First, 50.7% isn’t like, super likely. Even though it’s greater than 50%, if you look at a large number of collections of 23 people, each of whom is drawn randomly from a uniform distribution as far as birthdays are concerned, in almost half of them, people will not share a birthday, as you might expect giving the problem some cursory thought.

The reason it’s as high as 50.7% is that we haven’t specified what the birthday is. If the question is “how many people do you have to gather so that the probability that at least two of them share the birthday April 23 exceeds 50%”, then indeed that number is quite large: 613, considerably more than the number of days in a year. (It’s a nice little exercise to figure out how you’d get that answer.)

But here, while you’re looking for the relatively uncommon occurrence of two people sharing a birthday, there are 365 different ways in which that can happen, and you’re not picky about which one of the ways it is. So you’re not really looking at the probability of a single rare event; you’re looking at the probability that at least one of a bunch of different rare events happens. And that goes up very quickly as the number of different rare events you’re considering goes up.

As always, if you’re not convinced, simulation is a useful tool. Here’s an R function that simulates a bunch of “rooms” of [math]n[/math] different people each:

## Define bday function for simulation 
bday <- function(n, r) { 
  # Simulates r rooms of n people each where integers represent birthdays 
   
  # Sample from discrete uniform(1,365) distribution with replacement 
  birthdays <- sample(1:365, n*r, replace = TRUE) 
   
  # Divide into rooms 
  rooms_matrix <- matrix(birthdays, r, n) 
   
  return(rooms_matrix) 
} 

The output is an [math]r\times n[/math] matrix where each row corresponds to a room and columns correspond to individual people within each room. To numerically estimate the probability that any two people out of [math]n[/math] share a birthday, all you have to do is count the number of rows in which at least one element is duplicated, and then divide that by the number of rows in total.

Let’s loop through values of [math]n[/math] from 2 (since we know the probability is zero for [math]n=1[/math]) to 80 (at which point the probability estimates for a simulation are going to basically be one):

## Initialize probability vector 
probs_sim <- rep(0, 80) 
 
set.seed(2019) 
 
## Loop through values of n 
for(n in 2:80) { 
  birthdays_sim <- bday(n, 1e5) 
   
  # For each row of the matrix, check for duplicates 
  duplicated_bdays <- apply(birthdays_sim, 1, function(x) any(duplicated(x))) 
   
  # Probability is the mean of this vector 
  probs_sim[n] <- mean(duplicated_bdays) 
} 

We can plot the results:

Once you get to [math]n=10[/math] the probability starts rising very rapidly, and just eyeballing it, it does appear that the probability passes the 50% mark at just over 20 people.

And at exactly what number does it do that?

> min(which(probs_sim >= 0.5)) 
[1] 23 

Probably larger than you think.

Let’s make two simplifying assumptions: (1) birthdays are uniformly distributed throughout the year [they aren’t], and (2) nobody is born on leap-day [they are]. Under these conditions, the probability of any one person having any given birthday is [math]1/365.[/math]

Now, let’s re-frame the question: what is the probability that out of [math]21[/math] people, every one of them has a different birthday? It turns out we have a handy notation for this. If there is a list of [math]21[/math] people, the number of ways to assign them each a unique birthday (out of [math]365[/math]) is called the number of permutations* of [math]365[/math] taken [math]21[/math] at a time. This is denoted [math]P(365,21)[/math] or [math]_{365}P_{21},[/math] and is given by

[math]\qquad P(365,21)=\dfrac{365!}{(365-21)!}.[/math]

This turns out to be a rather large number, on the order of [math]3\times 10^{53}.[/math]

On the other hand, the total number of ways to assign birthdays to [math]21[/math] people (i.e. allowing repeats) is [math]365^{21} \approx 6\times 10^{53}.[/math]

So, the probability that each of the birthdays is unique is given by

[math]\qquad \dfrac{P(365,21)}{365^{21}}=\dfrac {365!}{344!\times 365^{21}}.[/math]

If you use a good calculator (e.g. WolframAlpha), you will get a value of about [math]0.556.[/math]

Now, to answer the original question: the probability that at least two do share a birthday is [math]1-0.556 = 0.444,[/math] or about [math]44\%.[/math]

*You might think that we should be using combinations rather than permutations here. You are wrong. Calculating this using combinations sets you up for a world of hurt; I will leave it to you to figure out why.

It means if you select 23 random people, the chance that at least 2 of them share a birthday is just above 50%.

As to how to solve that…

Before anything else, let's get out of the way that I'm excluding February 29.

