Cutting a stick twice

Consider three situations -
$1)$ A stick, placed at $[0,1]$ is cut at a point $x$ given by R.V $X$ uniformly distributed in $[0,1]$. Another cut is made in $[0,x]$ given by R.V $Y$, uniform in the support region.
$2)$ A stick, placed at $[0,1]$ is cut at a point $y$ given by R.V $Y$ uniformly distributed in $[0,1]$. Another cut is made in $[y,1]$ given by R.V $X$, uniform in the support region.
$3)$ A stick, placed at $[0,1]$ is cut at two points $x$ and $y$ uniformly distributed. The random variable $Y = min(x,y)$ and $X = max(x,y)$

Practically / Intuitively, all the situations seem the same, because you are basically making two cuts and all three have the same support region, i.e, the right lower triangle in the unit square. However, By calculating probability distribution, we get-
$1)f_{XY}(x,y)=\frac 1x$
$2)f_{XY}(x,y)=\frac 1{1-y}$
$3)f_{XY}(x,y)= 2$
Can someone explain as to how this is practically possible? I feel that every event in $1$ has an equally probable event in $2$ and $3$ ($2$ is just the same stick put the other way, from $[1,0]$ isn't it?)

• Note cases 1 and 2 are the same except for a change of variable $(x,y)\to(1-y,x)$. Case 3 is a very different situation from the others. – Macavity Mar 10 '16 at 15:51

In each of the situations there are random variables $U,V$ with $0<U<V<1$ a.s.

In situation 1) $V$ has uniform distribution on $[0,1]$ and $U$ has not.

In situation 2) $U$ has uniform distribution on $[0,1]$ and $V$ has not.

In situation 3) $U$ neither $V$ has uniform distribution on $[0,1]$.

So the situations are evidently not the same.

• You are right, but that's what I already know. I said that maths proved my intuition wrong. My argument goes like this - Every event, say $(x,y)$ in $1$ has the same probability of occurance as in $2$ since, you can just rotate the view $180^o$ and that's the same as case $1$.. isn't it? – Win Vineeth Mar 10 '16 at 14:55
• Events $\{U=x,V=y\}$ have probability $0$ hence do not provide information about the distribution. – drhab Mar 10 '16 at 15:07
• How come $(U_1,V_1)$ and $(V_2,U_2)$ have equal distribution? Their support regions are different at first. And, by $(x,y)$ I meant their pdf. – Win Vineeth Mar 10 '16 at 15:08
• Sorry, I meant to say that $(U_1,V_1)$ and $(1-V_2,1-U_2)$ have equal distribution. – drhab Mar 10 '16 at 15:40
• I understand somewhat, thank you. But It's not completely clear. – Win Vineeth Mar 10 '16 at 16:10

The regions over which case 1 and 2 are distributed are different. Sketch a picture in the xy-plane of both to see.