no, the number of private keys that will unlock any given public address is MUCH less than 2^96
I don't know the average number, somebody probably does though?
Why?
There are ~2^256 private keys
The address is some kind of hash function of private key
So we do a map from 2^256 values to 2^160 hashes
If the hash function is really good - we will receive the same hash value for ~2^96 different private keys (sometimes slightly more, sometimes slightly less)
Hmmmm... I see what you are saying, sorry for my prior misunderstanding amaclin
Of course! There are only 160 bits of public addresses space.
To understand with an extreme example: If there were only 2 bits of address space then only 4 possible public addresses, and of course no matter what hash func you apply to 256 bits, it would have to reduce down to 1 of those 4 public addresses. So in this example you could hash any random 256 bit privkey and ~ 1 in 4 times hit a winner.
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So I stand corrected:
On average ~2^96 private keys will unlock any single address. 
The effective privkey space one needs to search is "only" ~2^160 before you get a hit on average. Perhaps a few bits less due to collisions.
But it's not like we can just chop off 96 bits from privkey space before we search, valid privkeys will be scattered "randomly" throughout the 256 bit space.
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And lets think of some ways to increase the chances of unlocking some coin:
One could check against the list of the richest 2^16 (64k) addresses, further reducing the number of hashes required to 144 bits.
With average luck you would hit jackpot when ~50% through search, so 1 bit less = 143 bits.
To try 2^143 hashes at 10TH/s ~=
35,357,599,566,417,147,294,418 Years.
Universe age ~=
13,798,000,000 years
So still very safe against logical brute force attack.
