Probability of being born on a leap day?

Question

Given that today is a leap day, does anyone know the probability of being born on a leap day?

Answer 1

To accurately predict that probability using statistics, it would be helpful to know where the birth took place.

This page http://chmullig.com/2012/06/births-by-day-of-year/ has a graph showing a subset of the number of births per day (multiplying the 29th by 4, which is incorrect, and undesirable for this question, but it also links to the original data and gives a rough indication of what you can expect) in the United States. I would assume that this curve doesn't hold true for other countries, and especially not for other continents. In particular the southern hemisphere and equatorial region may show a substantial derivation from these results - assuming that climate is a determining factor.

Furthermore, there's the issue of "elective birth" (touched upon by the authors of http://bmjopen.bmj.com/content/3/8/e002920.full ) - in poorer regions of the globe, I would expect a different distribution of births, simply because (non-emergency-) cesarian sections or induced birth are rarer than in developed countries. This skews the final distribution of births.

Using the American data, assuming ~71 Million births (rough graphed mean * 366) and 46.000 births on February 29ths, not correcting for the distribution of leap years in the data, because the precise period is not indicated, I arrive at a probability of around ~0.000648. This is slightly below the value one would expect given a flat distribution of births, and thus in line with the general impression give by the graph.

I'll leave a significance test of this rough estimation to a motivated reader. But given that the 29th (though uncorrected - the year 2000 injects a below average bias into the data) scores low even for the already low February standards, I assume a relatively high confidence that the null-hypthosesis of equal distribution can be rejected.

Answer 2

Sure. See here for a more detailed explanation: http://www.public.iastate.edu/~mlamias/LeapYear.pdf.

But essentially the author concludes, "There are 485 leap years in 2 millennia. So, in 2 millennia, there are $485(366) + (2000-485)(365)= 730485$ total days. Of those days, February 29 occurs in 485 of them (the leap years), so the probability is $485/730485=0.0006639424$"

Answer 3

I think the answer to this question can only be empirical. Any theoretical answer would be flawed without accounting birthday selection phenomena, seasonality etc. These things are impossible to deal with theoretically.

The birthday data is hard to find in US for privacy reasons. There's one anonymous data set here. It's from insurance applications in USA. The difference from other reports, such as a popular often cited NYT article, is that it lists the frequency of births by date, instead of simple ranking of days in a year. The weak point is of course the sampling bias, since it comes from insurance: uninsured people are not included etc.

According to the data there were 325 births in Feb 29 of total 481040. According to the Roy Murphy, the sample spans from 1981 through 1994. It includes 3 leap years of total 14 years. Without any adjustments the probability would be 0.0675% of being born on Feb 29 between 1981 and 1994.

You can adjust the probability by accounting for the frequency of leap years, which is close to 1/4 (not exactly though), e.g. by multiplying this number by $14/12$ to arrive to 0.079% estimate. Here, the conditional probability $p$ of being born on Feb 29 in a leap year is linked to the observed frequency $F_o=325$ by the frequency $f_L=3$ of leap years in a sample: $$F_o=f_L/N\cdot F\cdot p,$$ where $N=14$ is the number of years in a sample, and $F=481040$ is the total frequency of births.

Normally, the probability of leap years is $p_L\approx 1/4$, hence, the long run average probability $P_L$ of being born on Feb 29 is: $$P_L=p_L\cdot p\approx \frac{p_L\cdot N}{f_L} \frac{F_o}{F} \approx 0.079\%$$

You might be interested in the conditional probability $p$ of being born on Feb 29 given that you were born on leap year: $$p= \frac{N}{f_L} \frac{F_o}{F}\approx 0.32\%$$

So, the link between $P_L$ and $p$ is based on some couple of assumptions, e.g. that the probability of being born in any given year is uniform, and doesn't change.

Of course, this discussion was US centric. Who knows what are the patterns in other countries.

UPDATE: We automatically assumed that OP is Gregorian calendar. It gets even more interesting if you consider different calendars such as lunar calendar Hijri, where the leap years are every 30 years or so.

