## Odds and Probability: Commonly Misused Terms in Statistics – An Illustrative Example in Baseball

Yesterday, all 15 home teams in Major League Baseball won on the same day – the first such occurrence in history.  CTV News published an article written by Mike Fitzpatrick from The Associated Press that reported on this event.  The article states, “Viewing every game as a 50-50 proposition independent of all others, STATS figured the odds of a home sweep on a night with a full major league schedule was 1 in 32,768.”  (Emphases added)

Screenshot captured at 5:35 pm Vancouver time on Wednesday, August 12, 2015.

Out of curiosity, I wanted to reproduce this result.  This event is an intersection of 15 independent Bernoulli random variables, all with the probability of the home team winning being 0.5.

$P[(\text{Winner}_1 = \text{Home Team}_1) \cap (\text{Winner}_2 = \text{Home Team}_2) \cap \ldots \cap (\text{Winner}_{15}= \text{Home Team}_{15})]$

Since all 15 games are assumed to be mutually independent, the probability of all 15 home teams winning is just

$P(\text{All 15 Home Teams Win}) = \prod_{n = 1}^{15} P(\text{Winner}_i = \text{Home Team}_i)$

$P(\text{All 15 Home Teams Win}) = 0.5^{15} = 0.00003051757$

Now, let’s connect this probability to odds.

It is important to note that

• odds is only applicable to Bernoulli random variables (i.e. binary events)
• odds is the ratio of the probability of success to the probability of failure

For our example,

$\text{Odds}(\text{All 15 Home Teams Win}) = P(\text{All 15 Home Teams Win}) \ \div \ P(\text{At least 1 Home Team Loses})$

$\text{Odds}(\text{All 15 Home Teams Win}) = 0.00003051757 \div (1 - 0.00003051757)$

$\text{Odds}(\text{All 15 Home Teams Win}) = 0.0000305185$

The above article states that the odds is 1 in 32,768.  The fraction 1/32768 is equal to 0.00003051757, which is NOT the odds as I just calculated.  Instead, 0.00003051757 is the probability of all 15 home teams winning.  Thus, the article incorrectly states 0.00003051757 as the odds rather than the probability.

This is an example of a common confusion between probability and odds that the media and the general public often make.  Probability and odds are two different concepts and are calculated differently, and my calculations above illustrate their differences.  Thus, exercise caution when reading statements about probability and odds, and make sure that the communicator of such statements knows exactly how they are calculated and which one is more applicable.

## Christian Robert Shows that the Sample Median Cannot Be a Sufficient Statistic

I am grateful to Christian Robert (Xi’an) for commenting on my recent Mathematical Statistics Lessons of the Day on sufficient statistics and minimally sufficient statistics.

In one of my earlier posts, he wisely commented that the sample median cannot be a sufficient statistic.  He has supplemented this by writing on his own blog to show that the median cannot be a sufficient statistic.

Thank you, Christian, for your continuing readership and contribution.  It’s a pleasure to learn from you!

## Statistics Lesson and Warning of the Day – Confusion Between the Median and the Average

Yesterday, I attended an interesting seminar called “Transforming Healthcare through Big Data” at the Providence Health Care Research Institute‘s 2014 Research Day.  The seminar was delivered by Martin Kohn from Jointly Health, and I enjoyed it overall.  However, I noticed a glaring error about basic statistics that needs correction.

Martin wanted to highlight the overconfidence that many doctors have about their abilities, and he quoted Vinod Kohsla, the co-founder of Sun Microsystems, who said, “50% of doctors are below average.”  Martin then presented a study showing an absurdly high percentage of doctors who think that they are “above average”.  A Twitter conversation between attendees of a TED conference in San Francisco and Vinod himself confirms this quotation.

The statement “50% of doctors are below average” is wrong in general.  By definition, 50% of any population is below the median, and the median is only equal to the average if the population is symmetric.  (Examples of symmetric probability distributions are the normal distribution and the Student’s t-distribution.)  Vinod meant to say that “50% of doctors are below the median”, and he confirmed this in the aforementioned Twitter conversation; I am disappointed that he justified this mistake by claiming that it would be less understood.  I think that a TED audience would know what “median” means, and those who don’t can easily search for its meaning online or in books on their own.

In communicating truth, let’s use the correct vocabulary.

## Exploratory Data Analysis: The 5-Number Summary – Two Different Methods in R

#### Introduction

Continuing my recent series on exploratory data analysis (EDA), today’s post focuses on 5-number summaries, which were previously mentioned in the post on descriptive statistics in this series.  I will define and calculate the 5-number summary in 2 different ways that are commonly used in R.  (It turns out that different methods arise from the lack of universal agreement among statisticians on how to calculate quantiles.)  I will show that the fivenum() function uses a simpler and more interpretable method to calculate the 5-number summary than the summary() function.  This post expands on a recent comment that I made to correct an error in the post on box plots.

> y = seq(1, 11, by = 2)
> y
[1]  1  3  5  7  9 11
> fivenum(y)
[1]  1  3  6  9 11
> summary(y)
Min.   1st Qu.   Median    Mean     3rd Qu.    Max.
1.0     3.5       6.0       6.0      8.5       11.0

Why do these 2 methods of calculating the 5–number summary in R give different results?  Read the rest of this post to find out the answer!

Previous posts in this series on EDA include

## Using the Golden Section Search Method to Minimize the Sum of Absolute Deviations

#### Introduction

Recently, I introduced the golden search method – a special way to save computation time by modifying the bisection method with the golden ratio – and I illustrated how to minimize a cusped function with this script.  I also wrote an R function to implement this method and an R script to apply this method with an example.  Today, I will use apply this method to a statistical topic: minimizing the sum of absolute deviations with the median.

While reading Page 148 (Section 6.3) in Michael Trosset’s “An Introduction to Statistical Inference and Its Applications”, I learned 2 basic, simple, yet interesting theorems.

If X is a random variable with a population mean $\mu$ and a population median $q_2$, then

a) $\mu$ minimizes the function $f(c) = E[(X - c)^2]$

b) $q_2$ minimizes the function $h(c) = E(|X - c|)$

I won’t prove these theorems in this blog post (perhaps later), but I want to use the golden section search method to show a result similar to b):

c) The sample median, $\tilde{m}$, minimizes the function

$g(c) = \sum_{i=1}^{n} |X_i - c|$.

This is not surprising, of course, since

$|X - c|$ is just a function of the random variable $X$

– by the law of large numbers,

$\lim_{n\to \infty}\sum_{i=1}^{n} |X_i - c| = E(|X - c|)$

Thus, if the median minimizes $E(|X - c|)$, then, intuitively, it minimizes $\lim_{n\to \infty}\sum_{i=1}^{n} |X_i - c|$.  Let’s show this with the golden section search method, and let’s explore any differences that may arise between odd-numbered and even-numbered data sets.