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I saw a which said that points were picked uniformly and independently. I have a feeling this is important for the solution but I am not sure what they mean by uniformly and independently. Any help would be appreciated. I will post the original question here.

Question: Four points are chosen uniformly and independently at random in the interior of a given circle. Find the probability that they are the vertices of a convex quadrilateral.

Note: I am not looking for a solution to the problem, just for clarification on what uniformly and independently mean.

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  • $\begingroup$ "Uniformly" means without applying any preference to a particular part of a region. So, if you were choosing uniformly amongst a discrete set of options, say the numbers $1$ through $10$, you're just as likely to choose $5$ as you are $2$: the probability of drawing any given number is $\frac{1}{10}$. "Independently" here really just means that the last point you chose has no impact on any future point you choose. So, say that $A_1$ is your first choice and $A_2$ your second choice. The probability of drawing both is $\Pr[A_1 \cap A_2] = \Pr[A_1] \cdot \Pr[A_2]$. $\endgroup$ – Matt.P Aug 11 '18 at 17:31
  • $\begingroup$ Uniformly in this case means the probability density is a constant $\frac{1}{\pi r^2}$. Independent means the choice of any one point has no effect on the choice of any other point. $\endgroup$ – herb steinberg Aug 11 '18 at 17:33
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    $\begingroup$ @Matt.P This should be an answer. $\endgroup$ – Arnaud Mortier Aug 11 '18 at 17:38
  • $\begingroup$ Thank you. That really clears it up. $\endgroup$ – John Wick Aug 11 '18 at 17:56
  • $\begingroup$ I know you weren't looking for the solution to the problem; but if you ever do look for it, you'll find it here: mathworld.wolfram.com/SylvestersFour-PointProblem.html $\endgroup$ – joriki Aug 11 '18 at 19:26
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The brief answer is that the four random points follow the "obvious" distribution — the distribution you would have assumed if you were the sort of person who didn't realize you were assuming a probability distribution.

A more precise statement is:

  • That $Q_i$ is chosen uniformly means that for any subregion $S$ of the interior of the circle, $\mathbf{P}(Q_i \in S)$ is proportional to the area of $S$
  • That the four points are chosen independently means that for any four regions $S_i$, you have $$ \mathbf{P}(Q_1 \in S_1 \text{ and } \ldots \text{ and } Q_4 \in S_4) = \mathbf{P}(Q_1 \in S_1) \cdot \ldots \cdot \mathbf{P}(Q_4 \in S_4) $$
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Sometimes a picture helps more than words and formulas, at least for the term "uniform":

"Independent" means that a single one of the random points does not care where the other points are, but its position is picked independently of the others, as if they were not there. E.g. to be not independently chosen, a point could try to avoid being placed at a position where many other points have been placed already. This can give a uniform pattern, but they are not chosen independently (now the order in which the points are chosen matters).


As Hurky said, when one (mostly a non-mathematician) says "random", what he means most of the time includes the properties uniform and independent. E.g. throwing a dice gives uniformly and independently chosen random numbers from $\{1,2,3,4,5,6\}$. Let's see what the same experiment might look like when one of the properties is missing:

  • not uniform: the dice is biased, e.g. it was manipulated to show the $6$ more often (on average) than any other number.
  • not independent: the dice "has a memory". E.g. it tries to not show the same number as on the last throw. Sometimes people think that because there was no $6$ for a long time, the probability that there will come a $6$ is now higher. This can only be true if the dice does not generate independent numbers, which is usually not the case.
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I'm reformulating an earlier comment I made as an answer.

"Uniformly" means without applying any preference to a particular part of a region. So, if you were choosing uniformly amongst a discrete set of options, say the numbers $1$ through $10$, you're just as likely to choose $5$ as you are $2$: the probability of drawing any given number is $1$ in $10$.

"Independently" here really just means that the last point you chose has no impact on any future point you choose. So, say that $A1$ is your first choice and $A2$ your second choice. The probability of drawing both is $\Pr[A1 \cap A2]= \Pr[A1] \cdot \Pr[A2]$.

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