Birthday Problem for 3 people

Question

I know that, in a room of 23 people, there is a 50-50 chance that two people have the same birthday. However, what I want to know is: How many people do you need to have a 50-50 chance that 3 people share the same birthday?

Note: Assume for this question, that birthdays are equally distributed

Answer 1

The question linked in the comments, while not your exact question, is very relevant. In particular, reading the answers posted there will tell you that an exact computation for triples is gross. However, Byron Schmuland's answer gives us a useful estimate: $$ P(\text{at least one triple with } N \text{ people}) \approx 1 - e^{-{N \choose 3}/365^2} $$ From this we can simply solve for what $N$ makes this value $\frac12$. We need \begin{align*} 1 - e^{-{N \choose 3}/365^2} &= \frac12 \\ e^{-{N \choose 3}/365^2} &= \frac12 \\ e^{{N \choose 3}/365^2} &= 2 \\ {N \choose 3} &= 365^2 \ln 2 \\ N(N-1)(N-2) &= 6 \cdot 365^2 \ln 2 \\ N^3 - 3N^2 + 2N - 6 \cdot 365^2 \ln 2 &= 0 \end{align*}

A computer calculation reveals that the only real root of this cubic polynomial is $N \approx 83.13$. Hence we would expect to need $83$ or $84$ people in the room before having a 50-50 chance of getting three people with the same birthday.

UPDATE: This is only an approximation, and it turns out it's off by more than I expected. To get an exact answer, I used the exact formula given here, and found that $N = 87$ is actually closest, with a probability of $.49945$ that three people have the same birthday.

Answer 2

Empirically the answer seems to be $87$ or $88$.

More precisely, the probability of at least three people sharing a birthday is very close to $0.50$ if there are $87$ people in total (making the standard assumption of an i.i.d. uniform distribution over $365$ days).

Using the following R code to test a million cases for $87$ people in a room

days   <- 365
people <- 87
cases  <- 1000000
set.seed(1)
maxhits <- function(x){ max(table(x)) }
eg <- matrix(sample(days, people*cases, replace = TRUE), nrow=cases)
table(apply(eg, 1, maxhits) )

the sample distribution for the most people sharing a single birthday was

 1      2      3      4      5      6      7 
13 500212 461918  36116   1688     50      3  

Answer 3

The answer for n people can be found on OEIS: https://oeis.org/A014088 (Minimal number of people to give a 50% probability of having at least n coincident birthdays in one year)

For 3 people is 88.

Answer 4

Algorithm leading to an exact answer:

First, you must find a closed form formula for $~f(n) = ~$ the probability that there is at least one occurrence of three people sharing the same birthday, given that there are exactly $~n~$ people in the room.

Then, it becomes a simple matter of using computer assistance to determine the smallest value of $~n,~$ such that $~f(n) \geq 1/2.$

So, the entire problem reduces to deriving the closed form formula for $~f(n).$

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$f(n)~$ will be expressed as

$$f(n) = 1 - \frac{N}{D} ~: ~D = 365^n,$$

where $~N~$ denotes the number of equally likely ways that you can have no occurrence of 3 people sharing the same birthday.

Given $~n~$ fixed, and $~\displaystyle p \in \left\{ ~0, ~1, ~2, ~\cdots, ~\left\lfloor \frac{n}{2} ~\right\rfloor ~\right\}, ~$
let $~g(n,p)~$ denote the number of ways that you can have exactly $~p~$ pairs of people, where the two people in each pair have the same birthday, but no one else in the room has that birthday.

Then

$$N = \sum_{p = 0}^{\left\lfloor \frac{n}{2} ~\right\rfloor} g(n,p),~$$

so the entire problem reduces to providing a closed form formula for $~g(n,p).$

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To compute $~g(n,p):$

• There are $~\displaystyle \frac{n!}{2^p \times (n-2p)! \times p!} ~$ distinct ways of forming the $~p~$ pairs of people.
  The denominator's $~p!~$ factor is used to accommodate the fact that the order that the pairs are formed is irrelevant.

