The Near-miss Birthday Problem. Gregory Quenell Plattsburgh State

Transcription

1 The Near-miss Birthday Problem Gregory Quenell Plattsburgh State 1

2 The Classic Birthday Problem Assuming birthdays are uniformly distributed over 365 days, find P (at least one shared birthday) in a random sample of n people. Solution: Use complementation. P (at least one shared birthday) = 1 P (no shared birthday) = 1 P (n different birthdays) 2

3 Finding P (n different birthdays) 3

4 Finding P (n different birthdays) Birthday of person days {}}{ 4

5 Finding P (n different birthdays) Birthday of person days {}}{ P (n different birthdays) = number of ways to place n 1 birthdays in 364 days with no collision ( ) total number of ways to place n 1 birthdays in 365 days 5

6 Finding P (n different birthdays) Birthday of person days {}}{ P (n different birthdays) = number of ways to place n 1 birthdays in 364 days with no collision ( ) total number of ways to place = = n 1 birthdays in 365 days ( ) 364 (n 1)! n 1 (365) n (364 (n 2)) = (364) n 1 (365) n 1 6

7 Some numbers P (at least one shared birthday) n P (shared birthday) = 1 (364) n 1 (365) n 1 7

8 The Near-miss Birthday Problem Assuming birthdays are uniformly distributed over 365 days, find at least one pair of birthdays P that are either coincident or adjacent in a random sample of n people. Solution: Use complementation. P (at least one near miss) = 1 P (no near misses) = 1 P (n isolated birthdays) 8

9 Finding P (n isolated birthdays) Birthday of person days {}}{ P (n isolated birthdays) = number of ways to place n 1 birthdays in 364 days with no collision and no two birthdays adjacent ( ) total number of ways to place n 1 birthdays in 365 days 9

10 Finding P (n isolated birthdays) Birthday of person days {}}{ P (n isolated birthdays) = This is still (365) n 1 number of ways to place n 1 birthdays in 364 days with no collision and no two birthdays adjacent ( ) total number of ways to place n 1 birthdays in 365 days 10

11 Finding P (n isolated birthdays) Birthday of person days {}}{ P (n isolated birthdays) = This is still (365) n 1 number of ways to place n 1 birthdays in 364 days with no collision and no two birthdays adjacent ( ) total number of ways to place n 1 birthdays in 365 days How do we count these? 11

12 Counting isolated birthdays n 1 isolated birthdays in 364 days {}}{ a 1 a 2 a 3 a 4 a n 1 a n Every arrangement of n 1 isolated birthdays corresponds to a gap sequence a 1, a 2,..., a n in which { a1 + a a n = 364 (n 1) a i 1 for all i 12

13 Aside on counting A sequence a 1, a 2,..., a n of positive integers such that a 1 + a a n = S is called an n-part composition of S. Theorem: The number of n-part compositions of S is ( ) S 1. n 1 Proof: Write down a string of S dots. Then there are S 1 inter-dot spaces. 13

14 Aside on counting A sequence a 1, a 2,..., a n of positive integers such that a 1 + a a n = S is called an n-part composition of S. Theorem: The number of n-part compositions of S is ( ) S 1. n 1 Proof: Write down a string of S dots. Then there are S 1 inter-dot spaces. 3 {}}{ {}}{ 3 {}}{ Placing bars in n 1 of these S 1 spaces determines an n-part composition of S, and conversely. 14

15 Application to birthdays n 1 isolated birthdays in 364 days {}}{ a 1 a 2 a 3 a 4 a n 1 a n There are ( ) [364 (n 1)] 1 n 1 possible gap sequences. = ( ) 364 n n 1 15

16 Application to birthdays n 1 isolated birthdays in 364 days {}}{ a 1 a 2 a 3 a 4 a n 1 a n There are ( ) [364 (n 1)] 1 n 1 possible gap sequences. = ( ) 364 n n 1 Result: ( number of ways to place n 1 isolated birthdays in 364 days ) = ( ) 364 n (n 1)! n 1 = (364 n) n 1 16

17 The near-miss birthday formula P (no near miss) = (364 n) n n 1 The probability of at least one near miss in a random sample of n people is 1 (364 n) n n 1. The least n for which this probability exceeds 0.5 is n = 14: P at least one pair of coincident or adjacent birthdays in a random sample of 14 people

18 More numbers P (shared birthday) = 1 (364) n 1 (365) n 1 n P (shared) P (near miss) P (near miss) = 1 (364 n) n 1 (365) n 1 18

19 Birthdays shared by k or more people Let (X 1, X 2,..., X 365 ) be a random vector in which X i is the number of people in a random sample of size n who were born on day i. Then (X 1, X 2,..., X 365 ) follows a multinomial distribution with n things, 365 bins, and constant probability 1/365. Thus P ((X 1, X 2,..., X 365 ) = (x 1, x 2,..., x 365 )) = ( ) n x 1 x 2 x 365 = n n! x 1! x 2! x 365! ( ) n We want P (max(x 1, X 2,..., X 365 ) k), the probability that some date is the birthday of k or more people in the sample. Again, we use complementation: P (max(x 1, X 2,..., X 365 ) k), = 1 P (max(x 1, X 2,..., X 365 ) k 1) = 1 P (X i k 1) for all i. 19

20 Finding P (X i k 1) for i = 1, 2,..., 365 We need n! 365 n k 1 k 1 k 1 x 1 =0 x 2 =0 x 365 =0 x 1 + x x 365 =n ( 1 x 1! 1 x 2! 1 ) x 365! Consider the product ( 1 0! + 1 ) ( 1! (k 1)! 0! + 1 ) 1! (k 1)! factor for x 1 factor for x 2 ( 1 0! + 1 ) 1! (k 1)! factor for x 365 To pick out the terms with x 1 + x x 365 = n, introduce a tracer variable: ( τ 0 0! + τ 1 1! + + τ ) ( k 1 τ 0 (k 1)! 0! + τ 1 1! + + τ ) ( k 1 τ 0 (k 1)! 0! + τ 1 1! + + τ ) k 1 (k 1)! The coefficient of τ n in this product is exactly the sum that we want. 20

21 The multiple-birthday formula We have P (X i k 1 i) = n! 365 n coeff of τ n in ( τ 0 0! + τ 1 1! + + τ k 1 ) 365 (k 1)! And so P (max(x i ) k) = 1 n! ( τ 365 [τ n 0 ] n 0! + τ 1 1! + + τ k 1 ) 365 (k 1)! 21

22 An example In a random sample of 100 people, what s the probability that there are six (or more) who share a birthday? It s 1 100! ( τ [τ ] 0! + τ 1 1! + τ 2 2! + τ 3 3! + τ 4 4! + τ 5 ) 365 5! Mathematica says the answer is / (This is about ) 22

23 Multiple birthday probabilities Probabilities of at least one date in the calendar being the shared birthday of k people for k = 3, 4, and 5. 23