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

Let’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.
Let’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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The answer should be D. 11/216.
There are 2 cases:
1. Both dices show the same number, so it's 6 out of 36. Then for the other player the probability to roll the same number is 1/6 x 1/6.
2. There are 30 out of 36 options for the dices to be different numbers. The probability for the other player to roll the same is 2/6 x 1/6.

Probability = 1/6 x 1/6 x 1/6 + 5/6 x 2/6 x 1/6 = 11/216

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The second player roll the first dice. It should be equal to one of the dices the first player rolled. It doesn't matter which one, so there are 2 options out of 6. For the second dice there is only 1 option left, which makes it: 2/6 * 1/6.



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If we are already considering (1,2) and (2,1) to be the same set while calculating probability then why do we have total number of pairs = 30? Shouldn't it be 15?

Hi All,

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:

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

We have to account for two possible situations (since Pam's roll dictates the probability that Robin's roll will match) : Pam rolls the same number on both dice or Pam rolls two different numbers. The probabilities that you highlighted in red are those two probabilities: there's a 1/6 chance that it's the same number twice and a 5/6 chance that it's two different numbers.

GMAT assassins aren't born, they're made,
Rich
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My way of doing this:

Total cases : 6^2 * 6^2= 36*36
Now required cases:
1. Both dices have same number for both the person. Example (1,1) and (1,1)
Such 6 cases.
2. Different number in 2 dice example (1,2) and (1,2)- This can have 4 arrangements, (1,2)(1,2), (1,2)(2,1),(2,1)(1,2),(2,1)(2,1)
and total number of ways to pick these types are 6C2.

so total cases = 6 + 6C2 * 4

Required probability = ( 6 + 6C2 * 4) / ( 36*36 ) = 11/216