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

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Bunuel
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink] 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] 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] 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] New post   30 May 2014, 01:31
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mrvora wrote:
Hi Bunuel, Can you please explain your answer of (1/6)^3 = 1/216 in case mary scores 17.


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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Re: Mary and Joe are to throw three dice each. The score is the [#permalink] 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] New post   07 Mar 2015, 05:35
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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] New post   06 May 2021, 05:19
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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Re: Mary and Joe are to throw three dice each. The score is the [#permalink] New post   06 May 2021, 10:25
Bunuel wrote:
bsaikrishna wrote:
Thanks for the reply.

What if the question is to find out the probability of the sum to be greater than 12?

How can we solve it using the same expected value approach?


Since we know that the probability of getting more than 10 (11, 12, ..., 18), is 1/2 then we should find the probability of getting 11 and 12 and then subtract these values from 1/2.

But you won't need this for the GMAT as there will be lengthy calculations involved:
11 can be obtained by combination of the following: 1-4-6, 1-5-5, 2-3-6, 2-4-5, 3-4-4, 3-3-5 --> 3!+3!/2!+3!+3!+3!/2!+3!/2!=27.
12 can be obtained by combination of the following: 1-5-6, 2-4-6, 2-5-5, 3-3-6, 3-4-5, 4-4-4 --> 3!+3!+3!/2!+3!/2!+3!+1=25.

P=1/2-(25+27)/6^3=7/27.

All combinations:
The sum of 3 - 1;
The sum of 4 - 3;
The sum of 5 - 6;
The sum of 6 - 10;
The sum of 7 - 15;
The sum of 8 - 21;
The sum of 9 - 25;
The sum of 10 - 27;
The sum of 11 - 27 (notice equals to the combinations of the sum of 10);
The sum of 12 - 25 (notice equals to the combinations of the sum of 9);
The sum of 13 - 21 (notice equals to the combinations of the sum of 8);
The sum of 14 - 15 (notice equals to the combinations of the sum of 7);
The sum of 15 - 10 (notice equals to the combinations of the sum of 6);
The sum of 16 - 6 (notice equals to the combinations of the sum of 5);
The sum of 17 - 3 (notice equals to the combinations of the sum of 4);
The sum of 18 - 1 (notice equals to the combinations of the sum of 3).
Total = 2*(1+3+6+10+15+21+25+27) = 216 = 6^3.

Hope it's clear.



11 can be obtained by combination of the following: 1-4-6, 1-5-5, 2-3-6, 2-4-5, 3-4-4, 3-3-5 --> 3!+3!/2!+3!+3!+3!/2!+3!/2!=27.
12 can be obtained by combination of the following: 1-5-6, 2-4-6, 2-5-5, 3-3-6, 3-4-5, 4-4-4 --> 3!+3!+3!/2!+3!/2!+3!+1=25.
can anyone explain this step please
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink] New post   31 Aug 2021, 06:58
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 [#permalink] New post   Updated on: 30 Nov 2023, 05:40
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testtakerstrategy wrote:
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



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

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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Re: Mary and Joe are to throw three dice each. The score is the [#permalink] New post   06 Oct 2021, 21:16
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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Re: Mary and Joe are to throw three dice each. The score is the [#permalink] New post   07 Oct 2021, 01:00
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

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Re: Mary and Joe are to throw three dice each. The score is the [#permalink] New post   23 Jan 2024, 05:57
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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