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

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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, 22:12
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stne wrote:
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.


I guess you are on the right track here. In case of 9, you give 1 to each of the 3 dice and you are left with 6. Now when you try to split 6 among the three groups, you will have 3 cases which look like this:
6, 0, 0
0, 6, 0
0, 0, 6

What you are saying here is that first die shows 7, second shows 1 and third shows 1. This case doesn't work, does it?
All these 3 cases don't work and you need to remove them.
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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 Oct 2012, 14:03
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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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New post 04 Oct 2012, 03:10
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wisc4llin wrote:
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.


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 Jan 2013, 06:56
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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!!!

Kudos for the Solution plz.... :)
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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, 23:10
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, 01:31
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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, 01:31
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, 05:35
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 22 Aug 2019, 22:13
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: PS: Probability  [#permalink]

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Re: PS: Probability   [#permalink] 20 Sep 2019, 05:17

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