I have $x$ darts that I randomly throw on a dartboard with $y$ slots (the v shaped slices like you get when you cut a cake). Now I select a particular slice on the dartboard . What's the probability that no dart would have hit that slice?
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4$\begingroup$ In my opinion this should go to Math.SE $\endgroup$– ABcDexterJun 14, 2016 at 14:36
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1$\begingroup$ Uniform, I.I.D. with a single boundary at the edge of the union of the slices I guess? That's some pretty accurate random throwing :) $\endgroup$– Jonathan AllanJun 14, 2016 at 14:42
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1$\begingroup$ Has the Board no Bulls eye (and half bull)? Does the outer border around the slices exist and also count? Are the wires between the slices excluded (bouncers)? Are there double and tripple fields at the board? Do you mean a real dart board, or a simplified abstract dart board? I know, I am maybe overcorrect, but I play darts ;) $\endgroup$– kl78Jun 14, 2016 at 14:56
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2 Answers
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$P = (1-\frac{1}{y})^x$ : One slice covers $\frac{1}{y}$ of the dartboard's surface, so one dart avoids one slice with probability $(1-\frac{1}{y})$, and over $x$ throws with probability $(1-\frac{1}{y})^x$
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It is that :
first dart misses the slice AND second also misses the slice AND .... AND last (xth) also misses the slice. which is equal to (y-1/y) * (y-1/y) * (y-1/y) * .... * (y-1/y) (x times) = (y-1)^x / y^x final result = (y-1)^x / y^x
