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Out of n people what is the probability that there four people having (four) consecutive birthdays?

I could isolate four people and then determine the generic case:

2/365 chance for the second person (given the previous fixed), then 2/365 chance for the third person (given the previous fixed), then 2/365 chance for the fourth person (given the previous fixed):


But now I have n choose 4 ways I can select, and I am asking for at least one of those instances having the probability above, which requires me to use inclusion-exclusion.

That's why I'm looking for an approximation that's simple.

Ignoring the special cases on Dec 31 and Jan 1 is also fine.

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up vote 2 down vote accepted

As usual, I'm ignoring leap years and assuming that each person's birthday is chosen independently from $\{1,\ldots,365\}$ with equal probability. For my calculations the possibility of "wrap-around" sequences (e.g. days 363, 364, 365 and 1) complicates things, so I'll ignore that. Let $A(n, d)$ be the probability that there is at least one consecutive $4$-tuple with $n$ independent "birthdays" chosen from $\{1,\ldots, d\}$. Let $B_1(n,d)$ be the corresponding probability given that day $0$ has been "pre-chosen" (so days 1,2,3 would make a $4$-tuple), and similarly $B_2(n,d)$ with days $-1$ and $0$ pre-chosen and $B_3(n,d)$ with days $-2,-1$ and $0$. Now condition on the number $X$ of times day $1$ is chosen (which has binomial distribution with parameters $n$ and $1/d$). We get the recurrence relations $$ \eqalign{A(n,d)& = P(X=0) A(n,d-1) + \sum_{x=1}^{n-3} P(X=x) B_1(n-x,d-1)\cr & = (1-1/d)^n A(n,d-1) + \sum_{x=1}^{n-3} {n \choose x} (1/d)^x (1-1/d)^{n-x} B_1(n-x,d-1)\cr B_1(n,d) &= P(X=0) A(n,d-1) + \sum_{x=1}^{n-2} P(X=x) B_2(n-x,d-1)\cr &= (1-1/d)^n A(n,d-1) + \sum_{x=1}^{n-2} {n \choose x} (1/d)^x (1-1/d)^{n-x} B_2(n-x,d-1)\cr B_2(n,d) &= P(X=0) A(n,d-1) + \sum_{x=1}^{n-1} P(X=x) B_3(n-x,d-1)\cr &= (1-1/d)^n A(n,d-1) + \sum_{x=1}^{n-1} {n \choose x} (1/d)^x (1-1/d)^{n-x} B_3(n-x,d-1)\cr B_3(n,d) &= P(X=0) A(n,d-1) + P(X \ge 1)\cr &= (1-1/d)^n A(n,d-1) + 1-(1 - 1/d)^n}$$ with $A(n,d) = 0$ if $d \le 3$ or $n \le 3$.

EDIT: And here are the results: a plot of $A(n,365)$ for $n$ from $1$ to $200$:

enter image description here

The first $n$ for which $A(n,365) \ge .5$ is $92$.

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There are ${365\choose 4}$ possible sets of four birthdays. How many sets of four consecutive birthdays are there??

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Birthdays need not be different. – André Nicolas Dec 7 '12 at 0:33
Then you have $365^4$ quadruples of birthdays. – ncmathsadist Dec 7 '12 at 1:30

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