55

For any one relationship, the odds of sharing the same month and day are approximately 1 in 365 (not exactly because of leap year and because births are not exactly evenly spaced within a year. If you add in year, it's probably something like 1 in 3000 or 4000 (most people have relationships with people relatively close in age). But that' a priori. That ...


38

As Peter pointed out, it is impossible to calculate coincidences after the fact. Your question got me thinking, and I realized my girlfriend and I also have a strange birthday coincidence. She was born exactly 432 days before me! And we are also in a successful relationship! I don't know what this probability is, but it is the exact same as yours!


21

Sorry, a bit new here so please excuse me if this doesn't help too much. The US Social Security Administration keeps records of births and deaths and has their information available for purchase (apparently for a hefty price): Here However I found a source that claims to have bought it and is offering it for free (as well as offering the data sorted by ...


18

Luckily someone has posted some genuine birthday data with a bit of discussion of a related question (is the distribution uniform). We can use this and resampling to show that the answer to your question is apparently 23 - the same as the theoretical answer. > x <- read.table("bdata.txt", header=T) > birthday <- data.frame(date=as.factor(x$date)...


13

Assuming all birthdays are equally likely and the birthdays are independent, the chance that $k+1$ aliens do not share a birthday is $$p(k;N) = 1\left(1-\frac{1}{N}\right)\left(1-\frac{2}{N}\right)\cdots\left(1-\frac{k}{N}\right).$$ Its logarithm can be summed asymptotically provided $k$ is much smaller than $N$: $$\log(p(k;N)) = -\frac{k(k+1)}{2N} - \...


11

We can be even more precise than @Mike Shi's data: the most dangerous of all birthdays is the very first one. The 1st day mortality rates reported there are around 0.2 % for industrialized countries and 0.8 % average for all countries. Which means that the risk of dying on the day of birth is at least as high as the risk of dying at any of the following ...


10

So, first of all, the odds of sharing some weird connection with any random person are probably quite high. From experience I'd guess around 20% or so, no way to seriously calculate that, but no matter what it exactly is, just want to be clear having a special weird connection means nothing (though it is fun). Then, something the other didn't take into ...


9

If it's an event specified before the fact, you can simply break it down: The chance that your boyfriend was born the same year as you is actually very high (especially given many situations tend to bring people of very similar age together); it's a very difficult probability to calculate, though, without data. If you had that probability, P(Same day and ...


9

Notice that your answer is never equal to $1$ regardless of how high $n$ is. However, obviously if $n=366$ then there must be two people with the same birthday. So basically, the correct answer captures the fact that for everyone to have a different birthday, you begin running out of dates the more people are in the room.


8

Yes, the independence assumption is invalid. Suppose you know that Alice and Bob have different birthdays, and you also know that Bob and Carol have different birthdays. What is the chance that Alice and Carol have different birthdays? Is it still 364/365? No. Suppose Bob was born at day X. Alice was NOT born at the day X. Since X is not the Carol's ...


8

Taking the question literally According to wikipedia, 33.2% of married couples in the United States differ in age by less than one year. Thus, a baseline estimate for sharing the same date of birth would be the above statistic divided by two (because it captures 2 years) for sharing the same year multiplied by the probability of sharing the same birthday: $...


8

I think the logic is wrong since the probability of two persons having different birthdays is dependent on the fact that they need to have different birthdays than all the others. A simple example birthday paradox for A,B and C not having birthday on the same weekday. Each of these pairs are 1/7 in a vacuum. But given A had birthday an a Monday and B and C ...


7

The $i^{th}$ iterm will be selected $\text{Binom}(m, \, 1/n)$ times. From this, you can find all the quantities you want, because, e.g., $$\mathbb{E}[\text{number of pairs}] = \sum_{i = 1}^n \mathbb{P}[i^{th} \text{ item appears twice}] $$ For example, the expected number of pairs is given by $$ n \cdot \mathbb{P}[\text{Binom}(m, \, 1/n) = 2]. $$ You can ...


6

It is possibly in the language used. In mathematics, language is very important, so one has to be careful how something is described or discussed. Also, these situations are a little different, though it is important not to conflate the concepts of order and replacement. Language aspect: A 'hand' in cards does not denote order. 'number of possible ...


5

Here's an argument why the probability of death on the birthday may be higher than on other days: Birthdays are emotionally charged days. More over, people tend to celebrate it somehow.. So there is an excess of factors (relative to the person's usual life style) that increase biological stress (excess emotions, excess drinking, excess eating, excess dancing,...


