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General question: Given a dartboard of unit radius, what's the probability that a dart randomly lands within a circle of radius 1/3 centered inside the dartboard?

Standard answer: The dart is thrown such that it hits each point with equal likelihood. The probability that it lands within the inner circle is the ratio of the areas of the two circles, which is 1/9.

Other formulation: Suppose that the dart is thrown directly at the center of the dartboard, but wind comes from a random direction. For any wind direction, the velocity of the wind pushes the dart some distance from the center of the dartboard to its perimeter, with each distance being equally likely. Each wind vector is equally likely and, for each vector, each distance is equally likely. The probability of being less than 1/3 units from the center of the dartboard is 1/3.

The definition of randomness is different for each formulation. In the standard answer, a random vector is chosen from the set ${(x,y)\colon \; x^2+y^2 \leq 1}$ and we ask the probability that $x^2 + y^2 \leq \frac{1}{9}$. In the other formulation, a random vector is chosen ${(x,y)\colon \; x^2+y^2 = 1}$ and, on this vector, a distance $d$ is chosen uniformly on $[0,1]$. We ask the probability that $d\leq\frac{1}{3}$.

I understand the math of solving this problem, but I don't understand intuitively why these different conceptions of randomness give two different answers. It seems that both are valid means for answering the general question. Intuitively, why do they give different answers?

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Intuitively, imagine modeling the second formulation as follows: randomly select an angle to the $x$-axis, calling it $\theta$, then model the location of the dart as falling uniformly in a very thin rectangle along the line $y = (\tan\theta) x$. Approximately, the dart is in the inner circle with probability $1/3$. However, when you consider the collection of all such thin rectangles (draw them, say), you will see that they have more overlapping area near the center of the dartboard, and less overlap towards the perimeter of the dartboard. This will be more obvious as you draw the rectangles larger and larger (though the approximation will be worse). As you make the rectangles thinner, the approximation gets better, but the same principle applies: you are putting more area around the center of the circle, which increases the probability of hitting the inner circle.

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    $\begingroup$ This is also why the area of a circle is not equal to $\int_0^{2\pi} \int_0^R 1 d\theta dr$. $\endgroup$
    – shabbychef
    Commented Aug 30, 2010 at 19:03
  • $\begingroup$ Thanks for this response. To follow-up: is the solution to the the "other formulation" (wind vectors), 1/3, correct given the issues that you raise? $\endgroup$
    – Charlie
    Commented Sep 3, 2010 at 21:56
  • $\begingroup$ yes, it is correct. Both formulations are a little bit fishy, though. Imagine if the dart board were 30 feet in diameter. I don't think either of these models would be appropriate in that case $\endgroup$
    – shabbychef
    Commented Sep 20, 2010 at 5:27
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It seems to me that the fundamental issue is that the two scenarios assume different data generating process for the position of a dart which results in different probabilities.

The first situation's data generating process looks like so: (a) Pick a $x \in U[-1,1]$ and (b) Pick a $y$ uniformly subject to the constraint that $x^2+y^2 \le 1$. Then the required probability is $P(x^2 + y^2 \le \frac{1}{9})= \frac{1}{9}$.

The second situation's data generating process is as described in the question: (a) Pick an angle $\theta \in [0,2\pi]$ and (b) Pick a point on the diameter that is at an angle $\theta$ to the x-axis. Under this data generating process the required probability is $\frac{1}{3}$ as mentioned in the question.

As articulated by mbq, the issue is that the phrase 'randomly lands on the dartboard' is not precise enough as it leaves the meaning of 'random' ambiguous. This is similar to asking what is the probability of coin landing heads on a random toss. The answer can be 0.5 if we assume that the coin is a fair coin but it can be anything else (say, 0.8) if the coin is biased towards heads.

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Think of the board as a filter -- it just converts the positions on board into an id of a field that dart hit. So that the output will be only a deterministically converted input -- and thus it is obvious that different realization of throwing darts will result in distribution of results.
The paradox itself is purely linguistic -- "random throwing" seems ok, while the true is that it misses crucial information about how the throwing is realized.

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