### Rare Events, Probability and Sample Size

```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,
expectedE ≈ 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
expectedE = 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 (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.
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.
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, 2nd D4)
(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.
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: (1st D8, 2nd D8)
(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 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.
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: (1st D10, 2nd D10)
(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 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.
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.
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.
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.
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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