Complementation Law;

[math]\mathbb{P}(A)^c=1-\mathbb{P}(A).[/math]

Using complementation law in Birthday Paradox.

[math]\mathbb{P}(atleast[/math] [math]2[/math] [math]have[/math] [math]same[/math] [math]Birthday)=1-\mathbb{P}(noone[/math] [math]have[/math] [math]same[/math] [math] Birthday).[/math]

Calculation of Probability of no [math]2[/math] having same Birthday.

[math]\mathbb{P}(1B)=(\frac{365}{365}).[/math]

[math]\mathbb{P}(2B)=(\frac{364}{365}).[/math]

[math]\mathbb{P}(3B)=(\frac{363}{365}).[/math]

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[math]\mathbb{P}(70B)=(\frac{365-69}{365}).[/math]

Since;[math]\mathbb{P}(noBirthday)=(\frac{365}{365})(\frac{364}{365})(\frac{363}{365})...(\frac{365-69}{365})[/math]

Therefore;

[math]\mathbb{P}(2Birthday)=1-\mathbb{P}(noBirthday).[/math]

Beautifying the equation with Factorial.

[math]\mathbb{P}(2Birthday)=1-(\frac{365!}{(365^{70})(365-70)!})=0.9999[/math]

Concept Required;Probability Theorem.

No. It is about 99.916%, a bit less than 99.99%.

[math]\displaystyle 1-\frac{365!}{(365-70)!365^{70}}=0.99916\ldots[/math]

This assumes a uniform distribution over 365 possible birth dates. There are actually 366, but one of them is significantly more rare than the others. If you allow 366 birth dates, the answer changes to roughly 99.914%.

The precise probability, taking into account the likelihood of being born on Feb 29, is somewhere between these two figures. Determining the exact value requires assumptions on the distribution of birth years among the attendees of the party.

Furthermore, birth dates are not uniformly distributed through the year

Such concentration effects will increase the likelihood of birthday collisions among any number of people, likely beyond the 99.99% figure for 70 people mentioned in the question.

Footnotes

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.

It’s much easier to work out the probability that nobody has same birthday as anybody else in the room first. Let’s say their birthday is independent

For the first person in the room,he or she’s birthday could be any day in a year,and the second person’s birthday could be any day but the first person’s birthday. etc.etc.

So the probability that nobody has the same birthday as anybody else in the room is

[math]\displaystyle\prod_{n=0}^{99} \dfrac{365-n}{365}[/math]

[math]=\displaystyle\prod_{n=0}^{99} \dfrac{365-n}{365} \dfrac{(365-100)!}{(365-100)!}=\dfrac{365!}{365^{100}(365-100)!}=\dfrac{365!}{365^{100}265!}[/math]

OK,modern solution:

And don’t forget that the final answer is 1 minus this number

So the answer is[math]8-3.0724892785158286e[/math]

Footnotes

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.

It’s not easy to give an exact answer because dates of birth are not evenly distributed. Also, what if you’re looking at a convention of “people who were drafted to serve in Vietnam in the first year of the draft lottery”? They’ll all be drawn from some restricted set of dates of birth, the dates that the lottery picked early.

But in the normal course of events, it’s a very high probability.

By the time you’re up to 60 people, say, if they all have different birthdays, then every one of the remaining 40 have to at any rate miss those 60 numbers. So the chance that they all do is in the ballpark of (5/6)^40 which is pretty small. Well less than 1 in 1000.

Think of any two people as a pairing.

