Rare Events Probability and Sample Size Rare Events

Rare Events, Probability and Sample Size

Rare Events An event E is rare if its probability is very small, that is, if Pr{E} ≈ 0. Rare events require large samples to ensure proper observation.

Minimal Sample Size as a Function of Probability Suppose that we have an Event, E, with probability PE. Then the quantity (1/ PE) represents the sample size with an expected count for E of 1. That is, expected. E ≈ 1 for n ≈ (1/ PE). When PE is small, (1/ PE) is large.

Rareness, Sample Size and Probability An event E is rare relative to sample size n when the expected count for E is less than 1. That is, when expected. E = n* PE < 1 n < 1/PE.

Pairs of Dice and the Pair (1, 1)

Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1, 1). Pair of Fair Dice, each with face values {1, 2, 3} per die (1 st D 3, 2 nd D 3) (1, 1)(2, 1)(3, 1) (1, 2)(2, 2)(3, 2) (1, 3)(2, 3)(3, 3) Pr{(1, 1) shows} = Pr{1 shows from 1 st D 3}*Pr{1 shows from 2 nd D 3} = (1/3)*(1/3) = 1/9 . 1111111 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/9) 11. 11111. The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/9) = 9 < 100.

Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1, 1). Pair of Fair Dice, one with face values {1, 2, 3, 4} per die and one with face values {1, 2, 3} per die: (1 st D 4, 2 nd D 3) (1, 1)(2, 1)(3, 1)(4, 1) (1, 2)(2, 2)(3, 2)(4, 2) (1, 3)(2, 3)(3, 3)(4, 3) Pr{(1, 1) shows} = Pr{1 shows from 1 st D 4}*Pr{1 shows from 2 nd D 3} = (1/4)*(1/3) = 1/12 . 0833 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/12) 8. 33 The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/12) = 12 < 100.

Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1, 1). Pair of Fair Dice, each with face values {1, 2, 3, 4} per die: (1 st D 4, 2 nd D 4) (1, 1)(2, 1)(3, 1)(4, 1) (1, 2)(2, 2)(3, 2)(4, 2) (1, 3)(2, 3)(3, 3)(4, 3) (1, 4)(2, 4)(3, 4)(4, 4) Pr{(1, 1) shows} = Pr{1 shows from 1 st D 4}*Pr{1 shows from 2 nd D 4} = (1/4)*(1/4) = 1/16 . 0625 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/16) 16 The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/16) = 16 < 100.

Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1, 1). Pair of Fair Dice, each with face values {1, 2, 3, 4, 5, 6, 7, 8} per die: (1 st D 8, 2 nd D 8) (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8) (2, 1)(3, 1)(4, 1)(5, 1)(6, 1) (2, 2)(3, 2)(4, 2)(5, 2)(6, 2) (2, 3)(3, 3)(4, 3)(5, 3)(6, 3) (2, 4)(3, 4)(4, 4)(5, 4)(6, 4) (2, 5)(3, 5)(4, 5)(5, 5)(6, 5) (2, 6)(3, 6)(4, 6)(5, 6)(6, 6) (2, 7)(3, 7)(4, 7)(5, 7)(6, 7) (2, 8)(3, 8)(4, 8)(5, 8)(6, 8) (7, 1)(8, 1) (7, 2)(8, 2) (7, 3)(8, 3) (7, 4)(8, 4) (7, 5)(8, 5) (7, 6)(8, 6) (7, 7)(8, 7) (7, 8)(8, 8) Pr{(1, 1) shows} = Pr{1 shows from 1 st D 8}*Pr{1 shows from 2 nd D 8} = (1/8)*(1/8) = 1/64 . 0156 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/64) 1. 56 The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/64) = 64 < 100.

Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1, 1). Pair of Fair Dice, each with face values {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} per die: (1 st D 10, 2 nd D 10) (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8) (1, 9) (1, 10) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8) (2, 9) (2, 10) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8) (3, 9) (3, 10) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (4, 7) (4, 8) (4, 9) (4, 10) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (5, 7) (5, 8) (5, 9) (5, 10) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (6, 7) (6, 8) (6, 9) (6, 10) (7, 1) (7, 2) (7, 3) (7, 4) (7, 5) (7, 6) (71, 7) (7, 8) (7, 9) (7, 10) (8, 1) (8, 2) (8, 3) (8, 4) (8, 5) (8, 6) (8, 7) (8, 8) (8, 9) (8, 10) (9, 1) (9, 2) (9, 3) (9, 4) (9, 5) (9, 6) (9, 7) (9, 8) (9, 9) (9, 10) (10, 1) (10, 2) (10, 3) (10, 4) (10, 5) (10, 6) (10, 7) (10, 8) (10, 9) (10, 10) Pr{(1, 1) shows} = Pr{1 shows from 1 st D 10}*Pr{1 shows from 2 nd D 10} = (1/10)*(1/10) = 1/100 . 01 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/100) =1 The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/100) = 100.

Pair of Fair Dice, one with face values {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} and one with face values {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Pr{(1, 1) shows} = Pr{1 shows from D 12}*Pr{1 shows from D 10} = (1/12)*(1/10) = 1/120 0. 00833 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/120) 0. 833. The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/120) = 120 > 100.

Pair of Fair Dice, each with face values {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}: (1 st D 12, 2 nd D 12) Pr{(1, 1) shows} = Pr{1 shows from 1 st D 12}*Pr{1 shows from 2 nd D 12} = (1/12)*(1/12) = 1/144 ≈ 0. 006944444 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/144) ≈. 6944444 The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/144) = 144 > 100.

Pair of Fair Dice, one with face values {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} and one with face values {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Pr{(1, 1) shows} = Pr{1 shows from 1 st D 20}*Pr{1 shows from 2 nd D 10} = (1/20)*(1/10) = 1/200 = 0. 005 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/200) =. 50 The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/200) = 200 > 100.

Pair of Fair Dice, one with face values {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24} and one with face values {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Pr{(1, 1) shows} = Pr{1 shows from 1 st D 24}*Pr{1 shows from 2 nd D 10} = (1/24)*(1/10) = 1/240 ≈ 0. 0042 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1, 1)} = 100*(1/240) ≈. 42 The smallest sample size for which we expect to observe one or more tosses showing the pair (1, 1) is 1/(1/240) = 240 > 100.