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m08#21

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Re: m08#21 [#permalink] New post 26 Aug 2011, 18:21
nightwing79 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) \frac{24}{64}
(B) \frac{32}{64}
(C) \frac{36}{64}
(D) \frac{40}{64}
(E) \frac{42}{64}

[Reveal] Spoiler: OA
B

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ok - I know the solution is long if I go the traditional route.
So - I switched to educated guessing and it was easier.
btw -Good question.
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Re: m08#21 [#permalink] New post 03 Apr 2012, 23:21
A very belated reply to an ancient problem. I kept this GMAT of the day problem in my inbox to see if I can come up with my own solution.

So, this is how I did it, I am not sure if it will be under 2 mins because I classify this as a tough problem so on the real test I wouldn't mind spending a little extra on this to give some lead away to the Prep.

Here goes:
I drew a 6x6 matrix with all possibile values on the 2nd and 3rd dice.
So, it would consist of 36 squares
1,1 1,2..... and so on
2,1 2,2..... and so on
finally
6,1 6,2..... and so on

So, lets vary the first dice to see a pattern of the number of possibilities greater than value of 10.
So, when the first dice is 1, you would have (1,6,6); (1, 6,5); (1,5,6) and so on. This represents a total of 3+2+1 = 6 possibilities (which physically represents the bottom right corner right triangle of the 6x6 matrix with a vertex at (6,6); (4,6); and (6,4).

Now continue with the first dice value of 2
total is 4+3+2+1 = 10

Now continue with the first dice value of 3
total is 5+4+3+2+1 = 15

You get the pattern?

so, the pattern here is 6, 10, 15, 21, 26, 30 for the 6 values of the first dice. The number of possibilities increases by one diagonal towards the left top corner vertex value of (1,). Notice: The last two additional diagonal drops in value because the 6x6 square starts to narrow down toward the top corner.

So, when we add all the values we get 108.

Total # of possibilities is 216. (6x6x6 which represents 6 possible values for each dice)

So, probability is 108/216 = 1/2 = 32/64.

Hope this helps!

This took me around 2:40 mins. I will take that on a GMAT for 700+ problem anyday!!

This method is both visual as well as uses the same concept used on the OG to solve one of the problems (I can't recall which one but I remember its a visual problem with all the dots).
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Re: m08#21 [#permalink] New post 21 Jun 2012, 03:18
To outscore Mary, Joe has to score in the range of 11-18. The probability to score 3 is the same as the probability to score 18 (1-1-1 combination against 6-6-6, if 1-1-1 is on the tops of the dice the 6-6-6 is on the bottoms). By the same logic, the probability to score x is the same as the probability to score 21 - x . Therefore, the probability to score in the range 11-18 equals the probability to score in the range of 3-10. As 3-18 covers all possible outcomes the probability to score in the range 11-18 is \frac{1}{2} or \frac{32}{64} .


In the above explanation, can someone please explain how Prob. to score x is the same as 21 -x ? Where do we get 21 from?
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Re: m08#21 [#permalink] New post 21 Jun 2012, 03:58
Expert's post
teal wrote:
To outscore Mary, Joe has to score in the range of 11-18. The probability to score 3 is the same as the probability to score 18 (1-1-1 combination against 6-6-6, if 1-1-1 is on the tops of the dice the 6-6-6 is on the bottoms). By the same logic, the probability to score x is the same as the probability to score 21 - x . Therefore, the probability to score in the range 11-18 equals the probability to score in the range of 3-10. As 3-18 covers all possible outcomes the probability to score in the range 11-18 is \frac{1}{2} or \frac{32}{64} .


In the above explanation, can someone please explain how Prob. to score x is the same as 21 -x ? Where do we get 21 from?


The lowest score is 3 (1+1+1) and the highest score is 18 (6+6+6).
The probability of getting 3 is the same as the probability of getting 18=21-3;
The probability of getting 4 is the same as the probability of getting 17=21-4;
...

As for the solution.

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.

Also discussed here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-86407.html and here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it's clear.
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Re: m08#21 [#permalink] New post 21 Jun 2012, 22:28
answer should be 1/2 because no. of possibilities of summations are 3-18 that is 16 sum possibilities, 3-10 makeup 8.
That is option B.
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Re: m08#21 [#permalink] New post 23 Jun 2012, 04:44
Bunuel wrote:
teal wrote:
To outscore Mary, Joe has to score in the range of 11-18. The probability to score 3 is the same as the probability to score 18 (1-1-1 combination against 6-6-6, if 1-1-1 is on the tops of the dice the 6-6-6 is on the bottoms). By the same logic, the probability to score x is the same as the probability to score 21 - x . Therefore, the probability to score in the range 11-18 equals the probability to score in the range of 3-10. As 3-18 covers all possible outcomes the probability to score in the range 11-18 is \frac{1}{2} or \frac{32}{64} .


