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# Mr. Smith throws three dice and scores 10 (for example 5+2+3

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Manager
Joined: 02 Jul 2004
Posts: 218
Mr. Smith throws three dice and scores 10 (for example 5+2+3 [#permalink]

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11 Aug 2004, 03:09
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Mr. Smith throws three dice and scores 10 (for example 5+2+3 or 4+4+2)
Mr. Brown throws dice next. What is the probability that Mr. Brown will outscore Mr. Smith?

(time - 1 minute)
Senior Manager
Joined: 22 Jun 2004
Posts: 392
Location: Bangalore, India

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11 Aug 2004, 05:32
I think it is 54/216.

I have taken all the permuations of the three DICE numbers exeeding 10 i.e. 11 to 18. Hope my counting is correct

OlegC wrote:
Mr. Smith throws three dice and scores 10 (for example 5+2+3 or 4+4+2)
Mr. Brown throws dice next. What is the probability that Mr. Brown will outscore Mr. Smith?

(time - 1 minute)

_________________

Awaiting response,

Thnx & Rgds,
Chandra

Manager
Joined: 02 Jul 2004
Posts: 218

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16 Aug 2004, 01:27
Anybody else wants to try this one? I posted this problem because I think it is very GMAT-like, looks tough but there is a shortcut that makes it easy
Senior Manager
Joined: 22 Jun 2004
Posts: 392
Location: Bangalore, India

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16 Aug 2004, 03:20

This is my mail and not guest's mail.

I wanna wonder how it can be solved in one minute. How many tricks do we have to remember??

Anonymous wrote:
mallelac wrote:
I think it is 114/216.

I have taken all the permuations of the three DICE numbers exeeding 10 i.e. 11 to 18. Hope my counting is correct

I edited this problem once. However, it did not get reflected. The first answer I gave was 54.

OlegC-awaiting the short-cut

OlegC wrote:
Mr. Smith throws three dice and scores 10 (for example 5+2+3 or 4+4+2)
Mr. Brown throws dice next. What is the probability that Mr. Brown will outscore Mr. Smith?

(time - 1 minute)

_________________

Awaiting response,

Thnx & Rgds,
Chandra

Manager
Joined: 02 Jul 2004
Posts: 218

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17 Aug 2004, 01:10
probability to score 18 = probability to score 3
probability to score 17 = probability to score 4
probability to score 16 = probability to score 5
............................
probability to score 11 = probability to score 10

therefore:
probability to score in the range 3-10 = probability to score in the range 11-18

therefore, probability to score in the range 11-18 = 1/2

Director
Joined: 20 Jul 2004
Posts: 592

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04 Sep 2004, 17:10
Wow! that really needs some out-of-the-box thinking.

1) Whenever there is pb problem that seems time consuming,
a. Look to see if it is solvable by "1 - pb(Event not happening)" rule.
OR
b. look for symmetry (like the one above) -- another example is "pb that A sits behind B = pb that A sits in front B = 1/2"

2) Just for the account,

For a two dice throw,
Pb of sum being 2 = Pb of sum being 12 = 1/36
Pb of sum being 3 = Pb of sum being 11 = 2/36
Pb of sum being 4 = Pb of sum being 10 = 3/36
Pb of sum being 5 = Pb of sum being 09 = 4/36
Pb of sum being 6 = Pb of sum being 08 = 5/36
Pb of sum being 7 = 6/36
So, here the half-half symmetry, as in three dice throw, does not work.
But, pb of sum 2-6 = pb of sum 8-12 = 15/36 = 5/12

For a three dice throw,
Pb of sum being 3 = Pb of sum being 18 = 1/216
Pb of sum being 4 = Pb of sum being 17 = 3/216
Pb of sum being 5 = Pb of sum being 16 = 6/216
Pb of sum being 6 = Pb of sum being 15 = 10/216
Pb of sum being 7 = Pb of sum being 14 = 15/216
Pb of sum being 8 = Pb of sum being 13 = 21/216
Pb of sum being 9 = Pb of sum being 12 = 25/216
Pb of sum being 10= Pb of sum being 11 = 27/216
So, here the half-half symmetry, pb of sum 3-10 = pb of sum 11-18 = 108/216 = 1/2
04 Sep 2004, 17:10
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