5

The events "$k$ people have a birthday on Jan 1" and "$k$ people have a birthday on Jan 2" are not mutually exclusive, and so the probability that either of them occurs is not equal to the sum of their individual probabilities. You would need to subtract the probability that both events occur, but this is likely to become extremely complex once you scale up ...


4

It actually does not. The birthday "problem" as usually stated (there's actually no problem...) only relies on days of the year (and an independence assumption). Most of the time intuition in statistics must be double-checked by calculation, in here the calculation (no matter how you do it), does not use the year, hence the problem is independent from the ...


4

There is not much difference between the two. In case of individuals the probability of two people sharing birthdays is $1/365$. In case of couples it is $1/365^2$, that's all. You just replace this number and it will do. You can think about it in terms of each couple having a sort of joint birthday, a total of $365^2$ possibilities instead of $365$.


4

Let's call our random variables $X_1$, ... $X_n$. I assume that you meant that they are integer valued, also look at the comments to your original question. Let's look at the case that there is a unique minimum which is $j$ and is assumed by $X_i$ (and only $X_i$). The probability for this case is $$p(X_i = j) \prod_{k \neq i} p(X_k > j) = (\frac{1}{x})(\...


4

Although the question is about birthdays, the "birthday paradox" isn't really relevant here. It's about how many random samples you need to take before you expect at least two samples among them to be equal (a collision). Your question is mostly about the probability of two samples being equal. If there were 30 people in your relationship then you'd expect ...


4

Indeed, the problem was already solved. It's a generalization of Birthday paradox. The formula for probability mass function is The probability of getting $k$ unique values from $[0, n)$ when choosing $m$ times is given by: $$P(V = k) = \binom{n}{k}\displaystyle\sum_{i=0}^k (-1)^i \binom{k}{i} \left(\frac{k-i}{n}\right)^m $$ where $V$ is a ...


4

If you have three days $x$, $y$, $z$, the events $x\ne y$, $x\ne z$, and $y\ne z$ are not independent, but you treat them as such.


3

The chances for this to happen.... two people having their birthday on the same day as explained by the other posters is 1/365 * 1/30 to be conservative here with the age ranges. To be in a relationship, a successful one multiply by maybe 1/2 or 1/3?! However, for you to be in a relationship, you first have to be here. For you to be here, your mom and dad ...


3

For a given value of $(n,m)$ [to be read as (days,draws)], one can compute exactly $$\mathbb{P}(V = k) = \binom{n}{k}\displaystyle\sum_{i=0}^k (-1)^i \binom{k}{i} \left(\frac{k-i}{n}\right)^m$$for all values of $k$ at a cost of $\text{O}(n²)$ and then simulate exactly from this distribution by generating $U(0,1)$ variates and comparing them with the ...


3

I'll answer question b) because it's more general, and question a) can just be thought of as a special case of b) where the adjacency matrix is simply the identity matrix. I'll give you the exact method, though approximate methods might be called for because the computation of the exact solution scales rapidly with number of people. I don't think there's a ...


2

I agree with Erik's answer in the case of a uniform distribution. Here I will assume a general distribution $P(X_i = j) \doteq p_j$ for all $j = 1,...,x$ (obviously $1 = \sum_{j=1}^xp_j$). $$P(\text{there is a unique minimum}) = \sum_{k=1}^nP(X_k\text{ is the unique minimum})$$ $$ = nP(X_1\text{ is the unique minimum})$$ $$ = n\sum_{j=1}^xP(X_1 = j,X_1\text{...


2

In the usual context of the birthday problem, if the first two values are not distinct the problem is solved and the third and subsequent values are irrelevant. What is confusing here is that you have written Explanations of the birthday problem state that if we sample a third time, the probability that this third sample is distinct from the first ...


2

Assuming you aim for the case that exactly (so not at least) 11 double birthdays and 1 triple birthdays occur: $$\begin{align}p(60,n_2=11,n_3=1) &= \frac{\text{possibilities with 11 double birthdays and 1 triple birthdays}}{\text{all possibilities}}\\ &= \frac{\frac{60!}{35!22!3!} 21!! \,\cdot \,365 \cdot 364 \cdot \, ... \, \cdot (365-n+1+1\...


2

When in doubt, simulate. (I'm sure that you can actually put together a formula, but it will likely be painful to look at.) I'll work with the criterion being "at least one triple birthday" and "at least eleven double birthdays" -- changing the code below to checking for exactly so many birthdays is not hard. n.sims <- 1e5 n.persons <- 60 counter <...


2

There's no birthday-paradox effect here. In the birthday paradox, the somewhat unintuitive solution comes from the fact that you have several people and the possible ways of choosing two increase quadratically with the total number of people in the room. In your problem, you just draw two random numbers from the range $[0,255]$ repeatedly (you could map that ...


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