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

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29  
Note that births are not distributed uniformly throughout the year, so the probability of a randomly-chosen day being a leap day is not the same as the probability of being born on one. – Ben Millwood Feb 29 at 13:04
16  
Of whom being born? All people in history? Today? All people alive? Prospectively into the future? Probabilities are meaningless unless the events to which they refer are well-defined. – whuber Feb 29 at 13:55
13  
100% of the people born today will be. Does that help? – aslum Feb 29 at 14:08
8  
Lots of parents don't want their kids to be born on leap day. So with the rise of scheduled C-sections, the probability will be lower than a random day. fivethirtyeight.com/features/… – James Lawruk Feb 29 at 17:11
2  
I agree with @whuber, that the question is ill-defined. Without proper definition of probability space the question cannot be answered. Hence the downvote. – mpiktas Mar 1 at 8:25
up vote 24 down vote accepted

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 $485/730485=0.0006639424$"

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9  
Why it cant be calculated as 1/(Number of days in 4 years) = 1/1461 = 0.00068 ? – Siddhesh Feb 29 at 8:42
20  
@Siddhesh There is a rule regarding the centuries. So e.g. 2100 is not a leap year – Floo0 Feb 29 at 8:46
7  
@Siddhesh, Unfortunately, it's not quite that simple. Leap years a bit more complex. An average year length is actually, on average, 365.2425 days not 365.25. As written on the wikipedia leap year page, "The Gregorian calendar . . . removes three leap days every 400 years, which is the length of its leap cycle. This is done by removing February 29 in the three century years (multiples of 100) that cannot be exactly divided by 400.[3] The years 2000 and 2400 are leap years, while 1800, 1900, 2100, 2200, 2300 and 2500 are common years." – StatsStudent Feb 29 at 9:09
10  
I don't see why you need to consider 2000 years; leap years are on a 400 year cycle, so why not just reduce to "there are 97 leap years in 400 years"? – Philip Kendall Feb 29 at 12:27
7  
Why should we consider such minor influences as "cancelled" leap days in century years not multiples of 400 but on the other hand not factor in external influences, such as almost all births being delayed past or introduced before the 29th of February, just to save the child from the inconvenience (or other reasons)? - At least here in Germany the probability for a birth to occur on any 29th of February is (estimated) nearly zero. – Alexander Kosubek Feb 29 at 13:02

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.

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The data set has Feb 30 and 31 dates of birth. That's funny. It needs a good clean up before the analysis, but it has a lot of data which is great. – Aksakal Mar 3 at 14:14

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

enter image description here

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3  
It's a useful analysis (+1). It makes me wonder about what connection there is, if any, between the frequencies you analyze and the (vaguely defined) probability asked for in the question. – whuber Feb 29 at 15:55
1  
@whuber, the probabilities in my answer are for cases like analysis of insurance applications or some user data. For instance, you have a web site and want to flag problematic user data. You could compare the frequency of Feb 29 birthdays with my probabilities. However, if you are planning a family and asking this question, then my numbers are useless, pretty much. The reason is that they don't account for factors such as when exactly the couple is actually copulating or fertility and period patterns of couples, for instance, which is the main determinant of the birth day date. – Aksakal Feb 29 at 16:53
    
Glad to see you did not start with the math before considering other factors, beyond pure stats – TheBlastOne Mar 1 at 7:02

The cover to my favorite book ever 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 eachother: 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 is 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.

enter image description here

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.

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2  
A brief description of the model used in also available in the blog post here: andrewgelman.com/2012/06/19/… – Sakari Cajanus Mar 2 at 14:15

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.

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7  
This is a nice analysis of the chance that a randomly chosen date would be February 29th. I believe most of the discussion in this thread focuses on the fact that this doesn't actually answer questions about chances of births, which do not really behave like random draws from the set of possible days. – whuber Feb 29 at 15:28
    
A much easier way is to say that there are 97 leap years per 400 years the way you have already worked out. Calculate the number of days in 400 normal years. 400 * 365 = 146000. Then you need to add the 97 leap days giving 146097. – CJ Dennis Mar 1 at 7:06

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.

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2  
Errr, I was ready to vote this up, and then I got to Well, for anyone already born that is easy. It is either 0 or 1. No. – mattdm Mar 2 at 3:05
    
I guess this hinges a bit on the interpretation of probabilities. Assume I have flipped a coin. I looked at it and know the outcome (e.g. heads). You are standing next to me, but have not seen the outcome. What is the probability of the coin showing heads (for you, for me, "objectively")? In the example above: for the given (born) person the probability is either 0 or 1 (assuming he knows on which date he is born). If you pick a person at random and you had to guess his birthday, the probability it being a Feb. 29 is than again an empirical question. – data_enthusiast Mar 3 at 8:53

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".

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2  
I don't think this answers the question being asked, because the leap day, Feb. 29, only exists in particular calendar systems. Those calendar systems have only been in use in particular societies during the recent historical eras. For example, this question is not intelligible to someone who reckons time using the Hebrew calendar, which has no "February" at all! Moreover, even if we assume a calendar with a leap day, it still doesn't resolve the indeterminacy surrounding the probability distribution of births to days. – General Abrial Mar 1 at 22:25
    
@user777, thats irrelevant. If you belong to a culture that doesn't recognize leap day, there would still be individuals that are born on our leap day. – Octopus Mar 2 at 17:59
    
@Octopus Not if they were born before October, 1582, the month that the Gregorian calendar was introduced. The question is not specific enough to permit one to discern which populations are under consideration, which is why my comment is critically relevant. – General Abrial Mar 2 at 18:05
    
@user777, you're splitting hairs. The point is the Gregorian calendar exists today and it can be used to place every single day in history, whether or not they observed it at that time. – Octopus Mar 2 at 18:27
    
@Octopus How do you know that's the point? – General Abrial Mar 2 at 18:43

I asked my sister, whose bithday is February 29, and she said, "The result of my own empirical study was that it is 1.00, obviously."

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