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3 darts are thrown (equal probability of landing anywhere on the dart board). What's the probability that they all land on a same half of the dart board?

Edit: I know the first 2 darts can land anywhere but don't know how to find the probability that the 3rd lands in an acceptable region.

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    $\begingroup$ Your posted question should say something about your own thoughts on the matter. In particular, if you don't understand the question, tell us at what point you get lost, or if you understand it but don't know what to do after than, tell us where your difficulty is. $\endgroup$ – Michael Hardy Sep 1 '20 at 20:06
  • $\begingroup$ I edited to include what I've thought about. $\endgroup$ – lollab Sep 1 '20 at 20:08
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    $\begingroup$ Two hints: (1) the distance from the centre is irrelevant, so you can restrict the problem to looking at angles; (2) without loss of generality, assume that the first dart hits the board on the positive $x$-axis (i.e., assume $\theta_1=0$ in polar co-ordinates). $\endgroup$ – Théophile Sep 1 '20 at 20:18
  • $\begingroup$ @Théophile I don't understand what you mean by your first hint. If the three darts land in a small triangle about the center, then they aren't in the same half of the board. If they fall in the same triangle near the edge, then they are in the same half. I know you must mean something else, but I don't see what. Please add another detail or two for me. $\endgroup$ – saulspatz Sep 1 '20 at 20:29
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    $\begingroup$ @Théophile In the case you are interested by an alternative solution, I have provided one. $\endgroup$ – Jean Marie Sep 2 '20 at 8:45
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Suppose rays are drawn from the center of the disk to the places where the first two darts landed. And suppose for example the angle between those two rays is $23^\circ.$ Say you rotate the ray to the first dart $23^\circ$ counterclockwise to get the ray to the second dart. If you go counterclockwise from that ray $180^\circ$ you will sweep out a part of the disk within which, if the third dart lands there, then all three will be on the same side of that line formed from that ray and the ray $180^\circ$ counterclockwise from it. But also, if you start with the ray to the second dart and go $180^\circ$ clockwise you will sweep out a region within which, if the thrid dart lands there, then all three will be in that half of the disk.

Thus you will fail to put all three on the same side of some line through the center only if the third dart lands in the region $23^\circ$ wide that is opposite the one bounded by the the two aforementioned rays. Thus the probability that they would be in a common half of the disk is $$ \frac{360^\circ-23^\circ}{360^\circ}. $$

All of that holds if the angle is $23^\circ.$ But the angle is uniformly distributed between $0^\circ$ and $180^\circ,$ so what you need is the expected value of $$ \frac{360^\circ - (\text{random angle})}{360^\circ}. $$ Since an angle uniformly distributed between $0^\circ$ and $180^\circ$ is on average $90^\circ,$ the random quantity above is on average $$ \frac{360^\circ-90^\circ}{360^\circ} = \frac 3 4. $$

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  • $\begingroup$ "But the angle is uniformly distributed between $0^\circ$ and $180^\circ$ ". Is this because it's a circle and so has constant radius, as opposed to, for example, and ellipse? $\endgroup$ – Adam Rubinson Sep 1 '20 at 20:42
  • $\begingroup$ @AdamRubinson : Certainly if it had some other shape then the distribution would be different. $\endgroup$ – Michael Hardy Sep 1 '20 at 20:44
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    $\begingroup$ Thought so. Makes sense. It should also be mentioned that one or more darts landing in the centre, or two darts landing on a diameter all have probability zero, and so these scenarios can be ignored. $\endgroup$ – Adam Rubinson Sep 1 '20 at 20:47
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    $\begingroup$ There is another way to obtain the result (see my answer). $\endgroup$ – Jean Marie Sep 1 '20 at 22:10
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Let us represent impact points by their polar coordinates $(r_k,\theta_k)$ on a unit radius disk. An essential preliminary remark is that we need only consider angles $\theta_k$ (see detailed explanations in the Edit below). Therefore, this question is geometricaly equivalent to the following one: being given a triangle with vertices on the unit circle, what is the probability that this triangle doesn't contain the origin ?

This issue has been treated here where several answers give $1/4$ for the probability of the complementary event (the triangle contains the origin). Therefore, the probability that the darts belong to a same half of the dartboard is $1-1/4=3/4$.

Remark: A somewhat connected issue can be found here.

Edit: (following a remark by @BlueRaja - Danny Pflughoeft) More precisely, on the geometrical side, saying that all impact points $I_k:=r_ke^{i\theta_k}$ are in the same half-plane is equivalent to say that all points on the unit circle $J_k:=e^{i\theta_k}$ are in the same half-plane.

[an even more precise reason is that "being in the same half-plane" is a property attached to the "convex hull" of all points $ae^{i\theta_1}+be^{i\theta_2}+ce^{i\theta_3}$ with positive values of $a,b,c$.]

On the probability side, using terms defined here, the new underlying σ-algebra is a quotient space of the initial one with a "canonical" transfer of the probability law.

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  • $\begingroup$ The question you linked is talking about points chosen on (the boundary of) the circle, not in the circle. You can show these problems are equivalent, but that should not just be assumed. $\endgroup$ – BlueRaja - Danny Pflughoeft Sep 2 '20 at 4:49
  • $\begingroup$ @BlueRaja - Danny Pflughoeft Yes, you are right. I should have explained that in this issue, only polar angle θ matters. I have taken this into account in an "Edit" to my answer. $\endgroup$ – Jean Marie Sep 2 '20 at 6:21

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