The Riddler Express for Friday, February 23 was a gambler's ruin problem, so I thought I would walk you through the solution and show you how the following problem can be solved. Suppose you have one quarter, and I have two quarters. Each round, a die is rolled. If a 1, 2, 3, or 4 … Continue reading Gambler’s Ruin: A Lesson in Probability

# Category: Problem Solving

# Aliquot Divisors

When the idea came to write about this post, I had never came across the term "aliquot" to my recollection. Let's quickly deconstruct that sentence and assign some probabilities. I've never came across the term "aliquot": probability = 0.01 I don't recall having come across the term "aliquot": probability = 0.99 From this, I hope … Continue reading Aliquot Divisors

# Shortest Line Segment

This week's Riddler Express has interested readers solving for the shortest possible distance between two diagonals in a cube. The solution in this post will involve some higher level mathematics toward the end. You've been warned. I've changed the coordinates from the original posed problem with the one below. On this cube, there is a … Continue reading Shortest Line Segment

# The Yanghui Triangle, Part I

Many of you know Yanghui's Triangle as Pascal's triangle. However, about 500 years earlier (according to Wolfram Mathworld), a Chinese mathematician Yanghui studied this triangle. In China, it is referred to as the Yanghui Triangle. How is it produced? In my previous blog post, Counting Can Be Really Tough, I introduced you to combinations. As a … Continue reading The Yanghui Triangle, Part I

# Jumping Lily Pads

A frog on the shore (think of the shore as lily pad zero) has a line of 20 lily pads, each numbered 1 to 20. The frog is able to jump one or two lily pads at a time. That is, starting from shore, the first jump can either be to lily pad 1 or … Continue reading Jumping Lily Pads