July 16, 2021
I have three dogs: Fatch, Fetch and Fitch. Yesterday, I found a brown 12-inch stick for them to play with. I marked the top and bottom of the stick and then threw it for Fatch. Fatch, a Dalmatian, bit it in a random spot — leaving a mark — and returned it to me. In her honor, I painted the stick black from the top to the bite and white from the bottom to the bite.
I subsequently threw the stick for Fetch and then for Fitch, each of whom retrieved the stick by biting a random spot. What is the probability that Fetch and Fitch both bit the same color (i.e., both black or both white)?
Probability: \(\frac{2}{3}\).
Explanation:
Without loss of generality (WLOG), let us assume the stick is of unit length. The original puzzle can be parsed into the following: pick a number uniformly on the number line between 0 and 1. Then subsequently pick two more numbers between 0 and 1 uniformly. What is the probability that the last two numbers picked fall on the same side of the number line as divided by the first number?
The answer to the above equivalent problem is simply the average of the probabilities given each realization of the first number parting the number line into two segments. For example, say the first number is 0.5, well then the probability of Fetch and Fitch picking the same side is simply one-half. On the other extreme, if the first number is very close to 0 or very close to 1, then the probability of the latter two dogs picking the same side of the stick would be very close to 1, since one of the halves is essentially the entire stick. The average of all those scenarios can be calculated using calculus.
So let's execute our plan. Let the first number be \(x\) for \(0 \leq x \leq 1\). Assuming Fitch and Fetch bite the stick indepenently of one another, the probability of them biting the same side would be:
\begin{align} &P(\text{Biting One Side} | x)\\ &= (1-x)^2 + x^2 \\ &= 2x^2 - 2x + 1 \\ \end{align}
Where \(x\) is chosen uniformly randomly by Fatch, the first dog. Hence, \(X \sim \mathbf{Unif}(0, 1)\) has the following uniform distribution pdf:
\begin{equation} f_X(x) = 1, \quad \text{for} \quad 0 \leq x \leq 1 \end{equation}
Putting everything together:
\begin{align} &P(\text{Biting One Side}) \\ &= \int_0^1 P(\text{Biting One Side} | x) f_X(x) \mathrm{d}x\\ &= \int_0^1 (2x^2 - 2x + 1) \times 1 \mathrm{d}x\\ &= \left. \frac{2x^3}{3} - x^2 + x \right|_0^1 \\ &= \boxed{\frac{2}{3}} \end{align}