread for this problem on the GMAT Club’s forum 🙂

Happy studying!

Hi,

My questions relates to the first problem. Isn’t (1/6)^3 equal 1/216 -? if so, then there is no right answer in the first problem. Please let me know if I am not correct.

Thank you,

Nadya

Hi Chris,

I somehow found the first one, the toughest among the lot, but after reading your explanation, I understood, where was i making the mistake. Thanks for such a nice explanation

I believe, the last problem could be solved in a bit easier way i.e. by just counting the number of instance.

Here’s my explanation, please correct me if I am wrong –

The possible options for the number on the dice in the left hand are : 3, 4, 5, 6

The possible options on the right hand are :

6 :- 10 (1 + 1 to 4; 2 + 1 to 3; 3 + 1 to 2; 4 + 1)

5 :- 6 (1 + 1 to 3; 2 + 1 to 2; 3 + 1)

4 :- 3 (1 + 1 to 2; 2 + 1)

3 :- 1 (1 + 1)

The total favorable outcomes will be 20

The total number of outcomes will be 6 * 6 * 6

Prob = 20 / (6 * 6 * 6)

Regards

Umang

Interesting, I actually hadn’t seen this problem in ages and I tried it again from scratch and actually solved it the way you did (also, I did this in my head, which I wouldn’t be able to do as easily–if at all–with the method explained above). So definitely an economical method (took me only about a minute to answer the question).

Thanks for pointing that out 🙂

For number 1, it can be solved way more easily. Think of a random process R which says the probability of getting 3 dice the same and 2 different. Since the dice are independent, we can throw them one at a time.

We do throw 1, and the probability of getting any number is 1.

We do throw again. The probability of getting the throw number of throw 1 is 1/6.

For throw 3, the probability of getting the same number again is 1/6.

Now for number 4, the probability of getting a number other than the repeated one is 5/6. Easy.

For throw 5, the probability of getting a new number is 4/6. We multiply and get 20/6^4, which I think is the correct answer. It’s approximately 1.5%.

For problem 2, same idea. 5/6 * 4/6 = 5/9

Problem 3: first off, any number above 7 beats the single die. It’s a simple check for those who like dice: out of 36 cases, 21 beat the first die.

Out of the other 6v6, the 1 always to any of the 15 cases. The 2 loses to 14 cases. The 3 loses 12 cases. The 4 loses to 11, the 5 to 6 and the 6 to none. Do the product and you get your answer. (This is easier done by multiplying everything and then checking denominators of the answers.)

Hi

Why Magoosh Team is not responding?

Hi, Pavan

We have a high number of comments on our blogs every day, so it may take us a few days to respond, but we will definitely get to them soon! 🙂 If you sign up for a premium plan, we answer all questions through our own Support system within 24 hours, so I would definitely recommend that if you have a lot of questions. :).

Best,

Margarette

Hi Chris

please explain this….

Nine highschool boys gather at the gym for a game of mini-volleyball. Three teams of 3 people each will be created. How many ways are there to create these 3 teams?

We are selecting three people for three different groups: 9C3 x 6C3 x 3C3 = 1680.

Hope that helps!

But the answer is 280

He’s wrong…partially…

its 9C3 X 6C3 X 3C3

but then we need to know that the order of choosing the teams doesnt matter

therefore we need to divide it by 3!

and you get 280

i think answer is 280