Probability of 4 parents sharing 2 birthdays

Question

I'm a math idiot and I'm trying to figure something out. My mother was born on April 10th as was my friend's mother. My father was born on July 4th as was my friend's father.

How do I calculate the probability of that happening?

EDIT: How would one calculate the probability of 2 pairs of shared birthdays? I think this is the answer I'm looking for

Halp.

-Brett

Answer 1

When you have a problem you find a bit difficult it is sometimes easier to try to solve a simplified example.

Instead of birthdays, let us say that your parents are throwing dice. Let's call your father $f_1$, your mother $m_1$, and you friend's parents $f_2$ and $m_2$. If $f_1$ throws a 2 and $m_1$ throws a 5, then $f_2$ has to throw a 2 and $m_2$ has to throw a 5. The probability of this happening is $$ \frac{1}{36}. $$

If only one of the parents have to match, then the probability is: the probability that $f_1$ matches $f_2$ and $m_1$ matches $m_2$ which is 1/36, PLUS the probability that $f_1$ mathes $m_2$ and $m_1$ matches $f_2$ which is also 1/36, MINUS the probability that they all match together (because this has already been counted). This will happen every sixth time, so it is 1/216. So the probability of either parents matching a throw of the die is: $$ \frac{2}{36} - \frac{1}{216}. $$

To verify this, you can simulate 1 million dice tosses and count the results. One of the things that make answering questions about statistics and probability easier than many questions in mathematics - you can brute force the answer in a computer and check what you calculated! :)

Here is some R-code for running the simulation.

# Throwing a die
NSIM = 1000000
hit = rep(0, NSIM)
for (i in 1:NSIM) {
  bd = sample(1:6, size=4, replace = TRUE)
  f1 = bd[1] 
  m1 = bd[2]
  f2 = bd[3]
  m2 = bd[4]
  if (f1 == f2 & m1 == m2) {
    hit[i] = 1
  } 
} 
sum(hit)/NSIM
1/36

# Father and friend's mother (or vice-versa) can share 'die'
hit2 = rep(0, NSIM)
for (i in 1:NSIM) {
  bd = sample(1:6, size=4, replace = TRUE)
  f1 = bd[1] 
  m1 = bd[2]
  f2 = bd[3]
  m2 = bd[4]
  if (f1 == f2 & m1 == m2 | f1 == m2 & m1 == f2) {
    hit2[i] = 1
  } 
} 
sum(hit2)/NSIM
2/36 - 1/216  # Remove occurences when they all throw the same number

My results:

> sum(hit)/NSIM
[1] 0.027855
> 1/36
[1] 0.02777778
> sum(hit2)/NSIM  
[1] 0.050728  
> 2/36 - 1/216  # Remove occurences when they all throw the same number  
[1] 0.05092593

By extending this to birthdays, we get: $$ \frac{1}{365^2} \approx 0.0000075 $$ when the birthdays have to match up. And when they can cross over: $$ \frac{2}{365^2} - \frac{1}{365^3} \approx 0.0000149 $$

Simulating in R again. Did 100 million this time - took a while!

# Birthday simulations - using 1 to 365
NSIM = 100000000 # 100M simulations
# Fathers match and mothers match
hit = rep(0, NSIM)
# Father and friend's mother (or vice-versa) can share birthday
hit2 = rep(0, NSIM)
# Simulation
bdsim = matrix(sample(1:365, size=4*NSIM, replace = TRUE), ncol=4)
for (i in 1:NSIM) {
  bd = bdsim[i, 1:4] 
  f1 = bd[1] 
  m1 = bd[2]
  f2 = bd[3]
  m2 = bd[4]
  if (f1 == f2 & m1 == m2) {
    hit[i] = 1
  } 
  if (f1 == f2 & m1 == m2 | f1 == m2 & m1 == f2) {
    hit2[i] = 1
  } 
} 
sum(hit)/NSIM
1/365^2
sum(hit2)/NSIM
2/365^2 - 1/365^3  # Remove occurences when they all throw the same number

And to verify the theoretical results with the simulated results:

> sum(hit)/NSIM
[1] 7.62e-06
> 1/365^2
[1] 7.506099e-06
> sum(hit2)/NSIM
[1] 1.485e-05
> 2/365^2 - 1/365^3  # Remove occurences when they all throw the same number
[1] 1.499163e-05

Answer 2

Since you don't care about overlaps (you said "I'm looking for the probability of all 4 falling on those exact dates."), you take the probability your mom's birthday is April 10th, and multiply that by the probability your friend's mom's birthday is April 10th, then you multiply that by the probability your dad was born on July 4th, and multiply that by the probability your friend's dad was born on July 4th.

This means $\frac{1}{365}*\frac{1}{365}*\frac{1}{365}*\frac{1}{365} = (\frac{1}{365})^4 = 5.63*10^{-11}$ very slim (THIS IS EXCLUDING LEAP YEARS). Multiply by 100 to get a percent: $5.63*10^{-9}$%

Bear in mind, that since you said you only care about birthdays falling on specific dates, the probability your mom's b-days are April 10th, and dad's are July 4th, is THE EXACT SAME PROBABILITY of your mom's birthday being January 28th, your friend's mom's Decmber 2nd, your Dad's April 11th, and your friend's dad's being October 30th.

If you only care about 2 pairs of the same birthday, the probability would actually be higher than for any select birthday assigned (the probability above). The probability of 2 pairs of shared birthdays gets more complicated as well...

for that, you can now see @CoveredInChocolate's answer

Fun fact: if there's a group of 23 people, there's over a 50% chance that 2 or more of them share a birthday.

Answer 3

The probability the moms share a birthday is $1/{365}$ (approximately; this ignores leap years).
The probability the dads share a birthday is $1/365$.
Since these two events are independent, the probability they both happen is $1/365^2$.