It is currently 19 Oct 2017, 09:54

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m08#21

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Manager
Status: On...
Joined: 16 Jan 2011
Posts: 184

Kudos [?]: 70 [0], given: 62

### Show Tags

26 Aug 2011, 19:21
nightwing79 wrote:
Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?

(A) $$\frac{24}{64}$$
(B) $$\frac{32}{64}$$
(C) $$\frac{36}{64}$$
(D) $$\frac{40}{64}$$
(E) $$\frac{42}{64}$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

ok - I know the solution is long if I go the traditional route.
So - I switched to educated guessing and it was easier.
btw -Good question.
_________________

Labor cost for typing this post >= Labor cost for pushing the Kudos Button
http://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html

Kudos [?]: 70 [0], given: 62

Senior Manager
Joined: 08 Jun 2010
Posts: 389

Kudos [?]: 100 [0], given: 13

Location: United States
Concentration: General Management, Finance
GMAT 1: 680 Q50 V32

### Show Tags

04 Apr 2012, 00:21
A very belated reply to an ancient problem. I kept this GMAT of the day problem in my inbox to see if I can come up with my own solution.

So, this is how I did it, I am not sure if it will be under 2 mins because I classify this as a tough problem so on the real test I wouldn't mind spending a little extra on this to give some lead away to the Prep.

Here goes:
I drew a 6x6 matrix with all possibile values on the 2nd and 3rd dice.
So, it would consist of 36 squares
1,1 1,2..... and so on
2,1 2,2..... and so on
finally
6,1 6,2..... and so on

So, lets vary the first dice to see a pattern of the number of possibilities greater than value of 10.
So, when the first dice is 1, you would have (1,6,6); (1, 6,5); (1,5,6) and so on. This represents a total of 3+2+1 = 6 possibilities (which physically represents the bottom right corner right triangle of the 6x6 matrix with a vertex at (6,6); (4,6); and (6,4).

Now continue with the first dice value of 2
total is 4+3+2+1 = 10

Now continue with the first dice value of 3
total is 5+4+3+2+1 = 15

You get the pattern?

so, the pattern here is 6, 10, 15, 21, 26, 30 for the 6 values of the first dice. The number of possibilities increases by one diagonal towards the left top corner vertex value of (1,). Notice: The last two additional diagonal drops in value because the 6x6 square starts to narrow down toward the top corner.

So, when we add all the values we get 108.

Total # of possibilities is 216. (6x6x6 which represents 6 possible values for each dice)

So, probability is 108/216 = 1/2 = 32/64.

Hope this helps!

This took me around 2:40 mins. I will take that on a GMAT for 700+ problem anyday!!

This method is both visual as well as uses the same concept used on the OG to solve one of the problems (I can't recall which one but I remember its a visual problem with all the dots).

Kudos [?]: 100 [0], given: 13

Manager
Joined: 13 May 2010
Posts: 122

Kudos [?]: 24 [0], given: 4

### Show Tags

21 Jun 2012, 04:18
To outscore Mary, Joe has to score in the range of 11-18. The probability to score 3 is the same as the probability to score 18 (1-1-1 combination against 6-6-6, if 1-1-1 is on the tops of the dice the 6-6-6 is on the bottoms). By the same logic, the probability to score $$x$$ is the same as the probability to score $$21 - x$$ . Therefore, the probability to score in the range 11-18 equals the probability to score in the range of 3-10. As 3-18 covers all possible outcomes the probability to score in the range 11-18 is $$\frac{1}{2}$$ or $$\frac{32}{64}$$ .

In the above explanation, can someone please explain how Prob. to score x is the same as 21 -x ? Where do we get 21 from?

Kudos [?]: 24 [0], given: 4

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128872 [0], given: 12183

### Show Tags

21 Jun 2012, 04:58
teal wrote:
To outscore Mary, Joe has to score in the range of 11-18. The probability to score 3 is the same as the probability to score 18 (1-1-1 combination against 6-6-6, if 1-1-1 is on the tops of the dice the 6-6-6 is on the bottoms). By the same logic, the probability to score $$x$$ is the same as the probability to score $$21 - x$$ . Therefore, the probability to score in the range 11-18 equals the probability to score in the range of 3-10. As 3-18 covers all possible outcomes the probability to score in the range 11-18 is $$\frac{1}{2}$$ or $$\frac{32}{64}$$ .

