RE: CBR – The Way to Split Coins (DRAFT)
We can assume I > B or better I -> infinity ( or n -> 0)
I'm not sure this is a safe assumption to make. It looks like B can be very large, e.g. if N = 10 and M = 100 then B = 64, and C = nB = 640, which is impossible since C is the number of coins cooling down, and cannot exceed N.
If we look at this iteratively, we start with N = nI coins. In the beginning, S = N = nI. Suppose n of those coins stake, meaning one UTXO stakes. Then S = N - n = nI - n = n(I - 1). If another n coins stake, then S = N - 2n = nI - 2n = n(I - 2). So if j is the number of staked blocks in the last 16 hours, then S = n(I - j). If I = j, i.e. if the number of staked blocks = number of UTXOs, this implies that all of the coins have staked and S = 0.
If it is the case that when you stake, all of the coins in a UTXO stake, then B should equal I, meaning that the maximum number of blocks you stake should equal the number of UTXOs you have.
If I understand your model correctly, it seems like there is an assumption that the probability of staking a block = number of your coins/number of total coins. Another way of saying this is if you have 10 coins in one UTXO out of 100, and your chance of staking is 10%, then someone else with 20 coins in one UTXO will have a 20% chance of staking. If this is correct, then by linearity of expectations, it shouldn't make a difference how many UTXOs you have - you could have one, or many, but the number of blocks you stake over a large enough time period should be the same. If however it is the case that you have a higher probability of staking a block by splitting coins in a single UTXO into two separate UTXOs, then an analysis can be made as to how much a person can exploit this.
Are you sure about: ...
As wallet =/= UTXO (or address =/= UTXO) a portion of a wallet will stake, but whole UTXO.
I'm not sure at all, I was wondering myself out of ignorance.
You can check list of UTXOs with a command
That's good to learn, thanks!
No, as sum(n) = N; usually we use n to count things, here I've used it (not a great choice) as a value of one utxo. To have a chance to stake 64 times (within 16h) you would have to split N = 10 into 64 chunks of n=0.15625 each (equal ns for simplicity).
@tomasbrod commented on splitting coins, but his message is not clear to me.
Also,
is not accurate enough
I meant
(now updated in the main text, thanks)
I'm not sure this contradicts my original point, as the formula for B still only contains N, M, and t.
This is what I'm trying to get at, that the maximum number of blocks a person can stake is equal to the number of UTXOs that person has. Theoretically, if one makes more UTXOs than there are blocks in some time period, then a person could theoretically stake all of those blocks, though the probability of that will be extremely low.