Hue's Counting?

In Search of the Elusive Blue

This question was given to me by Rudy Roggio¹ so I want to acknowledge that up front. Furthermore, according to a very well respected weather quantitative researcher in the Chicago area who uses the pseudonym ‘Midas’,

“Very slow if you can't do this basic math. How will you predict colour of sky? Makes for weak meteorologist if can't solve basic probability.”

It’s therefore a clear signal of one’s quant capabilities and important to be able to do these with ease.

Without further ado, the question was: Suppose that you have an urn with 9 red balls and 1 blue ball. You will keep on sampling balls from this urn (with replacement) until you have seen at least one of each. What do you expect the ratio of red balls to the total number of balls drawn to be?

1. < 0.9

2. = 0.9

3. > 0.9

Intuition

While the waiting times for each color might follow geometric-like distributions, the quantity R/N is not simply additive or linear. This is why you can’t “just add up” the expected times and divide—nonlinear operations on random variables (like taking a ratio) require more careful treatment than linear ones.

Early Blues Pull Down the Average. While late blue draws do contribute ratios close to 0.9, early blue draws (which are more impactful in pulling the average down) skew the overall expected ratio downward; we therefore expect the expected ratio to be < 0.9.

One can also look at this with Jensen’s Inequality² in mind and realize that

\(\mathbb{E}\left[\frac{R}{N}\right] \leq \frac{\mathbb{E}[R]}{\mathbb{E}[N]} \)

Solution

There are two main cases to consider here

1. The first draw is Red

2. The first draw is Blue

The total expectancy is the sum of these two outcomes weighted by their respective probabilities.

\(\mathbb{E}\left[\frac{R}{N}\right] = P(\text{First Red}) \cdot \mathbb{E}\left[\frac{R}{N} \mid \text{First Red}\right] + P(\text{First Blue}) \cdot \mathbb{E}\left[\frac{R}{N} \mid \text{First Blue}\right]\)

\(\mathbb{E}\left[\frac{R}{N}\right] = 0.9 \cdot \mathbb{E}\left[\frac{X}{1+X} \right] + 0.1 \cdot \mathbb{E}\left[\frac{Y}{1+Y}\right]\)

\(\mathbb{E}\left[\frac{X}{1+X}\right] = 0.1\cdot\sum_{x=1}^{\infty}\frac{x}{1+x}\cdot(0.9)^{x-1}\)

\(\mathbb{E}\left[\frac{Y}{1+Y}\right] = 0.9\cdot\sum_{y=1}^{\infty}\left(1 - \frac{y}{1+y}\right)\cdot(0.1)^{y-1}\)

\(\mathbb{E}\left[\frac{R}{N}\right] \approx 0.7924\)

1. Rudy Roggio, Quantitative Trader. LinkedIn Profile, https://www.linkedin.com/in/rudy-roggio/.

2. "Jensen's inequality." Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/wiki/Jensen%27s_inequality.