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:Indeed, both probabilities coincide with the left and right side of the Nicomachus identity, but the events are certainly not equivalent. In fact, one can only conclude their probabilities are equal ''because'' of the Nicomachus identity. There are no further references either, and I could not find anything else in google. I think this is an original idea someone posted. <!-- Template:Unsigned IP --><small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/61.127.95.156|61.127.95.156]] ([[User talk:61.127.95.156#top|talk]]) 13:08, 12 February 2018 (UTC)</small> <!--Autosigned by SineBot-->
:I agree; that section is at best confusing, and at worst wrong. I've added a "Confusing section" tag. [[User:PatricKiwi|PatricKiwi]] ([[User talk:PatricKiwi|talk]]) 06:54, 16 February 2021 (UTC)
::It needs a source, but there is a very natural four-dimensional polyhedral interpretation in which this equality of probabilities is analogous to the equality of polyhedral numbers. The LHS is the volume of the subset max(x,y,z)≤w of the unit 4-cube, which by [[Cavalieri's Principle]] in the w-direction is just the integral over all choices of w of the cube max(x,y,z)≤w in the 3-dimensional cross-section for that w, and the RHS is the volume of a Cartesian product of two isosceles right triangles 0≤x≤y≤1 and 0≤z≤w≤1 in the xy and zw planes. So it's the same principle, a sum (or this time integral) over cubes equals the square of a triangle. To me that's so similar as to make it obviously not coincidental. Presumably one could even prove these volumes/probabilities equal without calculating them first, as a limiting case of the Nicomachus identity —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 07:01, 16 February 2021 (UTC)
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