First, a shoutout to February’s winners, Gian Luca Degani and Daniel Jennings. Gian Luca used the approach outlined below to solve the main puzzle. For the bonus, Daniel used a probability tree, evaluated Jo’s win probability using a large number of terms, and binary searched over Jo’s accuracy until the probability was very close to $50\%$. While our approach sidesteps the heavy computation, approaches like Daniel’s are very useful in real-world problems with messy structure.

Main puzzle

Jo and Flo hit their targets with probabilities $p_j$ and $p_f$.

If Jo wins, it can happen in one of four ways:

• Jo hits Flo on the first shot, or
• Jo misses Flo, Flo misses her, then Jo hits Flo, or
• Jo and Flo miss each other twice, then Jo hits Flo, or
• Jo and Flo miss each other three times, then Jo hits Flo using her last snowball.

The probability that Jo wins on the first throw is just $p_j$. But if she misses, Flo also has to miss if Jo is going to have another chance. The chance of Jo and Flo both missing each other is ${(1-p_j)(1-p_f).}$

So, the total probability of Jo winning before she runs out of snowballs is

$$ \begin{align*}P(\text{Jo wins})= p_j &+ \overbrace{(1-p_j)(1-p_f)p_j}^\text{Jo wins on second throw} \\&+ \overbrace{(1-p_j)^2(1-p_f)^2p_j}^\text{Jo wins on third throw}\\ &+\overbrace{(1-p_j)^3(1-p_f)^3p_j}^\text{Jo wins on fourth throw}\end{align*} $$

which, for the accuracies given, is about $37.9\%$.

Bonus

If Jo and Flo have an endless supply of snowballs, there is no limit to the number of rounds.

Like before, the quickest way for Jo to win is to hit Flo on the first throw.

But if Jo and Flo both miss with their first snowball, then Jo finds herself back in the same situation: throwing first with an endless stack of snowballs. This means that the probability of winning after two misses is the same as the probability of winning before the two misses:

$$ \begin{align*}P(\text{Jo wins}) &=P(\text{Jo wins immediately}) + P(\text{Jo and Flo both miss})P(\text{Jo wins}) \\ &= p_j + (1-p_j)(1-p_f)P(\text{Jo wins}).\end{align*} $$

Isolating for $P(\text{Jo wins})$, we get

$$ P(\text{Jo wins})=\frac{p_j}{1-(1-p_j)(1-p_f)}. $$