First, the case of 1 random person. There are 365 possible birthdays for them. Of them, 365 of them yield a unique birthday, 0 yield a match. So the probability of all unique birthdays is [math]\frac{365}{365}.[/math]

Now, let's add a second random person. There are 365 possible birthdays. 1 yields a match, 364 a unique birthday; the probability of person #2 having a unique birthday assuming no prior matches is [math]\frac{364}{365}.[/math] That makes the probability of all unique birthdays [math]\frac{365\times 364}{365\times 365}.[/math]

Now, let's add a third random person. There are 365 possible birthdays. Assuming no prior matches, 2 yield a match, 363 a unique birthday. That's a probability that #3 has a unique birthday of [math]\frac{363}{365}.[/math] That makes the probability of all 3 having unique birthdays [math]\frac{365\times 364\times 363}{365\times 365\times 365}.[/math]

You'll notice that the probability of all unique birthdays is getting multiplied by smaller and smaller numbers, making it decrease. When you have 23 people, it's [math]\frac{365\times 364\times...\times 344\times 343}{365^{23}}.[/math] That number comes out to just under 50%- about 49.3%.

Since the probability of all unique birthdays drops below 50%, it means the probability of a match exceeds 50% (just barely).

Let’s first compute the probability that all 30 people have different birthdays. (I’ll assume 365 possible birthdays, ignoring leap years. And also that birthdays are equally distributed, which they are not).

Pick the first person. They have some birthday. The second person must have a different birthday than this and the probability of that is:

[math]\frac{364}{365}[/math].

The third person must have a different birthday than both of the first two and the probability of that is:

[math]\frac{363}{365}[/math].

Continue this until you do 30 people. Now multiply these fractions together to get the probability that all 30 have different birthdays:

[math]\frac{364}{365}*\frac{363}{365}*\frac{362}{365}*…*\frac{336}{365}=29.4\%[/math]. Note that there are 29 factors here.

If we subtract this from 1, we get the probability that at least 2 people have the same birthday is:

[math]100\%-29.4\%=70.6\%[/math].

The breakeven point is when there are 23 people in the room. This means that if there are 23 people in a room, there’s about a 50% chance that two or more will have the same birthday.

In many of my classes there are 60 students. When I check on birthdays, I’ve always had students with the same birthday. There’s less than a 1 in 200 chance that all 60 students have different birthdays.

How is it possible that there is 50% chance that two people in a room of 23 have the same birthday?

This is really easy to understand when you look at it the right way. Suppose we pick just two of the people in the room - what’s the probability that they have the same birthday? Let’s ignore the complication of 29th February birthdays as it doesn’t make much difference to the answer. Well, whatever birthday the first person has, there’s a 1 in 365 chance that the second person has this birthday. Or put another way there’s a [math]\frac{364}{365}[/math] chance that they’ll have different birthdays.

But now consider in a room of 23 people how many distinct pairs of people there are that could be chosen. The first person could be paired with any of the remaining 22, the second person with any of the remaining 21, and so on to the penultimate person being paired with the last person. So there are 22+21+20+….+2+1 = 253 distinct pairs of people.

So what’s the chance that none of these 253 pairs of people share the same birthday? Well we know that if we want to find the chance of an event happening n times then we need to raise the chance of the event happening once to the nth power. For example the chance of throwing a 6 on a die four times in a row is [math]\left(\frac{1}{6}\right)^4[/math].

Hence to find the chance of none of the 253 pairs having the same birthday we need to calculate [math]\left(\frac{364}{365}\right)^{253}[/math]. When you do this you get just under 50%. Hence the converse, ie: the chance that at least one of the pairs has the same birthday, is just over 50%.

In summary the point is that in a room of 23 people there are 253 chances of getting a pair with matching birthdays, and this raises smallish odds to become a more likely than not event. It’s counter-intuitive because our brain thinks in terms of just 23 people, without realising that this in fact represents 253 chances.

In a room full of 23 people there are [math]253=\frac{23 \times 22}{1\times 2}[/math] unique pairs of people, and the number of pairs is what matters when thinking about the probability that any pair of people have the same birthday.

This might help you to absorb the result, which is certainly counterintuitive at first sight.

Usually for simplicity real problems such as the fact that people can be born on February 29, people may be twins, and the distribution of birthdates among people in a given country is not uniform over the year are just ignored and you assume that the probability of any given birthday is equal and of course that there are 365 days in a year.

Then you can get the idea of the solution easily by just solving it for two people.

The chance that two people do not have the same birthday would be: [math]\frac{365}{365} \frac{364}{365}[/math], which is already a little less than one. This follows since there are only 364 possible birthdays to choose for the second person, such that he or she doesn't have the same birthday as the first one. The first person must have some birthday, so the probability that he or she does have some birthday is 1, accounting for the first factor.

