>Len,
>
>We could ask how much a 3rd ball would cost, and come up with an even better algorithm. <g>
>
>>>
>>>However, if we would not involve probability theory, Steve's solution is the most compact, though a little bit more complex, than Sergey's.
>>
>>I'm not disagreeing, Steve's is the best solution giving the lowest possible maximum number of drops, but it's not a guaranteed minimum number of steps for all instances.
>>
>>(Just me being pendantic on the wording of the question, I guess <g>)

Now if cost is no object, 100 balls simultaneously dropped, 1 from each floor & count the number unbroken, should get the required answer in the minimum time (assuming they don't bounce too far).