DonaldRauscher.com

A Blog About D4T4 & M47H

538 Riddler: Puzzle of the Monsters' Gems

26 May ’16

There are three ways that this game can end: slaying a rare monster (the most likely), slaying an uncommon monster, or slaying a common monster (the least likely). I began by thinking about the probability of each of these events happening. The probability of the game ending by slaying a rare monster can be expressed as the following:

Extending that same logic, we can calculate the probability that we end the game by slaying an uncommon and common monster as well:

...these sum to 1 and match what we would expect directionally.

Next, I added the expected number of common monsters slayed to each of these scenarios. If the game ends by slaying a common monster (which only happens 15% of the time), this is simply 1. In the other two scenarios, the number of common monsters slayed is a binomial random variable, conditioned on not all the previously slayed monsters being of a single type. Bayes helps us reduce the expression:

Putting this all together:

538 Riddler: Puzzle of the Overflowing Martini

19 May ’16

Here's my solution to this week's 538 Riddler:

If the liquid reaches fraction of the way up the glass when upright, then the liquid goes fraction of the way up the glass on the opposite side just before it begins to pour.

Dandelin spheres were key to my proof. They prove that the top of the liquid (a conic section) forms an ellipse. They also prove that a sphere tangent to the top of the liquid and the sides of the glass intersects that elliptical conic section and one of its foci! I derived two expressions for the volume of the liquid (one pouring, one upright) and set them equal to one another. Interestingly, the steepness of the glass ( in my proof), drops out of the expression!