Consider a race that is run by several people, including Alice, Bob and Charlie.

1) Is it possible that the probability that Alice finishes before Bob is strictly greater than \( \frac{1}{2} \) AND the probability that Bob finishes before Alice is strictly greater than \( \frac{1}{2} \)?

2) Is it possible that the probability that Alice finishes before Bob is strictly greater than \( \frac{1}{2} \) AND the probability that Bob finishes before Charlie is strictly greater than \( \frac{1}{2} \) AND the probability that Charlie finishes before Alice is strictly greater that \( \frac{1}{2} \)?

Can we generalize this to having \(n\) people?

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TopNewestIf Alice finishes before Charlie, and Charlie finishes before Bob, i will be denoting it as \(\textrm{ACB}\). I will be assuming that no two people can finish together.

1) Let's assume both \(P(\text{AB}), P(\text{BA})\), are both strictly greater than \(\dfrac{1}{2}\), then this means \(P(\text{AB}) + P(\text{BA}) > 1\).

However, we know that \(\text{AB }\)and \(\text{BA}\) are mutually exclusive events, and collectively exhaustive, therefore \(P(\text{AB}) + P(\text{BA}) =1\). This is a contradiction, therefore our initial assumption was wrong, which means, both \(P(\text{AB})\) and \(P(\text{BA})\) cannot be strictly greater than \(\dfrac{1}{2}\)

2) I am working on this right now.... – Pranshu Gaba · 3 years, 1 month ago

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I suppose this problem comes from nontransitive dice, although by making it a scenario of runners my thoughts of probability are suddenly shattered. Still figuring out how to formalize the probabilities as runners. – Ivan Koswara · 3 years, 1 month ago

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– Calvin Lin Staff · 3 years, 1 month ago

That is a good interpretation, and gives you a possible construction almost immediately.Log in to reply