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805+ Level|   Probability|      
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wisc4llin
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Mary and Joe are to throw three dice each. The score is the 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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Mary and Joe are to throw three dice each. The score is the 04 Oct 2012, 03:10
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wisc4llin
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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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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Mary and Joe are to throw three dice each. The score is the 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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Mary and Joe are to throw three dice each. The score is the 29 May 2014, 23:10
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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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Mary and Joe are to throw three dice each. The score is the 30 May 2014, 01:31
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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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If Mary scores 17, then for Joe to outscore her should get the score of 18 (max possible with three dice: 6+6+6=18). The probability of getting 18, so the probability of getting 6 on each of the three dice, is 1/6*1/6*1/6 = (1/6)^3 = 1/216.

Does this make sense?
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Mary and Joe are to throw three dice each. The score is the 07 Mar 2015, 01:31
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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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Mary and Joe are to throw three dice each. The score is the 07 Mar 2015, 05:35
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deepakaj
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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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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Mary and Joe are to throw three dice each. The score is the 06 May 2021, 05:19
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This is a good question. Counting the combinations soon becomes too tedious. Thats when I thought the probability distribution will be tree-like around the median, which happens to be precisely between 10 and 11. However, stupid mistake from me first when I took 18 as the highest value and 1 as the lowest value, getting a median of 9,5.
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Mary and Joe are to throw three dice each. The score is the 31 Aug 2021, 06:58
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VeritasKarishma Bunuel --- do you have similar problems you would recommend looking at? Where we have a varying standard distribution over the set of all possible outcomes? I am having trouble finding such questions for combinatorics/probability
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Mary and Joe are to throw three dice each. The score is the Updated on: 30 Nov 2023, 05:40
Originally posted by KarishmaB on 31 Aug 2021, 21:35.
Last edited by KarishmaB on 30 Nov 2023, 05:40, edited 1 time in total.
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testtakerstrategy
VeritasKarishma Bunuel --- do you have similar problems you would recommend looking at? Where we have a varying standard distribution over the set of all possible outcomes? I am having trouble finding such questions for combinatorics/probability
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Check these videos for basics of Combinatorics:
Video on Permutations: https://youtu.be/LFnLKx06EMU
Video on Combinations: https://youtu.be/tUPJhcUxllQ
Video on Probability: https://youtu.be/0BCqnD2r-kY
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Mary and Joe are to throw three dice each. The score is the 06 Oct 2021, 21:16
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My approach: Denominator= 6*6*6
Numerator=>(Find a pattern for above 10),i.e {1,4,6},{1,5,6},{1,6,6}. similarly for2,3,4,5,6. This position can be taken in 6 formats. Hence, the ans is (3*6*6)/(6*6*6) =1/2
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Mary and Joe are to throw three dice each. The score is the 07 Oct 2021, 01:00
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Took me most of a minute to see this, but then it became easier

To calculate the all the different SUMS and ways to obtain them that would beat Mary when 3 dice are rolled would take an astronomical amount of time. Hence, there has to be a shorter method.


The minimum sum that can be rolled is all 1s ——-> 1 + 1 + 1 = 3

There is 1 way to roll this

The maximum sum that can be rolled is all 6s ———> 6 + 6 + 6 = 18

There is 1 way to roll this


Then the next sum above the minimum is 4 ——> would need to roll: 1 - 1 - 2

There is 3!/2! = 3 ways to do this


The next sum less than the max is 17 ————> would need to roll: 6 - 6 - 5

There is 3!/2! = 3 ways to do this


On it goes, each consecutive SUM going up from the minimum and coming down from the maximum will have the same number of possibilities

The following sums will have an equal number of ways to obtain the sum via 3 dice:

Sum of 5 and Sum of 16

Sum of 6 and Sum of 15

Sum of 7 and Sum of 14

Sum of 8 and Sum of 13

Sum of 9 and sum of 12

Sum of 10 and Sum of 11


Based on the Symmetry, it will be just as likely to roll:

P(SUM is 10 or less) = P(Sum is 11 or more)


In order to roll a higher sum than Mary, he must roll an 11 or more.

The probability that he will do so is (1/2) or


32/64

Posted from my mobile device
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Mary and Joe are to throw three dice each. The score is the 15 Sep 2025, 07:11
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What if Mary scores any number like 9, 11, 12... not the exact half. How will we solve that?
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Official Solution:

Mary and Joe each roll three dice. The total score for each person is the sum of the numbers on their three dice. If Mary scores 10, what is the probability that Joe will have a higher score in his attempt?

A. \(\frac{24}{64}\)
B. \(\frac{32}{64}\)
C. \(\frac{36}{64}\)
D. \(\frac{40}{64}\)
E. \(\frac{42}{64}\)


To outscore Mary, Joe must achieve a score in the range of 11-18. The probability of scoring 3 is the same as the probability of scoring 18 (the 1-1-1 combination is opposite the 6-6-6 combination when considering the tops and bottoms of the dice). Similarly, the probability of scoring \(x\) is the same as the probability of scoring \(21 - x\). Consequently, the probability of scoring within the 11-18 range is equal to the probability of scoring within the 3-10 range. Since the range of 3-18 encompasses all possible outcomes, the probability of scoring within the 11-18 range is \(\frac{1}{2}\).

Alternative explanation

The expected value when rolling a single die is \(\frac{1}{6}*(1+2+3+4+5+6)=3.5\).

Therefore, the expected value when rolling three dice is \(3*3.5=10.5\).

Mary scored 10, so the probability of getting a sum greater than 10 (11, 12, 13, ..., 18), or greater than the average, is the same as getting a sum less than the average (10, 9, 8, ..., 3). Since these two events are complementary (adding to 1) and symmetrical, the probability of each is 1/2:

The probability of getting a sum of 3 (the minimum possible sum) = the probability of getting a sum of 18 (the maximum possible sum);

The probability of getting a sum of 4 = the probability of getting a sum of 17;

The probability of getting a sum of 5 = the probability of getting a sum of 16;

...

The probability of getting a sum of 10 = the probability of getting a sum of 11.

Therefore, the probability distribution is symmetrical, and thus the probability of getting a sum from 3 to 10 = the probability of getting a sum from 11 to 18 = 1/2.


Answer: B
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Mary and Joe are to throw three dice each. The score is the 15 Sep 2025, 07:21
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Saun2511
What if Mary scores any number like 9, 11, 12... not the exact half. How will we solve that?

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Then you'd need to count cases. Check this post: https://gmatclub.com/forum/mary-and-joe ... l#p1032531
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