In the above explanation, can someone please explain how Prob. to score x is the same as 21 -x ? Where do we get 21 from?


The lowest score is 3 (1+1+1) and the highest score is 18 (6+6+6).
The probability of getting 3 is the same as the probability of getting 18=21-3;
The probability of getting 4 is the same as the probability of getting 17=21-4;
...

As for the solution.

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.

Also discussed here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-86407.html and here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it's clear.


Hi Bunnel,

I didn't get these steps

Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.???

The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);????

and is he throwing dice once or thrice?

i am totally confused with the wordings of the question....can you please explain this one??
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Re: m08#21 [#permalink] New post 23 Jun 2012, 04:48
Expert's post
kotela wrote:
Hi Bunnel,

I didn't get these steps

Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.???

The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);????

and is he throwing dice once or thrice?

i am totally confused with the wordings of the question....can you please explain this one??


Please check the links provided above and also this post: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html#p1032531
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Re: m08#21 [#permalink] New post 23 Jun 2012, 05:32
I solved it, although it took me a while. One of the things that took my time is that i could not come to 64 in the denominator, since 3 attempts give us 216, so i did not even think that 32/64 is the same as 108/216. I think that on the real GMAT they will use the answer which comes from 108/216, e.g. 36/72, 18/36....or 1/2.
Bunuel don't you think so? or that is one of the tricks?
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Re: m08#21 [#permalink] New post 27 Aug 2012, 07:23
see. no ofDigit( sum) from 11-8===8
10-3===is also 8
as consequtive and equal.
Total p===1
so each 1/2.
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Re: m08#21 [#permalink] New post 05 Sep 2013, 02:34
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IMO 32/64 is the answer, Equal possibilities.
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Re: m08#21 [#permalink] New post 12 Aug 2014, 08:45
abhicoolmax wrote:
mercurialcc wrote:
i don't think there is a need to list all the combinations.

with 3 dices, the possible range of score is from 3 (3 dices of one) to 18 (3 dices of six). so there's a any number between 3 and 18 has a probability of:

1/(18-3+1) = 1/16


if a person has to roll 11 or over to win, the probability will be

P(rolling 11) + P(rolling 12) + ... + P(rolling 18) = 1/16 + ... + 1/16 = 8/16 = 1/2 = 32/64 ===> Ans. B

Alternatively:
all available scores is 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
the score can't be 1 or 2 because u have 3 dices

if mary rolled 10, to be higher than her, u can roll 11 and up
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18

bold occurrences / all occurrences = 8/16


Oops. You got the answer by fluke :) You made BIG mistake in your assumption (even you 2nd solution is INCORRECT! 8/16 is NOTHING!).... Be careful my friend. These are the careless mistakes that cost you bigtime.

P(3) is NOT equal to P(5). WHY?

Number of ways 3 can occur = {1,1,1} = 1
Probability of 3 occurring = 1/(6*6*6) = 1/216

Number of ways 4 can occur = {1,1,2}*3C1 = 3
Probability of 5 occurring = 3/(6*6*6) = 1/72

Number of ways 5 can occur = {1,1,3}*3C1 + {1,2,2}*3C2 = 6
Probability of 5 occurring = 6/(6*6*6) = 1/36

.....

The symmetry is across the 2 ends of the SUM and not across the entire set.

Please see my solution above to inculcate the right thought process!


Hello..
Could you please explain, how you arrived at following ? :
Number of ways 4 can occur = {1,1,2}*3C1 = 3
Number of ways 5 can occur = {1,1,3}*3C1 + {1,2,2}*3C2 = 6

I am sorry but I am very bad at probability and combinatorics !! :(
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m08#21 [#permalink] New post 12 Aug 2014, 19:36
Bunuel wrote:
This question was posted in PS forum as well. Here is my solution for it:

Expected value of a roll 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 have more then 10 (11, 12, 13, 14, 15, 16, 17, 18), or more then average, is the same as to have less than average=1/2.

P=1/2.

Think this approach is the easiest one.


Expected value is an amazing approach to solving this problem.

I find it strange that the fractions in the answer have no relation to the problem i.e. There is no way of solving this problem where you get 64 in the denominator. This is my first time seeing this. I wonder if the actual GMAT could have something like this. I usually inspect PS answer choices and try to get clues from them when I'm stumped. Until now.
m08#21   [#permalink] 12 Aug 2014, 19:36
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