Consider The Birthday Problem In This Question

Consider The Birthday Problem In This Problem We Calculate The Pro

1. Consider the Birthday Problem. In this problem, we calculate the probability that, in a group of n people, at least two have the same birthday. Let E be the event that at least two people share a birthday. In order to calculate P(E), we first need a sample space.

A possible sample space consists of n-tuples of integers from 1 to 365, representing each person’s birthday. (Leap years are not considered.)

(a) List or describe the sample space for n = 200. What is the size of the sample space?

(b) For n = 200, propose an algorithm for enumerating the number of tuples in the sample space where at least two people share a birthday. Describe it in words or pseudocode, not actual code.

(c) Given a supercomputer that performs 33.86 petaflops (33.86 × 10^15 floating point operations per second), and recognizing that scanning each tuple costs 1 floating point operation, how long would it take to process all tuples for n = 200? Calculate the time in seconds, days, and years.

2. The previous calculations show that direct enumeration is infeasible at n = 200. For smaller n such as 3, the sample space (approximately 49 million tuples) is manageable with enumeration. Calculate P(E) for n = 3 using counting principles, and verify it matches 1 minus the probability that all birthdays are distinct.

3. Repeat the calculation for n = 4. Comment on how the complexity of computing P(E) via counting principles grows as n increases.

4. Consider a variation: "What is the probability that in a group of n people, at least three share the same birthday?" Use complement probability to solve this problem for large n, as enumerating the sample space is infeasible.

Paper For Above instruction

Introduction

The Birthday Problem is a classic question in probability theory that explores the likelihood that, within a randomly selected group of people, at least two individuals share the same birthday. This problem highlights surprising results in probability and has applications ranging from cryptography to statistical sampling. In this essay, we delve into the formal structure of the problem, computational considerations, and extensions to related variations, focusing on large group sizes where brute-force calculations become impractical and probabilistic reasoning offers more efficient solutions.

Sample Space and Its Size

For a group of n = 200 people, each individual has a birthday uniformly distributed among 365 days (excluding leap years). The sample space comprises all possible tuples where each element represents an individual’s birthday. Specifically, the sample space is the set of all ordered sequences of length 200 with entries from the set {1, 2, ..., 365}. The total number of such tuples is 365 raised to the 200th power, reflecting the fact that each person’s birthday is independent and equally likely.

Mathematically, the sample space size is |Ω| = 365^{200}. This exponentiation results in an astronomically large number, approximately 3.84 × 10^{485} tuples, which significantly exceeds any feasible computational enumeration or storage capacity.

Algorithm for Enumerating Tuples with Shared Birthdays (n=200)

To enumerate tuples where at least two people share a birthday, one can propose the following algorithm:

1. Initialize a counter to zero for tuples with shared birthdays.
2. Iterate over all possible tuples in the sample space (from 1 to 365 for each of the 200 positions).
3. For each tuple:
   • Check if at least two entries are identical.
   • If this condition is true, increment the counter.

However, explicitly generating and checking all tuples is impossible for n=200. Instead, theoretical calculations or probabilistic approximations are used. The described algorithm serves as a conceptual framework—practical implementation is infeasible due to the sample space's vast size.

Computational Feasibility and Estimated Processing Time

The supercomputer's power is specified as 33.86 petaflops, or 33.86 × 10^{15} floating point operations per second. Assuming each tuple scan requires 1 floating point operation, the total number of operations needed to enumerate all tuples for n=200 is 365^{200}. The total processing time T in seconds can be estimated as:

T = Total operations / Flops per second = 365^{200} / (33.86 × 10^{15})

Given the size of 365^{200} (~10^{485}), T ≈ 10^{485} / 3.386 × 10^{16} ≈ 10^{469} seconds.

Converting seconds into days and years:

• Days: ≈ 10^{469} / 86,400 ≈ 10^{464} days.
• Years: ≈ 10^{464} / 365 ≈ 10^{461} years.

These estimates demonstrate that, even with the fastest supercomputers, exhaustive enumeration at n=200 is completely unfeasible due to astronomical computational time frames, reinforcing the need for probabilistic methods.

Counting Principles for Smaller n: n=3 and n=4

For smaller groups, counting principles enable direct calculation of the probability that at least two individuals share a birthday. For n=3, the total sample space is 365^{3} ≈ 4.91 × 10^{7} tuples. The probability that all three have distinct birthdays is:

P(\text{all distinct}) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} ≈ 0.9918

Thus, the probability that at least two share a birthday is:

P(\text{at least two}) = 1 - 0.9918 ≈ 0.0082

This matches the complement probability calculation, confirming consistency. For n=4, the probability all are distinct is:

P(\text{all distinct}) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} ≈ 0.9837

Consequently, the probability that at least two share a birthday is approximately 1 - 0.9837 = 0.0163.

As n increases, this calculation rapidly involves more multiplicative steps, illustrating the computational complexity. The approach remains straightforward for small n but becomes increasingly cumbersome, and at larger n, infeasible without approximation.

Variation: Probability at Least Three Share a Birthday

Extending the problem to "at least three sharing a birthday" involves more complex complement probability reasoning. Instead of focusing solely on cases with at least one pair, we analyze the probability that all birthdays are either unique or occur at most twice, then subtract from 1 to find the probability that at least one birthday occurs three or more times.

For large n, directly enumerating the sample space or applying counting principles becomes impossible. The solution employs the principle of inclusion-exclusion or Poisson approximation for the counts of repeated birthdays.

The probability that a specific birthday is shared by exactly three people (and not more) can be approximated via the Poisson distribution as n grows large, where the expected number of triples is small. Consequently, the probability that at least one birthday occurs three times is roughly:

P(\text{at least three}) ≈ 1 - P(\text{no triple birthdays})

where P(\text{no triple birthdays}) can be estimated using probabilistic models that assume independence and uniformity, providing computational feasibility for large n.

Conclusion

The Birthday Problem exemplifies the counterintuitive nature of probability and highlights the limitations of brute-force approaches to combinatorial problems involving large sample spaces. For small groups, explicit counting provides accessible solutions, but for large n, probabilistic and statistical methods are not only more practical but necessary. The extension to scenarios involving triples or higher repetitions further complicates direct enumeration, emphasizing the significance of probabilistic approximation techniques in complex combinatorial problems.

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