And this proceeds with the third person and the fourth person and so on, so that a product of strictly decreasing factors which are all less than or equal to one is generated. This product decreases to less than 50% more quickly than you expect, and in fact that happens already when there are only 23 factors in the product. You should work it out with a calculator. It doesn't take too long.

Since the case in which no two people share the same birthday and the case in which at least two people share the same birthday are clearly mutually exclusive and these two cases cover all the possibilities, the probability that two people do share the same birthday is then greater than 50% when there are 23 people, and you have the result.

It is an easy problem to simulate on a computer using Monte Carlo methods, too.

It seems strange, but it is true.

It seems low, doesn’t it? But it works like this:

Let there be one person in a room. They can have any day out of 365 as their birthday.

Another person enters. In order not to share the first person’s birthday they have 364 days to choose from. Obviously there is a 364/365 chance that they randomly manage this.

A third person enters. Reasoning similarly, there is a 363/365 chance that they don’t share either of the first two birthdays. So the chance that we randomly managed this was 1 x 364/365 x 363/365.

For each additional person that we add, we multiply the existing probability by an increasingly small fraction out of 365.

By the time you get to the 23rd person, it happens that 1 x 364/365 x … 343/365 is close to 1/2.

I’ve left leap years out of this to keep the maths simple. If you want to work out the numbers including leap years, be my guest.

This is a well known problem and most people are surprised by the answer (I remember I was). The way to approach this is to calculate the probability that everyone has a different birthday and subtract the answer from 1. the first guy can be born any day. The next guy can be born any other day so 364/365. The third guy has TWO days he can’t be born on so his is 363/365. This goes on in a similar manner until we get to the 23rd person who must avoid 22 days so 343/365. Lets multiply them all together and then subtract from 1. (364*363*363*…*344*343)/365^22 = .4927. Subtract that from one 1-.4927 = .5023 i.e. there is a better than 50% chance two people share the same birthday!

Let’s answer the question “What is the probability of two randomly-selected people out of a population of 24 having the same birthdate?”.

This way, we don’t have to deal with birth years and can focus on calendar dates alone.

Let’s also say, for the moment at least, that we’re not considering the date February 29; our putative “year” thus has 365 possible dates.

We can do this using a well-known technique called normalization: since something has to happen, the probabilities of all possible outcomes must sum to 1. This means that in an either-or situation (coin flipping, for instance), the probability that something doesn’t happen is 1 - the probability that it does.

So let’s deal with the inverse problem: what is the probability, as a function of the population size [math]n[/math], that no two people share a birthdate? Obviously, by the time [math]n[/math] has increased to 366, there must be at least one “matched pair” and that probability falls to zero (and the probability of a match therefore rises to 100%).

Similarly, there is a 100% probability of “no match” at [math]n = 1[/math].

Extending this, the second person can have one of 364 possible birthdays; the third one of 363, and so on. Mathematically, we can express it like this:

[math]\displaystyle P_{\text{no match}} = \prod_{n = 1}^{365} \dfrac{365 - n + 1}{365} \tag 1[/math]

As already stated,

[math]\displaystyle P_{\mathrm{match}} = 1 - P_{\text{no match}} = 1 - \prod_{n = 1}^{365} \dfrac{365 - n + 1}{365}[/math]

[math]= 1 - \dfrac{365!}{365\,^n \left(365 - n\right)!}[/math]

For the case [math]n = 24[/math], this comes out to about 53.8%. Some might claim that the desired probability is 53.7% if we allow 366-day years, but I disagree, on the grounds that that result is based on the faulty assumption that each birthdate is equally likely; as February 29 only occurs every fourth year (during our lifetimes, anyway), that birthdate is arguably only one-fourth as likely as the other 365.

The 23% comes from some dubious logic. It goes like this

For the first person there is a [math]\frac{22}{364}= 0.0604[/math] or 6.04% chance that they will not have a match. So there is a [math](1 - \frac{22}{364})^{23} = 0.238[/math] or 23.8% chance that they will have a match.

A better answer is to look at the probability that there is not a match, call it p(a)

[math]p(a) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \ldots \frac{343}{365} = 0.492703[/math]

so the probability of getting a match is [math]1 - p(a) = 1−0.492703=0.507297[/math]or 50.7%.

In fact the reason that 23 is often used as an answer in these problems is that it is the lowest number of people for which the probability of getting a match passes 50%.

Danny

“What is the probability of two people having the same birthday in a room full of people?”

Higher than you might think.

