Mary and Joe are to throw three dice each. The score is the : GMAT Problem Solving (PS) - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 14:43

### 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

# Mary and Joe are to throw three dice each. The score is the

Author Message
TAGS:

### Hide Tags

Moderator
Joined: 02 Jul 2012
Posts: 1231
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Followers: 115

Kudos [?]: 1385 [0], given: 116

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

04 Jul 2012, 08:35
ohh. Cant believe i missed that. Thanks a lot
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Intern
Joined: 26 Aug 2012
Posts: 1
Followers: 0

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

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

01 Oct 2012, 02:52
Bunuel wrote:
noboru 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?

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.

P=1/2.

Hi Bunel,
I used below appraoch to get to thr solution.
Let me know if this is correct.

This problem can be drilled down to find all possible ways to get sum 11 and above.

Dice 1 : 6 , Dice2 : 4 and Dice3 : {1,2,3,4,5,6}
This gives total 6 combinations(1 x 1 x 6).
Now these values can be interchanged on other dices so = 6 x 3 x 2 x 1.
So Combination 1 = 36 ways

Step2 :
Similarly lets try to get values 6,5 and {1,2,3,4,5,6}
Therefore Combination 2 = 36 ways

Step3 :
Similarly lets try to get values 6,6 and {1,2,3,4,5,6}
Therefore Combination 3 = 36 ways

Therefore in all we have 36 + 36 + 36 = 108 ways.

and total number of ways = 216

Therefore probablity = 108/216.
Intern
Joined: 12 Oct 2012
Posts: 11
Location: Singapore
GMAT 1: 710 Q49 V35
GPA: 3.65
Followers: 1

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

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

05 Nov 2012, 18:53
navo wrote:
Bunuel wrote:
noboru 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?

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.

P=1/2.

Hi Bunel,
I used below appraoch to get to thr solution.
Let me know if this is correct.

This problem can be drilled down to find all possible ways to get sum 11 and above.

Dice 1 : 6 , Dice2 : 4 and Dice3 : {1,2,3,4,5,6}
This gives total 6 combinations(1 x 1 x 6).
Now these values can be interchanged on other dices so = 6 x 3 x 2 x 1.
So Combination 1 = 36 ways

Step2 :
Similarly lets try to get values 6,5 and {1,2,3,4,5,6}
Therefore Combination 2 = 36 ways

Step3 :
Similarly lets try to get values 6,6 and {1,2,3,4,5,6}
Therefore Combination 3 = 36 ways

Therefore in all we have 36 + 36 + 36 = 108 ways.

and total number of ways = 216

Therefore probablity = 108/216.

It is not clear what you mean by "Now these values can be interchanged on other dices so = 6 x 3 x 2 x 1."
What about combinations {5,5,5}, {4,4,4} for example? I don't see they are counted in your approach.

I also tried to solve the task by counting combinations which would give 10 points or less with the plan to substract this number from total number of combinations. Then I realized that this value would be the same as if I would count number of combinations resulting in 11 points and higher.
Actually we dont even need to find average expected value.

There are 16 possible scores: 3,4,5,...16,17,18.
Work from opposite sides of this set. Possibility to get 3 points is the same as for 18, possibility of 4 is the same as of 17, and so on.
The last pair in the middle is 10 and 11.
So possibility of score from 3 to 10 = possibility 11 to 18.

The question is about getting score more than 11, i.e. second half of the set = exactly half of choices.
Manager
Joined: 28 Dec 2012
Posts: 115
Location: India
Concentration: Strategy, Finance
GMAT 1: Q V
WE: Engineering (Energy and Utilities)
Followers: 3

Kudos [?]: 66 [1] , given: 90

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

05 Jan 2013, 05:56
1
KUDOS
Joe should score 11,12... 17,18 to score over mary.
Now lets consider 1 by 1:

18: 6,6,6 combination ---> 1 arrangement 18 -> 1 possibility
17: 6,6,5 combination ---> 3 arrangement 17 -> 3 possibilities
16: 6,6,4 combination ---> 3 arrangement
6,5,5 combination ---> 3 arrangement 16 -> 6 possibilities
15: 6,6,3 combination ---> 1 arrangement
6,5,4 combination ---> 6 arrangement
5,5,4 combination ---> 3 arrangement 15 -> 10 possibilities

similarly for 14,13,12,and 11 we have 15,21,25,27 possibilities respectively.
Total favorable: 1+3+6+10+15+21+25+27 = 108 possibilities

Probability = 108/(6*6*6) = 1/2 or 32/64

Two important points... the solution is not the shortest but shows systematic listing method useful for other questions. Secondly,
It appeared as though a series was forming.. which is not the case!!!

