Could someone please explain me how it works for comparing uniform distributions?

I have three independent uniform distributions:

a from 0 to Ab from 0 to Bc from 0 to Cwith A<=B<=C

so why P(a is max(a,b,c)) is A^2/(6BC) and not P(a>b).P(a>c) ?( P(a>b) = A/2B )I know that for identical uniform distributions, the min, median and max tend to be evenly distributed in the interval at 1/4, 2/4, 3/4, each being in the middle of the space left when you have put the others.

what is the formula for P(a is min(a,b,c)) ?

This is not a school problem,  I'm trying to solve analytically some [0,1] toy poker games…

Textile garbling again is really annoying.

a from 0 to A

b from 0 to B

c from 0 to C

with A<=B<=C

Anyone knows how to write in the forum without lines being aggregated?

shouldn't it be A^2/(3BC)? (if A=B=C, shouldn't it be 1/3?)

It's not P(a>b)P(a>c) because those a>b and a>c aren't independent events. Both are saying “a is big”, so why “a is big” should be uncorrelated with “a is big”?

Your probability space is a brick in 3D space. The part with b > A or c > A contributes nothing to the probability in question, so we're only interested in the cube 0 <= a, b, c <= A. This cube's volume is A/B * A/C of all the brick's volume. And the a=max() set occupies 1/3 of the cube due to symmetry. There, 1/3 * A/B * A/C = A^2/(3BC).

Oh thanks alihv for this geometrical explanation, it seems so simple now!

Yes you are right, A^2/(6BC) was half this volume with added condition that b>c, so probability(a>b>c).

So I guess the formula for P(a=min(a,b,c)) is :

1-a(1/2b+1/2c-a/(3 b c))