The way you calculate this is a little counter-intuitive. You ask the reverse question: in a room with N people, where we can assume birthdays are scattered randomly, what are the odds that no one shares a birthday with anyone else?

(That will give you a probability, a number between 0 and 1. Subtract that number from 1 and you get the converse: the probability that at least one person shares a birthday with someone else.)

Start with one person. What are the odds this person does not share a birthday with anyone else in the room? The odds are 100%, or 1, because there's no one else in the room! Let's represent that as 365/365.

Now we add a second person. What are the odds the 2nd person doesn't share a birthday with the 1st one? Ignoring leap years, there are 364/365 chances that they don't share a birthday. The two birthdays are independent, so we multiply. Now we have 365/365 x 364/365.

When we add a third person, the same logic applies — this person shouldn't have the same birthday as either of the first two. So for the 3rd person, the probability is 363/365.

Keep multiplying elements of this series — 365/365 x 364/365 x 363/365 x 362/365 x 361/365… until you get what you want. For example you might want to know: how many random people need to be in a room, for the chances of a common birthday to be over 50%? The number of people turns out to be 23. (Check it and see.)

Assuming 365 days per year and that birth dates are randomly chosen, the likelihood that of n people none will have a common birthday is [math]\frac{365}{365}\frac{364}{365}\dots\frac{365-n+1}{365} = \frac{365!}{(365-n)!\,365^n}[/math]. The likelihood that at least two will share a birthday is thus one minus this probability. For n = 100, the likelihood of a shared birthday is ~0.99999969275, which is to all intents and purposes a certainty. If the people had genuinely been chosen at random you could essentially bet your life that at least two would share a birthday. Even with fifty people there’s a 97% chance of a common birthday.

There are [math]n={15 \choose 2}=105[/math] trails, each pair of people is a trial. The probability of a success, having the same birthday, on any trial is [math]{1\over 365}.[/math]

You want to find the probability there is at least one 1 success.

Since you've assumed that birthdays are all equally likely, you've ruled out twins or triplets etc in your 15 people.

You’re dealing with a random variable, [math]X,[/math] with the binomial distribution having [math]n=105[/math] and [math]p={1 \over 365}.[/math] And you want to find [math]P(X \ge 1)= 1-P(X=0). [/math]

[math]1-P(X=0)=1-{105 \choose 0}\bigl ( {1 \over 365} \bigr )^0 \bigl ( {364 \over 365} \bigr )^{105}=0.2503.[/math]

Sine [math]n[/math] is large and [math]p[/math] is small, the probability can be approximated using the Poisson distribution with parameter [math]\lambda =np= {105 \over 365}.[/math] The approximation is very good, it gives probability 0.2500.

It is possible, because the question asks for any two people in the room to be born on the same date, whilst you might interprete it to be about two specific people which check that condition. That sounds a bit weird but this does indeed change the situation a lot.

Think about it this way: How likely is it that you and your neighbour both win the lottery tomorrow (assuming the two of you take part, of course)? Quite close to zero, I’d guess. But how high are the chances, that two random people out of all the lottery participants do so? I bet you won’t be like “Oh my god! Two people have won the lottery!” if that happens.

It is the probability of at least 2 out of 22 having the same birthday which is governed by the Binomial Probability Law. It is obtained by summing the following expression for x =2 to 22, where n =22, p = 1/7, q = 6/7.

The probability that no two people share a birthday in a group of n people. Note that the vertical scale is logarithmic (each step down is 1020 times less likely).

It is not possible to answer this question as all people do not have their birth date being equal probable on all the days. We have seasons of marriages accordingly the number of births are clustered around certain months rather than in all possible months.

Calculation of probability is possible if the outcomes are known and they have equal probability like in the case of tossing of a coin or throwing of a dice.

In case we have some prior data where the individuals are divided according to their birth dates then based on the frequency observed there we can say which date has what probability?

See the birthday data taken from Google (the data is by Lane Sainty). You will observe that all days are not equal probable. But using this data, you may be able to say that for a given data, what are the chances of two people sharing the same birth date more so for the area for which these data is collected. For other areas, again, we cannot answer the question raised by you.

You answer these problems by reversing the question to ask what the chances are that none will have the same birthday. Then you subtract the answer you get from 1 and it gives you the probability that at least two people share the same birthday. So, if we want to find the probability of them NOT sharing a birthday, let’s imagine all 20 people line up outside the room and come in one-by-one and state their birthdays. The first person into the room has a 365/365 (or 100% chance of not announcing a birthday that anyone else has stated yet, since he is the first. The second guy has a 364/365 chance of not having the same birthday. If the second guy doesn’t have the same birthday as the first, there are now two days that are off limits. So, the third person has a 363/365 chance of not having a birthday that matches either of the first two. This continue for all 20 people, with the last one having a 346/365 chance of having a different birthday than everyone who came before, IF they were all different up to that point. Now we just multiply all the numbers together and we get (365•364•363•…•346)/(365^20) = 0.5853.

