> for the first 6 tosses, the chances of getting all heads are .5**6.

Agreed.

> But if you get 20 tosses, that's fourteen more tries at
> getting all six in a row.  So that would be 14 * .5**6,
> or is it 15 * .5**6 ?

Following that logic you would add the probabilities of each
try, not multiply, if my memory serves me -- which it hasn't
lately :[

But adding the individual probabilities means you reach 1
if you do 64 tosses, and we know that the probability of
getting 6 heads in a row in 64 tosses cannot be 1.  That
is why is figured the number of tosses does not matter.

Also, since a few asked, the order what you toss before
getting 6 heads in a row does not matter.  Here is a possible
solution "graph":
   HHHHHH
   THHHHHH
   xTHHHHHH
   xxTHHHHHH
   xxxTHHHHHH
   ...
   xxxxxxxxxxxxxTHHHHHH

where x could be a H or a T.

Kind of fun,
Jeff