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@satcat
stacking since: #84328longest cowboy streak: 10
50 sats \ 0 replies \ @satcat 1h \ parent \ on: Trudeau Promises to resign Politics_And_Law
He's fine with granting himself those
40 sats \ 1 reply \ @satcat OP 3 Jan \ parent \ on: Anyone have experience with LNMarkets? Stacker_Stocks
Awesome, LN login works fine on my Android. Their api seems to work pretty well too.
Oh awesome that's great to hear! Thanks, for the vote of confidence, and happy trading to you also :)
The logistics and economics of a long thin city just don't make sense, this is never going to happen, absolute utter waste of resources.
0 sats \ 0 replies \ @satcat 17 Dec 2024 \ parent \ on: Bitcoin in Europe Faces Narrative Shift news
Hopefully DOGE changes that
Pollieve is definitely already capitalizing on his weakness, calling for a non-confidence vote tonight
TL;DR: Canadians are frustrated with economic issues like housing, living costs, and inflation under Trudeau's leadership. His government has faced scandals, key resignations, and electoral losses. Unfulfilled promises, internal party unrest, and the NDP withdrawing support add to the challenges, potentially leading to an early election with unfavorable polls for the Liberals.
He won't step down in a million years, it's his life obsession to never lose at anything. Of anything he'll probably try to get another 4 years after his second term.
I copy pasted you're question into Claude 3.5, but none of the formulas were copied, nonetheless I think it got it right.
Response:
I notice some mathematical expressions might have been cut off in your message. From context, this appears to be about Bitcoin private key collisions and the birthday problem. Let me help fill in the details:
Bitcoin has 2^256 possible private keys (this is the total number of possible keys in the keyspace).
Let's call the number of randomly generated keys n.
Let's solve this step by step:
- This is a variation of the birthday problem. The probability of a collision can be calculated using:
P(collision) = 1 - P(no collision)
- The probability of no collision when choosing n keys from 2^256 possibilities is:
P(no collision) = (2^256)/(2^256) × (2^256-1)/(2^256) × ... × (2^256-(n-1))/(2^256)
This can be rewritten as:
P(no collision) = ∏(i=0 to n-1) (1 - i/2^256)
- For small probabilities relative to the size of the keyspace, we can use the approximation:
P(collision) ≈ n^2 / (2 × 2^256)
- For a 0.1% chance of collision, we want:
0.001 = n^2 / (2 × 2^256)
Solving for n:
n = √(0.001 × 2 × 2^256)
n ≈ 2^128 × √0.002
n ≈ 4.8 × 10^38 keys
- For 8 billion people: 4.8 × 10^38 ÷ 8 × 10^9 ≈ 6 × 10^28 keys per person
These calculations assume we're using the approximation formula, which is valid when the probability is small relative to the keyspace (which it is in this case).
This demonstrates why Bitcoin private key collisions are practically impossible - even with billions of people generating enormous numbers of keys, the probability of a collision remains infinitesimally small due to the massive keyspace of 2^256.
Note: I performed these calculations directly. While they should be accurate given the approximation formula used, for absolute precision with such large numbers, you might want to verify with specialized cryptographic software or mathematical tools.