Fountain and Snowballs

Snowy application of probability theory

Introduction

It was the end of exam period during winter holidays. There was a thin layer of snow everywhere. While walking through the city center I noticed a fountain covered with a thin layer of snow. I casually started throwing snowballs into the fountain. The snowballs were passing through the snow layer and creating holes. Then I thought of a game: what if I throw multiple snowballs to clear out holes, and then take one final throw to score into those previously made holes? Afterwards, I thought to myself: What is the probability to score into a hole with that final snowball?, and I realized this can be calculated using basic statistics!

Information

• area of the fountain
  • \(S_{f}=1,000\ \text{cm}^2\)
• average area of the hole created from a snowball
  • \(S_{E(s)}=10\ \text{cm}^2\)
• total number of snowballs thrown (\(x-1\) setup throws + \(1\) final target throw)
  • \(X\)
• probability of success on the final throw
  • \(P(S_x)\)
• probability of failure on the final throw
  • \(P(F_x)\)

Assumptions

• The person who throws the snowballs is throwing them completely randomly
• All throws will always be scored somewhere within the area of the fountain
• The first \(x-1\) setup throws each create a brand new, unique hole in the snow layer that does not overlap with any previously made holes
• The final (\(x\)-th) throw scores a success if it lands inside an existing hole, and a failure if it lands on remaining snow

Solution

The first \(x-1\) setup throws clear a total area equal to \(S_{E(s)} \cdot (x-1)\). Because the final throw lands completely randomly, the probability of it landing in a hole scales linearly with the total area cleared by the setup throws. Considering that the area of the fountain (\(S_f\)) is limited, after a certain number of setup throws the probability of success on the final throw will be \(100\%\).

• number of attempts to achieve maximum probability
  • \(\displaystyle X_{\text{max}}=\frac{S_{f }}{S_{E(s)}}+1=\frac{1,000 }{10 }+1=101\)
  • \(+1\) accounts for the final target throw
• probability function
  • \(\displaystyle\frac{S_{E(x)}\cdot(x-1)}{S_f}=\frac{10(x-1)}{1,000}=\frac{x }{100 }-\frac{1 }{100}=0.01x-0.01\)
  • \(p(x)=\begin{cases}0.01x-0.01 &x \in \{1,2,3,\dots,100\}\\ 1&x\ge 101\\0&\text{otherwise}\end{cases}\)
• example
  • What is the probability of scoring on the final throw if you throw \(20\) snowballs in total (\(19\) setup throws \(+\) \(1\) target throw)?
  • \(P(S_{20})=p(20)=0.19\)
  • \(P(F_{20})=1-p(20)=0.81\)

Of course, these assumptions are quite unrealistic for a real-world game, but they allow to explore this elegant probability puzzle!