In the above explanation, can someone please explain how Prob. to score x is the same as 21 -x ? Where do we get 21 from?

The lowest score is 3 (1+1+1) and the highest score is 18 (6+6+6).
The probability of getting 3 is the same as the probability of getting 18=21-3;
The probability of getting 4 is the same as the probability of getting 17=21-4;
...

As for the solution.

Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?
A. 24/64
B. 32/64
C. 36/64
D. 40/64
E. 42/64

Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.
Expected value of three dice is 3*3.5=10.5.

Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2 = 32/64.

That's because the probability distribution is symmetrical for this case:
The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);
The probability of getting the sum of 4 = the probability of getting the sum of 17;
The probability of getting the sum of 5 = the probability of getting the sum of 16;
...
The probability of getting the sum of 10 = the probability of getting the sum of 11;

Thus the probability of getting the sum from 3 to 10 = the probability of getting the sum from 11 to 18 = 1/2.

Also discussed here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-86407.html and here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it's clear.
_________________

Kudos [?]: 128872 [0], given: 12183

Intern
Joined: 12 Jun 2012
Posts: 3

Kudos [?]: [0], given: 0

### Show Tags

21 Jun 2012, 23:28
answer should be 1/2 because no. of possibilities of summations are 3-18 that is 16 sum possibilities, 3-10 makeup 8.
That is option B.

Kudos [?]: [0], given: 0

Director
Joined: 28 Jul 2011
Posts: 521

Kudos [?]: 297 [0], given: 16

Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE: Accounting (Commercial Banking)

### Show Tags

23 Jun 2012, 05:44
Bunuel wrote:
teal wrote:
To outscore Mary, Joe has to score in the range of 11-18. The probability to score 3 is the same as the probability to score 18 (1-1-1 combination against 6-6-6, if 1-1-1 is on the tops of the dice the 6-6-6 is on the bottoms). By the same logic, the probability to score $$x$$ is the same as the probability to score $$21 - x$$ . Therefore, the probability to score in the range 11-18 equals the probability to score in the range of 3-10. As 3-18 covers all possible outcomes the probability to score in the range 11-18 is $$\frac{1}{2}$$ or $$\frac{32}{64}$$ .

In the above explanation, can someone please explain how Prob. to score x is the same as 21 -x ? Where do we get 21 from?

The lowest score is 3 (1+1+1) and the highest score is 18 (6+6+6).
The probability of getting 3 is the same as the probability of getting 18=21-3;
The probability of getting 4 is the same as the probability of getting 17=21-4;
...

As for the solution.

Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?
A. 24/64
B. 32/64
C. 36/64
D. 40/64
E. 42/64

Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.
Expected value of three dice is 3*3.5=10.5.

Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2 = 32/64.

That's because the probability distribution is symmetrical for this case:
The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);
The probability of getting the sum of 4 = the probability of getting the sum of 17;
The probability of getting the sum of 5 = the probability of getting the sum of 16;
...
The probability of getting the sum of 10 = the probability of getting the sum of 11;

Thus the probability of getting the sum from 3 to 10 = the probability of getting the sum from 11 to 18 = 1/2.

Also discussed here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-86407.html and here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it's clear.

Hi Bunnel,

I didn't get these steps

Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.???

The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);????

and is he throwing dice once or thrice?

i am totally confused with the wordings of the question....can you please explain this one??
_________________

+1 Kudos If found helpful..

Kudos [?]: 297 [0], given: 16

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128872 [0], given: 12183

### Show Tags

23 Jun 2012, 05:48
kotela wrote:
Hi Bunnel,

I didn't get these steps

Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.???

The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);????

and is he throwing dice once or thrice?

i am totally confused with the wordings of the question....can you please explain this one??

