diff --git a/CHANGELOG.md b/CHANGELOG.md
index 7a196a9c..91669b28 100644
--- a/CHANGELOG.md
+++ b/CHANGELOG.md
@@ -20,6 +20,7 @@ Change Log / Ray Tracing in One Weekend
 
 ### The Rest of Your Life
   - Fix    -- Fix typo of "arbitrary" (#1589)
+  - New    -- Added a bit more explanation of Buffon's needle problem (#1529)
 
 
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diff --git a/books/RayTracingTheRestOfYourLife.html b/books/RayTracingTheRestOfYourLife.html
index de7f9af0..888cc33e 100644
--- a/books/RayTracingTheRestOfYourLife.html
+++ b/books/RayTracingTheRestOfYourLife.html
@@ -94,8 +94,13 @@
 Estimating Pi
 --------------
 The canonical example of a Monte Carlo algorithm is estimating $\pi$, so let's do that. There are
-many ways to estimate $\pi$, with the Buffon Needle problem being a classic case study. We’ll do a
-variation inspired by this method. Suppose you have a circle inscribed inside a square:
+many ways to estimate $\pi$, with _Buffon's needle problem_ being a classic case study. In Buffon's
+needle problem, one is presented with a floor made of parallel strips of floor board, each of the
+same width. If a needle is randomly dropped onto the floor, what is the probability that the needle
+will lie across two boards? (You can find more information on this problem with a simple Internet
+search.)
+
+We’ll do a variation inspired by this method. Suppose you have a circle inscribed inside a square:
 
   ![Figure [circ-square]: Estimating $\pi$ with a circle inside a square
   ](../images/fig-3.01-circ-square.jpg)
@@ -449,8 +454,8 @@
 
 One Dimensional Monte Carlo Integration
 ====================================================================================================
-Our Buffon Needle example is a way of calculating $\pi$ by solving for the ratio of the area of the
-circle and the area of the circumscribed square:
+Our variation of Buffon's needle problem is a way of calculating $\pi$ by solving for the ratio of
+the area of the circle and the area of the circumscribed square:
 
     $$ \frac{\operatorname{area}(\mathit{circle})}{\operatorname{area}(\mathit{square})}
        = \frac{\pi}{4}