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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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New post 05 Nov 2009, 14:41
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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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Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 05 Nov 2009, 14:59
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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

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.

Answer: B.
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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, 12:17
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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, 23:20
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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, 05:37
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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, 04:01
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, 15:55
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, 23:05
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 06 Oct 2010, 00:26
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, 03:46
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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, 17:23
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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 22 Jan 2012, 04:32
Thanks for the reply.

What if the question is to find out the probability of the sum to be greater than 12?

How can we solve it using the same expected value approach?
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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 22 Jan 2012, 05:30
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bsaikrishna wrote:
Thanks for the reply.

What if the question is to find out the probability of the sum to be greater than 12?

How can we solve it using the same expected value approach?


Since we know that the probability of getting more than 10 (11, 12, ..., 18), is 1/2 then we should find the probability of getting 11 and 12 and then subtract these values from 1/2.

But you won't need this for the GMAT as there will be lengthy calculations involved:
11 can be obtained by combination of the following: 1-4-6, 1-5-5, 2-3-6, 2-4-5, 3-4-4, 3-3-5 --> 3!+3!/2!+3!+3!+3!/2!+3!/2!=27.
12 can be obtained by combination of the following: 1-5-6, 2-4-6, 2-5-5, 3-3-6, 3-4-5, 4-4-4 --> 3!+3!+3!/2!+3!/2!+3!+1=25.

P=1/2-(25+27)/6^3=7/27.

All combinations:
The sum of 3 - 1;
The sum of 4 - 3;
The sum of 5 - 6;
The sum of 6 - 10;
The sum of 7 - 15;
The sum of 8 - 21;
The sum of 9 - 25;
The sum of 10 - 27;
The sum of 11 - 27 (notice equals to the combinations of the sum of 10);
The sum of 12 - 25 (notice equals to the combinations of the sum of 9);
The sum of 13 - 21 (notice equals to the combinations of the sum of 8);
The sum of 14 - 15 (notice equals to the combinations of the sum of 7);
The sum of 15 - 10 (notice equals to the combinations of the sum of 6);
The sum of 16 - 6 (notice equals to the combinations of the sum of 5);
The sum of 17 - 3 (notice equals to the combinations of the sum of 4);
The sum of 18 - 1 (notice equals to the combinations of the sum of 3).
Total = 2*(1+3+6+10+15+21+25+27) = 216 = 6^3.

Hope it's clear.
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Re: Mary and Joe are to throw three dice each  [#permalink]

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New post 06 May 2012, 03:04
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P(3) + P(4) + P(5)..........+P(10) = P(11) + P(12) + P(13).......+P(18)

and P(3) + P(4) + ....... + P(18) = 1

Therefore P(11) + P(12) + P(13).......+P(18) = 1/2 or 32/64

Option (B)
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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, 00:05
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, 01:11
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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 23 Sep 2012, 08:17
Bunuel wrote:
bsaikrishna wrote:
Thanks for the reply.

What if the question is to find out the probability of the sum to be greater than 12?

How can we solve it using the same expected value approach?


Since we know that the probability of getting more than 10 (11, 12, ..., 18), is 1/2 then we should find the probability of getting 11 and 12 and then subtract these values from 1/2.

But you won't need this for the GMAT as there will be lengthy calculations involved:
11 can be obtained by combination of the following: 1-4-6, 1-5-5, 2-3-6, 2-4-5, 3-4-4, 3-3-5 --> 3!+3!/2!+3!+3!+3!/2!+3!/2!=27.
12 can be obtained by combination of the following: 1-5-6, 2-4-6, 2-5-5, 3-3-6, 3-4-5, 4-4-4 --> 3!+3!+3!/2!+3!/2!+3!+1=25.

P=1/2-(25+27)/6^3=7/27.

All combinations:
The sum of 3 - 1;
The sum of 4 - 3;
The sum of 5 - 6;
The sum of 6 - 10;
The sum of 7 - 15;
The sum of 8 - 21;
The sum of 9 - 25;
The sum of 10 - 27;
The sum of 11 - 27 (notice equals to the combinations of the sum of 10);
The sum of 12 - 25 (notice equals to the combinations of the sum of 9);
The sum of 13 - 21 (notice equals to the combinations of the sum of 8);
The sum of 14 - 15 (notice equals to the combinations of the sum of 7);
The sum of 15 - 10 (notice equals to the combinations of the sum of 6);
The sum of 16 - 6 (notice equals to the combinations of the sum of 5);
The sum of 17 - 3 (notice equals to the combinations of the sum of 4);
The sum of 18 - 1 (notice equals to the combinations of the sum of 3).
Total = 2*(1+3+6+10+15+21+25+27) = 216 = 6^3.

Hope it's clear.


Bunuel I know this can be a lengthy,but can you show how the different combinations of 9 is equal to 25 I am getting 28
also the different combinations of 10 I am getting 36 well as it should be 27

the combinations from 3 to 8 matches with yours but for 9 and 10 I am getting a different answer.

Thank you
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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 Sep 2012, 22:52
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stne wrote:
Bunuel I know this can be a lengthy,but can you show how the different combinations of 9 is equal to 25 I am getting 28
also the different combinations of 10 I am getting 36 well as it should be 27

the combinations from 3 to 8 matches with yours but for 9 and 10 I am getting a different answer.

Thank you


Responding to a pm:

The method you are using is not correct. It is fine for the sum till 8. It fails for 9, 10, 11 and 12.
If you enumerate, you will get the same numbers as Bunuel.

First let me point out that when you decide to use a particular method, you should fully understand the method. First go through this post to understand why you can use 7C5 or 7C2 to get a sum of 8 (and to get the smaller sums too).

http://www.veritasprep.com/blog/2011/12 ... 93-part-1/

Focus on method II of question 2.

Notice how you divide n identical objects among m distinct groups. Let’s take the example of a sum of 7. You have to divide 7 among 3 dice such that each die must have at least 1 (no die face can show 0). First step is to take 3 out of the 7 and give one each to the three dice. Now you have 4 left to distribute among 3 distinct groups such that it is possible that some groups may get none of the four. Think of partitioning 4 in 3 groups. This can be done in (4+2)!/4!*2! = 6C2 ways (check out the given link if you do not understand this)

This is how you obtain 6C2 (which is the same as 6C4) for the sum of 7.

The concept works perfectly till the sum of 8. Thereafter it fails. Now that you know why this method works for some values, can you guess why it fails for others?
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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 01 Oct 2012, 09:44
maybe something to do with the minimum and maximum value of a dice? Till 8 we can have minimum 1 and maximum 6 for a dice , 1 1 6 * \(\frac {3!}{2}\) , 3 2 3 *\(\frac {3!}{2}\) ,
5 2 1 * 3!,2 2 4 *\(\frac {3!}{2}\) , 1 3 4*3! = 3 +3 + 6 + 3 + 6 = 21

but for 9 we have only 1 6 2 * 3! , 1 3 5 *3!, 1 4 4 *\(\frac {3!}{2}\), 2 2 5 *\(\frac {3!}{2}\) ,
3 3 3 , 2 4 3 * 3! = 6+6+3+3+1+6 = 25

but when we use 8C6 or 8C2 for the sum of 9 , we are getting 3 extra cases, what are those extra cases ? If we can find those cases I think we can find the answer to your question.
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