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## Pam and Robin each roll a pair of fair

This topic has 3 expert replies and 4 member replies

### Top Member

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#### Pam and Robin each roll a pair of fair

Wed Sep 20, 2017 3:09 am
Pam and Robin each roll a pair of fair, six-sided dice. What is the probability that Pam and Robin will both roll the same set of two numbers?

A. 1/216

B. 1/36

C. 5/108

D. 11/216

E. 1/18

Can experts teach me the formulas here? Thanks

OA D

### GMAT/MBA Expert

Jeff@TargetTestPrep GMAT Instructor
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Tue Dec 12, 2017 5:17 pm
lheiannie07 wrote:
Pam and Robin each roll a pair of fair, six-sided dice. What is the probability that Pam and Robin will both roll the same set of two numbers?

A. 1/216

B. 1/36

C. 5/108

D. 11/216

E. 1/18
We need to determine the probability that when Pam and Robin each rolls a pair of fair, six-sided dice, they both roll the same set of numbers. There are two scenarios: when Pam and Robin both roll the same two numbers and when they roll two distinct numbers.

Scenario 1: When the two numbers on the dice are the same

Letâ€™s say they both roll 1s. That is, Pam rolls (1, 1) and Robin rolls (1, 1). The probability of this happening is

1/6 x 1/6 x 1/6 x 1/6 = 1/(6^4)

Since the probability is the same for all 6 pairs of numbers, the probability of their rolling the same numbers is 6 x 1/(6^4) = 1/(6^3) = 1/216.

Scenario 2: When the two numbers on the dice are distinct

There are 6 x 5 = 30 ways to roll two distinct numbers when rolling two dice.
Letâ€™s say Pam rolls (1, 2) and Robin also rolls (1, 2). The probability of this happening is:

1/6 x 1/6 x 1/6 x 1/6 = 1/(6^4)

However, if Pam rolls (1, 2) and Robine rolls (2,1), those are still considered the same set of numbers, and the probability of that occurring is also 1/(6^4).

Therefore, for each pair of distinct numbers rolled, the probability is 2 x 1/(6^4) = 2/(6^4). Since there are 30 such pairs, the overall probability is 30 x 2/(6^4) = 60/(^4) = 10/(6^3) = 10/216.

Finally, since the events in option 1 and those in option 2 are mutually exclusive, we use the addition rule of probability. That is, the probability that Pam and Robin will both roll the same set of two numbers is:

1/216 + 10/216 = 11/216

Answer: D

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### Top Member

regor60 Master | Next Rank: 500 Posts
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Wed Sep 20, 2017 8:19 am
If this were a matching one die to one die question, it could be simplified by having the first person roll and "select" the number to be matched. In which case, the second person has a 1/6 chance of matching.

There are 6x6 or 36 pairs of numbers the first person can roll. 6 of them are identical, 11,22,33,44,55,66. So there are 30 other pairs. Recognize that 12 is the same as 21, so 15 unique pairs.

So, 30/36 of the rolls will generate nonidentical pairs and 6/36 will generate identical numbers.

Breaking it down. The first person rolls nonidentical pair
30/36 = 5/6 probability

In order to match this pair, the second person must roll a pair from the nonidentical pair bucket, also with 5/6 probability. Further, he has to match that pair. There are 30 pairs but there are two ways of matching 12 = 21 for example.

So the overall probability is 5/6 (first person) x 5/6(second person)x2/30(second person) = 5/108

Continuing to the identical pairs.

First person rolls identical pair 6/36 = 1/6 probability

Second person rolls identical pair, also 1/6 probability, but also has to match the one rolled by the first person , 1/6 probability.

So combining: 1/6 (first person) x 1/6(second person) x 1/6 (second person again) =1/216

So the combined probability of matching pairs needs to add the probabilities above:

5/108 = 10/216 + 1/216 =D, 11/216

### GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
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Wed Sep 20, 2017 9:38 am
Hi lheiannie07,

This is a tougher probability question than average (and you likely will not see this exact situation on Test Day). This is meant to say that you shouldn't be too concerned about this prompt until you're picking up points in all the other 'gettable' areas first.

That having been said, the 'quirk' with this question is that you have to account for the probability of rolling two dice and getting the same number on both dice vs. getting two different numbers.

When rolling two dice, there are 36 possible outcomes:
-6 outcomes have the same number twice (1-1, 2-2, etc.)
-30 outcomes have two different numbers (1-4, 3-2, 5-1, etc.)

Thus, 6/36 = 1/6 of the outcomes are the same number twice
30/36 = 5/6 of the outcomes are two different numbers

IF Pam rolled 5-5, then Robin would have to also roll 5-5. The probability of that occurring would be: (1/6)(1/6) = 1/36.

IF Pam rolled 2-6, then Robin has two different ways to match up (2-6 OR 6-2). The probability of that occurring would be (2/6)(1/6) = 2/36 = 1/18

Accounting for all possible outcomes, the probability of Pam and Robin rolling the same result would be:
(1/6)(1/36) + (5/6)(1/18) =
1/216 + 5/108 =
1/216 + 10/216 =
11/216

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

Mo2men Legendary Member
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Thu Sep 21, 2017 12:47 pm
Rich.C@EMPOWERgmat.com wrote:
Hi lheiannie07,

This is a tougher probability question than average (and you likely will not see this exact situation on Test Day). This is meant to say that you shouldn't be too concerned about this prompt until you're picking up points in all the other 'gettable' areas first.

