I don’t have any background in geometric probability, so may I ask for forgiveness if the following is wrong or doesn’t make sense:

**Assumptions:**

- no three sampled points are collinear
- the (naive) geometric probability in euclidean space is equal to the ratio of lengths or areas
- the probabilites do not change under isometric transformations and/or uniform scaling
- the probability of a set $ P4:=\lbrace p_1,\ p_2,\ p_3,\ p_4\rbrace$ of four points being in convex configuration in the euclidean plane is independent of the order in which the elements are drawn by the sampling process; that in turn means that we assume that
- $ \lbrace p_1,\ p_2,\ p_3\rbrace$ resembles the corners of $ T_{max}$ , the triangle of largest area
- $ \lbrace p_1,\ p_2,\ p_3\rbrace$ have been drawn uniformly from the boundary of their circumcircle, which has implications on the probability of encountering $ T_{max}$ with specific values for the pair of smallest central angles.
- the smallest circle enclosing $ P4$ is the unit circle centered at the origin
- the longest side of $ T_{max}$ is bisected by the non-negative part of the x-axis

Under the above assumptions the probability of encountering a convex quadrilateral equals the blue area divided by the blue plus red area in the images below:

the sampling area in case of acute $ T_{max}$ equals the entire unit disk

the sampling area in case of acute $ T_{max}$ equals the unit disk with a notch

The notch in the case of obtuse $ T_{max}$ is owed to the assumption that the first three points resemble $ T_{max}$ which implies that points outside $ T_{max}$ that generate a deltoid configuration would contradict the maximality of the area of $ T_{max}$

If the above makes sense, then the probability of four points being in convex configuration could be calculated by integrating over the ratios of the blue areas over the entire sampling area as defined by the $ T_{max}$ multiplied with the probability of $ T_{max}$ as a consequence of uniform sampling on the boundary of the unit circle.

Questions:

- have similar ways of defining the sample spaces for Sylvester’s Four Point Problem already been described or investigated?
- what are objections against the proposed against the proposed definition of the available sample space on basis of $ T_{max}$ ?

**Remark:**

in case of obtuse $ T_{max}$ the area of the sample space can be calculated on basis of angles $ \alpha$ and $ \beta$ that are adjacent to the longest side of $ T_{max}$ as follows, keeping in mind that that longest edge is the diameter of the unit circle:

- the area of the lower blue half-disk equals $ \ \frac{\pi}{2}\ $
- the area $ \ A_{\alpha}\ $ of the union of $ \ T_{max}\ $ with the blue region opposite to angle $ \ \alpha$ equals $ \ \alpha+\sin(\alpha)cos(\alpha)\ $ and analogous $ \ A_{\beta}\ $ for angle $ \ \beta\ $
- the area of $ \ A_{T_{max}}\ $ of $ \ T_{max}\ $ equals $ \ \frac{1}{\cot(\alpha)+\cot(\beta)}\ $

The area of the sampling area for obtuse $ \ T_{max}\ $ then equals $ \ \frac{\pi}{2}+A_{\alpha}+A_{\beta}-A_{T_{max}}$