Topic: [XMR] Monero Speculation - page 1753. (Read 3313576 times)

Been buying a couple more coins.
Any idea where will be the next level of resistance? 

Good Morning all!
I bought a few hundred more.
It is a good way to start a rainy autumn day by buying a few hundred Moneros.
If you guys will not do it, I will buy even more in front of you.  
Remember what Risto said earlier: It is easy to sell large amounts of Moneros but hard to rebuy them cheaper. Personally I've noticed too much fishing will not take me anywhere and the emission is now pretty easy to swallow even if the price is higher. We are talking here only about thousands of dollars per day of emission.

P.S. I have a full confirmation of identity on Poloniex. It enables me to withdraw funds whenever I buy and the withdrawal is recommneded so that you do not fall in the trap of selling too early (this is a general notice).  

Lie to the other pirates and tell them that the magical gold coins have an invisible drawing of a new, more powerful god invisibly drawn on them?

bravo. 

 
  
There are 7 billion pirates (sorry, I meant n pirates) who all are using silver coins with different Gods drawn on them to transact with.  Pirate #1 wants all n-pirates to adopt new and improved gold coins with no Gods drawn on them.  Luckily, he is an idealist and isn't necessarily looking for maximum profitability in this game; his only goal is to convert as many of the pirates as possible to the magical gold coin standard.  Right.... I forgot to mention that these are magic gold coins which can be instantly teleported to anyone on the planet, and no one has to know about it.  Despite the advantages though, this might be an uphill battle.  Most pirates are deeply religious and deeply value their silver with Gods drawn on it.  These new magical gold coins might be superior, but they also use witchcraft (which is untrusted and considered dangerous by many).  
  
Overnight millions of magical golden coins begin to appear in Pirate #1's lair.  This is because he is the first believer, and after all: they have to go somewhere.  The power of these coins is such that anyone who makes a statement of faith: "I believe in the power of the magical gold coin." will also be able to share in the miraculous appearance of gold coins in their lair (although the total amount that appears will slowly decrease over time to a minimum).  To compound the problem, it is possible to compete for a greater share of the magical gold coins by building ornate altars in your lair, the materials of which can only initially be bought with the holy-stamped silver (because no one accepts the magical gold as money yet).  
  
Assuming that Pirate #1 has the ability to immediately contact up to 1% of the pirates (of varying dispositions towards the witchcraft that these coins use, and who also have varying connections to eventually contact the other 99% of pirates), and his goal is to convince as many as possible to value the magical golden coins over the holy silver coins, what is his ideal strategy?  Historical precedent?  
  

We make the following assumptions:

1) That M > N
2) If there is an even number of pirates left the senior pirate votes twice once in the regular vote and once to break a tie if needed
3) The coins cannot be broken down into smaller pieces of gold
4) There is zero cost for the hangings (For example the cost of the rope is zero)
5) Each pirate acts strictly in his best financial interest.

The solution is the senior pirate distributes one coin to each priate, except the second most senior pirate gets no coins and he keeps M+2-N coins for himself.

We start with N=2. In this secnario the senior pirate keeps all the coins and uses his tie vote to outvote the junior pirate
N = 3 The senior pirate only needs the vote of one pirate. So he pays the cheapest cost by giving one coin to the most junior and keeping the rest etc.

solution: the 2 youngest always reject so they can divide the M coins between themselves?
The second youngest will probably end up with most of the coins, but the youngest will demand at least round_up(M/N) coins
that is my guess :p

edit:
this works if N = 3

Let me think...
If N = 4, the second oldest don't want to die I guess, so he will bribe the 2 youngest
so he gets zero, and the other 2 get N/2, oldest one dies

I need to think about this some more, very interesting puzzle!

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edit2, very interesting...

N=1: (obvious)
=> P1 gets M

N=2: P1 breaks the ties, so he can just take it all
 => P1 gets M, P2 gets 0

N=3: P1 wants to live, so he need at least one of the pirates to accept his proposal.
The only way he can achieve that is by colluding with P2 or P3 and giving him all the coins.
No idea who would collude, but I am guessing P2 and P3 will start to lower their bid to make sure P1 picks them, it's a race to the bottom
I don't have a solution for this... Maybe P1 ends up with M-1 and P2 or P3 gets one coin
On the other hand, P2 can bribe P3 to hang P1 so he has a better chance to get some coins. He promises to give P2 some coins to collude with him so doesn't end up with zero coins (which would be the normal case for N=2)
But in that case, P1 will offer P3 all his coins, if he wants to live. It is also possible P1 offers P2 all of his coins, but this seems less likely, because P2 would get all the coins anyway for the case N=2, so he would demand additional coins or a service from P1.
=> It will be cheaper for P1 to just give all his coins to P3 (I think)

N=4: P1 wants to live, but P2, P3 and P4 can collude to take him out. they need all 3 votes though, so everybody needs to gain
P2 just doesn't want to die in N=3, so he will offer P3 and P4 an equal amount of coins (M/2), I think.
another option is that P4 bribes 2 other people to collude with him, possible P2 (because he would get zero coins in all other scenarios) and P3 or P4.
although I think in the end, P3 or P4 would again end up with all of the coins, because they think about the situation for N=3  and due to the same reasoning, I think P4 will end up with all the coins
=> P1 colludes with P2 and P4 to support him, P4 gets all the coins

same for N=5, 6, ...

