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Five darts are thrown at a dart board, and each has an equal probability of hitting any point on the board, what is the probability that all of the darts land on the same half of the board (that is, we could draw a line through the center of the bullseye such that all of the darts are on one side of the line)?

My thought process was as follows, P(of landing in one specific half of the board) = (0.5 * πr^2)/(πr^2) = half circle's area/the full circle's area = 1/2. And to calculate the probability that all of the 5 darts land on that half = (1/2)^5 = 1/32.

I think my solution is wrong, would anyone please tell me what is it wrong and what would be the correct way to solve it?

Moaz Nasem
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    Your solution is indeed incorrect, for a pretty straightforward reason. What you've calculated is the probability that the five darts land in a specific half of the dartboard: say, the left hand side of it. But that's distinct from the probability that there's some half of the dartboard that contains all of them. For a simple example of why this is so, consider the case of two darts: your approach would claim that they have probability 1/4 of being in the same half, but in fact for any two darts on a dartboard you can find a diameter they're both on the same side of. – Steven Stadnicki Nov 17 '23 at 20:57
  • @StevenStadnicki Thanks for your answer. yeah, that make sense, now I understand why my solution is wrong. How to approach this question then? – Moaz Nasem Nov 17 '23 at 21:04
  • If each point is equally likely, I think each angle is equally likely, and the distance from the center doesn't affect it being in a specific half. Calling each point's angle $a,b,c,d,e \in [0^{\circ}, 360^{\circ})$ then the problem should reduce to calculating the probability that $max-min \leq 180^{\circ}$ – dgeyfman Nov 17 '23 at 21:10
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    Is the line dividing the board into half drawn beforehand (eg,say, vertically) or can it be drawn after the darts have been thrown ? – true blue anil Nov 17 '23 at 21:12
  • Not a solution but a possible lower bound: The first dart determines a line through the origin (unless it's a bull which has zero likelihood). Then we want all four remaining darts on one side of that line, probability 2 times 1/16 or 1/8. So if this vague approach works the probability you seek is at least 1/8. – coffeemath Nov 17 '23 at 21:13
  • @trueblueanil If the line is predrawn prob is $2\cdot (1/2)^5.$ – coffeemath Nov 17 '23 at 21:16
  • @dgeyfman Thanks for your reply. can you please illustrate your answer furthermore? – Moaz Nasem Nov 17 '23 at 21:16
  • @coffeemath: I don't think that is right. The first dart determines a line... sure. So what? I can easily have two darts slightly to the right (clockwise) and two darts slightly to the left (counterclockwise) and yet have all five in the same final half of the board. – David G. Stork Nov 17 '23 at 21:16
  • @DavidG.Stork I was only going for a lower bound on the probability. – coffeemath Nov 17 '23 at 21:17
  • @dgeyfman Does this mean the probability will always be (max-min)/360 regardless of how many darts we have? the biggest angle between the first two darts is at max a 180 degrees, so on average it's 90 degrees. so the problem reduces to (360 - 90)/360 = 3/4? Is this the correct answer regardless of how many darts I have? – Moaz Nasem Nov 17 '23 at 21:24
  • This will be a duplicate of another question, but essentially you want $n-1$ darts to be in the half clockwise from a special dart, and that has probability $\dfrac{1}{2^{n-1}}$, but any of the darts could be the special one, making the overall probability $\dfrac{n}{2^{n-1}}$ since the $n$ different cases are mutually exclusive. https://math.stackexchange.com/questions/3811030/probability-that-3-darts-land-in-a-same-half-of-a-dart-board and https://math.stackexchange.com/questions/268635/what-is-the-probability-that-the-center-of-the-circle-is-contained-within-a-tria deal with $3$ darts – Henry Nov 17 '23 at 21:27
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    I agree with the max-min approach in the comment of @dgeyfman, with one exception. You have to use clock arithmetic. For example, if one of the angles is $~345^\circ,~$ and one of the angles is $~15^\circ,~$ you have to interpret the $~15^\circ~$ angle as $~(360 + 15)^\circ~$ to perform the calculation. This seriously complicates the problem. – user2661923 Nov 17 '23 at 21:30
  • @Henry Thanks for your reply, how did you calculate the 1/2^n−1 – Moaz Nasem Nov 17 '23 at 21:32
  • @MoazNasem You have a special dart and you want the other $n-1$ to be in the semicircle clockwise from the special dart. Each independently has probability $\frac12$ so just take the product – Henry Nov 17 '23 at 21:34
  • @Henry Thanks, that makes sense. so in a way the first dart a throw is just to show me where the targeted semicircle will be from the point it hit and 180 degrees from that. then the rest of the darts had each a probability of 1/2 to land in that half that explains the 2^n-1. – Moaz Nasem Nov 17 '23 at 21:39

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