First Question: You walk into a room filled with random people. You want to find another person in that room who has the same birthday as you. For example, June 15. How many people would need to be in the room? How do you go about solving this question?

Second Question: Same as above. However, you want to find another person in that room who has the same birth date as you. For example, June 15, 1985. How many people would need to be in the room? How do you go about solving this question?

Thanks, 32.209.69.24 (talk) 08:08, 17 January 2025 (UTC)[reply]

These are both different from the well-known birthday problem.

For the first, let's ignore the possibility of people born on February 29 in a leap year, so there are only 365 possible birthdays. Let us also assume that all 365 birthdays are equally likely, so for any fixed day D of the year, such as January 17, the probability ${\displaystyle p}$that a randomly selected person's birthday falls on that very same day is equal to ${\displaystyle {\tfrac {1}{365}}.}$The probability that this person's birthday falls on a different day is then equal to the complement ${\displaystyle 1{-}p={\tfrac {364}{365}}.}$

It is easier now to consider the complementary question: What is the probability ${\displaystyle q_{N}}$that none among ${\displaystyle N}$randomly selected persons has a given birthday D. The answer to the original question is then given by its complement, ${\displaystyle 1-q_{N}.}$

If ${\displaystyle N=0,}$there is no one whose birthday could be D, so ${\displaystyle q_{0}=1.}$If ${\displaystyle N=1,}$with just one other (randomly selected) person present, ${\displaystyle q_{1}}$is just the probability that this person's birthday is D, so ${\displaystyle q_{1}=1{-}p.}$Now suppose we already know ${\displaystyle q_{n}}$for some value of ${\displaystyle n.}$Then we can determine ${\displaystyle q_{n{+}1}}$by considering that the joint probability of two independent events co-occurring is equal to the product of their individual probabilities. Therefore ${\displaystyle q_{n{+}1}=q_{n}\times q_{1}.}$We can conclude that in general

${\displaystyle q_{n}=q_{1}^{n}=(1{-}p)^{n}=({\tfrac {364}{365}})^{n}.}$

The probability of the same birthday as yours among a random selection of ${\displaystyle N}$people is therefore ${\displaystyle 1-({\tfrac {364}{365}})^{N}.}$

Now note that as ${\displaystyle n}$gets larger and larger, the value of ${\displaystyle q_{n}}$gets smaller and smaller, but it never reaches zero exactly. Even if ${\displaystyle N=365,}$we find that ${\displaystyle 1-({\tfrac {364}{365}})^{365}=0.632625...~.}$To get to 99%, ${\displaystyle N}$should be at least ${\displaystyle 1679}$; ${\displaystyle 1-({\tfrac {364}{365}})^{1678}=0.98998...}$falls still short, but ${\displaystyle 1-({\tfrac {364}{365}})^{1679}=0.99001...}$reaches the target.

The approach assumes that the possible birthdays are uniformly distributed over the population, which is not the case in reality. However, to account for this, you only need to know the real value of ${\displaystyle p}$for day D and not for any other day.

To find a somewhat realistic answer to the second question is harder. In reality, the people in a room will not be a random sample from the total population. People below the age of 3 and over the age of 97 will be underrepresented, so if your own birthdate is January 17, 1925, the likelihood of today finding someone present to jointly celebrate your 100th birthday with is much smaller than that of finding a co-celebrant for your 35th birthday if your birthdate is January 17, 1990. The notion of "random selection" is not clearly applicable. You need to know at least the distribution of birthyears among the population from which the people in the room are selected, accounting both for the actual population pyramid and for age-based selection bias. When you have determined ${\displaystyle p_{\text{Y}},}$the probability that a person randomly selection from those present in the room has the same birthyear Y as you, instead of ${\displaystyle p=1/365}$you can use ${\displaystyle p=p_{\text{Y}}/365}$and proceed as above.  --Lambiam 11:44, 17 January 2025 (UTC)[reply]

Wow. Very thorough, detailed, and comprehensive. You certainly have a gift for math. Thanks! Let me read this over and process it all. I'll need a day or two. Thanks so much. 32.209.69.24 (talk) 08:39, 18 January 2025 (UTC)[reply]