• You can construe each of the $~p~$ pairs of people as one fused unit, that will be assigned one birthday. So, instead of having $~n~$ units, you have $~n - p~$ units, each of which are assigned a birthday.
  
  The number of distinct assignments is therefore $~\displaystyle \frac{365!}{(365 + p - n)!}.$

Therefore,

$$g(n,p) = \frac{n!}{2^p \times (n-2p)! \times p!} \times \frac{365!}{(365 + p - n)!}.$$

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$$\underline{\text{Final Summary}}$$

Use computer assistance to determine the smallest value of $~n~$ such that $~f(n) \geq \dfrac{1}{2}.$

$$f(n) = 1 - \frac{N}{D} ~: ~D = 365^n.$$

$$N = \sum_{p = 0}^{\left\lfloor \frac{n}{2} ~\right\rfloor} g(n,p).$$

$$g(n,p) = \frac{n!}{2^p \times (n-2p)! \times p!} \times \frac{365!}{(365 + p - n)!}.$$

Answer 5

There are 2 approaches to this kind of question (3 people with the same birthday...) You either 1) just want the answer. 2) want the satisfaction and understanding that comes from figuring it out from basic probability theory. (This can be a long hard road with no success guaranteed no matter how creative you are or how hard you work.)

If you just want the answer, the best way to get it is to write a little program that gets the answer by experiment. Using a random number generator to provide the data, and repeating the experiment a large number of times (for instance, 1,000,000 times) and seeing the result. You can easily get the answer to the birthday problem for any number of same birthdays (Sames), and also get answers to related questions- like, On the average how many pairs (OneLesses) of people had the same birthday at the time a new person had the same as one of the pairs making 3 with the same bday.
For this problem, Sames=3, OneLesses=number of pairs

the C code is below. It was created and run in the Pelles C compiler for windows, a free program.

In the case of the 3 person birthday problem, the results for a million runs the output of the program is-

" Avg # of tosses to get 3 Sames from 365 birthdays = 88.6971 Avg # of OneLesses, i.e. pairs, before 3 Sames from 365 birthdays = 10.1309"

So the average number of people in a room before there being 3 with the same birthday is 88.7 and at the time that happened, there were, on the average, 10 pairs of people with the same birthday already in the room.

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C code->

  Over=0;

while (!Over) {

             printf(" enter NumTrials,   ");
                 scanf("%d",&NumTrials);

                    printf(" enter number of outcomes,   ");
                 scanf("%d",&Outcomes);

                 printf(" enter number of Sames,   ");
                 scanf("%d",&Sames);

TotalTosses=0;
TotalOneLesses=0;
  MaxCount=0;

   for (i=1; i <= NumTrials; i++)
   {
    //  FOR NumTrials loop
             // initialize
                   for(j=0; j<=Outcomes; j++)  Results[j]=0;
                   Count=0;
                   OneLess=0;

          Done=0;       
           while(!Done)
          { 
          toss= rand () % (Cards) ;
                Count++;
                Results[toss]++;

                if ( Results[toss]== (Sames-1) )  OneLess++;

                if ( Results[toss]==Sames) Done=1;

            }  // End of While!Done


              Tosses[Count]++;


            TotalTosses+= Count;
             TotalOneLesses+= OneLess;

        }// for (NumTrials)

           AvgTosses= (double)TotalTosses /NumTrials ;
           printf( "Avg # of tosses to get %d Sames from %d Outcomes = %.4f \n ", Sames, Outcomes, AvgTosses);

           AvgOneLesses= (double)TotalOneLesses /NumTrials ;
           printf( "Avg # of OneLesses before  %d Sames from %d Outcomes = %.4f \n ", Sames, Outcomes, AvgOneLesses);

} // END While (!Over)

                   //quit program