Kudos for the Solution plz....
_________________

Impossibility is a relative concept!!

Manager
Joined: 28 Dec 2012
Posts: 115
Location: India
Concentration: Strategy, Finance
GMAT 1: Q V
WE: Engineering (Energy and Utilities)
Followers: 3

Kudos [?]: 66 [0], given: 90

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

05 Jan 2013, 06:57
Short-cut solution:

3 is minimum and 18 is maximum as the sum of three die.
Now, 18 will have same possibilities as 3 (1 possibility)
17 will have same possibilities as 4 (3 possibilites)
16 will have same possibilities as 5 (6 possibilites)
.
.
.
10 will have same possibilities as 11 (27 possibilites)

NOTE: 18+3 = 21, 17 + 4 = 21.... this helps to generate the series.

Thus all possibilites are half - half divided between 3-10 and 11-18. Thus if Joe has to get something from 11 to 18, his probability is 1/2

Kudos for this insight plz
_________________

Impossibility is a relative concept!!

Intern
Joined: 19 Apr 2012
Posts: 27
Followers: 0

Kudos [?]: 3 [0], given: 8

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

04 Apr 2013, 13:21

He needs a minimum score of 11:

The minimum throw must be: Dice 1: 4 or greater Dice 2: 4 or greater Dice 3: 3 or greater

3*( 3/6 * 3/6 * 4/6) = 1/2
Manager
Joined: 06 Feb 2013
Posts: 59
Followers: 1

Kudos [?]: 39 [0], given: 35

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

14 Apr 2013, 18:19
Quote:
Can someone please explain what mistake i'm doing:

Total No. Of Possible Outcomes = 216

Outcomes where Joe scores 10 or less:

111 ---> 1
222 ---> 1
333 ---> 1
112 ---> 3, 113 ---> 3, 114 ---> 3, 115 ---> 3, 116 ---> 3,
221 ---> 3, 223 ---> 3, 224 ---> 3, 225 ---> 3, 226 ---> 3,
331 ---> 3, 332 ---> 3, 334 ---> 3,
441 ---> 3, 442 ---> 3,

Outcomes where Joe scores more than 10 = 216 - 48 = 168

Probability = 168/216 = 7/9

Quote:
You are missing some cases:
123 - 6 ways;
124 - 6 ways;
125 - 6 ways;
126 - 6 ways;
134 - 6 ways;
135 - 6 ways;
136 - 6 ways;
145 - 6 ways;
234 - 6 ways;
235 - 6 ways.

So, total of 60 scenarios were missing. Together with the 48 cases you counted we would have 48+60=108 ways to get the sum of 10 or less, so the probability is 1-108/216=1/2.

Hope it helps.

My brain works in terms of combinatorics better than the "fast" approach. I feel it is much more intuitive. It is tough to come with the fast one immediately - it is good (My hats off to Bunuel and others who explain, get, and use this approach) but in my view it is more of a "you either come up with it or nothing" and therefore, completely unreliable (and just remembering solution to this problem will not get you the right answer once the similar problem is tweaked in a slightly different manner, as is the case with most GMAT problems). Combinations is a more conceptual and a step by step way that will and should eventually get you there. Now, I was wondering can a combinations way be summarized for this problem? Or what can be the most concise comb way to approach this? Ironically, to me this is one of the instances when shorter does not necessarily mean better, not to offend anyone here.
_________________

There are times when I do not mind kudos...I do enjoy giving some for help

Intern
Joined: 29 Jul 2012
Posts: 29
GMAT 1: Q V
Followers: 0

Kudos [?]: 14 [0], given: 7

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

16 May 2013, 16:24
Any suggested reading to refresh on "Expected Values"? I've been reading the Manhattan GMAT books and don't recall this "Expected Value" technique.
Current Student
Joined: 29 May 2013
Posts: 24
Concentration: Marketing, Entrepreneurship
Schools: Molson '17 (M)
GMAT Date: 08-20-2014
GPA: 3.34
Followers: 0

Kudos [?]: 21 [0], given: 8

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

29 Mar 2014, 21:53
Bunuel wrote:
noboru 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?

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.

P=1/2.

@Bunuel Plz help me with this.
My method to solve was as below. Where did I go wrong?

