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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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closed as off-topic by Ivo Beckers, JMP, Roland, manshu, Gareth McCaughan Jun 14 '16 at 15:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Ivo Beckers, JMP, Roland, manshu, Gareth McCaughan
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ In my opinion this should go to Math.SE $\endgroup$ – ABcDexter Jun 14 '16 at 14:36
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    $\begingroup$ 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 Allan Jun 14 '16 at 14:42
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    $\begingroup$ 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 ;) $\endgroup$ – kl78 Jun 14 '16 at 14:56
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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

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