First, thanks qmech, I think most of I wanted to write is not neccessary any more (or yet, lets see how the thread goes).
A "uniform probability distribution" means that given a point in the deck each card has a 1/52 probability of appearing there. Another example of a uniform probability distribution is a non-weighted die, which has a uniform probability distribution for all its numbers (1/6). This is a fully randomized deck.
It's even stricter. You don't only want to each card to have it's propor position, you want to have each of the 52!=8x10^67 possible permutations of the deck is equally likely. And 10^67 is a large number, the sun has 10^55 atoms (accorting to Wolfram Alpha). So that's nearly a trillion times the number of atoms in the sun.
Do you really come near this distribution? I don't know, it's hard to tell, it's really hard to generate significantly more than 10^67 shuffles to compare the outcome with the uniform distribution. At least if you are limited by the lifetime of the universe...
The good thing is that, when playing most card games, you are not really interested in the exact permutation. You care for what cards come to your hand. So when you distribute your 52card deck to 4 persons, each one getting 13 cards, you don't care in which cards they get first, what matter is which one come to their hand. That mean out the 8x10^67 permutations, you map (13!x13!x13!x13!)=1.3x10^39 permutations to one and the same hands.
And as described above, you can either use this large contraction to try to cancle out any distance from the uniform distribution of all
hands, or you you can leave it. Typical shuffles tend to leave cards that where close before the shuffle also close together after the shuffle. Which means, the probability of permutations where these two cards have a smaller distance from each other is 'a bit' larger than it should be, and the ones where they have a large distance is 'a bit' smaller. Where 'a bit' depends on how often you shuffle, but is always larger than 0.
So you want to give each player cards from all parts of the deck to compensate for this behaviour. If you don't, and A and B where next to each other before the shuffle, and he has card A, it's 'highly' likely that he also has B. At least more likely than it should be. If you do, it's significantly lower. See example (with extreme numbers for demonstration effect) above.