Joe should score more than 10 .Possible cases {4,3,4}* 3!/2!(ways of arrangement) , {3,4,5}*3! , {4,4,4}*1 ,{5,5,5}*1 ,{6,6,6}*1
Probability of getting any of the above case is (1/6)^3 ..
Thus, [(1/6)^3 * 3!] + [(1/6)^3 * 3!/2!] + [(1/6)^3 * 1] + [(1/6)^3 * 1] +[(1/6)^3 * 1] = 1/18 ..

Plz help !!!
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93028 [0], given: 10541

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

30 Mar 2014, 10:39
sachinwar wrote:
Bunuel wrote:
noboru 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?

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.

P=1/2.

@Bunuel Plz help me with this.
My method to solve was as below. Where did I go wrong?

Joe should score more than 10 .Possible cases {4,3,4}* 3!/2!(ways of arrangement) , {3,4,5}*3! , {4,4,4}*1 ,{5,5,5}*1 ,{6,6,6}*1
Probability of getting any of the above case is (1/6)^3 ..
Thus, [(1/6)^3 * 3!] + [(1/6)^3 * 3!/2!] + [(1/6)^3 * 1] + [(1/6)^3 * 1] +[(1/6)^3 * 1] = 1/18 ..

Plz help !!!

You are missing several other cases:
{6, 6, 5}
{6, 6, 4}
{6, 6, 3}
{6, 6, 2}
...
_________________
Intern
Status: GMAT_BOOOOOOM.............. Failure is not an Option
Joined: 22 Jul 2013
Posts: 10
Location: India
Concentration: Strategy, General Management
GMAT 1: 510 Q38 V22
GPA: 3.5
WE: Information Technology (Consulting)
Followers: 0

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

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

27 Apr 2014, 10:46
Bunuel wrote:
mariyea wrote:
How were you able to come up with (1+2+3+4+5+6)? I understand that one outcome out of six occurs when Joe rolls the dice but the other part... a bit puzzling???

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

How did you come up with multiplied by 1 , 2 and so on ..

To Roll a die .. possible outcome for each will be 1/6 .. but you multiplied this by 1 then 2 then 3 and so on .. means 1/6*1 .what is multiplied by 1 denotes here ?

Thanks
_________________

Kudos will be appreciated if it was helpful.

Cheers!!!!
Sit Tight and Enjoy !!!!!!!

Senior Manager
Joined: 28 Apr 2014
Posts: 291
Followers: 1

Kudos [?]: 35 [0], given: 46

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

28 Apr 2014, 02:48
MacFauz wrote:
Bunuel wrote:
noboru 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?

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.

P=1/2.

Can someone please explain what mistake i'm doing:

Total No. Of Possible Outcomes = 216

Outcomes where Joe scores 10 or less:

111 ---> 1
222 ---> 1
333 ---> 1
112 ---> 3, 113 ---> 3, 114 ---> 3, 115 ---> 3, 116 ---> 3,
221 ---> 3, 223 ---> 3, 224 ---> 3, 225 ---> 3, 226 ---> 3,
331 ---> 3, 332 ---> 3, 334 ---> 3,
441 ---> 3, 442 ---> 3,

Outcomes where Joe scores more than 10 = 216 - 48 = 168

Probability = 168/216 = 7/9

I was also thinking on similar lines!!
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93028 [0], given: 10541

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

28 Apr 2014, 02:50
himanshujovi wrote:
MacFauz wrote:
Bunuel 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?

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.

P=1/2.

Can someone please explain what mistake i'm doing:

Total No. Of Possible Outcomes = 216

Outcomes where Joe scores 10 or less:

111 ---> 1
222 ---> 1
333 ---> 1
112 ---> 3, 113 ---> 3, 114 ---> 3, 115 ---> 3, 116 ---> 3,
221 ---> 3, 223 ---> 3, 224 ---> 3, 225 ---> 3, 226 ---> 3,
331 ---> 3, 332 ---> 3, 334 ---> 3,
441 ---> 3, 442 ---> 3,

Outcomes where Joe scores more than 10 = 216 - 48 = 168

Probability = 168/216 = 7/9

I was also thinking on similar lines!!

Flaw in this solution is explained here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-86407.html#p1101519

Hope it helps.
_________________
Intern
Joined: 29 Apr 2014
Posts: 3
Followers: 0

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

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

29 May 2014, 22:10
1
This post was
BOOKMARKED
Hi Bunuel, Can you please explain your answer of (1/6)^3 = 1/216 in case mary scores 17.
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93028 [1] , given: 10541

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

30 May 2014, 00:31
1
KUDOS
Expert's post
mrvora wrote:
Hi Bunuel, Can you please explain your answer of (1/6)^3 = 1/216 in case mary scores 17.

