probability

ritula · Posted: Tue Sep 23, 2008 3:06 am Post subject: probability

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

sid1979 · Posted: Wed Oct 08, 2008 8:38 pm Post subject:

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.

chase4meg · Posted: Thu Oct 09, 2008 2:04 am Post subject:

Can any of the moderators/instructors help?

Thanks!

Ian Stewart · Posted: Thu Oct 09, 2008 6:16 am Post subject:

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.

forgetneoiamtheone · Posted: Fri Oct 10, 2008 5:44 pm Post subject:

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.

Ian Stewart · Posted: Fri Oct 10, 2008 7:10 pm Post subject:

forgetneoiamtheone · Posted: Sat Oct 11, 2008 4:04 pm Post subject:

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