abhirup1711 wrote: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?
A. 24/64
B. 32/64
C. 36/64
D. 40/64
E. 42/64

Here's one approach.

First consider rolling 1 die.
If you were to roll a die millions of times, what would be the average value rolled?
Well, since each outcome (1,2,3,4,5 and 6) are all equally likely, the average of the outcomes will be 3.5 (since (1+2+3+4+5+6)/6 = 3.5)
Of course, it's impossible to roll 3.5, but notice that the 3.5 divides the outcomes into two parts. We have the numbers less than 3.5 (that is 1,2,3) and the numbers greater than 3.5 (that is 4,5,6).

Also notice that, in one roll, P(rolling less than 3.5) = 1/2, and P(rolling more than 3.5) = 1/2

Now consider rolling 3 dice.
If the average expected outcome is 3.5 when one die is rolled, the average expected sum will be 10.5 when three dice are rolled (since 3.5 + 3.5 + 3.5 = 10.5)

IMPORTANT: If 10.5 is the average expected sums, then half of all sum will be less than 10.5 and half will be greater than 10.5. In other words, P(sum is less than 10.5) = 1/2 and P(sum is greater than 10.5) = 1/2

The question asks us to find P(sum is greater than 10). This is the same as P(sum is greater than 10.5), which means this probability = [spoiler]1/2 = B[/spoiler]

Cheers,
Brent

mrvora wrote:Hi Brent,
I understood your explanation. However, what would be answer if Mary scores 17 in her attempt? Would it be 1/16. I think NO.

We want P(Joe's sum > 17)
Since 18 is the greatest sum of 3 dice, we want P(sum is 18)
P(sum is 18) = P(6 on 1st die AND 6 on 2nd die AND 6 on 3rd die)
= P(6 on 1st die) X P(6 on 2nd die) X P(6 on 3rd die)
= 1/6 X 1/6 X 1/6
= 1/216

Cheers,
Brent