Probability each table leg was in each spot. I have a fold up table at home with six legs and three areas. Each area takes 2 legs to keep the table up, but when stored three legs are positioned on the left and three on the right.
Everytime I setup the table I randomly take the three legs on the left, put two on the left side and one in the middle. Then I take the three legs from the right and put two on the right and also one in the middle.
When folding it up I randomly pick one leg in the middle to go left and one to go right.
So basically a leg that started on the left when folded, could end up in the middle when setup and end on the right when folded again.
I've setup this table maybe a hundred times and everytime I wonder "Has each leg been in every spot (six spots when setup) at least once?"
I guess there is a 1/3 change that a leg ends up in the middle and then a 1/2 change it ends up on the other side, which by my uneducated mind results in a 16.6% probability it changes sides, but I have no idea how to go from there.
If you perfectly rotate the legs around, you could get them in each spot in 6 times. But I don't know that is even relevant at all.
So, what is the probability each leg has been in every spot?
 A: Suppose the legs are numbered and ordered in a list, for example [1,2,3,4,5,6]. When packed away the first three legs in the list are on the left, the last three are on the right. When erected the first two are on the left, the middle two in the middle and the last two on the right. 
Let's consider the possible permutations that occur between a table being packed away and then re-erected. If we start with a table with the order [1,2,3,4,5,6], the possible orders after being packed are [1,2,3,4,5,6] or [1,2,4,3,5,6]. After being reassembled the 18 possible permutations are:
P = {   [1,2,3,4,5,6],
        [2,3,1,4,5,6],
        [2,3,1,4,5,6],
        [1,2,3,5,4,6],
        [1,3,2,5,4,6],
        [2,3,1,5,4,6],
        [1,2,3,6,4,5],
        [1,3,2,6,4,5],
        [2,3,1,6,4,5],
        [1,2,4,3,5,6],
        [1,4,2,3,5,6],
        [2,4,1,3,5,6],
        [1,2,4,5,3,6],
        [1,4,2,5,3,6],
        [2,4,1,5,3,6],
        [1,2,4,6,3,5],
        [1,4,2,6,3,5],
        [2,4,1,6,3,5]} 
As some commenters mentioned, working out precisely the expectation that all legs end up in all spots is (probably) very hard, but I think this is a great example of a situation where running randomised simulations in a computer should give a pretty good real world answer. 
For a given number of disassemblies and erections between 1 and 100, I ran 10000 simulations and counted for how many simulations each of the legs occurred in each of the positions. The results:

So, if you have setup the table 100 times I would say it's very likely indeed (> 99%) that all the legs would have been in all positions!
A: Solution for one specific leg only.  (Note: It is not obvious, at least for me, how to generalize from one leg to six legs, coz the legs are dependent.  In other words, @JMP probably used code that has way more states than this solution.)
The following Markov chain will model moments when the leg is stored.  The state is $(x,y,z,b)$ where:


*

*$x,y,z \in \{0,1,2\}$ represent how many spots it has visited in the Left, Middle, Right pair of holes respectively, and 

*$b \in \{L,R\}$ represents which side its being stored.

*We start in state $\{0,0,0,L\}$
Consider a generic state $(x,y,z,L)$ (the $R$ case is symmetric).  


*

*There is $2/3$ prob it will visit one of the Left holes.  Conditioned on this happening, the next $b = L$, and $x$ will increment or stay the same:


*

*$x: 0\to 1$ with prob $1$

*$x: 1\to 2$ with prob $1/2$

*$x: 2$ cannot increment

*We can summarize the above by saying $Prob(x \to x+1 \,\,\text{i.e. increments}) = 1 - x/2$, and similarly $Prob(x \to x \,\,\text{i.e. stays constant}) = x/2$.


*There is $1/3$ prob it will visit one of the Middle holes.  Conditioned on this happening:


*

*The probabilities for $y$ to increment are analogous to above: $Prob(y \to y+1) = 1-y/2$

*In this case the next $b\in \{L, R\}$ with $1/2$ prob each.
So collecting everything together in terms of transition probabilities:


*

*$P((x,y,z,L) \to (x,y,z,R)) = \frac13 \frac12 {y \over 2}$

*$P((x,y,z,L) \to (x,y+1,z,R)) = \frac13 \frac12 (1 - {y \over 2})$

*$P((x,y,z,L) \to (x,y+1,z,L)) = \frac13 \frac12 (1 - {y \over 2})$

*$P((x,y,z,L) \to (x+1,y,z,L)) = \frac23 (1- {x \over 2})$

*$P((x,y,z,L) \to (x,y,z,L)) = \frac23 {x \over 2} + \frac13 \frac12 {y \over 2}$ 


*

*Note: the first term is the sub-case of visiting an already-visited Left hole, and the second term is the sub-case of visiting an already-visited Middle hole.


*Sanity check: the above $5$ probabilities sum to $1$, whew!  :)
So we have a Markov chain with $3 \times 3 \times 3 \times 2 = 54$ states, where each state typically can reach $5$ states (including itself), but sometimes fewer (if some of $x,y,z$ have maxed out).  What you want is the time $T$ (i.e. no. of steps) to go from $(0,0,0,L)$ to $(2,2,2,b)$ where you don't care whether $b=L$ or $R$.  You specific question is the value of $Prob(T \le 100)$.  
This kind of problem is well studied (I've seen it called "first visit", or "time to absorption" if you combine $(2,2,2,L)$ and $(2,2,2,R)$ into a single absorbing state).  IIRC closed form solution is possible...?  But of course the issue is that, without further insightful tricks, the transition matrix itself is complicated, so even a closed form (based on the matrix) will still be complicated.
