Everyone at bloomfield knoble knows not to ask me STEM (science, technology, engineering, math)-related questions, because I will not give a simple answer. I seize the opportunity to fill an entire whiteboard with formulas and computations as if I were giving a lecture at MIT. In my defense, I don’t try to be like that, it’s just so many STEM-related answers are based on knowing the answers to a bunch of other questions. For example, Chi square (calculating the relationship between two variables to determine if they are related) is pretty common in marketing. However, calculating Chi square means constructing the observed values table using the original dataset; using the f^e formula to construct the expected values table; using the Chi square former to calculate the Chi square value; using the df formula and the Chi square table to discover if that x^2 value is significant; and drawing a conclusion about the relationship between the two variables. So, yeah, ask me a question and I’m going to walk through the entire process to deliver the answer.
Sorry – got a bit off topic. See! I just used a paragraph to explain why people don’t ask me STEM-related questions.
Anyway, it turns out that Clark (associate creative director here at bloomfield knoble) and I have the same birthdate. One of the interns, who doesn’t know better, asked what the chances are that two people in a relatively-small office would have the same birthday, and the topic for this week’s blog was born. Let the whiteboard explanation (followed by the reason it matters in advertising/marketing) commence:
Let’s exclude February 29th because those people, like Gingers, are born without souls, so that a year has 365 days. Let’s also assume that all days are equally likely birthdays for a randomly chosen person. So how many people do you need to ask to be at least 50% certain that at least two of them have the same birthday? What’s your guess? Many people answer 183, which is about half of 365. This is a fairly well-known problem, so you might already know the answer is 23.
We arrive at the answer by computing the probability that everyone has a different birthday and then subtract this from 1. Start with just two people. The first can have any birthday and the second person must avoid this day, which has a probability of 364/365. The probability that two people share a birthday is thus 1 – 364/365 or about 0.003. Add another person. His or her birthday must avoid both previously taken birthdays, which has probability of 363/365. The probability that all three people have different birthday is 364/365 x 363/365 and the probability that there is some common birthday in a group of three is P(some common birthday) = 1 – 364/365 x 363/365 about 0.01. We keep doing this over and over. At 10 people, the the probability already exceeds 0.1 and at 22 people it is 0.48 and at 23 people the probability of some common birthday is 0.51. Thus, only 23 people are needed to be at least 50% certain that there is some common birthday.
Remember, this isn’t the same as the probability that somebody shares a particular birthday, which is how I’m going to spin this math lesson back to marketing and advertising.
There are, generally, two types of campaigns. There is the campaign where you are trying to reach a very specific audience and influence them all; and there is the type of campaign where you are trying to reach everyone and then influence some. The first campaign is like two people sharing a particular birthday – you have very specific criteria in mind and you determine the reach and frequency based on those criteria. These are, in my opinion, the best kind of campaigns and thanks to the willingness of people to give up their private information in return for cat pictures, very easy to accomplish. The second campaign is a bit trickier. This is the “maybe I should get a billboard” campaign. It’s become quite popular to dismiss these kinds of campaigns, simply because we – as ad people – don’t feel like we’ll reach the target audience or that they are simply too expensive to have an effective return on investment. But much of that “feeling” isn’t always based in true numbers.
Like the birthday problem, the number that seems correct (183 to hit 50%) isn’t actually the number. The same is true in different types of campaigns. It is easy to dismiss campaign elements like Digital Out of Home, or billboards, etc. as a “waste of money” because the length of time required to be seen by enough people may seem like too low of an ROI. However, a little statistical analysis may reveal that we don’t have to spend as much as we thought to be effective. I could show you the math behind that thinking, but I’ve run out of space on the whiteboard.
- another viral math puzzle
- I know that you know that I know that you know…., CSI Undergraduate Conference on Research, Scholarship, and Performance, April 2015
- Why the Cheryl birthday problem turned into the maths version of #TheDress
A STEM (Science / Technology / Engineering / Math) graduate and COO of bloomfield knoble, Thomas exemplifies the view that advertising is becoming an engineering discipline. He leads the integrated insights and strategic planning group in a way consistent with bloomfield knoble’s goal of bringing a strong analytical foundation to uncover fresh and innovative insights and business opportunities.