1–0.5853 = 0.4147, or 41.47% that at least two people in the room will share the same birthday.

There are 23 students in MTH105-online Fall 17 class. What is the probability that there is at least two sharing a birthday among these 23 people?

Assumption: Let’s assume that none of the students in this class have a birthday during leap year, thus we are assuming 365 possible birthdays per student.

There are [math]365^{23}[/math] total ways that the birthday calendar can be arranged in the class.

Next, there are [math]\frac{365!}{(365–22)!}=\frac{365!}{343!}[/math] arrangements where no two students have the same birthday (ie - the first student could have any birthday, the second could have any except for the same as the first, etc).

So the probability of no students having the same birthday is [math]\frac{365!/343!}{365^{23}}[/math] and thus the probability that some (at least two) have the same birthday is [math]1-\frac{365!/343!}{365^{23}}[/math].

The probability is:

[math]\displaystyle 1 - \left(\frac{364}{365}\right)^{\binom{n}2}[/math]

where [math]n[/math] is the number of persons in the room.

Why is that? Because the probability that at least two people in the room have the same birthday is [math]1[/math] minus the probability that no pair to two persons in the room have the same birthday.

The probability that a pair of two person is not born the same day is exactly

[math]\displaystyle 1 - \frac1{365} = \frac{364}{365}[/math]

In addition, in a room of [math]n[/math] persons, there are exactly [math]\binom{n}2[/math] possible pairs of persons, which means the probability that no pair of two persons in the room have the same birthday is

[math]\displaystyle \left(\frac{364}{365}\right)^{\binom{n}2}[/math]

We simply have to take [math]1[/math] minus that to get the actual probability.

Incidentally, for [math]n = 30[/math], the value is:

[math]\boxed{p \approx 0.696816 \approx 69.68\%}[/math]

There are 365 day in a year (let us assume we don't have a leap-year).

Two persons have their birthday on the same day, the remaining 11 on different days of the year.

So we pick one day for the common birthday: there are 365 choices. Then let us pick the two persons: there are [math]{n \choose 2}[/math] such choices. Now the remaining [math](n-2)[/math] persons are distributed over the remaining 364 days, so that no further duplication occurs:

There are [math]364*363*\dots *354[/math] such choices.

Finally you have 353 days with no birthday of the attending people.

The number of good cases is then

[math]365 {n \choose 2} 364*363*\dots 354 = {n \choose 2} \frac{365!}{353!}[/math]

which gives you a probability of

[math]{n \choose 2} \frac{365!}{353!*365^n}=0.178[/math]

for exactly 2 persons having same birthday.

Remember: the probability that 13 persons have all different birthdays is

[math]\frac{365}{365}\frac{364}{365}\cdots \frac{354}{365}\frac{353}{365}=0.8055[/math]

which gives the slightly higher probability of [math]0.1945[/math] for at least 2 having the same birthday.

Assuming the other 365 possible birthdays are equally likely, and the various events that person x has birthday y are all independent, the probability is [math]1 - 1 \cdot \frac{364}{365} \cdot \frac{363}{365} \cdots \frac{336}{365}[/math], or [math]1 - \frac{365 \cdot 364 \cdots 336}{365^{30}}[/math], which is approximately [math].7063162427[/math].

Under the same assumptions, suppose the number of people is [math]n[/math] (with [math]1 \le n \le 365[/math]). If we number the people from [math]1[/math] to [math]n[/math], then there are [math]365^n[/math] possible assignments of birthdays to these people, which we are assuming to be equally likely with one another. The number of possible assignments where all their birthdays are different is [math]365 \cdot 364 \cdot 363 \cdots (365-n+1)[/math]. So the probability of all birthdays different is [math]\frac{365 \cdot 364 \cdots (365-n+1)}{365^n}[/math], and hence the probability of at least two birthdays the same is [math]1-\frac{365 \cdot 364 \cdots (365-n+1)}{365^n}[/math]. If [math]n=30[/math] this turns out to be approximately [math].7063162427[/math], as indicated above. For [math]n=22[/math] it is approximately [math].4756953077[/math], and for [math]n=23[/math] it is approximately [math].5072972343[/math]. So [math]n=23[/math] is the crossover point where two birthdays the same becomes more likely than not.