Two people, one pairing:

Three people, three pairings:

Four people, six pairings:

Five people, ten pairings:

Generally speaking, any n people make (n)(n - 1)/2 pairings (also known as triangular numbers). And any one pairing has 1/365 chance to have the same birthday. So if you have, for example, 5 people, the chances that at least two of them have the same birthday is [math]1 - (364/365)^{10} = 1 - 0.973 = 0.027[/math]

That’s pretty low probability, but as you add more people, it goes up fast. You need only 23 people for the probability of one pair of same birthdays to be over 50%:

[math]23*22/2 = 253[/math]

[math]1 - (364/365)^{253} = 1 - 0.4995 = 0.5005[/math]

75th triangular number is 75*74/2 = 2775. So the chances of at least two out of 75 have the same birthday is [math]1 - (364/365)^{2775} = 1 - 0.00049 = 0.99951[/math]

“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.)

Its quite straight forward. Assume there are n random dudes under consideration. The probability we are interested is
P(n) = the probability of at least 2 dudes among the n having the same birthday.
But its complement P'(n) is much more easier to calculate.

P'(n) = No two dudes among the n dudes has same birthday.
= [math]\frac{365}{365}*\frac{365-1}{365}*\frac{365-2}{365}*..*\frac{365-(n-1)}{365}[/math]
= [math]\frac{365!}{(365-n)!*365^n}[/math]

Of course. P(n) = 1-P'(n)
In your case,p(42) = 1 - 0.086
= 0.914

A quick intuitive approach to see why this probability is so high.

Note: this is a quick approximation.

There are [math]{75 \choose 2} = \frac{75 \times 74}{2} = 2775[/math] pairs of people. For each pair of people, there is a [math]\frac{1}{366}[/math] chance that they have the same birthday. Or, for each pair there is a [math]\frac{365}{366}[/math] chance that they do not share a birthday and we look at [math]\left(\frac{365}{366}\right)^{2775} \approxeq 0.0005[/math].

Now, that was just an approximation. The actual probability of people sharing a birthday will be higher than that. Why? Well, let’s consider the case where there are 400 people. Now it is absolutely certain there will be people sharing a birthday, but the above reasoning would give a value just a tad bit below [math]100\%[/math].

That’s because we did not take into account that, if the first two people we look at do not share a birthday, there’s only [math]364[/math] possible days left for the other people. The probability of no two people sharing a birthday becomes: [math]\frac{365}{366} \times \frac{364}{366} \times \frac{363}{366} \times \cdots \times \frac{366 - 74}{366}[/math]. This gives a probability of about [math]0.0003[/math] that no two people share a birthday.

The simplest way to think of it is that, if there were 367 people, it would be impossible for them not to have the same birthday—there are only 366 birthdays to go around.

If there are 366 people, it’s almost impossible—each one has to be born on a separate day. Once you’d found 365 of them all had different birthdays, there’s only a very tiny chance (one in 366) that the last person had the right birthday for them all to have different birthdays.

But even the 364th person had only a two in 366 chance to be different.

Even halfway through the list, after you’d found 183 people, all with different birthdays, the next person already has a 50/50 chance of breaking the streak.

So, the farther you get into the list, the smaller chances of keeping the streak alive. The 94th person has a one in four chance of breaking the streak, but you’ve got more than 250 people to go, all of whom have a better than one in four chance of breaking the streak.

Obviously, as you work backward, from 94 to 75, the odds of the last person in the lists breaking the streak get better and better, but the list has to be pretty short in order for none of them to break the streak.

Suppose you have 22 people, none of whom share the same birthday. The next random person has about a 94% chance of keeping the streak alive—but to get to that point, you’ve already had twenty two tries at breaking the streak, which gets harder at each step.

It turns out the 23rd person is where you hit the break-even point. Yes, given that you already have 22 who don’t share the same birthday, there’s only a 94% chance that the 23rd will match one of them, but it’s hard to get to 22—in around two in five times, someone in the first 22 will already have broken the streak.

It’s that interplay between the work needed to get to stage n with the streak still alive, and the diminishing chances that the [math](n+1)[/math]th person can keep the streak alive that makes the break even point so surprisingly low.

There is not as far as I know because it belongs to another field that science just ignores and considers silly.

I am not even sure if science would even BOTHER to look at that stuff and keep an open mind.