So, I think that, as long as N =/= 2, the youngest will end up with all the coins and nobody dies
If N = 2 the oldest one gets them all


edit3:
found this on wikipedia:
This changes things a lot

Yup, that's right.
The likelihood to go up is bigger than to go down since in order to go down you need to dump pretty much over 100 btc's worth of coins.
Try to rebuy such an amount for lower price taking account the competition there is around 0.002-0.0022 levels.
Therefore I think the wall at 0.0023 will eventually be eaten or pulled.

I think most XMR traders are used to big walls by now :-P Or at least the medium size (10-20k). Also, if someone nibbles a bit off that, the rest will follow.

Looks like nobody wants to buy or sell.
If you sell, you might end up selling too low as there is incentive to start building up higher walls.
If you buy only a small amounts of coins, there is a sell wall at 0.0023 waiting which might scare the traders not to execute any buys.

Interesting situation - we will see how this will play out.  

The good thing is, there is now finally a good number of buy orders.

You can't hang users (or miners) in Bitcoin, they are anonymous (for practical purposes here) and new entry is permissionless. In fact there isn't even a "most senior" miner or user at all. So this analogy doesn't apply.

It is an interesting puzzle though, although like most of these puzzles the correct solution can depend on subtle differences in interpretation of the problem statement. Another one like that being the Monte Hall problem.

N pirates are dividing the loot of M gold coins. They don't trust each other etc. and have the following rules (only):
- most senior one starts by proposing a distribution of the coins between robbers
- a vote is held whether this is accepted, ties are broken by the senior guy
- if it is rejected, the senior guy is hanged and new proposal by the second-in-command.

Monero looks pretty tight now.
A big sell and big buy wall only less than 5 % spread.

Was frontrunning Warz' bid, asked him to let me buy 5k first, wall now:



 

No the mempool doesn't matter, because you still need an incentive for people to pay a non-negligible fee. Let's say the most profitable transaction in the entire pool pays some negligible fee so you include it (better than nothing right?). Then the next transition need only compensate you for the potential loss of your infinitesimal profit on the first transaction, which in turn is even more infinitesimal. Everyone transacting understands this so they understand that the fee needed to get into a block is virtually nothing. Because the fee necessary to get into the block at all is so small, even the highest fee transaction in the entire mempool won't pay a lot (unless someone fat fingers of course). So this all collapses. That's what the math is telling you.


You're going to choose the combination that gives you the best outcome. People transacting know that and will only pay something extremely negligible, not something that allows you to make a profit and actually consume electricity hashing.

Only in that single edge case. So, ok, if there's basically zero demand for transactions the orphan-risk analysis breaks down. But if the mempool contains more than one fee-containing tx, the orphan risk analysis seems to holds.

So again, put yourself in the shoes of that miner. How would you select transactions? Seems unlikely that you, being rational, would just shove all the transactions in a giant block without making sure that your giant block has a good-enough (relative to the fee-rev you're earning) chance of not being orphaned.

My fat fingers bought 648 XMR some time ago..... It was accident - I hoped just to nibble a bit and waite until there will be someone willing to sell me lower...

There is no equilibrium where people pay a significant fee. Given a choice between mining an empty block (zero reward) and mining anything else any rational miner will choose "anything else" even if that is a 100 terabyte transaction paying a fee of one satoshi.

You can likewise view a total of one terabytes worth of multiple transactions paying a total fee of 1 satoshi the same way. There is no pricing pressure that would cause people creating transactions to want to pay more.

Still not seeing how it breaks with no coinbase tx and only fees. The cost of each tx is the extra orphan risk caused by the extra X bytes needed to describe the tx, multiplied by the tx fee included in the tx. How is that not the case?

Now with current block reward, and for the foreseeable future, coinbase rewards will be much higher than avg fees, so the cost of orphaning a block is mostly going to be associated with that coinbase tx. But ultimately, why is the above invalid? There's *still* some cost to adding a tx to a block, so a rational miner will not do so unless the tx includes enough fee to cover that cost (again, the delta on orphan risk due to the additional bytes).

Sorry if I'm being thick on this, but I haven't yet seen a sufficient explanation.





Put yourself in the shoes of a bitcoin miner in the year 2141 (no block reward). There's demand for transaction confirmations, so you have a mempool with a bunch of transactions which include fees ranging from 0 to some positive number. It takes time for blocks to propagate, and orphan risk increases exponentially with the quantity of data you're transmitting. How do you select which transactions to include?

I'd chart orphan-risk per byte against the sets of transactions I can build and the resultant aggregate blocksizes and fees......Hmmm...that actually feels like it might an NP-complete problem (an instance of the knapsack problem?). Meh, will think about this later. Anyways, there are practical-enough solutions to those problems, even for largish N.

No you can just get a small incremental update to your scratchpad on each new block from the pool (that is actually what is done I think). The cost of doing that is so much lower than running a full node it does't provide a huge incentive, instead barely one at all.