UPDATE 2:

What's surprising is that estimated probability $p$ leads to the expected occurrence of birthdays in Feb 29 for this sample: $F\cdot p=1,527$. This is only lower than Jan 1 and Dec 25, which is consistent with NYT's ranking above! They don't describe the source of data, referring to only Amitabh Chandra, Harvard University, but it's either the same or the finding is robust.

Now, how likely it is that very peculiar days in Gregorian calendar: Jan 1, Dec 25 and Deb 29 would come randomly as the most popular birthdays? I say it's highly unlikely a random occurrence. Hence, it's even more interesting to see what's going on in other calendars such as Hijri.

UPDATE 3:

Note that both $P_L,p$ are higher than naive theoretical estimates:

$$\hat p\approx 1/366\approx 0.27%$$ $$\hat P_L\approx p\cdot\frac{366}{365*4+1}\approx 0.068%$$

UPDATE 4:

Ben Millwood made comment that the distribution of births by day of year is non-uniform. Can we test this statement? Using my data set we can run $\chi^2$ test on the theoretical distribution with a null hypothesis that the distribution is uniform. The result is the rejection, i.e. the distribution doesn't seem to be uniform.

The theoretical distribution is built like this. We assume that the birth frequency is uniform across all calendar days, i.e. in 14 years span across $14*365+3$ days. Then we roll up the days into days of year, which is 366. Obviously only 3 leap days were encountered and 14 non-leap days. Below is my MATLAB code and the distribution plot for comparison of theoretical and empiricals.

d=[0101 1482
...
1231 1352];
%%
tc = sum(d(:,2)); % total obs

idL = 60; % index of Feb 29

% theor frequency, assuming uniform
ny = 1994 - 1981 + 1; % num of years
nL = 3; % # of leap years: 1984, 1988, 1992
nd = 365*ny + nL; % total # of days

fc = tc/nd; % expected freq for calendar date in sample
td = ones(366,1)*fc*ny; % roll the dates into day of year
td(idL) = fc*nL;

fprintf(1,'non-leap day expected freq: %f\n',td(end))
fprintf(1,'leap day expected freq: %f\n',td(idL))
fprintf(1,'non-leap day average freq: %f\n',mean(d([1:idL-1 idL+1:end],2)))
fprintf(1,'non-leap day freq std dev: %f\n',std(d([1:idL-1 idL+1:end],2)))
fprintf(1,'leap day observed freq: %f\n',d(idL,2))

% plots
bar(d(:,2))
hold on
plot(td,'r')
legend('empirical','theoretical')
title('Distribution of birth dates 1981-1994')
set(gca,'XTick',1:30:366)
set(gca,'XTickLabels',[num2str(floor(d(1:30:366,1)/100)) repmat('/',13,1) num2str(rem(d(1:30:366,1),100))])
grid on

% chi^2 test
[h p]=chi2gof(d(:,2),'Expected',td)

OUTPUT:

non-leap day expected freq: 1317.144534
leap day expected freq: 282.245257
non-leap day average freq: 1317.027397
non-leap day freq std dev: 69.960227
leap day observed freq: 325.000000

h =

     1


p =

     0

Answer 4

My all-time favorite cover to a book provides some highly relevant evidence against the assumption of a uniform allocation of births to dates. Specifically that births in the US since 1970 exhibit several trends superimposed on each other: a long, multi-decade trend, a non-periodic trend, day-of-week trends, day-of-year trends, holiday trends (because procedures like Cesarean section allow one to effectively schedule the birthdate, and doctors often don't do those on holidays). The result is that the probability of being born on a randomly-chosen day in a year is not uniform, and because birth rate varies between years, not all years are equally likely, either. So the answer that just checks how many leap years there are in some interval and reckons from the calendar is making very strong assumptions which have little utility in describing the real world in any reasonable way!

This also provides evidence that Asksal's solution, while a very strong contender, is also incomplete. A small number of leap days will be "contaminated" by all off the effects at play here, so Asksal's estimate is also capturing (quite by accident) the effect of day-of-week and long-term trends along with the Feb. 29 effect. Which effects are and are not appropriate to include are not clearly defined by your question.

