POW with ROW who has the higher chance to mine a block? A group of $N$ miners trying to mine a block, I am one of them. In one scenario they are using a POW(proof of work) algorithm and in another, they use a ROW(race of work) algorithm.
The mining process is to execute the function $m()$. It returns a random hash number between $0 - 2^{32}$. In case of POW the first miner the find a hash smaller than some value $L$ wins. In case of ROW, the miner with the smallest hash after $T$ minutes wins.
In both scenarios I have 10% of the mining power, I manage to execute $m()$ 10% of the total time it is executed by all the miners until someone wins.
In what scenario do I have more chances to win the block, POW or ROW?
Also does the answer is changing if instead of 10% I have $p$ percent?
 A: Assume for simplicity no ties are possible, neither in values hashed, nor in timing (i.e. two miners cannot produce a hash "at the same time", so it's always obvious who won POW). Note that in both cases, the winning miner is the one with the lowest hash: this is obvious in ROW, but also in POW since the first hash below L is less than every hash that came before it. Then, if at the time the winner is declared you computed a fraction $f$ of all the hashes, for symmetry reasons the probability that the lowest hash is among those you computed is $f$.
A: Suppose you split your resources into 10 separate mining nodes and your opponent splits his resources into 90 separate nodes. You then have 100 identically powered nodes between you, and you run the competition between them, and look for who owns the winner. It should then be clear that you have a 10% chance of winning.
However, notice that "split your power into 90 nodes and let them run independently" for both modes of the game means that you're actually doing the same thing as "mine with all of your power at once". It's implicit in your description of the rules that each hashing attempt is independent of all the others, so whether you choose to artificially distribute the attempts you have the power to make among 90 imagined sub-miners or not does not change whether one of the attempts is the eventually winning one.
(In practice, $2^{32}$ is probably not a large enough search space to justify the assumption of independence. Actual coin mining uses a vastly larger space to approximate that assumption much better).
