Question

Predicting coincidence. Your statistical mechanics class has 25 students. What is the probability that at least two classmates have the same birthday?

Short answer

The probability that at least two classmates have the same birthday is approximately 0.569 (or 56.9%).

Step-by-step solution

- Understand the problem

We need to calculate the probability that at least two students in a class of 25 have the same birthday. This will involve principles of probability and the birthday paradox.

- Calculate total possible combinations

There are 365 days in a year, and each student can have any of those days as their birthday. Therefore, the total number of possible birthday combinations for 25 students is \[ 365^{25} \]

- Calculate the probability no two students have the same birthday

Start by calculating the probability that 25 students all have different birthdays. For the first student, there are 365 choices, for the second student, there are 364 choices, for the third student 363, and so on until 25 choices for the 25th student. The probability that all have different birthdays is:\[ P_{\text{distinct}} = \frac{365}{365} \times \frac{364}{365} \times \cdots \times \frac{341}{365} \]

- Simplify the expression

Simplify the above expression to:\[ P_{\text{distinct}} = \frac{365!}{(365-25)! \, 365^{25}} \]

- Calculate the probability of at least one shared birthday

The probability that at least two students share the same birthday is the complement of all students having distinct birthdays. So:\[ P_{\text{shared}} = 1 - P_{\text{distinct}} \]

- Calculate the numerical value

Using a calculator or a computational tool, calculate:\[ P_{\text{distinct}} \approx 0.4313 \]Therefore:\[ P_{\text{shared}} = 1 - 0.4313 = 0.5687 \]

Key concepts

Probability Theory

In this exercise, we explore a fascinating concept called the Birthday Paradox, which is rooted in probability theory. Probability theory is a branch of mathematics that deals with the chances of different outcomes occurring. When we say the probability of an event happening, we are talking about how likely it is to happen.

For instance, if you roll a fair six-sided die, each number (1 through 6) has an equal probability of 1/6 of being rolled. In our exercise, we want to find out the probability that at least two students share the same birthday in a class of 25 students. This specific problem uses the principles of probability to compute the likelihood of shared birthdays.

We started by understanding the total possible combinations of birthdays for the students. There are 365 days in a year, so for 25 students, the total number of possible combinations is given by: \[ 365^{25} \]

Next, we computed the probability that no two students share the same birthday. This was done by calculating the probability that each subsequent student has a different birthday from the ones before them, leading us to our formula: \[ P_{\text{\text{distinct}}} = \frac{365!}{(365-25)! \, 365^{25}} \]

Finally, we derived the probability of at least one shared birthday by subtracting the probability of no shared birthdays from 1. This gave us: \[ P_{\text{\text{shared}}} = 1 - P_{\text{\text{distinct}}} \]

Statistical Mechanics

Statistical mechanics is a theoretical framework that uses statistics to describe the behavior of systems with many components, like gases or collections of particles. While it might seem far removed from our problem, there's an interesting connection. Both birthday paradox and statistical mechanics deal with the behavior of systems under random circumstances.

In statistical mechanics, we often examine the distribution of particles in different states, just as we look at the distribution of birthdays among students. The idea is to compute the probabilities of different configurations or outcomes. For example, in the birthday problem, we're interested in the configuration where at least two students share the same birthday.

By analyzing runs of different particles (or in our case, birthdays), statistical mechanics helps us understand the average behavior of large systems. Even though this exercise involves only 25 students, the principles of statistical mechanics underpin the calculations, especially when we extrapolate to larger numbers and complex systems.

So, while the birthday paradox is simpler, the same statistical methods are often used in more complex scenarios within statistical mechanics to predict phenomena and behaviors.

Combinatorics

Combinatorics is another branch of mathematics closely related to our problem. It deals with counting, arrangement, and combination of objects. In the birthday paradox, we used combinatorial principles to calculate the number of ways students can have distinct or shared birthdays.

To find out how many ways 25 students can each have a unique birthday, we applied principles from combinatorics to calculate the number of valid arrangements, given that no two students share the same day. The formulaic approach: \[ P_{\text{\text{distinct}}} = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \text{\text{ ... }} \times \frac{341}{365} \] simplifies through combinatorial rules.

Utilizing factorial notation simplified our work. Factorials are products of all positive integers up to a number and denote the number of possible arrangements. That’s why we had: \[ \frac{365!}{(365-25)!\times 365^{25}} \]

Combinatorics not only helped us count the number of possible arrangements but also guided us in understanding probabilities in multi-step scenarios. This is a small application of combinatorics, but it showcases its power in solving real-world problems, making it easier to grasp how seemingly complex problems can be broken down into simpler, manageable parts.