What is the probability that you and a friend have different birthdays? (For simplicity, let a year have 365 days.) What is the probability that three people have three different birthdays? Show that the probability that \(n\) people have \(n\) different birthdays is $$ p=\left(1-\frac{1}{365}\right)\left(1-\frac{2}{365}\right)\left(1-\frac{3}{365}\right) \cdots\left(1-\frac{n-1}{365}\right) $$ Estimate this for \(n \notin 365\) by calculating in \(p\) [recall that \(\ln (1+x)\) is approximately \(x\) for \(x<1]\). Find the smallest (integral) \(n\) for which \(p<\frac{1}{2}\). Hence show that for a group of 23 people or more, the probability is greater than \(\frac{1}{2}\) that two of them have the same birthdit? (Try it with a group of friends or a list of people such as the presidents of the United States.)

Step by step solution

01

Identify the Total Number of Possible Birthdays

There are 365 possible days in a year, so each person's birthday can be any of these 365 days.

02

Calculate Probability for Two Different Birthdays

For you and a friend to have different birthdays, your friend's birthday must be on any of the remaining 364 days after your birthday is fixed. The probability is therefore: \[ P(\text{different birthdays}) = 1 - \frac{1}{365} \]

03

Calculate Probability for Three Different Birthdays

To have three different birthdays, the second person's birthday must differ from yours and the third person's birthday must differ from both. This gives: \[ P(\text{3 different birthdays}) = \frac{364}{365} \times \frac{363}{365} \]

04

Generalize for n People

The probability that n people each have different birthdays follows the general form: \[ P(\text{n different birthdays}) = \frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \text{...} \times \frac{365-(n-1)}{365} \]

05

Simplify Using Logarithms

Using the approximation \( \text{ln}(1-x) \text{ is approximately } x \) for small x, the product can be rewritten using logarithms: \[ \text{ln}(P) = \text{ln} \bigg(1 - \frac{1}{365}\bigg) + \text{ln} \bigg(1 - \frac{2}{365} \bigg) + \text{...} \] This approximates to: \[ \text{ln}(P) \backsimeq -\frac{1}{365}-\frac{2}{365}-\frac{3}{365} \text{...} -\frac{(n-1)}{365} \]

06

Summing the Series

Summing the series: \[ \text{ln}(P) \backsimeq - \frac{n(n-1)}{2 \times 365} \] Therefore: \[ P \backsimeq \text{exp}(- \frac{n(n-1)}{2 \times 365}) \]

07

Finding Smallest n for P < 1/2

We seek the smallest n such that: \[ \text{exp}(-\frac{n(n-1)}{730}) < \frac{1}{2} \] which simplifies to: \( \frac{n(n-1)}{730} \text{ln}(2) \approx 0.693. \)

08

Solving for n

Solving for n: \[ \frac{n(n-1)}{730} > \text{ln}(2) \] approximates at \( n \backsimeq 23 \) when plugged and solved numerically. Thus for n=23, probability falls below 1/2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory

Probability theory is a branch of mathematics that deals with the likelihood of different outcomes. It helps in quantifying the chance of an event occurring, expressed as a number between 0 and 1. If the probability is 0, the event will not happen at all. If it's 1, the event is certain to happen. For instance, in the birthday problem, we calculate the chance that no two people share the same birthday among a group.

Basic principles include:

• Understanding random variables and outcomes.
• Calculating probabilities using rules like the addition and multiplication rules.
• Concepts of independent and mutually exclusive events.

These principles are applied to estimate the probability of different birthdays in a group by making assumptions and calculations.

Combinatorics

Combinatorics is the area of mathematics focused on counting, arrangement, and combination of objects. It plays a crucial role in solving the birthday problem, where we deal with the different ways people can have unique birthdays. By understanding the fundamentals of combinatorics, you can calculate the total number of possible arrangements and their probabilities.

Key concepts include:

• Permutations: Different ways to order a set of items.
• Combinations: Different ways to choose items from a set without regard to order.
• Binomial coefficients, which help in counting the combinations.

In the birthday problem, we use permutations to figure out the number of unique birthday assignments.
Understanding combinatorics helps generalize the probability for n people to have different birthdays.

Birthday Paradox

The birthday paradox is a famous problem in probability theory. Despite seemingly low odds, it shows that in a group of just 23 people, there's a more than 50% chance that at least two people share the same birthday. This surprises many because it defies common intuition. To understand it:

• Calculate the probability that each new person added to a group has a unique birthday not shared with the others already in the group.
• Multiply these probabilities for a general formula.

The probability that all n people have different birthdays is: \[ P(n \text{ different birthdays}) = \frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \text{...} \times \frac{365-(n-1)}{365} \] As n increases, the chance of unique birthdays dramatically drops.
It illustrates how human perception can misjudge probabilities in large combinations.

Logarithmic Approximation

Logarithmic approximations simplify complex multiplicative problems into additive ones. It is invaluable in probability theory and combinatorics, especially for large numbers. For the birthday problem, using natural logarithms helps estimate probabilities.

Basic principles include:

• \text{ln}(1+x) approximately x, for small x values.
• Using this to convert a product of probabilities into a sum, making calculations easier.

For the birthday problem, the formula for n people's unique birthdays gets rewritten using logarithms for simplicity: \[ \text{ln}(P) \backsimeq -\frac{1}{365}-\frac{2}{365}-\frac{3}{365} \text{...} -\frac{(n-1)}{365} \] Summing these series and considering the approximation makes the problem manageable. It's also easier to compute the threshold (like p < 0.5) where probabilities change significantly. It turns the problem into simple arithmetic, making complex calculations accessible.