M08-21

Question

Mary and Joe each roll three dice. The total score for each person is the sum of the numbers on their three dice. If Mary scores 10, what is the probability that Joe will have a higher score in his attempt?

A.  \frac{24}{64}  
B.  \frac{32}{64}  
C.  \frac{36}{64}  
D.  \frac{40}{64}  
E.  \frac{42}{64}

Bunuel · Accepted Answer

Official Solution:

Mary and Joe each roll three dice. The total score for each person is the sum of the numbers on their three dice. If Mary scores 10, what is the probability that Joe will have a higher score in his attempt?

A.  \frac{24}{64}  
B.  \frac{32}{64}  
C.  \frac{36}{64}  
D.  \frac{40}{64}  
E.  \frac{42}{64}

To outscore Mary, Joe must achieve a score in the range of 11-18. The probability of scoring 3 is the same as the probability of scoring 18 (the 1-1-1 combination is opposite the 6-6-6 combination when considering the tops and bottoms of the dice). Similarly, the probability of scoring  x  is the same as the probability of scoring  21 - x . Consequently, the probability of scoring within the 11-18 range is equal to the probability of scoring within the 3-10 range. Since the range of 3-18 encompasses all possible outcomes, the probability of scoring within the 11-18 range is  \frac{1}{2} .
 
 Alternative explanation 
 
The expected value when rolling a single die is  \frac{1}{6}*(1+2+3+4+5+6)=3.5 .
 
Therefore, the expected value when rolling three dice is  3*3.5=10.5 .
 
Mary scored 10, so the probability of getting a sum greater than 10 (11, 12, 13, ..., 18), or greater than the average, is the same as getting a sum less than the average (10, 9, 8, ..., 3). Since these two events are complementary (adding to 1) and symmetrical, the probability of each is 1/2:
 
The probability of getting a sum of 3 (the minimum possible sum) = the probability of getting a sum of 18 (the maximum possible sum);
 
The probability of getting a sum of 4 = the probability of getting a sum of 17;
 
The probability of getting a sum of 5 = the probability of getting a sum of 16;
 
...
 
The probability of getting a sum of 10 = the probability of getting a sum of 11.
 
Therefore, the probability distribution is symmetrical, and thus the probability of getting a sum from 3 to 10 = the probability of getting a sum from 11 to 18 = 1/2.

Answer: B

ankushbagwale · Answer

Bunuel  can you please how the number 21 is derived in the original explanation?

Also, here we know that the average is 10.5 & not 10 so how are we deducing that half the times sum will be less than 10 ( & not 10.5 ) & half the time ( sum will be more than 10.5) .

I can understand since number showing up on the dice being integers we will never get 10.5 ( it will be either 10 or 11 but never 10.5). Nonetheless, technically speaking do we need to say that half the times sum will be greater than average & half the time it will be lesser.

Bunuel · Answer

The probability of scoring 1 + 1 + 1 = 3 is the same as the probability of scoring 6 + 6 + 6 = 18 = 21 - 3.
The probability of scoring 2 + 1 + 1 = 4 is the same as the probability of scoring 6 + 6 + 5 = 17 = 21 - 4.
...

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

Talayva · Answer

I think this is a  high-quality  question and I agree with explanation.

spetznaz · Answer

Great question

Could you please tell the source ??

namanhsuit · Answer

I think this is a  high-quality  question and I agree with explanation.

Shantanu1996 · Answer

High quality question. Hands off to the author!

RenanBragion · Answer

I think this is a  high-quality  question and I agree with explanation.

rchawla17 · Answer

I think this is a  high-quality  question and I agree with explanation.

SDW2 · Answer

Is there any other approach to this question?  VeritasKarishma   IanStewart   egmat

IanStewart · Answer

The fastest way to solve (and I'm really just saying what Bunuel said, but in different words) is to notice:

• when we roll one die, we're just as likely to get an above-average roll as a below-average one. That's what the earlier posts mean when they say dice outcomes are "symmetric". The same is true when we roll three dice.

• when we roll one die, the average roll is 3.5 (it's the average of the equally spaced list 1, 2, 3, 4, 5, 6, so it is the median of that list). So when we roll three dice, the average roll is (3)(3.5) = 10.5. Since we can't get exactly 10.5, since that's not an integer, and since we're as likely to get an above average sum as a below average one, we must get a sum greater than 10.5 exactly half the time, and a sum less than 10.5 the other half of the time. So we get 11 or greater 1/2 the time.

There's an obvious alternative approach, but it takes a long time: you can just list all of the ways to get a sum of 18 (we must get 6-6-6), and a sum of 17 (we can get 5-6-6, 6-5-6 or 6-6-5), and so on. Using certain counting shortcuts makes that perhaps practical to do in two minutes, but it's long, and it's too tedious to be the only available method in a real GMAT question. There are other ways one could solve that are similar -- e.g. you could divide the problem into cases where you imagine getting a '6' on the first die, then a '5' on the first die, and so on, but they aren't any faster. I don't see an alternative solution to the first one I suggest above that is at all fast.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Vikramaditya00 · Answer

I think this is a  high-quality  question and I agree with explanation.

A_ · Answer

I like the solution - it’s helpful.

quidemid · Answer

From 11-18 there are 8 outcomes, and total for 3 dice rolled is between 3 and 18 inclusive which is 16 outcomes. P=1/2. Does this work?

Bunuel · Answer

Check  HERE .

M08-21

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