Probabilistic Race Results

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 12 \frac{1}{2} AND the probability that Bob finishes before Charlie is strictly greater than 12 \frac{1}{2} AND the probability that Charlie finishes before Alice is strictly greater that 12 \frac{1}{2} ?

Can we generalize this to having nn people?

#Contradiction #Probability

Note by Calvin Lin
7 years, 2 months ago

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Comments

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 - 7 years, 2 months ago

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That is a good interpretation, and gives you a possible construction almost immediately.

Calvin Lin Staff - 7 years, 2 months ago

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

1) Let's assume both P(AB),P(BA)P(\text{AB}), P(\text{BA}), are both strictly greater than 12\dfrac{1}{2}, then this means P(AB)+P(BA)>1P(\text{AB}) + P(\text{BA}) > 1.

However, we know that AB \text{AB }and BA\text{BA} are mutually exclusive events, and collectively exhaustive, therefore P(AB)+P(BA)=1P(\text{AB}) + P(\text{BA}) =1. This is a contradiction, therefore our initial assumption was wrong, which means, both P(AB)P(\text{AB}) and P(BA)P(\text{BA}) cannot be strictly greater than 12\dfrac{1}{2}

2) I am working on this right now....

Pranshu Gaba - 7 years, 2 months ago
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