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

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

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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

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

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

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New post 24 Sep 2010, 11:17
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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)

Edit: NOTE THE ABOVE SOLUTION IS NOT CORRECT. SEE POST BELOW
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 24 Sep 2010, 22:20
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6
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.
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 25 Sep 2010, 04:37
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.


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

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New post 27 Sep 2010, 03:01
Is there any alternate approach to solve this problme?
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 27 Sep 2010, 14:55
Yes, but alternative approaches revolve around the same idea.

I can tell you how to reduce this problem to that of a multinomial expansion if you want, but the technique is beyond the scope of GMAT. The answer presented here is the simplest possible
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 05 Oct 2010, 22:05
How did you get the possible scores i.e 16 and so the probablity is 1/16

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

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New post 05 Oct 2010, 23:26
sanober1985 wrote:
How did you get the possible scores i.e 16 and so the probablity is 1/16

Bunuel wrote:
imania wrote:
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.


The possible scores are {3,4,5,...,18} which is 16 distinct numbers

But probability is NOT 1/16. The outcomes are not equally likely
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 06 Oct 2010, 02:46
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sanober1985 wrote:
How did you get the possible scores i.e 16 and so the probablity is 1/16

Bunuel wrote:
imania wrote:
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.


When you roll 3 dice you can have the following sums: 3 (min possible 1+1+1), 4, 5, 6, ...., 18 (max possible 6+6+6), so total of 16 possible sums. But as you can see in my previous post (the one you quote) the probability of these score are not equal, so it's not 1/16 for each.


devashish 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.


Amazing explanation, but is this a GMAT type question, if yes then I doubt I will ever be able to solve such questions in Real GMAT Time and space. It is too far fetched for me to even think I can crack such a question in normal finite time, forget GMAT Time !!!


Don't worry, you won't see such kind of question on GMAT.
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 31 Jan 2011, 16:23
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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???
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 31 Jan 2011, 16:36
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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.
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 03 Jul 2012, 23:05
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,

Adding everything up = 48

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

Probability = 168/216 = 7/9
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 04 Jul 2012, 00:11
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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,

Adding everything up = 48

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

Probability = 168/216 = 7/9


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

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New post 05 Jan 2013, 05:56
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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!!!

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

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New post 29 May 2014, 22:10
Hi Bunuel, Can you please explain your answer of (1/6)^3 = 1/216 in case mary scores 17.
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 30 May 2014, 00:31
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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?
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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

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New post 07 Mar 2015, 04:35
deepakaj wrote:
What would be the approach to be followed if the question asked is to find the probability of Joe hitting more than 15??


More than 15 means 16 (6-6-4, 6-5-5), 17 (6-6-5) and 18 (6-6-6):

6-6-4 can occur in 3!/2!=3 ways each having the probability of 1/6^3.
6-5-5 can occur in 3!/2!=3 ways each having the probability of 1/6^3.

6-6-5 can occur in 3!/2!=3 ways each having the probability of 1/6^3.

6-6-6 can occur in only 1 way with the probability of 1/6^3.

3*1/6^3 + 3*1/6^3 + 3*1/6^3 + 1/6^3.
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 28 Oct 2018, 04:39
Bunuel

The way I did this:

values greater than 10 are 11,12,13,14,15,16,17,18. i.e 8 values
so, 8/18 (18 is the maximum Value of all three dices), so 8/18 *4/4 = 32/64.

Is this correct ?
GMAT Club Bot
Re: Mary and Joe are to throw three dice each. The score is the &nbs [#permalink] 28 Oct 2018, 04:39

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