What is the probability that in a randomly chosen group of r people at least…

Solution overview: compute the probability that at least three people share a birthday by using the complement (no day has three or more people).

Step 1 (complement):

• We count arrangements where every calendar day is used by at most two people. Such an arrangement consists of k disjoint pairs and r-2k singles, for some k = 0,1,..., floor(r/2).

Step 2 (count arrangements for fixed k):

• Choose which 2k of the r people form the pairs: C(r,2k).

• Partition those 2k people into k unordered pairs: number of pairings = (2k)! / (2!^k k!).

• Assign distinct days to the k pairs: falling factorial (365)_k = 365*364*...*(365-k+1).

• Assign distinct days to the r-2k singles from the remaining 365-k days: (365-k)_{r-2k}.

Combining these counts, the number of favorable birthday assignments (with no day having 3+ people) for fixed k is:

C(r,2k) * (2k)! / (2!^k k!) * (365)_k * (365-k)_{r-2k}.

Note that C(r,2k)*(2k)! = r!/(r-2k)!, so this simplifies to

r! / ((r-2k)! 2^k k!) * (365)_k * (365-k)_{r-2k}.

Step 3 (probability): divide by the total number of possible birthday assignments 365^r and sum over all possible k.

Probability(no day has 3+ people) = sum_{k=0}^{floor(r/2)} [ r! / ((r-2k)! 2^k k!) * (365)_k * (365-k)_{r-2k} ] / 365^r

Final answer: Probability(at least three people share a birthday) = 1 - sum_{k=0}^{floor(r/2)} [ r! / ((r-2k)! 2^k k!) * (365)_k * (365-k)_{r-2k} ] / 365^r.

Remarks: For k=0 this recovers the all-distinct case (365)_r / 365^r; for k=1 it gives the exactly one pair case. The full sum includes multiple disjoint pairs but excludes any configuration with a triple or higher multiplicity.