Please check the links provided above and also this post: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html#p1032531
_________________

Kudos [?]: 128872 [0], given: 12183

Manager
Joined: 28 Feb 2012
Posts: 115

Kudos [?]: 52 [0], given: 17

Concentration: Strategy, International Business
Schools: INSEAD Jan '13
GPA: 3.9
WE: Marketing (Other)

### Show Tags

23 Jun 2012, 06:32
I solved it, although it took me a while. One of the things that took my time is that i could not come to 64 in the denominator, since 3 attempts give us 216, so i did not even think that 32/64 is the same as 108/216. I think that on the real GMAT they will use the answer which comes from 108/216, e.g. 36/72, 18/36....or 1/2.
Bunuel don't you think so? or that is one of the tricks?
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Kudos [?]: 52 [0], given: 17

Intern
Joined: 02 Jul 2012
Posts: 2

Kudos [?]: 1 [0], given: 0

### Show Tags

27 Aug 2012, 08:23
see. no ofDigit( sum) from 11-8===8
10-3===is also 8
as consequtive and equal.
Total p===1
so each 1/2.

Kudos [?]: 1 [0], given: 0

Intern
Joined: 18 Jun 2012
Posts: 2

Kudos [?]: 7 [1], given: 2

### Show Tags

05 Sep 2013, 03:34
1
KUDOS
IMO 32/64 is the answer, Equal possibilities.

Kudos [?]: 7 [1], given: 2

Intern
Joined: 21 Jul 2014
Posts: 23

Kudos [?]: 2 [0], given: 13

### Show Tags

12 Aug 2014, 09:45
abhicoolmax wrote:
mercurialcc wrote:
i don't think there is a need to list all the combinations.

with 3 dices, the possible range of score is from 3 (3 dices of one) to 18 (3 dices of six). so there's a any number between 3 and 18 has a probability of:

1/(18-3+1) = 1/16

if a person has to roll 11 or over to win, the probability will be

P(rolling 11) + P(rolling 12) + ... + P(rolling 18) = 1/16 + ... + 1/16 = 8/16 = 1/2 = 32/64 ===> Ans. B

Alternatively:
all available scores is 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
the score can't be 1 or 2 because u have 3 dices

if mary rolled 10, to be higher than her, u can roll 11 and up
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18

bold occurrences / all occurrences = 8/16

Oops. You got the answer by fluke You made BIG mistake in your assumption (even you 2nd solution is INCORRECT! 8/16 is NOTHING!).... Be careful my friend. These are the careless mistakes that cost you bigtime.

P(3) is NOT equal to P(5). WHY?

Number of ways 3 can occur = {1,1,1} = 1
Probability of 3 occurring = 1/(6*6*6) = 1/216

Number of ways 4 can occur = {1,1,2}*3C1 = 3
Probability of 5 occurring = 3/(6*6*6) = 1/72

Number of ways 5 can occur = {1,1,3}*3C1 + {1,2,2}*3C2 = 6
Probability of 5 occurring = 6/(6*6*6) = 1/36

.....

The symmetry is across the 2 ends of the SUM and not across the entire set.

Please see my solution above to inculcate the right thought process!

Hello..
Could you please explain, how you arrived at following ? :
Number of ways 4 can occur = {1,1,2}*3C1 = 3
Number of ways 5 can occur = {1,1,3}*3C1 + {1,2,2}*3C2 = 6

I am sorry but I am very bad at probability and combinatorics !!

Kudos [?]: 2 [0], given: 13

Intern
Joined: 03 Oct 2011
Posts: 17

Kudos [?]: 10 [0], given: 4

### Show Tags

12 Aug 2014, 20:36
Bunuel wrote:
This question was posted in PS forum as well. Here is my solution for it:

Expected value of a roll of one die is 1/6(1+2+3+4+5+6)=3.5.
Expected value of three dice is 3*3.5=10.5.

Mary scored 10, so the probability to have more then 10 (11, 12, 13, 14, 15, 16, 17, 18), or more then average, is the same as to have less than average=1/2.

P=1/2.

Think this approach is the easiest one.

Expected value is an amazing approach to solving this problem.

I find it strange that the fractions in the answer have no relation to the problem i.e. There is no way of solving this problem where you get 64 in the denominator. This is my first time seeing this. I wonder if the actual GMAT could have something like this. I usually inspect PS answer choices and try to get clues from them when I'm stumped. Until now.

Kudos [?]: 10 [0], given: 4

m08#21   [#permalink] 12 Aug 2014, 20:36

Go to page   Previous    1   2   [ 32 posts ]

Display posts from previous: Sort by

# m08#21

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.