Pam and Robin each roll a pair of fair, six-sided dice. What is the pr

I'm not sure if my approach is right.

6/36 x 6/36 = 1/6 x 1 /6 = 1/36. I think the answer is B. I am not sure though.

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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Dear kolodits, How do you able to get the value for \(\frac{2}{6}\)?

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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At first I misread and thought it was asking for the same sum of the pairs of dice which would yield a different answer.

There are two cases. The first is when the first person rolls two different numbers. This happens 5/6 of the time. The second is when the first person rolls the same number on both die. This happens 1/6 of the time.

In the first case, the second person has a 2/6 chance to roll one of the two numbers that the first person rolled with the first die. If the first die is one of the two numbers, then the second person only has a 1/6 chance for the second die to match up. This means that in the first case, the second person has a 2/6 * 1/6 = 1/18 chance to have the same roll. This applies 5/6 of the time.

In the second case, both dice must match, so the second person has a 1/6 chance to roll the same number with each of the two die. This means the total probability is 1/6 * 1/6 = 1/36. This applies 1/6 of the time.

This means that the total probability is 5/6*1/18 + 1/6*1/36 = 5/108 + 1/216 = 10/216 + 1/216 = 11/216. Answer is D

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?

1/6 of the time? 5/6 of the time? How are you saying that? can you please explain?

Please clarify why you included additional probabilities (highlighted in red).

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

\(? = {{\# \,\,{\rm{favorables}}} \over {\# \,\,{\rm{total}}\,\,\left( {{\rm{equiprobables}}} \right)}}\)

\(\# \,\,{\rm{total}}\,\,\left( {{\rm{equiprobables}}} \right) = {6^2} \cdot {6^2}\,\,\,\,\left( {{\rm{taking}}\,\,{\rm{results}}\,\,{\rm{in}}\,\,{\rm{order,}}\,\,{\rm{for}}\,\,{\rm{both}}\,\,{\rm{players}}} \right)\)

\(\# \,\,{\rm{favorables}}\,\,\, = \,\,\,\left\{ \matrix{\\
\,6\,\,::\,\,{\rm{first}}\,\,{\rm{player}}\,\,{\rm{gets}}\,\,{\rm{same}}\,\,{\rm{number}}\,\,{\rm{twice,}}\,\,{\rm{second}}\,\,{\rm{player}}\,\,{\rm{gets}}\,\,{\rm{same}}\,\,{\rm{ones}} \hfill \cr \\
\,\,\, + \,\,\,\,\,\left( {{\rm{mutually}}\,\,{\rm{exclusive}}\,{\rm{!}}} \right) \hfill \cr \\
\,\left( {36 - 6} \right) \cdot 2!\,\,\,::\,\,\,{\rm{first}}\,\,{\rm{gets}}\,\,\left( {x,y} \right)\,\,{\rm{with}}\,\,x \ne y\,\,,\,\,{\rm{second}}\,\,{\rm{gets}}\,\,\left( {x,y} \right)\,\,{\rm{or}}\,\,\left( {y,x} \right) \hfill \cr} \right.\)

\(? = {{66} \over {{6^4}}} = {{11} \over {{6^3}}} = {{11} \over {216}}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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

Two cases, P(picking both balls different) = (6/6* 5/6* 1/6*1/6) * 2 - since Pam can get (1,2 or 2,1) and Sam can get (2,1) & (1,2)
One number P(both balls same) - 6/6*1/6*1/6*1/6

Total = 11/216
Ans:C

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.
For this part, shouldn't it be 15 pairs ? Because Pam (1,2) and Robin (2,1) can also be P(2,1) and R(1,2) so aren't these similar pairs ? Hence shouldn't the total pairs be 15 ?

Probability of pairs of same numbers = 6/36 = 1/6
So, probability of same pairs of same numbers = (1/6)*(1/36) = 1/216
Now,
Probability of same pairs of different numbers = (1-1/6)*(2/36) = 10/216

Here, 2/36 because if pam rolls (1,6) then robin has 2 options. So, 2*(1/36) = 2/36

So, answer = 11/216

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Official Explanation:

This is a pairs probability question (our goal is to match outcomes, rather than to get any particular numbers). So we will be using our "am I still happy?" approach. However, it is critical to note that the first two rolls do make a difference, even though we are happy with any first two rolls. Specifically, (if we assume that Pam rolls first) it makes a difference whether Pam rolls a pair or not. If she does, then it will be a bit harder for Robin to match Pam's rolls.

Pam's probability of rolling a pair is 1/6 ; any first value rolled by Pam is totally fine (probability 6/6=1 that we will be happy with Pam's progress towards some pair), but only one of Pam's six possible second rolls will make us happy if a pair is our goal. Conversely, Pam's probability of not rolling a pair is 1−1/6=5/6

Now, if Pam rolled a pair, Robin will have to roll one particular number twice in a row to match Pam. For instance, if Pam's rolls were a pair of threes, then Robin would have to roll a three and then a three again to match. Robin's probability to match Pam would be 1/6∗1/6=1/36

If Pam did not roll a pair, Robin would simply have to match either of Pam's numbers on the first roll and the other on the second. For instance, if Pam's rolls were a four and a five, then Robin could roll either a four or a five on his first roll, and subsequently would need to roll the other value on his second roll. That is, Robin's probability to match Pam would be 2/6∗1/6=2/36.

Putting it all together, there is a 5/6 chance for Robin's probability to be 2/36 or a 1/6 chance for Robin's probability to be 1/36. Combining these probabilities gives 5/6∗2/36+1/6∗1/36=11/216. Answer D is correct.

What worked for me:

Case 1: The 1st 2 dice have same numbers.

The 1st die has 6 possible values out of 6. The 2nd die has only 1 possible value which will be same to the 1st die, similarly the 3rd and 4th die only have 1 possible value.

Total possible outcomes: 6 x 1 x 1 x 1

Case 2: The 1st 2 dice have different numbers.

The 1st die has 6 options, the 2nd die has 5 possible outcomes (as 1 number used by 1st die). Now the 3rd die has 2 options - it can take either of the 2 numbers which has come up on the 1st 2 dice. And the 4th die has only 1 option - the 1 number that remains after the 3rd die has been allotted 1 number.

Total possible outcomes: 6 x 5 x 2 x 1

Sum of all favourable outcomes = Case 1 + Case 2 = 6 + 6 x 5 x 2 = 6 + 60 = 66
All possible outcomes = 6 x 6 x 6 x 6 = 1296

Required probability = \(\frac{66}{1296} = \frac{11}{216}\)
Answer D.