That having been said, the 'quirk' with this question is that you have to account for the probability of rolling two dice and getting the same number on both dice vs. getting two different numbers.

When rolling two dice, there are 36 possible outcomes:
-6 outcomes have the same number twice (1-1, 2-2, etc.)
-30 outcomes have two different numbers (1-4, 3-2, 5-1, etc.)

Thus, 6/36 = 1/6 of the outcomes are the same number twice
30/36 = 5/6 of the outcomes are two different numbers

IF Pam rolled 5-5, then Robin would have to also roll 5-5. The probability of that occurring would be: (1/6)(1/6) = 1/36.

IF Pam rolled 2-6, then Robin has two different ways to match up (2-6 OR 6-2). The probability of that occurring would be (2/6)(1/6) = 2/36 = 1/18

Accounting for all possible outcomes, the probability of Pam and Robin rolling the same result would be:
(1/6)(1/36) + (5/6)(1/18) =
1/216 + 5/108 =
1/216 + 10/216 =
11/216

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Hi Rich,

can you please detail the highlighted part in red? why Robin has 2 way to match up?

Thanks for your support

### GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
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Thu Sep 21, 2017 7:31 pm
Hi Mo2men,

The prompt asks for the probability that both women will roll the same two numbers - but no emphasis is placed on the ORDER of the numbers. So if Pam rolls 2-6, then EITHER of Robin's dice could be the 2 and the other would then have to be the 6 (thus 2-6 or 6-2 would be a 'match').

Imagine if we rolled one die at a time. The probability of rolling a 2 OR a 6 on the first die is 2/6.

Assuming that first roll is a 'match' (either the 2 or the 6), then the second die would have to be the other number - and the probability of THAT happening would be 1/6.

(2/6)(1/6) = 2/36 = 1/18

GMAT assassins aren't born, they're made,
Rich

_________________
Contact Rich at Rich.C@empowergmat.com

Mo2men Legendary Member
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Thu Sep 21, 2017 10:14 pm
Rich.C@EMPOWERgmat.com wrote:
Hi Mo2men,

The prompt asks for the probability that both women will roll the same two numbers - but no emphasis is placed on the ORDER of the numbers. So if Pam rolls 2-6, then EITHER of Robin's dice could be the 2 and the other would then have to be the 6 (thus 2-6 or 6-2 would be a 'match').

Imagine if we rolled one die at a time. The probability of rolling a 2 OR a 6 on the first die is 2/6.

Assuming that first roll is a 'match' (either the 2 or the 6), then the second die would have to be the other number - and the probability of THAT happening would be 1/6.

(2/6)(1/6) = 2/36 = 1/18

GMAT assassins aren't born, they're made,
Rich
Dear Rich,

Thanks for your reply. However, it proved that either the question wording is unclear or my assumptions are wrong.

When you rolling pair of dice, do you consider rolling them one by one or both in same time. The reason of question is the that PAM rolls pair of dice in one shot and gets 2-6. Then, Robin rolls her pair of dice in one shot. So it is nonsensical to assume if she get 2-6 or 6-2 as both are same. At the end, pair of dice is thrown at SAME time.
The case you describe happens when Pam a die by die then Robin rolls die by die so in first she can get 2 or 6 and the other die would be the other number. Now it makes sense if that the case in question.
How can you detect which scenario from my above explanation, if I'm correct.
I hope you understand my confusion and help.

Thanks in advance for your suppport

### Top Member

regor60 Master | Next Rank: 500 Posts
Joined
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261 messages
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Fri Sep 22, 2017 5:23 am
Mo2men wrote:
Rich.C@EMPOWERgmat.com wrote:
Hi Mo2men,

The prompt asks for the probability that both women will roll the same two numbers - but no emphasis is placed on the ORDER of the numbers. So if Pam rolls 2-6, then EITHER of Robin's dice could be the 2 and the other would then have to be the 6 (thus 2-6 or 6-2 would be a 'match').

Imagine if we rolled one die at a time. The probability of rolling a 2 OR a 6 on the first die is 2/6.

Assuming that first roll is a 'match' (either the 2 or the 6), then the second die would have to be the other number - and the probability of THAT happening would be 1/6.

(2/6)(1/6) = 2/36 = 1/18

GMAT assassins aren't born, they're made,
Rich
Dear Rich,

Thanks for your reply. However, it proved that either the question wording is unclear or my assumptions are wrong.

When you rolling pair of dice, do you consider rolling them one by one or both in same time. The reason of question is the that PAM rolls pair of dice in one shot and gets 2-6. Then, Robin rolls her pair of dice in one shot. So it is nonsensical to assume if she get 2-6 or 6-2 as both are same. At the end, pair of dice is thrown at SAME time.
The case you describe happens when Pam a die by die then Robin rolls die by die so in first she can get 2 or 6 and the other die would be the other number. Now it makes sense if that the case in question.
How can you detect which scenario from my above explanation, if I'm correct.
I hope you understand my confusion and help.

Thanks in advance for your suppport
You can consider a 2-6 as equal to 6-2, but you have to recognize that there are 2 ways to achieve this, so it doesn't make any difference if you view them as distinct or identical.

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