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 \(n \ll 365\) by calculating \(\ln p\) [recall that \(\ln (1+x)\) is approximately \(x\) for \(x \ll 1]\). 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 birthday. (Try it with a group of friends or a list of people such as the presidents of the United States.)

Short Answer

Expert verified
For two people, the probability is \[ \frac{364}{365} \]. For three, it's \[ \frac{364}{365} \times \frac{363}{365} \]. For \ n \ people, the probability is approximated by \[ p \approx e^{-\frac{n(n-1)}{730}} \]. For \ n = 23, the probability of a shared birthday > \ \frac{1}{2} \.

Step by step solution

01

Probability that two people have different birthdays

To find the probability that you and a friend have different birthdays, we start with the total number of possible birthdays: 365. Since one person's birthday does not affect the other's, there are 365 possibilities for the first person, and 364 possibilities for the second person to have a different birthday.
02

Calculate the probability

The probability that you and a friend have different birthdays is given by the ratio of the number of favorable outcomes to the total number of outcomes. This ratio is \[\frac{364}{365}\]. Therefore, the probability that two people have different birthdays is: \[\frac{364}{365} \approx 0.9973\].
03

Probability that three people have different birthdays

Similarly, to find the probability that three people have three different birthdays, we extend the process. The third person can have any of the remaining 363 days: \[\frac{363}{365}\]. Therefore, the combined probability that three people have three different birthdays is: \[\frac{364}{365} \times \frac{363}{365} \].
04

Generalize to n people

For n people to have n different birthdays, we generalize the process. The probability is given by the product: \[ p = \left(1-\frac{1}{365}\right)\left(1-\frac{2}{365}\right)\left(1-\frac{3}{365}\right)\cdots\left(1-\frac{n-1}{365}\right) \].
05

Estimate the probability for small n

For \( n \ll 365\), we can use the approximation \[ \ln(1-x) \approx -x \] for \[ x \ll 1 \]. Therefore, \[ \ln p \approx -\left(\frac{1}{365} + \frac{2}{365} + \frac{3}{365} + \cdots + \frac{n-1}{365}\right) = -\frac{1}{365} \sum_{k=1}^{n-1} k = -\frac{1}{365} \times \frac{(n-1)n}{2} \]. Converting back, \[ p \approx e^{-\frac{n(n-1)}{730}} \].
06

Find the smallest n for p < 1/2

To find the smallest integral n for which \[ p < \frac{1}{2} \], we set up the inequality \[ e^{-\frac{n(n-1)}{730}} < \frac{1}{2} \]. Taking the natural logarithm of both sides, \[ -\frac{n(n-1)}{730} < \ln(\frac{1}{2}) \] which simplifies to \[ \frac{n(n-1)}{730} > \ln(2) \]. Solving this inequality yields \[ n(n-1) > 730\ln(2) = 504.42 \]. The smallest integer n satisfying this is n = 23.
07

Probability for group of 23 people

Therefore, with a group of 23 people or more, the probability that at least two people share the same birthday is greater than \( \frac{1}{2} \). Test this theory with a group of friends to see the results.

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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 the branch of mathematics dealing with the analysis of random phenomena. In this context, the *birthday problem* is a classic example used to explain and understand probability.
To calculate the probability that you and a friend have different birthdays, consider there are 365 possible days for a birthday. If we assume your friend's birthday doesn't affect yours, there are 365 options for you and 364 options for your friend.
The formula for the probability of different birthdays is thus:\[ \frac{364}{365} \]. This principle extends to calculate for more people by multiplying fractions for subsequent birthdays.
Combinatorics
Combinatorics is the field of mathematics focused on counting, combination, and permutation. It is vital in solving the birthday problem in understanding how combinations of birthdays can occur.
The probability that three people all have different birthdays is calculated by:\[ \frac{364}{365} \times \frac{363}{365} \].
By generalizing this approach, we can express the probability that n people have distinct birthdays as:\[ p = \left(1-\frac{1}{365}\right) \left(1-\frac{2}{365}\right) \cdots \left(1-\frac{n-1}{365}\right) \]. This combinatorial approach gives insight into the decreasing probability as the group size increases.
Exponential Approximation
Exponential approximation allows simplifying complex probability calculations. It is particularly useful when n is much smaller than 365.
Using natural logarithms to approximate, we start with \( \ln (1-x) \approx -x \) for small values of x. Then, calculating the natural log of the product gives us:\[ \ln p \approx -\frac{1}{365} \sum_{k=1}^{n-1} k = -\frac{1}{365} \times \frac{(n-1)n}{2} \].
Exponentiating the result, we obtain:\[ p \approx e^{-\frac{n(n-1)}{730}} \]. This allows for quick estimation of the probability with less computational effort.
Group Theory
Group theory, though more commonly applied in algebra, also provides insight into an organized structure of elements. In the context of the birthday problem, it can represent the grouping of people and their possible shared birthdays.
This theory helps in understanding why with a group of 23 people or more, the probability that at least two people share the same birthday becomes greater than one-half. It’s fascinating to observe this probabilistic behavior within group combinations and arrangements.
Birthday Paradox
The birthday paradox, often surprisingly counterintuitive, explains that in a group of just 23 people, there’s more than a 50% chance that at least two individuals share a birthday.
Despite there being 365 days of the year, the paradox highlights the rapid increase in the probability of shared birthdays with more people. This can be practically tested with groups of friends or historical records of individuals, like presidents.
In mathematical terms, for n = 23, we solve:\[ \frac{23 \times 22}{2} / 365 \approx 0.5 \]. This peculiar yet fascinating result demonstrates the power and sometimes mind-boggling nature of probability theory and combinatorics.

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Most popular questions from this chapter

The following game was being played on a busy street: Observe the last two digits on each license plate. What is the probability of observing at least two cars with the same last two digits among the first 5 cars? 10 cars? 15 cars? How many cars must you observe in order for the probability to be greater than \(\frac{1}{2}\) of observing two with the same last two digits?

A bit (meaning binary digit) is 0 or \(1 .\) An ordered array of eight bits (such as 01101001) is a byte. How many different bytes are there? If you select a byte at random, what is the probability that you select \(11000010 ?\) What is the probability that you select a byte containing three 1 's and five 0 's?

(a) A weighted coin has probability of \(\frac{2}{3}\) of showing heads and \(\frac{1}{3}\) of showing tails. Find the probabilities of \(h h, h t, t h\) and \(t t\) in two tosses of the coin. Set up the sample space and the associated probabilities. Do the probabilities add to 1 as they should? What is the probability of at least one head? What is the probability of two heads if you know there was at least one head? (b) For the coin in (a), set up the sample space for three tosses, find the associated probabilities, and use it to answer the questions in Problem 2.12 .

There are 3 red and 2 white balls in one box and 4 red and 5 white in the second box. You select a box at random and from it pick a ball at random. If the ball is red, what is the probability that it came from the second box?

Two decks of cards are "matched," that is, the order of the cards in the decks is compared by turning the cards over one by one from the two decks simultaneously; a "match" means that the two cards are identical. Show that the probability of at least one match is nearly \(1-1 / e.\)

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