Online Course Discussion Forum

math challenge II-A Combinatorics 9.30

 
 
Picture of Areteem Professor
Re: math challenge II-A 9.30
by Areteem Professor - Monday, 8 June 2020, 9:04 PM
 

Correct. Let $p_1$, $p_2$, and $p_3$ be the probabilities that the first, second, or third person win the game, respectively. Since one of them wins the game, the probability that someone wins is $1$. That is, $p_1 + p_2 + p_3 = 1$. On the other hand, labeling the people "first", "second", and "third" is rather arbitrary, since they all play at the same time by the exact same rules. Thus $p_1 = p_2 = p_3$. This argument is what we are referring to in the solution when we say "by symmetry, the answer should be $\dfrac{1}{3}$".