```Good morning Dave, et al.,

> >  Myopic Miners: This bribery attack relies on all miners
> >
> >
> > being rational, hence considering their utility at game conclu-
> > sion instead of myopically optimizing for the next block. If
> > a portion of the miners are myopic and any of them gets to
> > create a block during the first T − 1 rounds, that miner would
> > include Alice’s transaction and Bob’s bribery attempt would
> > have failed.
> > In such scenarios the attack succeeds only with a certain
> > probability – only if a myopic miner does not create a block
> > in the first T − 1 rounds. The success probability therefore
> > decreases exponentially in T . Hence, to incentivize miners
> > to support the attack, Bob has to increase his offered bribe
> > exponentially in T .
>
> This is a good abstract description, but I think it might be useful for
> readers of this list who are wondering about the impact of this attack
> to put it in concrete terms. I'm bad at statistics, but I think the
> probability of bribery failing (even if Bob offers a bribe with an
> appropriately high feerate) is 1-exp(-b*h) where `b` is the number of
> blocks until timeout and `h` is a percentage of the hashrate controlled
> by so-called myopic miners. Given that, here's a table of attack
> failure probabilities:
>
> "Myopic" hashrate
> B 1% 10% 33% 50%
> l +-
> o 6 | 5.82% 45.12% 86.19% 95.02%
> c 36 | 30.23% 97.27% 100.00% 100.00%
> k 144 | 76.31% 100.00% 100.00% 100.00%
> s 288 | 94.39% 100.00% 100.00% 100.00%
>
> So, if I understand correctly, even a small amount of "myopic" hashrate
> and long timeouts---or modest amounts of hashrate and short
> timeouts---makes this attack unlikely to succeed (and, even in the cases
> where it does succeed, Bob will have to offer a very large bribe to
> compensate "rational" miners for their high chance of losing out on
> gaining any transaction fees).
>
> Additionally, I think there's the problem of measuring the distribution
> of "myopic" hashrate versus "rational" hashrate. "Rational" miners need
> to do this in order to ensure they only accept Bob's timelocked bribe if
> it pays a sufficiently high fee. However, different miners who try to
> track what bribes were relayed versus what transactions got mined may
> come to different conclusions about the relative hashrate of "myopic"
> miners, leading some of them to require higher bribes, which may lead
> those those who estimated a lower relative hash rate to assume the rate
> of "myopic" mining in increasing, producing a feedback loop that makes
> other miners think the rate of "myopic" miners is increasing. (And that
> assumes none of the miners is deliberately juking the stats to mislead
> its competitors into leaving money on the table.)

A thought occurs to me, that we should not be so hasty to call non-myopic
strategy "rational".
Let us consider instead "myopic" and "non-myopic" strategies in a population of
miners.

I contend that in a mixed population of "myopic" and "non-myopic" miners, the
myopic strategy is dominant in the game-theoretic sense, i.e. it might earn
less if all miners were myopic, but if most miners were non-myopic and a small
sub-population were myopic and there was no easy way for non-myopic miners to
punish myopic miners, then the myopic miners will end up earning more (at the
expense of the non-myopic miners) and dominate over non-myopic miners.
Such dominant result should prevent non-myopic miners from arising in the first
place.

The dominance results from the fact that by accepting the Alice transaction,
myopic miners are effectively deducting the fees earned by non-myopic miners by
preventing the Bob transaction from being confirmable.
On the other hand, even if the non-myopic miners successfully defer the Alice
transaction, the myopic miner still has a chance equal to its hashrate of
getting the Bob transaction and its attached fee.
Thus, myopic miners impose costs on their non-myopic competitors that
non-myopic miners cannot impose their myopic competitors.
If even one myopic miner successfully gets the Alice transaction confirmed, all
the non-myopic miners lose out on the Bob bribe fee.

So I think the myopic strategy will be dominant and non-myopic miners will not
arise in the first place.

Regards,
ZmnSCPxj
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```On Sun, Jun 28, 2020 at 2:16 PM David A. Harding via bitcoin-dev <
bitcoin-dev@lists.linuxfoundation.org> wrote:

> So, if I understand correctly, even a small amount of "myopic" hashrate
> and long timeouts---or modest amounts of hashrate and short
> timeouts---makes this attack unlikely to succeed (and, even in the cases
> where it does succeed, Bob will have to offer a very large bribe to
> compensate "rational" miners for their high chance of losing out on
> gaining any transaction fees).
>

We were separately working on a similar problem, and wrote a paper as well:
https://eprint.iacr.org/2020/774 *

We look at the Alice's-Fees/Bob's-Bribe ratio. We also look at "strong" and
"weak" miners in this context. If a miner is weak, their hash-rate is lower
than this fees/bribe ratio. If they are strong, their hash rate is more
than this fees/bribe ratio. In this setting, it turns out that if there are
only strong miners, Bob will win. If there is at least one weak miner,
Alice has to win, given a reasonable timeout value. We found it awesome
that lightning has a parameter called "channel-reserve_satoshis", which
directly helps in countering this bribe by giving Alice some leeway in fees.

* Ph.D students want to write papers, unfortunately.
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