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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Mary and Joe are to throw three dice each. The score is the

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Retired Moderator Joined: 02 Sep 2010
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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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.

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

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Bunuel wrote:
bsaikrishna wrote:

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

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1
Joe's possible outcomes range from 3 (if he throws all 1) to 18 (if he throws all 6).

Each option, from 3-18, is equally probable if we assume a fair dice. There are 8 outcomes (3, 4, 5, 6, 7, 8, 9, 10) where Joe gets 10 or less, thereby underscoring or tying Mary. There are 8 outcomes (11, 12... 18) in which Joe outscores Mary.

Because there are 8 ways for Joe to outscore, and 8 ways for Joe to underscore or tie and they are all equally probable, there is a 50% chance Joe will outscore Mary.
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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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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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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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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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?

A. 24/64
B. 32/64
C. 36/64
D. 40/64
E. 42/64

To outscore Mary, joe has to score 11 or more.
first die-6
second die-4
third die- 1
if he gets 6 on the first die then he has no more options on first throw but in second die he may get 4 or 5 or 6 so total 3 outcomes
on the third die he has 6 options(1,2,3,4,5,6).
so total outcomes= 1*3*6=18
it can be done in !3 ways
so total outcome= !3*18
Desired outcome= !3*18/6*6*6
= 1/2
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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Bunuel Hello,
How exactly are we expected to figure out that, in this case, the distribution of the scores of three dices is symmetrical about the mean without having to compute it?
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Mary and Joe are to throw three dice each. The score is the  [#permalink]

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Bunuel

Can you please explain the significance of the following two statements?

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.

What is the need to find the expected value of one die? And how did we come up with (1+2+3+4+5+6)?
I'm trying hard to understand when we should multiply the probability of an outcome with its value?
For each dice, 1 through 6 are the possible outcomes. And you have multiplied each outcome with its probability of occurrence i.e. 1/6 Mary and Joe are to throw three dice each. The score is the   [#permalink] 19 Jun 2020, 12:36

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