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m08#21

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01 Jun 2009, 22:56
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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) $$\frac{24}{64}$$
(B) $$\frac{32}{64}$$
(C) $$\frac{36}{64}$$
(D) $$\frac{40}{64}$$
(E) $$\frac{42}{64}$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

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 $$\frac{32}{64}$$

i understand the logic above. however I was trying to do it differently and am no getting the same solution. where am i going wrong this my logic:

the minimum grouping of numbers that need to appear on the dice is 3,3,4. There one dice must be 3,4,5,6 another must be 3,4,5,6 and one must be 4,5,6.

so you get (4)(4)(3)/((6)(6)(6))

where am i going wrong. I think my logic flow is right but I am missing something.
therefore there is
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04 Jun 2009, 08:08
You might want to consider group with dice results smaller than 3 such as 1,3,6; 2,2,6; 2,3,5

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20 Aug 2009, 11:22
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Here is the long way of doing it for those who are too stupid to realize the simplicity of the problem (i.e. me)

Total combinations are 6x6x6 = 216.

Now, p(eleven or more) = 1- p(10 or less)

Count the number of ways in which you can get 10 or less

If you roll a 6 the first time, (xyz = roll x, roll y, roll z)
then if you roll 6,1, there are 3 ways to roll 10 or less (611,612,613) (notice the sum of xyz <= 10)

so
61..3 ways
62..2 ways
63..1 way
Total = 6

51..4 ways
52..3
53..2
54..1
Total 10

41..5
424
433
442
451
Total 15

316
325
334
343
352
361
Total 21

216
226
235
244
253
262
Total 26

116
126
136
145
154
163
Total 30

Total = 108/216 = 1/2
1-1/2 = 1/2

For rolling 2 1st and 1st, the pattern isnt as obviously so you have to manually count them.

I suck

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21 Aug 2009, 21:09
bipolarbear wrote:
Here is the long way of doing it for those who are too stupid to realize the simplicity of the problem (i.e. me)

Total combinations are 6x6x6 = 216.

Now, p(eleven or more) = 1- p(10 or less)

Count the number of ways in which you can get 10 or less

If you roll a 6 the first time, (xyz = roll x, roll y, roll z)
then if you roll 6,1, there are 3 ways to roll 10 or less (611,612,613) (notice the sum of xyz <= 10)

so
61..3 ways
62..2 ways
63..1 way
Total = 6

51..4 ways
52..3
53..2
54..1
Total 10

41..5
424
433
442
451
Total 15

316
325
334
343
352
361
Total 21

216
226
235
244
253
262
Total 26

116
126
136
145
154
163
Total 30

Total = 108/216 = 1/2
1-1/2 = 1/2

For rolling 2 1st and 1st, the pattern isnt as obviously so you have to manually count them.

I suck

Total combinations are 6x6x6 = 216., incorrect!
when the dice are threw, the sum of the points S=1+2+3 is the same as S=3+2+1
So the total combinations are less than 216.
Haven't figured out the rest yet.
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21 Aug 2009, 21:33
The total combination of the sum could be 56 possibilities.

and the ones >10 would have 28

Can't figure out the formulated solution so far.
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03 Feb 2010, 23:07
Economist wrote:
I cannot figure out how 21-x is generalized !!!

Quote:
The probability to score X is the same as the probability to score 21-X.

If X=3 (which is 1+1+1 and which is min value in our case) than the prob 21-X=18 (which is 6+6+6 and which is max in our case).
If X=4, 21-X=17 and so on.
Finally,
If X=18 (6+6+6 and max), 21-X=3 (1+1+1 and min)
We cannot take any other number:
if we take 22-X, then 22-3=19 (but we have only 6+6+6=18!)
by the same logic, 20-X=2 when X=18.

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04 Feb 2010, 05:21
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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.
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02 Jun 2010, 06:19
Question statement looks misleading to me. Where it is mentioned that Mary scored from 3 to 10?

I thought the question says that she scored 10:
145, 136
226, 235, 244, 253
334

So, these 7 combinations can be arranged itself in 3! ways. So, total ways = 42
Therefore, required probability = 42/216 = 7/36

Please tell me where I am missing
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20 Aug 2010, 08:09
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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

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20 Aug 2010, 22:17
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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.

