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

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?

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.

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.

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

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

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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