If we start with the probability that they all have different birthdays we get the probability we want by subtracting that from 1.

Assume 366 possible birthdates, the probability that the second persons is different from the first is 365/366.

The probability that the third persons if different from the first two is 364/366

We carry on until we get to the probability that the 30th persons is different from the other 29 which is 337/366

We need the probability that all of these are true so we need the product of all of them which, using a spreadsheet is 0.27054 so the probability we want, that two of them will have the same Birthday is 1 - 0.27054 = 0.72946.

Basically this is happening because we are multiplying 30 numbers together that are all less than 1 and that gets small fast.

Danny

If there are 30 people in a room, what is the probability that 2 of them have the same birthday?

When I first came across this question, years ago, it was in the form of an entertaining party game…

But it does illustrate some great principles of Probability!

Assumptions

• There are 365 days in the year being considered.
• I don't count a Leap Year in my calculations.
• There are no twins (or triplets or quadruplets, for that matter) in the room
• Everyone knows the date of their own birthday (yes it sounds silly but I recall one party I attended where someone claimed not to know their birthdate). Maybe their loss of memory had something to do with their alcohol flow! I couldn't possibly comment!
• Everyone in the room is sober, or at least sober enough to remember their own birthdate.
• I've assumed an even distribution of birthdates, an assumption which is almost certainly wrong, but it makes the maths a lot easier.

Analysis

• Everyone in the room, all 30 of them, have an equal chance of being born on any particular date of the year.
• I recall reading some years ago that nine months after a major power failure in New York (I think it was), there was a significant increase in the birthrate.
• I suppose that if the nights were cold and dark, going to bed and hoping that your partner would keep you warm is a reasonable action!
• Equally there might be a significant increase in birthrates, 9 months after Christmas or New Year's Days, which would suggest a higher birth rate in September!.
• Notwithstanding the possibly higher birtrates due to the above factors, my analysis below, presumes that NO DATE has any higher or lower probability than any other date for the purposes of this question. An assumption that is almost certainly incorrect.
• Therefore for the 30 people in the room each of them has any of 365 days which could be their birthdate.
• Therefore all the 30 people in the room have 365 * 365 * 365 •••
• ( 30 times in total ) or ( 365 )³⁰ possible birthdates.
• That's rather a large number, ( 365 )³⁰ which by any stretch of the imagination, is an extremely large number.
• Rather than give myself eyestrain and a headache playing with a lot of large numbers, I can either proceed directly to work out the probability of two of the people in the room having the same birthdate, or deduce the probability that everyone has a unique birthday and deduct that probability from 100 percent, which will give me the probability that at least 2 of the people in the room have the same birthdate.
• That latter option seems far more sensible! (and time saving)

Key Facts

1. Everyone of the 30 people in the room, has to have been born on one of 365 days of a year.
2. Consequently the total combinations of possible probabilities is (365 )³⁰
3. Already my eyes are going blurry and my head is spinning.
4. In case you're wondering, ( 365 )³⁰ = 7.392408x10⁷⁶, (rather lot of zeroes!)
5. It seems to me to be easier to proceed on the assumption that of all the 30 people in the room each has their own unique birthdate(which they don't share with anyone else in the room), and subtract that probability from 1, which means that at least one person in the room does share a birthdate with someone else in the room.
6. Consider Person 1 in the room. They have 365 days which could be their birthdate, which is not shared with anyone else in the room of 30 people
7. We can therefore state that that Person 1 has a birthdaye of ====== =( 365/365 ) days in the year to chose from.
8. Person 2 has 364 days on which their birthdate might fall, as Person 1 has 1 of 365 days that could be their birthday, without sharing with anyone else in the room.
9. Therefore the Probability that Person 1 and Person 2 have different birthdates is ( 365/365 )( 364/375 )
10. Person 3 now enters the room. They must not have a borthdate on the same dates as Person 1 or Person 2.
11. So far the probability that all three people have unique birthdates is ===( 365/365 )(364/365 )( 363/365 )
12. When Person 4 enters the calculation is ( 365/365 )(364/365 )( 363/365 )( 362/365 )
13. The above process continues for each of the 30 people.on the room.
14. More and more people.enter the room, until the 30th person arrives.
15. The 30th person has (365 - 29 = 336) possible days to choose which could be their birthdate and is unique.
16. At this point we can state quite clearly that the total number of combinations of days in which none of the 30 people in the room have a birthdate in common is (365)*(364)* (363)*(362)*(361)*(360)*(359)*(358)*(357)*(356)*(355 *(354)*(353)*(352)*(351)*(350)*(349)*(348)*(347)*( 346 )*( 345 )*( 344 ) * ((343 )*(342)*(341)*(340)*(339)*(338)*(337)*(336)
17. We can show this calculation using factorial noation, or we can use
18. [(2.5104129675x10⁷⁷⁹)/(1.15632337x10⁷⁰²)] / [ 365 ]³⁰
19. At this point we can determine the Probability that out of 365 days, we have [ ( 365 )!/( 335 )! ]/[ 365 ]³⁰ where the denominator is the maximum number of combinations.

