The arithmetic behind hybrids is rather simple, like
that of mixing drinks. In all cases we start with one s orbital (think of it
as 100 mL of rum) and three p orbitals (three times 100mL of pepsi).
Now what remains is to make some mixtures (hybrids or drinks). At the
end, there always has to be four of them (since we have started with four
orbitals), although some p may remain unmixed (depending on hybridization
type).

Let's start with some basics. The angle between two bonds formed by two hybrids with hybridization indices of m_{i}
and m_{j }is quantitatively defined by these indices through
the cosine formula on the left. The formula is based on simple
geometrical considerations (which we are going to skip here). The
important point is that the bond angles can be measured experimentally
and provide information about specific hybrids used to make bonds or
to hold lone pairs. If the bonds are equivalent, the situation is even
simpler. It leads, for example, to the familiar angles of 109.5^{o}
between two sp^{3 }hybrids, 120^{o} between two sp^{2}
hybrids, and 180^{o} between two sp hybrids. 
For any individual hybrid orbital the fraction of s (f_{s})
and the fraction of p (f_{p}) must add up to 1 (i.e. whole orbital),
and–by definition–the
ratio of p to s is equal to the hybridization index, m. A simple
rearrangement of these two conditions tells us how to calculate f_{s}
and f_{p} when m is known. On a given atom, all fractions of
s in all hybrid orbitals must add up to one, and all fractions of p
must add up to the number of p orbitals used in a given hybridization type
(3 for sp^{3}, 2 for sp^{2} and 1 for sp). In other
words, we have to deal with "full" orbitals, and no orbital, or even its
piece, can be "lost" or "gained".
Let's consider a simple example of water molecule. The oxygen
atom is sp^{3} hybridized and forms two equivalent bonds to
hydrogen atoms using sp^{m1}hybrids (m_{1} to be
solved for). The two equivalent lone pairs reside in sp^{m2}
hybrids (m_{2 }to be solved for). From the
experimental observation the H–O–H
bond angle is 104.5^{o}. Using the cosine formula for two
equivalent hybrids we can find that cos(104.5^{o}) =
–1/m_{1}= –0.25, or m_{1}
= 4.0.
The table below summarizes all the arithmetic manipulation
needed to calculate all the relevant fractions. The calculation can
be performed for f_{s}'s or f_{p}'s (or both if
internal check is desired). 

Hybrid 
hybridization index 
f_{s} 
f_{p} 
O–H bond 
m_{1} = 4.0 
1/(m_{1}+ 1) = 1/5 = 0.20 
m_{1}/(1+ m_{1}) = 4/5 = 0.80 
O–H bond 
m_{1} = 4.0 
1/(m_{1}+ 1) = 1/5 = 0.20 
m_{1}/(1+ m_{1}) = 4/5 = 0.80 
Total used for bonds: 

2/(m_{1}+ 1) = 2/5 = 0.40 
2m_{1}/(1+ m_{1}) = 8/5 = 1.60 
Remaining per lonepair hybrid: 

(1  0.40)/2 = 0.30 
(3  1.60)/2 = 0.70 
lonepair hybrid 
m_{2} = 0.70/0.30 = 2.3 
0.30 
0.70 
lonepair hybrid 
m_{2} = 0.70/0.30 = 2.3 
0.30 
0.70 
Total used for all hybrids: 

2 x 0.2 + 2 x 0.3 =
1.0 
2 x 0.8 + 2 x 0.7 =
3.0 
In conclusion,
the scharacter is withdrawn from the O–H bonds
and increased in the lonepair hybrids. This transfer helps to stabilize the
lone pairs by bringing them closer to the oxygen's nucleus (on average). The
bonding electrons get stabilization from both nuclei (O and H) and of course
are in a bond built out in part of H's s orbital.
