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Reflections on Probability Problem#2

Here's another problem that I actually got correct, but just to compare my reasoning with the "view result" I clicked there and the explanation is so weird. Here's the problem and it's solution follows:

Two people, Alice and Bill, each roll a fair 20-sided die. What is the probability that Alice's roll is higher than Bill's roll?

Solution: The probability that they roll the same number is \(\frac{1}{20},\) so the probability that they do not roll the same number is \(\frac{19}{20}.\) By symmetry, if they do not roll the same number, each of Alice and Bill is equally likely to have rolled the higher number. The answer is thus \[\frac{1}{2} \cdot \frac{19}{20} = \frac{19}{40}.\]

What I was thinking: Bill either rolls 1, or 2, or,...., or 19. Then Alice beats Bill if she gets any of the next 19 numbers, or the next 18 numbers, or,....,or the last number 20. Then using a formula, Alice can beat Bill in \(\frac{19(19+1)}{2}\) ways which give same result.

My question: Alice and Bill having equally likely chances to roll higher means that we're like flipping a coin? How will I be able to spot this strategy in other problems? If I were to write this in words, does this mean that we are calculating P(Alice wins) = P(They do not roll same and Bill loses) = P(They do not roll same)P(Bill loses) = P(They do not roll same and Alice wins)?

Reflections: I guess I do understand the technique. My previous paragraph kind of Illustrates what is meant by "symmetry" and equal likelihood. So I'm making this note more like a reflexion kind of thing than a question.

Note by Jay B
4 months, 1 week ago

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