Why does the chain yield in nuclear fission sum up to 200%? The table of nuclides states the fission yield, or chain yield $Y$, the percentage of decays of $^{235}$U that lead to an isotope with mass number $A$. Why do they sum up to 200%?
Wikipedia just states 

Yield is usually stated as percentage per fission, so that the total yield percentages sum to 200%. Less often, it is stated as percentage of all fission products, so that the percentages sum to 100%. Ternary fission, about 0.2% to 0.4% of fissions, also produces a third light nucleus such as helium-4 (90%) or tritium (7%).

 A: 1 reaction will produce 2 daughter nuclei.
So percentage per reaction sums up to  200%.
In your comment, you mention making it 0.5 and 0.5. This also a correct way because you are taking percentage per total product which will sum upto 100%.
A: The question that you would like the fission yield tables to answer is

This isotope makes up some fraction of all the fission products; what is it?

Those fractions should add up to 100%.
The question that's being answered by these tables is slightly different:

What fraction of fission events produce this isotope?

Since fission usually produces two heavy fragments, the sum over all the fission fragments is approximately 200%.  This is easier to measure, because you can tell how many fissions have occurred by looking only at the amount of uranium remaining in the fuel. Ensuring that you have accurately measured all of the myriad fission products is a lot trickier; for instance, some fraction of the fission products are noble gases, which might escape from the fuel assembly.
This approach is consistent with normalizations of convenience in other branching-ratio tabulations with multiple particles in the final state. For a specific example, consider 
$$
\rm n + {}^{110}Cd \to {}^{111}Cd + \gamma
$$
which is some fraction of what happens when you put thermal neutrons on natural cadmium.  The gamma rays carry away a total energy of 7.0 MeV, which is the "neutron separation energy" for Cd-111.  However the average energy of these photons is one or two MeV, and the Cd-111 has a couple of low-lying excitations through which most of these gamma-ray cascades pass.  If we could count all of the gamma rays, the probabilities per neutron capture would add up to a few hundred percent.
But in this particular case it's not possible to reliably measure the energies of all of the very soft gamma rays from the closely-spaced high-energy states, so the branching ratios in my reference are instead normalized to one of the more prominent transitions from a low-energy state.
This confuses me every time I look it up (once every few years), because the calibration photon is assigned a relative intensity of 100%, but a different common photon is given a relative intensity of 274%.
A: A toy model:


*

*consider a heavy isotope e.g. $^{235}$U + neutron. $^{236}$U state undergoes the fission

*in the simplified view, $^{236}$U creates only (let's forget those important neutrons) two fragments (lighter and heavier) and if the first fragment has A=100, the other must have A=136 - see the two peaks in the isobar plot from wiki.



Nuclei produced in fission in graph by their mass A.


*

*the yield as defined in wiki, describes the proportion of the produced fragments (isobars) to the progenitor ($^{236}$U here). Following possible $\beta$ decays do not change A, so the isobar plot nicely conserves the original picture.

*If all $^{235}$U+n convert to the A=100 nucleus (but also one A=136 nucleus), then A=100 and also A=136 have both 100% yield.

*In case of exclusively ternary fission (three fragments), three nuclei would have 100% yield.
Conclusion: 
The wiki text seems slightly perplexing. While the idea is evident and very practical, the other processes involved (I could guess an example) make the 200% not extremely precise. 
A: It took my probably longer than necessary, but now I understand why it sums up to 200%. An example illustrated it to me. Let me enlighten you with a nuclear power plant. But first, the answer:
Let us assume that $^{235}$U would only split into 1 specific isotope. So when it splits, 2 nuclei are created. $2=200\%$

So, the fission yield does not state the percentage a certain isotope represents of possible splits of Uranium, it rather counts how many nuclei you would have at the end, if you split 100 Uranium nuclei!
This is a helpful information, e.g. if you liked to know the production rate of $^{131}$ I (iodine) of a nuclear power plant. Let us assume we have a powerful power plant with an energy production of 1 GW, and an efficiency of 20%. Let us further assume that the energy released in the fission of one $^{235}$U atom is 200 MeV. The fission yield of $^{131}$ I is given in the table of nuclides: $Y_{^{131}I}=2.885%$, as well as the half-life: $T_{1/2}=8.02$ d. Let us sum up everything we know in tabular form:


*

*$P_{el}=1$ GW

*$\eta=0.2$

*$\delta E=200$ MeV

*$Y_{^{131}I}=2.885\%$

*$T_{1/2}=8.02$ d $\Rightarrow\lambda=1\cdot 10^{-6}$ s$^{-1}$
From the electrical power production $P_{el}$ and the efficiency $\eta$ we can determine the thermal power production $P_{th}=\frac{P_{el}}{\eta}=5$ GW.
We know that one decay of uranium yields 200 MeV, therefore the number of $^{235}$U has to change according to: $\frac{\mathrm{d}N_{^{235}U}}{\mathrm{d}t}=\frac{P_{th}}{\delta E}$.
With one decay of $^{235}$U, $Y_{^{131}I}=2.885\%$ are created (on average), which means the production of $^{131}$I per second is given by
$$
P_{^{131}I} = Y_{^{131}I}\frac{\mathrm{d}N_{^{235}U}}{\mathrm{d}t}.
$$

To sum up: The fission yield states how many daughter nuclei are produced under the decay of 100 $^{235}$U isotopes. As the mother splits into two different daughters, the total number of the daughters sum up to 200. This is then broken down to percentage values, called the fission yield. This can be used to calculate the production of daughter nuclides. 

