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probability

by ritula » Tue Sep 23, 2008 3:06 am
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?

Ans: 32/64
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by sid1979 » Wed Oct 08, 2008 8:38 pm
Hi,

I am trying to solve the problem but not sure whether i am correct or not. Also it is not matching with your answer. But my way is very time consuming also. Basically the question is asking how many triplets are there whose sum is > 10. Then only Joe will win. So what i found is how many ways we can score 10. Then probabiltiy of scoring >10 will be 1-p(scoring <=10). So the following is the list of triplets which will add up to 10.
163,136,154,145,262,226,254,245,244,361,316,352,325,343,334,451,415,442,424,433,541,514,532,523,631,613,622. So it comes out to be 27 cases. So Probabibilty of Joe winning is 1-27/216 = 61/72. Note 216 is the total no of cases if we throw 3 dice. (6*6*6). Please let me know my approach. But in the real test if there is no other approach for this problem then better to guess & move ahead.

Sid.
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by chase4meg » Thu Oct 09, 2008 2:04 am
Can any of the moderators/instructors help?

Thanks!
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by Ian Stewart » Thu Oct 09, 2008 6:16 am
Note that an 'average' roll on one die is 3.5, so the average sum on three dice is 3*3.5 = 10.5. So the question is asking for the probability that you roll an 'above average' sum (11 or greater). The probability of an 'above average' sum when you roll dice is equal to the probability of a 'below average' sum; that is, the probability is 1/2.

I have no idea why they've written "32/64" instead of 1/2 as the correct answer; even if you do the calculation the long way (case by case), you won't get 64 in the denominator.
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by forgetneoiamtheone » Fri Oct 10, 2008 5:44 pm
sum can range from 3 to 18,
So no. of possibilties=16

number of possibilties greater than 10= (18-11)+1=8

so, prob=1/2.
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by Ian Stewart » Fri Oct 10, 2008 7:10 pm
forgetneoiamtheone wrote:sum can range from 3 to 18,
So no. of possibilties=16

number of possibilties greater than 10= (18-11)+1=8

so, prob=1/2.
Unfortunately things aren't so simple. The outcomes are not equally likely- you are much more likely to have a sum of 10 than to have a sum of 3, for example. By the logic above, I imagine you might conclude that the probability is 1/16 that you will get a sum of 3, but it's not; to get a sum of 3, you need to roll a 1 on each die, so the probability is (1/6)*(1/6)*(1/6) = 1/216.
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by forgetneoiamtheone » Sat Oct 11, 2008 4:04 pm
Thanks Ian for pointing out the flawed assumption.
The likelihood of each individual outcome didnt cross my mind.

Luckily, the symmetery of the distribution of outcomes around mean(the logic u used above) turned out to be saviour.

p[3]=p[18]
p[4]=p[17]
..and so on
therefore, p(3)+..+p(10)= p(11)+...p[18]=1/2
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by Cumulonimbus » Sun May 26, 2013 5:33 pm
Ian Stewart wrote:Note that an 'average' roll on one die is 3.5, so the average sum on three dice is 3*3.5 = 10.5. So the question is asking for the probability that you roll an 'above average' sum (11 or greater). The probability of an 'above average' sum when you roll dice is equal to the probability of a 'below average' sum; that is, the probability is 1/2.

I have no idea why they've written "32/64" instead of 1/2 as the correct answer; even if you do the calculation the long way (case by case), you won't get 64 in the denominator.
Hi Ian,

What will happen if the number of dies are changed to 4:
total possible sums - 4, 5,....24
Here mid term is 13,

I understand getting a P(sum of >=14) = P(sum<=12), is this correct?
What will be the prob of Sum = 13.

Also is this a GMAT worthy question?
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by info2 » Thu Oct 06, 2016 11:19 am
Ian Stewart wrote:Note that an 'average' roll on one die is 3.5, so the average sum on three dice is 3*3.5 = 10.5. So the question is asking for the probability that you roll an 'above average' sum (11 or greater). The probability of an 'above average' sum when you roll dice is equal to the probability of a 'below average' sum; that is, the probability is 1/2.

I have no idea why they've written "32/64" instead of 1/2 as the correct answer; even if you do the calculation the long way (case by case), you won't get 64 in the denominator.
Hello Ian

Can you give some tips on how to make events equally likely in such cases?

Thanks
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