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joe and moe throw 3 dice each. toe is the sum of points on all 3 dice. If toe of moe is 10, what is the probability that joe will have more toe than moe ?
a) 24/64
b) 32/64
c)36/64
d)40/64
e)42/64
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The possible ‘toe’ (sum) values for the 3 dices are 3 to 18.
Now, there is 1 way of getting toe = 3 ( 1,1,1)
3 ways of getting toe = 4 (1,1,2 : arranged in 3 ways for 3 dices)
6 ways of getting toe = 5 (1,2,2; 1,1,3 : arranged in 3 ways each)
:
3 ways of getting toe = 17 (6,6,5 : arranged in 3 ways)
1 way of getting toe = 18 (6,6,6)
If you notice the number of ways of getting toes is increasing in geometric progression. Similarly if you go in descending order from 18 , the number of ways again increase in geometric progression. And it seems the maximum ways of getting any sum is 18.
So the possible ways of getting toes = 3 to 18 are as follows:
In order to win. Joe needs to get the sum on 3 dices between 11 to 18 (both inclusive). And as you can see there are 50% ways of doing that.
But I think there should be a better way of doing it, because it took me 20 minutes to just figure out the pattern for the number of ways of getting all toe values.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.