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>)