If they have the same key, they would have access to your bitcoin..
But the odds of this ever happening is very slim.
Topic: What if someone generates the same address/key as you? (Read 2230 times)
This is a doubt which pop up every week or so, generating the same key is out of the question and there are math proofs for this.
Probably now you have understood that it's almost impossible to generate an existing address therefore don't worry about that 
I think moat people understand it will take billions of years to brute force an address, but it is still an extremely thin chance someone else can generate a key to your address. It just makes you feel insecure.
If you understand how incredibly low the chance is, you shouldn't feel insecure.
Here is what DannyHamilton responded to a very similar question a few months ago.
It is also possible that all the air in the room will spontaneously collect in one corner suffocating everyone in the room. The chance is extremely low, so low it is like impossible, but extremely low chance does not equal impossible.
Now ask any lay person if they think that it is possible for all the air in the room to suddenly collect tightly in a corner leaving everyone to suffocate, and we'll see if "impossible" in general use means what you think it means.
Now ask any lay person if they think that it is possible for all the air in the room to suddenly collect tightly in a corner leaving everyone to suffocate, and we'll see if "impossible" in general use means what you think it means.
EDIT: Oops, DannyHamilton mentioned that analogy faster than me.
Interestingly I asked the question then...
That is not as much fun. I can also just put 2256 into my calculator, still not a much fun.
Math is fun.
Agreed.
However, if you trust Google to be able to handle basic mathematics properly, you can also just enter:
2^256
into the Google search text box and click the "Google Search" button.
First of all, does every public key have a corresponding private key or can more than one public key correspond to just one private key?
Every public key is calculated directly from its corresponding private key. There are slightly less than 2256 possible public/private key pairs.yes I don't think it will happen anytime soon because that is 1 big fuckin number of possibilities, something like
1.157920893×10^77. Well that's what a mate told me it was and he is pretty smart. I can't claim I worked that out.
Can't see it happening in my lifetime.
10x = 2256
ln(10x) = ln(2256)
x(ln(10)) = 256(ln(2))
x = 256(ln(2))/(ln(10))
x = 77.06367888997918597471715704947
2256 = 1077.06367888997918597471715704947
= 10(0.06367888997918597471715704947 + 77)
= (100.06367888997918597471715704947)(1077)
= 1.1579208923731619542357098500863 * 1077
Math is fun.
Can't see it happening in my lifetime.
As long as the private keys are randomly generated with proper entropy, I can't see if happening in the lifetime of any human being that will ever exist on the face of the earth.
I dont get it, you mean there is a chance the private key would be same?
Only if the wallet is using a faulty random number generator. As long as the private key is randomly generated with sufficient entropy, it won't ever happen.
If there is a bug in some wallet software, then it is possible that it will generate an insufficiently random number resulting in another copy of that software generating the same random number for someone else.
I dont get it, you mean there is a chance the private key would be same?
Absolutely, but it's millions times more likely your dad's actually your maternal grandpa.
I dont get it, you mean there is a chance the private key would be same?
First of all, does every public key have a corresponding private key or can more than one public key correspond to just one private key?
Every public key is calculated directly from its corresponding private key. There are slightly less than 2256 possible public/private key pairs.yes I don't think it will happen anytime soon because that is 1 big fuckin number of possibilities, something like
1.157920893×10^77. Well that's what a mate told me it was and he is pretty smart. I can't claim I worked that out.
Can't see it happening in my lifetime.
I can get an account cracked... in, according to my computer, E-988985950 years.
So you have more chances at winning the lottery?
I suppose that depends on the lottery. Different lotteries have different odds. However, any lottery that has the same odds of winning as the odds of generating the same bitcoin address as someone else would be a lottery that nobody would play (since nobody would ever win).
As an example, lets assume that it's after the year 2140 and all the bitcoins that will ever exist have already been mined.
To demonstrate how unlikely it is, lets assume that nobody ever lost any private keys and no bitcoins have ever been permanently lost, so ALL bitcoins are accessible. Then lets assume that all the bitcoins are spread out into the smallest possible outputs. That would result in 2,099,999,997,690,000 addresses each with exactly 0.00000001 BTC in them. This gives you the largest possible set of bitcoin addresses that could ever exist with bitcoins in them.
