Attn Staticians/Mathmeticians: Real Life, Really Hard Problem. . .?

Question

Can someone please explain step by step how to calculate the probability that 3 siblings would each have their "Golden Birthday" in the same year?

"Golden Birthday"= The year in which you turn "X" years old if your birthday is the "Xth" day of any given month. (My birthday is Feb 4th 1980 so my "golden birthday" year was 1984, when I turned 4. If your birthday is the 15th of the month, it would be whatever year you turned 15, so on & so forth. . . )

Here are the facts:

Oldest was born the 24th of September,
Middle was born the 21st of October,
Youngest was born the 16th of December

AND ALL IN THE SAME YEAR:
Oldest turned 24yrs old,
Middle turned 21yrs old, AND
Youngest turned 16yrs old.

What is the statistical likelihood of this happening in any given family in any given year? What are the odds of this happening when 3 siblings are involved?

Answer 1

For the first part of the the problem, you are asking the probability that a family has a Golden Birthday (I will call it GB) in any particular year.

Assume we do not know the ages of any of the family members, and we are selecting a year at random.

Every person will have one GB in their life, and if we assume a lifespan of 80 years, the probability that it falls in any one randomly chosen year is 1/80, or 1.25%.

This percentage will hold for each family member, so you add this percentage for each one. Therefore, for a family with four members the probability is 4*(1.25%) = 5%.

This problem changes if you know the ages of the family members, because the GB will fall within the first 31 years of their lives. So for anyone over 31, the probabilty that their GB falls in a randomly chosen year is zero.


The second problem asks what happens when three siblings are involved. We make the same assumptions as before, and add a new one -- that none of the siblings are twins born on the same day, and the three are not triplets born on the same day. I take out the triplets and the twins because then cannot make the assumption of independent bithdates, which I need for the analysis below.

In probability, if any three events are independent of each other, the probability of them all happening at once is the product of the three.


In the first problem, we assumed a lifespan of 80 years. so the probabilty of one sibling having a GB in a given year was (1.25%). Here we have to be a little more careful, because the lifespans of the three siblings may not eactly overlap. Let's assume that the three siblings were born at most 20 years apart. Then all will have lived in a 100 year timespan.

So now the probability that one sibling has a GB in this 100 year timeframe is 1/100, or 1%. Then the probability of all three having a GB in the same given year is (1%)*(1%)*(1%) = .0001% or roughly 1 in 10,000.

This is a low probability, but it is not like one in millions. The reason is that everyone must have a GB at some point in their lifetime, and it must occur within a limited number of years. The range of possibilities is large, but it is not vast. We are assuming for this part of the problem that we are only considering families with three siblings.

We have unfinished business with the triplets and twins. If the three siblings are triplets, then the probability that their GB is in the same year is 1, and assuming a lifespan of 80 years, the probability that this falls in any particular year is just 1*(1/80) = 1.25%.

If two of the siblings are twins and one is not, then the probability of their GB being in the same year is 2/3, and assuming that the twins and the other sibling are born no more than 20 years apart (so that their combined lifespans cover 100 years), the probability that all three GB's fall in the smae year is (2/3)*(1/100)
or 0.67%.


Note that the solutions to this problem do not use the details of the calculation of Golden Birthday, they use the fact that each person gets a GB in exactly one year of their life. We could be talking about any event that happens once in a lifetime, and if we are just looking at the year in which it happens, the answer is the same.

While we are on the topic, the probability that someone has a Golden Birthday that is four years after thier birthdate, as you do, is (factoring in leap year):

(12/365)*(3/4) + (12/366) *(1/4) = 3.3%.

Finally, although it will not be your golden one, happy birthday.