How to assess the strength of the acid? Which is the stronger acid in the given pairs?
  
  
  $\ce{HClO3}$ and $\ce{HBrO3}$
  $\ce{HClO2}$ and $\ce{HClO}$
  $\ce{H2Se}$ and $\ce{H2S}$
  



given that the number of oxygen atoms is the same, the most electronegative atom should result in a stronger acid, thus $\ce{HClO3}$ is stronger.
$\ce{HClO2}$ is stronger since it has more oxygen atoms.
Here's where my logic fails, since the structure is the same (both have $\ce{2H+}$) then it boils down to 2 things to account for: electronegativity and the size of the atom. Going with electronegativity, sulfur is more electronegative thus stronger acid  which was my original answer. But, the correct answer is $\ce{H2Se}$ is the stronger acid which implies in this case atomic size ($\ce{Se}$ is larger than $\ce{S}$) outweighs electronegativity. I do not understand why in this case $\ce{H2Se}$ is stronger than $\ce{H2S}$ given sulfur is more electronegative, and most importantly why the same logic used in 1. can't be applied here and how do I discern where to account for atomic size specifically or when to account for electronegativity?


Any detailed-dumbed down explanations  on the logic are highly appreciated.
Let's start with $\ce{HXO_3}$.

In $\ce{HXO_3}$, after removal of proton, an $\ce{XO_3-}$ ion is formed which is resonance stabilised. For example, if X was $\ce{Cl}$,



Now, this shows that the $X$ needs to have a tendency to pull the electrons towards itself which decreases in the order :$\ce{Cl&gt;Br&gt;I}$. So more resonance is present in $\ce{ClO_3^-}$ than in $\ce{BrO_3^-}$ making the conjugate base of $\ce{HClO_3}$ more stable. Hence it is more acidic. 
So acidic strength decreases as: $\ce{HClO_3&gt;HBrO_3&gt;HIO_3}$

Now let's compare $\ce{HXO_3}$ and $\ce{HXO_2}$. You should now know that the acidic nature depends on the stability of the resulting conjugate base. Here, the number of resonance structure for $\ce{HXO_3}$ is more. Hence, the acid strength decreases in the order:   $\ce{HClO_3&gt;HClO_2&gt;HClO}$. So its actually the number of resonance structures that affects the acidic strength.

Finally, let's come to hydrides of 16th group. Let me represent any 16th group element as $Y$. Now, as size of $Y$ increases, the $\ce{H-Y}$ bond length increases. This means, the $\ce{H-Y}$ bond becomes weaker and removal of $\ce{H+}$ ion becomes easier. The size of atom decreases in the order :$\ce{O&gt;S&gt;SE&gt;Te}$.

Hence, the acidic strength decreases as : $\ce{H_2Te&gt;H_2Se&gt;H_2S&gt;H_2O}$ 

This shows that electronegative is not the only reason. For example, if you considered $\ce{HF}$ and $\ce{HCl}$, then you may think that $\ce{HF}$ is more acidic because $\ce{F}$ is more electronegative and hence will be happy to get a negative charge. But, charge on a small atom is unfavourable. As atomic size increases, the charge can spread over a larger area. Hence, $\ce{Cl-}$ is more stable. 

The same trend is seen in case of hydrides of 16th group. The charge in $\ce{Se-}$ is spread over a larger area. Hence, its more stable.