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

would you love to see how attacked it?
if Joe is expected to outscore his friend, he should get these sums, 11,12,13...18
all possibilities are from 3 to 18
so : prob =8/16 equal to 1/2
PS. If you are wondering how I came to 3 as min because 1+1+1
and likewise 18 is max (6+6+6)

Fantastic explanation!

How did you get the possible scores i.e 16 and so the probablity is 1/16

The possible scores are {3,4,5,...,18} which is 16 distinct numbers

But probability is NOT 1/16. The outcomes are not equally likely

When you roll 3 dice you can have the following sums: 3 (min possible 1+1+1), 4, 5, 6, ...., 18 (max possible 6+6+6), so total of 16 possible sums. But as you can see in my previous post (the one you quote) the probability of these score are not equal, so it's not 1/16 for each.




Don't worry, you won't see such kind of question on GMAT.

How were you able to come up with (1+2+3+4+5+6)? I understand that one outcome out of six occurs when Joe rolls the dice but the other part... a bit puzzling???

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

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.

P(3) + P(4) + P(5)..........+P(10) = P(11) + P(12) + P(13).......+P(18)

and P(3) + P(4) + ....... + P(18) = 1

Therefore P(11) + P(12) + P(13).......+P(18) = 1/2 or 32/64

Option (B)

Can someone please explain what mistake i'm doing:

Total No. Of Possible Outcomes = 216

Outcomes where Joe scores 10 or less:

111 ---> 1
222 ---> 1
333 ---> 1
112 ---> 3, 113 ---> 3, 114 ---> 3, 115 ---> 3, 116 ---> 3,
221 ---> 3, 223 ---> 3, 224 ---> 3, 225 ---> 3, 226 ---> 3,
331 ---> 3, 332 ---> 3, 334 ---> 3,
441 ---> 3, 442 ---> 3,

Adding everything up = 48

Outcomes where Joe scores more than 10 = 216 - 48 = 168

Probability = 168/216 = 7/9

You are missing some cases:
123 - 6 ways;
124 - 6 ways;
125 - 6 ways;
126 - 6 ways;
134 - 6 ways;
135 - 6 ways;
136 - 6 ways;
145 - 6 ways;
234 - 6 ways;
235 - 6 ways.

So, total of 60 scenarios were missing. Together with the 48 cases you counted we would have 48+60=108 ways to get the sum of 10 or less, so the probability is 1-108/216=1/2.

Hope it helps.

Bunuel I know this can be a lengthy,but can you show how the different combinations of 9 is equal to 25 I am getting 28
also the different combinations of 10 I am getting 36 well as it should be 27

the combinations from 3 to 8 matches with yours but for 9 and 10 I am getting a different answer.

Thank you

Responding to a pm:

The method you are using is not correct. It is fine for the sum till 8. It fails for 9, 10, 11 and 12.
If you enumerate, you will get the same numbers as Bunuel.

First let me point out that when you decide to use a particular method, you should fully understand the method. First go through this post to understand why you can use 7C5 or 7C2 to get a sum of 8 (and to get the smaller sums too).



Notice how you divide n identical objects among m distinct groups. Let’s take the example of a sum of 7. You have to divide 7 among 3 dice such that each die must have at least 1 (no die face can show 0). First step is to take 3 out of the 7 and give one each to the three dice. Now you have 4 left to distribute among 3 distinct groups such that it is possible that some groups may get none of the four. Think of partitioning 4 in 3 groups. This can be done in (4+2)!/4!*2! = 6C2 ways (check out the given link if you do not understand this)

This is how you obtain 6C2 (which is the same as 6C4) for the sum of 7.

The concept works perfectly till the sum of 8. Thereafter it fails. Now that you know why this method works for some values, can you guess why it fails for others?

maybe something to do with the minimum and maximum value of a dice? Till 8 we can have minimum 1 and maximum 6 for a dice , 1 1 6 * \(\frac {3!}{2}\) , 3 2 3 *\(\frac {3!}{2}\) ,
5 2 1 * 3!,2 2 4 *\(\frac {3!}{2}\) , 1 3 4*3! = 3 +3 + 6 + 3 + 6 = 21

but for 9 we have only 1 6 2 * 3! , 1 3 5 *3!, 1 4 4 *\(\frac {3!}{2}\), 2 2 5 *\(\frac {3!}{2}\) ,
3 3 3 , 2 4 3 * 3! = 6+6+3+3+1+6 = 25

but when we use 8C6 or 8C2 for the sum of 9 , we are getting 3 extra cases, what are those extra cases ? If we can find those cases I think we can find the answer to your question.

I guess you are on the right track here. In case of 9, you give 1 to each of the 3 dice and you are left with 6. Now when you try to split 6 among the three groups, you will have 3 cases which look like this:
6, 0, 0
0, 6, 0
0, 0, 6

What you are saying here is that first die shows 7, second shows 1 and third shows 1. This case doesn't work, does it?
All these 3 cases don't work and you need to remove them.

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.