Unencrypted Wallet Question

[LanYu]

If there is an address with bitcoins, where the wallet is unencrypted, would it be possible for somebody to generate that address and then lay claim to the coins at that address? If not, what mechanism prevents this?

[Remember remember the 5th of November]

If there is an address with bitcoins, where the wallet is unencrypted, would it be possible for somebody to generate that address and then lay claim to the coins at that address? If not, what mechanism prevents this?

Wallet encryption encrypts your whole wallet in the event that it's stolen. It is possible that a user independently generates an address that collides with yours i.e a collision and spend whatever coins the address has, but since we are talking about ~2^256 possible combinations, it's extremely unlikely, so much that you have a batter chance of winning the lottery a dozen of times.

[LanYu]

There are thousands of addresses that contain bitcoins. Isn't it possible that by generating a large amount of addresses, you will eventually stumble on one that already contains coins?

Maybe I am misunderstanding...

[BookLover]

Yes it is possible, but 2^256 is about 1.157920892 * 10^77 or a 1 in 115,792,089,200,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 chance of it happening.
It is quite a bit more likely for you and all your family to die by meteorite.  So moral of the story, buy meteorite insurance before worrying about address collisions.

Probability of death by meteorite found at :http://blogs.discovermagazine.com/badastronomy/2008/10/13/death-by-meteorite/#.UWhFucjLcTs it is 1 in 700,000 according to the site.

P.S.  Encrypting your wallet does nothing to prevent address collisions.

[LanYu]

Loud and clear...

I guess sometimes I underestimate the key size

[ar9]

Yes it is possible, but 2^256 is about1.157920892 * 10^77 or a 1 in 115,792,089,200,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 chance of it happening.
It is quite a bit more likely for you and all your family to die by meteorite.  So moral of the story, buy meteorite insurance before worrying about address collisions.

Probability of death by meteorite found at :http://blogs.discovermagazine.com/badastronomy/2008/10/13/death-by-meteorite/#.UWhFucjLcTs it is 1 in 700,000 according to the site.

P.S.  Encrypting your wallet does nothing to prevent address collisions.

What if someone decided to turn on a super rig, and churn out addresses, with a script to determine if the addresses are empty or not?
Anything stopping this?

[Remember remember the 5th of November]

Yes it is possible, but 2^256 is about1.157920892 * 10^77 or a 1 in 115,792,089,200,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 chance of it happening.
It is quite a bit more likely for you and all your family to die by meteorite.  So moral of the story, buy meteorite insurance before worrying about address collisions.

Probability of death by meteorite found at :http://blogs.discovermagazine.com/badastronomy/2008/10/13/death-by-meteorite/#.UWhFucjLcTs it is 1 in 700,000 according to the site.

P.S.  Encrypting your wallet does nothing to prevent address collisions.

What if someone decided to turn on a super rig, and churn out addresses, with a script to determine if the addresses are empty or not?
Anything stopping this?

He would need a really really powerful supercomputer. And maybe even then it would still take millions/billions of years. Quantum computers are a different story, but not yet possible.

[LanYu]

Aye, I did the math to back it up, and.... you guys are right.

Even if every single satoshi had its own unique address, then we have only used up 8.6361686e-62 percent of the keyspace. That means that you would have to generate about 5.513909e+61 addresses to even have a good chance on stumbling on a single address. Assuming you can generate a single address in 64 * 10^-9 seconds, it would take you 6.7140444e+48 years to find a single collision.

I am guessing that even a highly parallelized algorithm wouldn't put much of a dent in that, even with a super powerful computer.