To solve for the total probability that, out of the original ( 365 )³⁰ days, there are 365!/335! of them where no one shares a birthday, we simply divide the latter by the former, that is (365!/335!)/(365³⁰)

If you do this immense calculation, (using either a computer or an advanced calculator, we can determine that the probability that none of the 30 people will share a birthday is 29.36837573%. (Obviously rounded!)

If we subtract this from 100%, we get the probability that a minimum of two people do share a birthday, which is 70.63162427% or very slightly more than 70%

One way of understanding this process is recognising that, for every additional person, we have to calculate the probability that their birthday does not match with any of the others’ birthdays.

For numbers going past a few people, this number does grow quickly, as each person has a certain number of people with whom they can’t share a birthday, and this is true for every one of that certain number of people.

Summary and Conclusion

1. I started with determining the Probability that each person has a different birthdate.
2. I obtain the Probability I want by subtracting that from probability of 100% or 1.
3. There are 365 possible birthdates,
4. The probability that the second person is different from the first person is ( 364/365 ).
5. By the same reasoning, the probability that the third persons if different from the first two is 363/365
6. This process continues until I reached the point where the 30th persons birthdate is different lfrom the previous 29 people in the room, which is 336/365.
7. At this point, I have identified the probability that all of these people have a different birthdate.
8. To complete the analysis I needed to understand that each of the 30 people has 365 days of the year that could be their birthday.
9. On that basis, the total combination of all the possible probabilities of birthdays for all of the 30 people is =( 365 )( 365 )( 365 ) … 30 times or expressed as, ( 365 )³⁰.
10. In retrospect the best way to answer this question is to revise the question by determining whether anyone in the room shares a birthdate with anyone else in the room.
11. At this point, if we deduct the probability that no-one in the room has the same birthday from 1 or 100,%, we have in effect deduced the complement, that is, the probability that at least 2 people have the same birthdate .
12. What I have shown by this process is that the easiest way to answer this question is to answer a marginally different question, then deduct from 100% to get the complement.
13. The question asked if there were 30 people in the room, what is the probability that 2 of them have the same birthfaye. The analysis above shows that the required percentage is ⁷0.61% ( rounded to 4 significant
14. A Final Thought and Summary

Pick the first person.

They will have some birthday.

The second person must have a different birthday than this and the probability of that is:[ ( 364 )/( 365 ) ]

The third person must have a different birthday than both of the first two and the probability of that is:[ ( 363 )/( 365 ) ]

Continue this until you do 30 people.

1. Now multiply these fractions together to get the probability that all 30 have different birthdays:

( 364/365 )∗( 363 )/( 365 )∗( 362 )/( 365 )∗…∗( 336 )/( 365 ) = ( 29.4% ) …

Note that there are 29 factors here.

If we subtract this percentage from 1, we get the probability that at least 2 people have the same birthday, that is:100% − 29.4% = 70.6

In the earlier analysis, there is a fundamental assumption, that is is that there is a uniform distribution of birthdays in the population.

It is not hard to show that this assumption is false.]

The breakeven point occurs when there are 23 people in the room.

This means that if there are 23 people in a room, there’s about a 50% chance that two or more will have the same birthday ( assuming a uniform distribution of birthdays in the population, which is far from correct. )

Probably not surprisingly, there are higher birth rates around 9 months after New Year’s Eve and Christmas Eve

We can of course use 365 days and ignore leap years for the sake of simplicity and sanity (It doesn’t change the final result much, as the numbers are so large)

If there are 2 people in a room with 2 birthdays each randomly selected between 1 and 365, then the chance of no “birthday collision” is ( 364 )/( 365.)

Information that wasn't asked for but interesting just the same!

The 10 least common birthdays.

In a world of scheduled c-sections and inductions, doctors and delivery staff prefer to be home with their families on the big holidays rather than helping expand someone else’s families.

Another day where births are less common?

The 13th. I guess some people are still superstitious about the “unlucky”day

What day are the most babies born?

The most popular day for babies to make their entrance in 2019 was Tuesday, followed by Thursday.

Sunday was the slowest day, followed by Saturday.

