Regarding probability and the birthday paradox

combinatoricspermutationsprobabilityprobability theory

I have a question which states that
"In a group of 23 people what is the probability that there are two people with the same birthday? Assume there are 365 days in a year. Ignore leap years and such complications. Assume there is an equal probability of a person being born on each day of the year.". I solved it using the complement. I first computed the number of ways in which we can assign the birthdays to 23 people out of 365 days (without replacement). That gave 365 * (365-1).. (365-k+1). Then I divided this by 365^k. Then I subtracted the result from 1. But, the probability which I have now got may also contain 3 people having the same birthday or 4 people having the same birthday, etc. I want to know the probability of exactly two people having the same birthday. In short, what I have computer is, " what's the probability that AT LEAST TWO PEOPLE HAVE SAME BIRTHDAY" and what I'm looking for is "WHAT IS THE PROBABILITY OF EXACTLY TWO PEOPLE HAVING THE SAME BIRTHDAY".How do I compute that probability?

Best Answer

If by exaclty you mean any two peaople having same birthday and all the rest having different birthday, then we can do it for ingeneral having $i$ peaople out of $k$ with same birthday as follows:

For each fixed day of 365 days there are $\binom{k}{i}$ ways of chosing $i$ people having that day as birthday, by wich there are $364(364-1)(364-2)\ldots(364-k+i+1)$ ways of the rest $k-i$ people having different birthday, so the probability of exactly $i$ having the same birthday is $$P(i)=\frac{365\binom{k}{i}364(364-1)(364-2)\ldots(364-k+i+1)}{365^k}$$

Related Solutions

[Math] Birthday problem with 3 people

To solve this problem, take it step by step.

What is the probability of Annie having a birthday on a particular day of the year: 1/365

What is the probability of Boris having a birthday on the same day = (1/365) * (1/365)

Similarly, probability of Charlie having a birthday on the same day = (1/365)^3

The above answer is for a specific day in a year. Since we are fine with any day in the year, multiply the answer with 365 (total number of days in the year).

So, probability of all three having a birthday on the same day in the year = (1/365)^2

[Math] probability that at least $2$ people have same birthday from group of $N$ people

Let's rephrase the question.

Find the probability that any $2$ people have their birthday in the same day of the year in a group $4n$ people, of which $n$ born were in the leap year and $3n$ people were born in the regular year.

The complement of the event "any 2 people have their birthday" is "all people have birthday on different days of the year".

Consider two mutually exclusive possibilites: $\mathcal{A}_1$ - one of these birthdays falls on "Feb. 29", and $\mathcal{A}_2$ -- none of these birthdays fall on the leap day.

The number of configurations of the second kind is $$ N\left(\mathcal{A}_2\right) = \prod_{k=1}^{4 n} \left( 365 - (k-1) \right) $$ and the number of configurations of the first kind is $$ N\left(\mathcal{A}_1\right) = n \prod_{k=1}^{4 n-1} \left( 365 - (k-1) \right) $$ The total configuration size is $N_T=366^n 365^{3n}$, hence the probability of having no birthday duplicates is

$$ p_c = \frac{ N\left(\mathcal{A}_1\right) + N\left(\mathcal{A}_2\right)}{N_T} $$

Here is a confirmation in Mathematica:

(* computed probability expression *)
p[n_Integer] = (Product[365 - (k - 1), {k, 1, 4 n}] + 
     n Product[365 - (k - 1), {k, 1, 4 n - 1}])/(366^n 365^(3 n));

(* simulation result *) 
In[45]:= With[{n=7,reps=10^8},
   Sum[Boole[DuplicateFreeQ[Join[
       RandomChoice[Range[1,366],n],
       RandomChoice[Range[1,365],3n]]]],{k,1,reps}]/reps]//N

Out[45]= 0.34599694

In[53]:= N[p[7]]

Out[53]= 0.3460041650022472

(* verify that computed value belongs to the sampling 
    confidence interval *)
In[55]:= With[{pSample = Out[45], pExpected = p[7]},
 IntervalMemberQ[
  Interval[pSample + {-1, 1} Sqrt[(pSample (1 - pSample))/10^8]], 
  pExpected]]

Out[55]= True

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