# Volume of the probability simplex?

I think I might be misunderstanding the concept of a simplex and its volume.

Take the 2-dimensional simplex (a triangle) embedded in 3-dimensional space with vertices $$(1,0,0)$$, $$(0,1,0)$$, and $$(0,0,1)$$. When I calculate its area, I get $$\frac{\sqrt{3}}{2}$$. However, I've come across information suggesting that the volume (area in this case) of this probability simplex should be $$\frac{1}{2}$$.

Similarly, the 3-dimensional tetrahedron (a 3-dimensional simplex) embedded in 4-dimensional space with vertices $$(1,0,0,0)$$, $$(0,1,0,0)$$, $$(0,0,1,0)$$, and $$(0,0,0,1)$$. When I calculate its volume, I obtain $$\frac{1}{3}$$, but I've read that the volume of this 4-dimensional probability simplex is $$\frac{1}{6}$$.

Can someone explain the meaning of this discrepancy in the volume?

• Maybe you're thinking of $\{(x, y) \,:\, 0\leq x \leq y \leq 1\}$, which has volume 1/2, and $\{(x,y,z)\in\mathbb{R}^3\,:\,0\leq x \leq y \leq z \leq 1\}$, which has volume 1/6. Commented Oct 17, 2023 at 15:43

The area of the 2-simplex is indeed $$\sqrt{3}/2$$. There's a different 2-simplex, with vertices $$(0,0), (1, 0), (0, 1)$$ in the plane whose area is $$1/2$$; the corresponding 3-simplex in 3-space (with vertices at the origin and at all points with exactly one non-zero coordinate, which is 1, i.e. $$(1,0,0), (0,1,0), (0,0,1)$$) has volume $$1/6$$. In dimension $$n$$, that simplex has volume $$1/n!$$.