If Mary scores 17, then for Joe to outscore her should get the score of 18 (max possible with three dice: 6+6+6=18). The probability of getting 18, so the probability of getting 6 on each of the three dice, is 1/6*1/6*1/6 = (1/6)^3 = 1/216.

Does this make sense?
_________________
Senior Manager
Joined: 17 Sep 2013
Posts: 394
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Followers: 19

Kudos [?]: 270 [0], given: 139

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

08 Jun 2014, 21:19

For greater than 10- U need: 3 or more on 1 dice,4 or more on second,4 or more on third

Thus total no of possibilities are 4*3*3=36
Now this case is three times as likely because u can get a 3 or more on the other 2 dies as well..

36+36+36= 108
Total - 216

P- 1/2
_________________

Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..

Intern
Joined: 08 Jun 2014
Posts: 3
Followers: 0

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

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

08 Jun 2014, 22:38
If the problem were to be changed slightly and said that none of the first three digits repeat themselves, then the answer would become 3x2x2 = 12 going the the original method posted by Bunuel.
Is that correct?

Posted from my mobile device
Manager
Joined: 22 Feb 2009
Posts: 229
Followers: 5

Kudos [?]: 132 [0], given: 148

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

22 Jul 2014, 14:37
Bunuel wrote:
noboru 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?

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.

P=1/2.

I think it is the only way to solve the problem in less than 2 minutes. When I tried to calculate the total probability of 11, 12, ...18, it took me more than 10 minutes.
_________________

.........................................................................
+1 Kudos please, if you like my post

Intern
Joined: 01 Oct 2013
Posts: 1
Followers: 0

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

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

20 Aug 2014, 02:01
Bunuel wrote:
imania wrote:
would you love to see how attacked it?
if Joe is expected to outscore his friend, he should get these sums, 11,12,13...18
all possibilities are from 3 to 18
so : prob =8/16 equal to 1/2
PS. If you are wondering how I came to 3 as min because 1+1+1
and likewise 18 is max (6+6+6)

Unfortunately this approach is not right though for this particular case it gave a correct answer.

Consider this: if it were that Mary scored not 10 but 17 then Joe to outscore Mary should get only 18 and according to your approach as there are total of 16 scores possible then the probability of Joe getting 18 would be 1/16. But this is not correct, probability of 18 is (1/6)^3=1/216 not 1/16.

This is because not all scores from 3 to 18 have equal # of ways to occur: you can get 10 in many ways but 3 or 18 only in one way (3=1+1+1 and 18=6+6+6).

Hope it's clear.

========
Thanks Bunuel for the explanation however my 2 pence are as follows , The original explanation is perfectly sound, kindly correct me if I am wrong.

P of Joe getting more than 10 is 8 ways / 16 ways =1/2 (yay!)
8= 11, 12,....18; 16= 3,4....18;

Range of Summation of scores is 16 ; Max =18 and Min = 3 [ 3=1+1+1; 18=6+6+6 ] Prob of each =1/16.
lets consider an Outcome of 15 [ 3+6+6, 4+5+6............] ,

Question 1, why should be the permutations of numbers within the group be considered as the outcome is always 15 , per my example and as far as the verbiage of the question goes.... so aint these are identical groups of three since the outcome is 15 always. So each of the Outcomes i.e 16 of them [per question] shud have an equal probability i.e 1 /16.

Question 2, why are we considering the expected value when the question asks of Probability of Joe getting certain digits as outcomes.
Regards..
Intern
Joined: 24 Jan 2015
Posts: 1
Followers: 0

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

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

### Show Tags

07 Mar 2015, 00:31
What would be the approach to be followed if the question asked is to find the probability of Joe hitting more than 15??
Re: Mary and Joe are to throw three dice each. The score is the   [#permalink] 07 Mar 2015, 00:31

Go to page   Previous    1   2   3    Next  [ 44 posts ]

Similar topics Replies Last post
Similar
Topics:
2 Joey throws a dice 3 times and decides to invite 5 23 Apr 2015, 05:59
8 A and B in turns, throw a dice. If A gets a sum of 8 before 3 08 Aug 2013, 06:27
37 Mary and Joe are to throw three dice each. The score is the 13 22 Jan 2012, 02:07
23 Crowan throws 3 dice and records the product of the numbers 16 12 Feb 2011, 04:45
19 Mary and Joe are to throw three dice each. The score is the 5 23 Mar 2008, 12:23
Display posts from previous: Sort by