Multiply tries chance theory.

[Feldherren]

Short answer: yes, it's more likely to roll a 6 with two rolls of a d6 than it is with a single d6.

Long answer: ...but that doesn't mean you WILL roll a 6 even with six rolls.

When you want a result of 6 on a d6, you have a 1/6 chance of rolling it, assuming the die isn't loaded or unfair in some other way. When you roll 2d6, you have 32 possible results; 1 and 1, 1 and 2, 1 and ... and so forth.
When you roll 2d6, there are 11 possible combinations that include a 6. So you have 11/36 chance, or rather close to a third, of rolling one 6.

However, that said, you could roll 100d6 and still not turn up any 6s. Or roll all 6s. It's pretty unlikely, with the latter being unlikelier, barring loaded dice, but it can happen. Probabilities are used when there are no certainties, and I got a pearl on the very first mussel I boiled.
You can be somewhat sure you'll eventually get a pearl, as after x tries is the point at which probability flips around and says 'it's more likely you'll have gotten a pearl by this point than not', but unless the game biases the calculation in your favour based on 'this person has opened y mussels since they last got a pearl', each individual attempt is still exactly the same probability of getting a pearl.

[GreenScape]

http://en.wikipedia.org/wiki/Normal_distribution

there is chance after 1 000 000 000 rolls not to get 6. but because the chance is so fucking low( [5/6]^1000000000 = 0.<79181247 zeros here>896...) it will never happen in the world's history.