Question: The below algebraic solution is simple enough. But is there a way to "see" the answer using a clever trick or intuition? Given the algebraic solution, I feel like there should be. I just can't come up with anything.
My Attempt: If $Z \sim \mathrm{Binomial}(N,p)$ then $\mathrm{E}[Z] = Np$ and $\mathrm{Median}[Z] = \mathrm{Round}[Np]$. With $Y \sim \mathrm{Binomial}(3,0.5)$, $\mathrm{E}[Y] = 1.5$ and $\mathrm{Median}[Y] = 2$. If $Y=2$, then you win Game 2 below. So this suggests $p=0.5$ is a critical point of choice between Game 1 and 2.
Comment: My attempt is very rough. It relies on the Binomial density being symmetric about $p=0.5$. And it relies on an algebraic result for the expectation and median of Binomial random variable. I imagine there's a more intuitive approach out there.
Problem: You have a basketball hoop and someone says that you can play one of two games.
Game 1: You get one shot to make the hoop.
Game 2: You get three shots and you have to make two of three shots.
If $p$ is the probability of making a particular shot, for which values of $p$ should you pick one game or the other?
Algebraic Solution: Let $X \sim \mathrm{Binomial}(1,p)$ and $Y \sim \mathrm{Binomial}(3,p)$. Find values of $p$ for which $\Pr[X=1] \geq \Pr[Y \geq 2]$ which reduces to $p \geq -2p^3 + 3p^2$.
\begin{align*} \Pr[X=1] &\geq \Pr[Y \geq 2]\\ p &\geq -2p^3 + 3p^2\\ p &\leq 0.5 \end{align*}
And so you want to play Game 1 if $p \leq 0.5$