Official Solution: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. \(\frac{24}{64}\)
B. \(\frac{32}{64}\)
C. \(\frac{36}{64}\)
D. \(\frac{40}{64}\)
E. \(\frac{42}{64}\)
To outscore Mary, Joe has to score in the range of 11-18. The probability to score 3 is the same as the probability to score 18 (1-1-1 combination against 6-6-6, if 1-1-1 is on the tops of the dice the 6-6-6 is on the bottoms). By the same logic, the probability to score \(x\) is the same as the probability to score \(21 - x\). Therefore, the probability to score in the range 11-18 equals the probability to score in the range of 3-10. As 3-18 covers all possible outcomes the probability to score in the range 11-18 is \(\frac{1}{2}\) or \(\frac{32}{64}\).
Alternative Explanation Expected value of one die is \(\frac{1}{6}*(1+2+3+4+5+6)=3.5\).
Expected value of three dice is \(3*3.5=10.5\).
Mary scored 10 so the probability to get the sum more than 10 (11, 12, 13, ..., 18), or more than the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2 = 32/64.
That's because the probability distribution is symmetrical for this case:
The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);
The probability of getting the sum of 4 = the probability of getting the sum of 17;
The probability of getting the sum of 5 = the probability of getting the sum of 16;
...
The probability of getting the sum of 10 = the probability of getting the sum of 11;
Thus the probability of getting the sum from 3 to 10 = the probability of getting the sum from 11 to 18 = 1/2.
Answer: B
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