To calculate the probabilities, let’s break the problem into two parts:

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Part 1: Probability that at least one other guest shares your birthday

Assume there are $n = 13$ people at the dinner party, including yourself. Each person’s birthday is randomly distributed over $365$ days.

Complementary Approach:

1. The complement of "at least one other person shares your birthday" is "no one else shares your birthday."
2. If no one else shares your birthday, the remaining $12$ people must all have birthdays on days other than yours.

The probability that a single person does not share your birthday is: $P_{\text{not shared}} = \frac{364}{365}.$

For all $12$ other people, the probability that none share your birthday is: $P_{\text{none share}} = \left(\frac{364}{365}\right)^{12}.$

Now, the probability that at least one person shares your birthday is: $P_{\text{at least one shares}} = 1 - P_{\text{none share}}.$

Calculation:

$P_{\text{at least one shares}} = 1 - \left(\frac{364}{365}\right)^{12}.$ $P_{\text{at least one shares}} = 1 - \left(0.997260\right)^{12}.$ $P_{\text{at least one shares}} \approx 1 - 0.967692.$ $P_{\text{at least one shares}} \approx 0.0323.$

Thus, the probability is approximately 0.0323.

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Part 2: Probability that some pair of guests share the same birthday

This is the classic "birthday problem."

Complementary Approach:

1. The complement of "some pair of guests share the same birthday" is "no two guests share the same birthday."
2. If no two people share a birthday, the first person can have any of the $365$ days, the second person $364$ days, the third person $363$ days, and so on.

The total number of ways to assign unique birthdays to $13$ people is: $P_{\text{unique}} = \frac{365 \cdot 364 \cdot 363 \cdots (365 - 12)}{365^{13}}.$

The probability that at least one pair shares a birthday is: $P_{\text{pair shares}} = 1 - P_{\text{unique}}.$

Calculation:

The numerator (number of ways to assign unique birthdays): $365 \cdot 364 \cdot 363 \cdots 353 = \frac{365!}{(365 - 13)!}.$

The denominator (total possible birthday assignments): $365^{13}.$

So: $P_{\text{unique}} = \frac{365!}{(365 - 13)! \cdot 365^{13}}.$

Approximating this: $P_{\text{pair shares}} \approx 1 - 0.8264.$ $P_{\text{pair shares}} \approx 0.1736.$

Thus, the probability is approximately 0.1736.

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Final Results:

1. Probability that at least one other guest shares your birthday: 0.0323.
2. Probability that some pair of guests share the same birthday: 0.1736.

Would you like to see these calculations explained further or in more detail?

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Related Questions:

1. What is the probability for $n = 20$ people instead of $n = 13$?
2. How does the probability change if the year has $366$ days (leap year)?
3. What is the probability that exactly two people share a birthday in this group?
4. How would the probability change if there were $50$ guests instead of $13$?
5. How can these probabilities be approximated using logarithms?

Tip:

The "birthday paradox" often surprises people because probabilities grow faster than intuition expects!