equation for probability stumper? Jane says she was in town for two consecutive days last week (7 days), but won't say more. For each given day, what was the probability she was in town that day? I know there are 6 possible realizations, so it's 1/6 for the first and last days, and if I diagram it I can see readily that days 2 through 6 are 1/3.
But I wonder if there's an equation (or algorithm) that could supply an answer for any combination of days-in-duration.
 A: Sure...it's not complicated but it is badly error prone.  In particular, it is prone to the so-called off-by-one, or "fencepost" error (so-called because of the error one tends to make in counting the number of fenceposts needed to border  $N$ yards if you want to separate each by $1$ yard). I think that what follows is correct, but I advise testing it.
Let's say you had $N$ total days, numbered {$1,2,...,N$} and we know your friend stayed for $k$ consecutive days but we don't know which.  Of course we assume that any $k$ consecutive days are equally probable. In your case, $N=7$ and $k=2$.
How many blocks of $k$ consecutive days are there?  Assuming that $k<N$ we could start on any day up to day $N-k+1$.  Thus, in your case, the block could start on any of days $1, ..., 6$.  Thus there are $N-k+1$ equally probably blocks your friend might have selected.
Pick a day, $i$.  How many blocks is $i$ in?  Well, suppose $j$ is the start of a block containing $i$.  What do we know?  Well we need $1≤j≤i≤j+k-1≤N$.  Thus the smallest $j$ that works is $j_{min}=Max(1,i-k+1)$ and the largest $j$ that works is $j_{max}=Min(i,N-k+1)$. The total number of possible days is $j_{max}-j_{min}+1$  So the probability that your friend was here on day $i$ is $$\frac {Min(i,N-k+1)-Max(1,i-k+1)+1}{N-k+1}$$
Let's check that in your example.   if $i=1$ then this expression gives 
$\frac {1-1+1}{6}=\frac 16$ as desired.  If $i=4$ then we get $\frac {4-3+1}{6}=\frac 26$ as desired.  And if $i=7$ we get $\frac {6-6+1}{6}=\frac 16$ as desired.
