Sudoku candidate probability Supose a Sudoku puzzle row has four empty cells. The candidates for each cells is as follows:


*

*$3, 6, 9$    

*$7, 9$  

*$3, 6, 7$

*$3, 7$


Looking at the possible cells for $3$ (cells $1$, $3$ and $4$), the probability for $3$ in each cell could be:
a) $\frac13$, $\frac13$, $\frac13$: if we only look at the possible choices for $3$
b) $\frac13$, $\frac13$, $\frac12$: if we figure the total candidates for each cell.
Both the assumptions above are wrong.  
What is the probability of each cell being a $3$ in cells $1, 3$ and $4$?
 A: Whenever you talk about the probability of an event, you have to talk about the sample space (or equivalently, the measure) with respect to which you're taking the probability; otherwise, you run into potential paradoxes such as the Bertrand Paradox.  Since you're talking about a Sudoku puzzle, the (IMHO) most natural sample space for determining your probability is the space of correctly filled grids; by this measure, the probability of 3 being in e.g. the first place is just the number of correctly filled grids with 3 in that space divided by the total number of correctly-filled grids.
Unfortunately, because of the constraints involved, I don't believe there's any cleaner way of doing this than 'the hard way': enumerate all of the possibilities and then tally.  For convenience, I'll write the assignments in the form abcd, where a is the value in the first cell, b is the value in the second, etc.


*

*if the first cell is a 9, then by elimination (remove 9 from the possibilities for the second cell, then 7 from the third and fourth cells, then 3 from the third) the only arrangement is 9763.

*If the first cell is not a 9, then the second cell must be a 9; this leaves us with (36)9(367)(37).  Now, if the first cell is a 3 then the arrangement collapses (similarly to the above) into 3967; if the first cell is a 6 then we have one final ambiguity in the last two cells, giving us the configurations 6937 and 6973.


Thus, there are a total of four possible configurations for the puzzle: 9763, 3967, 6937 and 6973.  This gives probabilities of $\frac14$, $0$, $\frac14$, and $\frac12$ for 3 being in each of the four cells, respectively.
A: I'm going to map your cells as $1 = A, 2 = B, 3 = C$ and $4 = D$ so they won't get confused with the numbers to come.
Cell $B$ can be ignored as $3$ is not a possibility there.
Cell $A$ has $3$ possibilities, one of which is $3$.
Cell $C$ has $3$ possibilities, one of which is $3$.
Cell $D$ has $2$ possibilities, one of which is $3$.
It turns out that the the second approach is on the right track. It just needs to be scaled down to $100%$, or 1 total. $\frac13 + \frac13 + \frac12 = \frac76$. To turn that into a 1 you need to multiply by $\frac67$. This is done to the probability for each cell.

Cell $A = \frac13 \cdot \frac67 = \frac{6}{21} = \frac27 \approx 28.57%$
Cell $C = \frac13 \cdot \frac67 = \frac{6}{21} = \frac27 \approx 28.57%$
Cell $D = \frac12 \cdot \frac67 = \frac{6}{14} = \frac37 \approx 42.86%$

Note that this approach works for cases where several elements can be a particular possibility, but only one of them is permitted to be. Add up the individual probabilities, then multiply each by the reciprocal of the sum.