And this analysis only has bearing on the US, which has demographic trends which might be quite different from other nations or populations. Japan's birth rate has been declining for decades, for example. China's birth rate was regulated by the state, with some consequences for its nation's gender composition and hence birth rates in subsequent generations.

Likewise, Gelman's analysis only describes several recent decades, and it's not necessarily clear that this is even the era of interest to your question.

For those who get excited about this kind of thing, the material in the cover is discussed at length in the chapter on Gaussian processes.

Answer 5

February 29th is a date that occurs each year that is a multiple of 4.

However years that are a multiple of 100 but aren't one of 400, are not considered as leap years (E.g: 1900 is not a leap year while 2000 or 1600 are). Therefore, nowadays, it is the same pattern every 400 years.

So let's do the maths on a [0;400[ interval:

On a 400 years period, there is exactly 4 x 25 = 100 years that are a multiple of 4. But we have to subtract 3 (years multiple of 100 but not of 400) from 100, and we get 100 - 3 = 97 years.

Now we have to multiply 97 by 366 , 97 x 366 = 35502 (number of days in a leap year in a 400 years period), it remains (365 x (400-97)) = 110 595 (number of days that aren't in a leap year in a 400 years period).

Then we just have to add these two numbers in order to know the total number of days in a 400 years period: 110 595 + 35502 = 146 097.

To finish, our probability is the number of February 29th in a 400 years period so 97 given that there is 97 leap years divided by the total number of days of our interval:

p = 97 / 146097 ≈ 0,0006639424492

Hope this is right and clear.

Answer 6

I believe there are two questions being mixed up here. The one is "What is the probability of any given day being a Feb. 29th?". The second one is (and the one actually asked) "What is the probability of being born on a leap day?"

The approach of simply counting days seems to be misleading as Aksakal is pointing it. Counting days and calculating frequencies of Feb. 29th occuring addresses the question: "What is the probability that any given day is a Feb. 29th?" (Imagine waking up after a coma, no clue what day it is. The probability of it being a Feb. 29th is as pointed out above $p = \frac{97}{146097}\approx 0,00066394$).

Following Aksakal's answer, the probability can just be based on empirical studies of the distribution of births across the days of the year. Different data sets will come to different conclusions (e.g. due to effects of seasonality, long-term trends in birth rates, cultural differences). Aksakal pointed out a study (One comment: to account for the unrepresentative occurence of a leap year in the mentioned data (i.e. $\frac{3}{14}$) compared to the long-term frequency of leap year occurences (i.e. $\frac{97}{400}$) you would have to multiply the frequency of birth on Feb. 29th from the sample by $\frac{97}{400}\cdot\frac{14}{3} = \frac{679}{600} \approx 1.131667$).

Finally, there is a third possible interpretation of the question, which I believe was not intended though: "What is the probability of a specific person being born on a leap day?" Well, for anyone already born that is easy. It is either $0$ or $1$. For anyone not born but already conceived it also can be estimated using empirical studies on the length of pregnancy (see Wikipedia for an overview). For anyone not conceived yet, see above.

Answer 7

I've noticed that most of the answers above work this out by calculating the number of leap days in a particular period. There is a simpler way to get the answer, 100% accurately, by definition:

We use leap years to adjust the regular (365 day) calendar to the mean tropical year (aka mean solar year). The mean tropical year "is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth" (Wikipedia). The tropical year varies slightly, but the mean (average) tropical year is ABOUT 365.24667.

If out leap days are correct, then the chance of a randomly selected day being a leap day, is ((tropical year) - (non-leap-year)) / tropical year

Pluging in the approximate number we have, it's (365.24667-365)/365.24667, or 0.24667/365.24667, or 675 per million (0.0675%).

This, however, is for a randomly selected day. I imagine that this is substantially skewed by parents who would rather not have to explain to their kids, "your actual birthday only comes once per 4 years".