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Rare Events, Probability and Sample Size

Rare events occur when their probability is very small, highlighting the need for large sample sizes to observe them effectively. This article discusses the concept of minimal sample size required to expect at least one occurrence of a rare event, using the example of pairs of fair dice. It illustrates how the expected count for rare events relates to the sample size and probability, offering calculations for different die configurations. Understanding these principles is crucial for valid statistical analysis when dealing with rare occurrences.

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Rare Events, Probability and Sample Size

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  1. Rare Events, Probability and Sample Size

  2. 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.

  3. 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, expectedE≈ 1 for n ≈ (1/ PE). When PE is small, (1/ PE) is large.

  4. 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 expectedE= n* PE < 1 n < 1/PE.

  5. Pairs of Dice and the Pair (1,1)

  6. 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 (1st D3, 2nd D3) (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 1st D3}*Pr{1 shows from 2nd D3} = (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.

  7. 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: (1st D4, 2nd D3) (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 1st D4}*Pr{1 shows from 2nd D3} = (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.

  8. 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: (1st D4, 2ndD4) (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 1st D4}*Pr{1 shows from 2nd D4} = (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.

  9. 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: (1stD8, 2ndD8) (1,1) (2,1)(3,1)(4,1)(5,1)(6,1) (7,1)(8,1) (1,2) (2,2)(3,2)(4,2)(5,2)(6,2) (7,2)(8,2) (1,3) (2,3)(3,3)(4,3)(5,3)(6,3) (7,3)(8,3) (1,4) (2,4)(3,4)(4,4)(5,4)(6,4) (7,4)(8,4) (1,5) (2,5)(3,5)(4,5)(5,5)(6,5) (7,5)(8,5) (1,6) (2,6)(3,6)(4,6)(5,6)(6,6) (7,6)(8,6) (1,7) (2,7)(3,7)(4,7)(5,7)(6,7) (7,7)(8,7) (1,8) (2,8)(3,8)(4,8)(5,8)(6,8) (7,8)(8,8) Pr{(1,1) shows} = Pr{1 shows from 1st D8}*Pr{1 shows from 2nd D8} = (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.

  10. 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: (1stD10, 2ndD10) (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (7,1) (8,1) (9,1) (10,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (7,2) (8,2) (9,2) (10,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (7,3) (8,3) (9,3) (10,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (7,4) (8,4) (9,4) (10,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (7,5) (8,5) (9,5) (10,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) (7,6) (8,6) (9,6) (10,6) (1,7) (2,7) (3,7) (4,7) (5,7) (6,7) (71,7) (8,7) (9,7) (10,7) (1,8) (2,8) (3,8) (4,8) (5,8) (6,8) (7,8) (8,8) (9,8) (10,8) (1,9) (2,9) (3,9) (4,9) (5,9) (6,9) (7,9) (8,9) (9,9) (10,9) (1,10) (2,10) (3,10) (4,10) (5,10) (6,10) (7,10) (8,10) (9,10) (10,10) Pr{(1,1) shows} = Pr{1 shows from 1st D10}*Pr{1 shows from 2nd D10} = (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.

  11. 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 D12}*Pr{1 shows from D10} = (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.

  12. Pair of Fair Dice, each with face values {1,2,3,4,5,6,7,8,9,10,11,12}: (1st D12, 2nd D12) Pr{(1,1) shows} = Pr{1 shows from 1st D12}*Pr{1 shows from 2nd D12} = (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.

  13. 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 1st D20}*Pr{1 shows from 2nd D10} = (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.

  14. 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 1st D24}*Pr{1 shows from 2nd D10} = (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.

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