Scheduled c-sections and induced labours have a big influence on the reason why far fewer babies are born on the weekend, but spontaneous (unscheduled) deliveries occur less

When it comes to pregnancy and fertility in the United States, there are many statistics and facts to unpack.

How many babies are born in a day?

Worldwide, around 385,000 babies are born each day.

In the United States in 2019, about 10,267 babies were born each day. That's 1 percent less than in 2018 and the fifth year in a row that the number of births has declined.

Preliminary data for 2020 show another decline in birth numbers – this time of 4 percent, which will be the lowest number of births since 1979.

Experts are predicting a continued drop in 2021, too, perhaps as a result of women postponing pregnancies in light of the COVID-19 pandemic.

There was an 8 percent decline in babies conceived in December 2020 compared to the previous December.

What day are the most babies born?

The most popular day for babies to make their entrance in 2019 was Tuesday, followed by Thursday.

Sunday was the slowest day, followed by Saturday.

What time of day are the most babies born?

The highest percentages of births in 2019 occurred during the midday hours (noon to 3), followed by the morning hours of 9 to noon.

The fewest babies were born between midnight and 6:59 am.

What month are the most babies born?

In 2019, more newborns arrived in August than in any other month.

The second, third, and fourth most popular birthday months were July, September, and October, in that order. The least busy month for birth days was February.

According to preliminary data, the number of births for the U.S. declined for each month in 2020 and larger declines in births were seen in the second half of 2020 compared with the first half of the year.

Just in Case you Wanted to Know!!

PB

Let’s start by figuring out the probability that nobody has the same birthday as anybody else, then subtract that probability from [math]1[/math].

(By the way, I’m ignoring leap years and assuming there are [math]365[/math] unique birthdays).

For [math]n[/math] people in a room, the total number of combinations of birthdays without repetition is given by:

[math]V_{nr} = \frac{365!}{(365-n)!}[/math]

While the total number of combinations with repetition is given by:

[math]V_{r} = 365^n[/math]

And the probability that no two people have the same birthday in a group of [math]n[/math] people is given by:

[math]P(A) = \frac{V_{nr}}{V_{r}}[/math]

The probability that two people do share a birthday is therefore:

[math]P(B) = 1 - P(A)[/math]

You were asking about [math]100[/math] people:

[math]V_{nr} = \frac{365!}{(365-100)!}[/math]

This number is unspeakably huge, so we’re not going to calculate it outright.

[math]V_{nr} = 365^{100}[/math]

Another huge number.

[math]P(A) = \dfrac{\frac{365!}{265!}}{365^{100}} \approx 3.1\times10^{-7}[/math]

(Thanks WolframAlpha!)

[math]P(B) = 1 - (3.1\times10^{-7}) = 0.99999969 = 99.999969\text{%}[/math]

It might be surprising that the probability is so high. Here’s an analogy: imagine that you are asked to throw 100 darts — blindfolded — at a board divided into 365 wedges. Chances are extremely good that you’re going to hit the same wedge twice.

This is the explanation on the web.

Hope i help!

Footnotes

My answer is slightly different from the ones below. Among the 24 people in the room, there are 276 comparisons to be made (1+2+3+4…..+22+23). In each comparison, the probability that the birthdays are different is 364/365. This to the 276th power is 0.46898, which leaves a 53.1% chance that there is at least one match.

While the answer isn’t going to be vastly different if we include Feb 29, the setup would be as it turns it into a conditional probability problem. If someone’s birthday is Feb 29, the probability of someone having a different birthday is 1460/1461 rather than 1457/1461. But we can’t build that into the problem too easily since we don’t know the birthday of the first person with whom we start (or whether ANY of them were born on Feb 29). If 1460/1461 were needing to be used more than once, it would mean there are at least two people with that birthday, which would automatically increase the answer to 100%.

But we could specify a probability for the question, assuming one and only one person has the birthday of Feb 29. That would be 1-(364/365)^253. In this scenario, the Leap Day baby plays no real role. It would be the same as answering the question “What is the probability that in a room of twenty-THREE people, at least one pair has the same birthday, assuming no one was born on Feb 29?”

Others have answered this well in the case of a uniform distribution of birthdays. Assuming a uniform distribution between all the dates of the year gives the lowest probability for birthday coincidences to occur. At the other extreme is the distribution that would result if all people were born on the same day. Of course, this is the probability of coincidence (2 or more), not of precisely 2. If all were born on the same day, the probability of exactly 2 would be 0.

In any case, birthdays are not evenly distributed. Read more at this Quora link: What is the distribution of birthdays throughout the course of year?