I used a variation of this. The lowest sum is 3 (1+1+1), the highest is 18 (6+6+6). The mean of these is 10.5, meaning about half the digits sum should be above, half should be below. Roughly half should be greater than 10 then, so I just used elimination to see what what was close to half. This isn't a "clean" way to do it, but if you have two mins I think it's best to just guess and move on.
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22 Aug 2010, 05:06
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The easier approach for me is the following:
Because they dice 3 times, and count the sum of it, there are 16 possible results (from 3 to 18).
Mary scored 10. The question asks the probability to score HIGHER than Mary.
From 3 to 10 there are 8 possible results, and from 11 to 18 there are another 8 possible results.
It doesn't matter how many combination are there for every score. We can easily calculate anyways that there are a total of 216 (6x6x6) but that is irrelevant. We only need to know that scoring 3 and scoring 18 have the same number of combination, scoring 4 and 17 have the same number of combination again, etc.
So, to score more than 10, there are 8 possible results, which means 50% of the possible results (and combination), or 1/2, or 4/8, or 10/20, or 30/60, or 6/12, or 32/64, or whatever you want that is equivalent to 1/2 (or whatever answer choice you find that is equivalent).

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10 Sep 2010, 01:29
tt11234 wrote:
can anyone tell me what the probilities that this kind of questions will appear on the actual gmat?

I would say that the probability in this particular case equals to 1/2.
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01 Oct 2010, 07:26
In these kind of questions, one has to remember that we are choosing distinct nos. i.e. 8 distinct totals out of 16 total probabilities. Hence it would be a complete waste of time to manually list down each possibility. The questions clearly asks to "calculate" the required outcome, not to "arrange" the required outcomes.

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05 Jun 2011, 01:41
Pkit wrote:
tt11234 wrote:
can anyone tell me what the probilities that this kind of questions will appear on the actual gmat?

I would say that the probability in this particular case equals to 1/2.

the probablilty that we will be able to solve these type of questions in 2 mins is .00001
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25 Aug 2011, 05:52
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I solved it in < 30 secs. This is how I thought in my head.

Joe could win if 11-18, and loose if 3-10.

In other words, find number of ways x+y+z=A, such that A ={11, ... , 18}, and x,y,z={1, ... , 6}

Min(x+y+z)=1+1+1=3.
Max(x+y+z)=6+6+6=18.

P(3)=P(18), where P=Probability or number of ways, please think whatever you like. I like to think in terms of probability.
P(4)=P(17): why? one of the dices can take 2 values, while others constant. This is the key to understand the problem.

Extrapolate: This is the key to generalize the problem.
P(5)=P(16)
.
.
.
P(10)=P(11)

=> P(3)+P(4)+....+P(10) = P(18)+P(17)+....+P(11)
=> P(3, 4, ... , 10) = P(18, 17, ... , 11)
This is the key. After this think however you want to think. One of the ways:

=> P(3, ... , 18) = 1 = 2*P(18, ... , 11)
=> P(18, ... , 11) = 1/2 = 32/64

Of-course, you don't need all this math on paper - it was just to illustrate.

Tip: Try to find a pattern in combinatorial problems and try to generalize - problem becomes a piece of cake!

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25 Aug 2011, 06:04
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!

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26 Aug 2011, 14:02
As someone wrote before, is as simple as:

The probability to score between 3 and 10 is the same to score between 11 and 18, which is 1/2 <=> 32/64.

The tramp is to loose your time with long operations...

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26 Aug 2011, 14:20
CamiloCarellan wrote:
As someone wrote before, is as simple as:

The probability to score between 3 and 10 is the same to score between 11 and 18, which is 1/2 <=> 32/64.

The tramp is to loose your time with long operations...

Agreed, and that's what I also wrote too :p. But, I think it more important to understand the theory behind the solution - why P(3 ... 10) = P(18 ... 11). This is an easy question, but what if you were asked > 11 instead of > 10?

Answer = P(18, .... , 11) - P(11) = 1/2 - P(11) = 1/2 - P(see my next post)

Last edited by abhicoolmax on 26 Aug 2011, 15:27, edited 1 time in total.

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26 Aug 2011, 14:54
Well, so just go through the same process:

the probability of score between 3 and 11 is 9/16, the probability of score between 12 and 18 is 7/16. Correct?

Edit: I like your way, but to me this is faster, once you see the tramp

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26 Aug 2011, 15:25
CamiloCarellan wrote:
Well, so just go through the same process:

the probability of score between 3 and 11 is 9/16, the probability of score between 12 and 18 is 7/16. Correct?

Edit: I like your way, but to me this is faster, once you see the tramp

Sorry, that's WRONG

P(12 ... 18)
= P(11 ... 18) - P (11)
= 1/2 - [P(1,5,5)*3C1*2C2 + P(2,4,5)*3C1*2C1*1C1 + P(2,3,6)*3C1*2C1*1C1 + ................ )
= Definitely NOT equal to 7/16 :D

Please see my original 2 posts on why? Make sure you understand! You guys are making WRONG ASSUMPTION. Symmetry is across the 2 ends of the SUM, and NOT across the entire set.

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