Like attracts like whether they realise it or not.

Birthdays..OK...let's talk astrology here a bit and not laugh.

Ex: I was in the travel industry..the amount of people with sagittarius suns, or moons or prominent Jupiter ascendants were very noticeable. Jupiter/Sag very prominent. More than normal.

Somebody I knew : in his industry..Gosh...you are all LEOS ! the Leo people tended to like and employ those they liked best in terms of leadership, qualities...they employed their own..

I have seen that scenario again and again and again..

Once in an office ..out of 10 people, 8 of us were sagittarius and the 2 others .scorpios.

Totally defying the odds. And yes I did have same birthday with one.

I am not going to go further because statistic minded people will tell you ..heck of course it is a game of chance ..

not to me..

All right. So, let’s spell this out. 365 days — we asked that anyone born on February 29th stays home — and we assume that there’s even odds that anyone was born on the same day.

For 2 people, Alice and Bob, there’s 365^2 possibilities — Alice and Bob born on 1/1, Alice born on 1/1 and Bob born on 1/2, etc. But there are 365 ways Alice and Bob can share a birthday. So that’s 1/365 of a chance.

For 3 people (Alice, Bob and Carol), there’s 365^3 possibilities. There are 365*364 ways for Alice to share a birthday with Bob and not Carrol. There are also 365*364 ways for Alice to share a birthday with Carol and not Bob. There are 365*354 ways for Bob and Carol to share a birthday, but not Alice. And there’s 365 ways for Alice, Bob and Carol to all have the same birthday.

So let’s write that mathematically to find the odds.

[math]\frac{3\times365\times364 + 365}{365^3} = \frac{3\times 365 - 2}{365^2} \approx \frac{3}{365}[/math]

Now, what this means is that because we compare three pairs (and one trio), we have more combinations than one pair, even though there’s also more total possibilities. So the fraction goes up.

With four people, we have six pairs (Alice-Bob, Alice-Carol, Alice-Dave, Bob-Carol, Bob-Dave, Carol-Dave), four trios (Alice-Bob-Carol, Alice-Bob-Dave, Alice-Carol-Dave, Bob-Carol-Dave), and the foursome. We include the trios and foursome to make sure we don’t double-count, but consider that the number of pairs we have goes up faster than the number of people. Instead of 2, 3, 4, 5 we have 1, 3, 6, 10… this makes the number of successes increase faster than the number of all combinations.

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.

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 

As written, one cannot. (I used to write and review items for a well-known testing company.)

A school has 15 classes, okay. With 26 students. Umm, 26 total? In each class? 15 classes meaning five each trimester? Can students be in more than one class?

What is meant here by “celebrate”? Likely some percentage of students don't “celebrate” their birthdays.

The only real answer would be “hire enough private eyes to follow all students all the time, and observe until two students hold birthday celebrations on the same day.”

Now, to answer the intended question, look up the “pigeon hole principle” and figure out how it applies to your homework.

The approximation in this case would probably not be as good for 2 reasons:

1. The distribution of births year to year is not as uniform as the distribution of births day to day. Therefore, if we estimate with a uniform distribution, we will probably get a number of people required that is greater than the true number required.
2. Leap years become another problem, because on average there would be 2 or 3 leap years within a 10 year period. I’ll treat it as if there are exactly 2 leap years in my calculation. I’ll act as if you have instructed people to count their birthday as the leap day if they are born on the leap day. Some people born on leap days count their birthday as one of the non-leap days.

The number of possible birthdays in 10 years would be 3652 in that case.

If you treat the birth days as uniform, that means the calculation would be like so:

[math]1-\dfrac{_{3652}P_{n}}{3652^n}[/math]

That calculation first exceeds 0.50 when [math]n=72[/math]

Because of reason #1, the true number of people required for a 50% probability would likely be less than that, but how much less is the question. Similarly, the probability of a match is probably greater than the one given by the formula, but how much greater is unknown.

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).