Given that you are trying to find any one of 2,099,999,997,690,000 different addresses out of a total of approximately 1.46 X 1048 possible addresses, your odds of "winning" on any given attempt are approximately 1 in 6.96 X 1032
Looking at the Mega Millions lottery that is popular in the United States, the odds of winning the jackpot with a single set of 6 numbers are:
1 in 258,890,850
If I haven't messed up my math, the odds of winning the Mega Millions lottery jackpot twice in a row with that same set of numbers are therefore:
1 in 6.7 X 1016
The odds of winning the Mega Millions lottery jackpot three times in a row with that same set of numbers are then:
1.7 X 1025
We finally have better odds of finding a bitcoin address that has a balance if we compare to the odds of winning the Mega Millions lottery jackpot four times in a row with the same exact set of 6 numbers every time: 4.5 X 1033
Of course there aren't actually 2,099,999,997,690,000 different addresses that are holding a bitcoin balance (and there never will be due to lost bitcoins and the fact that people will generally store more than 0.00000001 BTC per adddress). If we use a more realistic number (perhaps something like 2,100,000,000 addresses with a balance) we find that:
You have better odds of winning the Mega Millions lottery jackpot all four times on your next four attempts with the exact same set of 6 numbers each time than you will ever have of have of generating a bitcoin address that already exists in a single attempt.
So you have more chances at winning the lottery?
So by what i understood, another reason BTC is superior to credit cards 
All these math are making my brain hurt, thanks.
The Possibility is always zero to get the same key, it is not impossible but just wait millions of years to get it .
.
Every now and then someone ask this question but the answer is written everywere, also in the bitcoin wiki.
Somewhere, I made calculations on this, but the gist is that it's many multitudes more likely for someone running one slow program to correctly guess the name, ID#, and CW2 of a credit card with an active LoC than it is for hundreds of botnets of hundreds of thousands of computers all running hyper-efficient software generating addresses to find a funded address.
Like Burt says, it's basically impossible for an incidental collision, and it's also basically impossible to intentionally find a used or funded BTC address. -But to answer your question, they'd be able to spend and send bitcoin in/to the address just like you.
ETA:
Here are the chances:
for credit card:
Like Burt says, it's basically impossible for an incidental collision, and it's also basically impossible to intentionally find a used or funded BTC address. -But to answer your question, they'd be able to spend and send bitcoin in/to the address just like you.
ETA:
Here are the chances:
Assuming 500K addresses which either have funds or will have funds within next three months,
0.0000000000000000000000000000000000000000034211388289180104270598866779539% chance per check.
Assuming average computer can do 250 optimized DSV-like checks per second,
0.00000000000000000000000000000000000000085528470722950260676497166948848% chance per second.
Assuming a "theft pool" is formed, and 500 of these computers averaging the above click/s,
0.00000000000000000000000000000000000042764235361475130338248583474424% chance per second.
Assuming each theft pool is one botnet, and 20 botnets, exactly the same, exist in these theft pools,
0.0000000000000000000000000000000000085528470722950260676497166948848% chance per second.
Per minute,
0.00000000000000000000000000000000051317082433770156405898300169309%
Per hour,
0.000000000000000000000000000000030790249460262093843538980101585%
Per day,
0.00000000000000000000000000000073896598704629025224493552243804%
Per month (30D),
0.000000000000000000000000000022168979611388707567348065673141%
Per year,
0.00000000000000000000000000026972258527189594206940146568988%
Assume worst-case scenario, DSV-like software can check 5000 addresses (10M of which are funded or will be funded within 6 months) per second, and 100 botnets of 50,000 computers each...
Per day,
0.000000000000000000000000014779319740925805044898710448761%
Per month,
0.00000000000000000000000044337959222777415134696131346283%
Per year,
0.0000000000000000000000053944517054379188413880293137978%
Per century,
0.00000000000000000000053944517054379188413880293137978%
Are we done, now?
ETA: Worse-than-worst case scenario. NSA can check 1T addresses per second, 1T addresses are funded.
Per century,
0.0000000000000021577806821751675365552117255191%
Per billion centuries,
0.0000021577806821751675365552117255191%
0.0000000000000000000000000000000000000000034211388289180104270598866779539% chance per check.
Assuming average computer can do 250 optimized DSV-like checks per second,
0.00000000000000000000000000000000000000085528470722950260676497166948848% chance per second.
Assuming a "theft pool" is formed, and 500 of these computers averaging the above click/s,
0.00000000000000000000000000000000000042764235361475130338248583474424% chance per second.
Assuming each theft pool is one botnet, and 20 botnets, exactly the same, exist in these theft pools,
0.0000000000000000000000000000000000085528470722950260676497166948848% chance per second.
Per minute,
0.00000000000000000000000000000000051317082433770156405898300169309%
Per hour,
0.000000000000000000000000000000030790249460262093843538980101585%
Per day,
0.00000000000000000000000000000073896598704629025224493552243804%
Per month (30D),
0.000000000000000000000000000022168979611388707567348065673141%
Per year,
0.00000000000000000000000000026972258527189594206940146568988%
Assume worst-case scenario, DSV-like software can check 5000 addresses (10M of which are funded or will be funded within 6 months) per second, and 100 botnets of 50,000 computers each...
