L9supp - N 10 20 21 22 23 30 40 50 p N .12 .41 .44 .47 .503...

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Example: - 19 × 19 square Go board (Japanese board game) - throw dart repeatedly and at random - mark square hit 1
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1. P (specific square hit) = 1 19 × 1 19 = 1 361 2. How many throws, until high probability of hitting some square again? 3. After N throws, what is prob- ability of hitting some square again? 2
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N throws P (all different) = 361 361 · 360 361 · ... · 361 - N + 1 361 P (some square at least twice) = 1 - P (all different) = 1 - 361 361 · 360 361 · ... · 361 - N + 1 361 = p N Fill in table using calculator. N 10 20 21 22 23 30 40 50 p N 3
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Or try nice approximation: 1 - x e - x p N = 1 - N - 1 Y i =0 361 - i 361 ! = 1 - N - 1 Y i =0 1 - i 361 ! 1 - N - 1 Y i =0 exp - i 361 ! = 1 - exp - N - 1 X i =0 i 361 = 1 - exp - ( N - 1) N 2(361)
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Unformatted text preview: N 10 20 21 22 23 30 40 50 p N .12 .41 .44 .47 .503 .70 .88 .97 4 Probability says that with only 23 darts, there is a better than 50% chance that one of the 361 squares will be hit more than once. Most people nd this surprising because they expect that the 23 darts would be spread out more (think bombs instead of darts for a stronger image) and so many more darts would be needed to achieve 50%. If you nd this surprising, try simulating on a computer. Time permitting we will do this in class. 5...
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This note was uploaded on 01/03/2012 for the course EE 1244 taught by Professor Drera during the Fall '10 term at Conestoga.

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L9supp - N 10 20 21 22 23 30 40 50 p N .12 .41 .44 .47 .503...

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