18 Heuristics and Decision Making
Textbook Chapters (and similar texts)
Demos
*******************************************************
Inquiry-based Activity: Birthday paradox
Note: Some versions of this activity have been described elsewhere – this is not a novel approach, but a hands-on guide.
Introduction: In this activity, students will explore the birthday paradox. The birthday paradox is demonstrated when you ask people, “how large would a group of people need to be in order for the odds of two people sharing a birthday to be 50%?” Many people answer this question with an answer that approximates 182, or close to 365 (the number of days in the year) divided by two. The actual answer is that the group would only need to contain 23 people for the odds of a shared birthday to be at 50%. In pairs, students will publicly-available data on athletic teams in order to test their ideas about the birthday paradox. This activity provides students with the opportunity to confront and think through their cognitive biases in a fun and easy-to-relate-to example.
Question to pose to students: In a group of 25, what are the odds that at least two people share a birthday?
Note: Year is not important, only the birth date
Note: 25 does not need to reflect your class size – the number 25 comes from the size of a Major League Baseball (MLB) team roster, which is used for this activity.
Students form a hypothesis: Students guess the answer to the question. You can specifically ask them to think through the numbers, or instruct them to “go with their guts,” depending on time/statistical background.
Students test their hypotheses: Students pair up. Assign them each to several rosters of MLB teams (e.g., Group A is assigned to the Orioles, Mets, and Blue Jays, Group B is assigned to the Mariners, Athletics, and Marlins, etc.). Official team rosters from the MLB website include birthdays. Have each group look up the rosters for their assigned teams and figure out whether there is a shared birthday. They need only note whether or not a birthday is shared on the roster, not how many people share a birthday. That is, they can consider themselves done with the roster when they find a match, or when they exhaust all players/birthdates and do not see a match. As a class, have students report out their findings.
Do the students’ hypotheses hold up?: Most students tend to underestimate what the odds of shared birthdays are. For the 2018 season, approximately 60% of MLB teams had shared birthdays on their roster, while average student guesses are somewhere around 15%. Discuss why this underestimation occurs. How does this type of error translate to real-world probability errors? Can they think of other areas where they may make similar errors?
The math behind the results –
In a group of 25 random people, the odds are 55.56% that two people will share the same birthday.
When considering a single pair (i.e., Person A having the same birthday as Person B), the odds are very high that they will NOT share the same birthday (p = .9973). If Person A and Person B share the same birthday, there is 1 in 365 days that is shared between these two people, that is p = (1/365) = .0027, and thus a p = 1.00–.0027 = .9973 chance that these two people do NOT share the same birthday.
But, if every possible pair is considered (i.e., Person A and B, Person A and C, Person A and D, etc., Person B and C, Person B and D, etc. where for Person A there are 24 unique pairs, for Person B there are 23 unique pairs, for Person C there are 22 unique pairs, etc.) there would be 24 + 23 + 22+ … = 300 separate pairs, each with the same likelihood of not matching. Therefore, if you compute the probability that there are NO matching birthdays across 300 separate pairs, there is a p = (.9973)300 = .4444 that there will be no matching pairs, and thus a p = 1.00 – .4444 = .5556 or 55.56% chance there will be a matching pair of birthdays in a group of 25 random people.
In this demonstration, we used sports team rosters, but you can use this demonstration with the students in your class, or any other groups. The only change to the computations above is the exponent used to find the probability of no matching pairs.