Per day,
0.000000000000000000000000014779319740925805044898710448761%
Per month,
0.00000000000000000000000044337959222777415134696131346283%
Per year,
0.0000000000000000000000053944517054379188413880293137978%
Per century,
0.00000000000000000000053944517054379188413880293137978%
Are we done, now?
ETA: Worse-than-worst case scenario. NSA can check 1T addresses per second, 1T addresses are funded.
Per century,
0.0000000000000021577806821751675365552117255191%
Per billion centuries,
0.0000021577806821751675365552117255191%
for credit card:
What is the probability (per check) of finding a valid, activated credit card number and selecting the correct CVV, cardholder's name, and expiration date, assuming the expiration date is not beyond five years into the future?
Maybe just for a US Capital One MasterCard credit card, to keep things simple. Uses MOD 10 algorithm. http://en.wikipedia.org/wiki/Mod_10 The first six digits of these cards are 517805. Digits 7-15 are unknown - the account ID #. Digit 16 is the MOD 10 checksum number.
This means there is a total pool of 99,999,999 accounts. Assume 10,000,000 are activated.
Expiration date is simple. Most (all?) aren't valid for more than 5 years. That gives a 1/60 chance of getting only the expiration date correct per check.
Cardholder's name is more of a clusterfuck. Let's assume only looking at "black" and "white" names (we're looking for a US account, remember) gives you 85% of all total active accounts. Let's assume common names make up 60% of all total active accounts, and that there are 50,000 common name combinations.
CVV is easy, and we'll assume we don't know how Capital One comes up with these numbers, so it's a simple 1/999 chance.
So. We need to successfully correct all of them in one go, and we have one ~1/10 chance (account #), one 1/60 chance (exp. date), one ~1/83333 chance (cardholder name), one 1/999 chance (CVV).
I think the per-check probability of all this comes to .000000000020020100300621424707921073926538% (low confidence, someone smart should check this because I originally posted this post as a question but ended up giving enough data where I thought I could solve it).
Keep in mind, per-check chance of finding funded bitcoin address is ~0.0000000000000000000000000000000000000000034211388289180104270598866779539%.
To make the numbers a little easier to grasp, here is %chance of finding bitcoin address if "DSV-like software can check 5000 addresses (10M of which are funded or will be funded within 6 months) per second, and 100 botnets of 50,000 computers each":
Per century,
0.00000000000000000000053944517054379188413880293137978%
If 100 botnets of 50,000 computers could check only 1,000 addresses per second (5x slower than above stats for bitcoin), the chance of correctly guessing info on an activated credit card is:
Per century,
10.01005015031071235396%
ETA: I left out PIN number. Point still stands.
Maybe just for a US Capital One MasterCard credit card, to keep things simple. Uses MOD 10 algorithm. http://en.wikipedia.org/wiki/Mod_10 The first six digits of these cards are 517805. Digits 7-15 are unknown - the account ID #. Digit 16 is the MOD 10 checksum number.
This means there is a total pool of 99,999,999 accounts. Assume 10,000,000 are activated.
Expiration date is simple. Most (all?) aren't valid for more than 5 years. That gives a 1/60 chance of getting only the expiration date correct per check.
Cardholder's name is more of a clusterfuck. Let's assume only looking at "black" and "white" names (we're looking for a US account, remember) gives you 85% of all total active accounts. Let's assume common names make up 60% of all total active accounts, and that there are 50,000 common name combinations.
CVV is easy, and we'll assume we don't know how Capital One comes up with these numbers, so it's a simple 1/999 chance.
So. We need to successfully correct all of them in one go, and we have one ~1/10 chance (account #), one 1/60 chance (exp. date), one ~1/83333 chance (cardholder name), one 1/999 chance (CVV).
I think the per-check probability of all this comes to .000000000020020100300621424707921073926538% (low confidence, someone smart should check this because I originally posted this post as a question but ended up giving enough data where I thought I could solve it).
Keep in mind, per-check chance of finding funded bitcoin address is ~0.0000000000000000000000000000000000000000034211388289180104270598866779539%.
To make the numbers a little easier to grasp, here is %chance of finding bitcoin address if "DSV-like software can check 5000 addresses (10M of which are funded or will be funded within 6 months) per second, and 100 botnets of 50,000 computers each":
Per century,
0.00000000000000000000053944517054379188413880293137978%
If 100 botnets of 50,000 computers could check only 1,000 addresses per second (5x slower than above stats for bitcoin), the chance of correctly guessing info on an activated credit card is:
Per century,
10.01005015031071235396%
ETA: I left out PIN number. Point still stands.
At that time maybe cryptocurrency not exist
who knows
its just not possible
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