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I know that mining pool shares have no value on their own, but it should be possible to find an average value based on the pool's average payout per share. This site describes the expected payout per share as ([block reward] - [pool operator's cut]) / [difficulty].

It seems to me that a share's "value" would necessarily be related to the difficulty as set by the pool--a pool with shares that are harder to find will have its users find fewer shares, making each share "worth" more. However, the equation above clearly doesn't work with the pool's difficulty, as most pools set difficulty to 1 and a share isn't worth ~48 BTC. How can I reconcile these issues to calculate the expected payout per share of a pool without knowing how many total shares are found?

Edit:

Ah, I think I've figured it out. The difficulty in the equation can be the network difficulty, because that's what makes it more or less likely that a given successful hash (earning a share) in a pool solves the block and earns the block reward. Thus, the network difficulty indirectly limits the value of a share. Is this correct?

Edit Again:

And now I'm back to being confused. If Pool A sets a higher difficulty than Pool B, Pool A's shares are worth more than Pool B's (assuming the same number of miners at a set hash rate) because they're rarer so each share will earn more at payout. How is this taken into account in the equation?

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2 Answers

up vote 2 down vote accepted

If the global network difficulty is D and the difficulty of shares in the pool is d, then the probability that a share will lead to a valid block is d/D. The reward in this case is B, so the average reward per share is B*(d/D). If the operator's average fee is f (so for example f=0.01 means 1% fee), the average reward miners get per share submitted is (1-f) * B * (d/D).

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Thanks for the answer! Can't believe I couldn't figure that out. –  ConstableJoe Nov 27 '12 at 14:19
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Variable difficulty mining:

If you are mining at a pool with a higher difficulty, then each share (proof-of-work) is worth more on average. Proving that you worked very hard is worth more than proving that you worked a little bit. Some pools now use variable difficulty, which means the difficulty can change over time. One share at difficulty X has the same average expected return as X shares of difficulty 1. It may be easiest to think of everything in terms of difficulty 1 shares.

Average value of a share (proof-of-work):

Basically each (difficulty 1) share's value is the block reward divided by the network difficulty. That's the average income of a block multiplied by the chance that the share creates a block (one divided by network difficulty). But keep in mind that transaction fees are part of the block reward. That's the bitcoin value of each share. If the pool uses merged mining the share would also have a value for each merged coin. That should also be included.

Example:

As I write this the new coins minted per block is about to drop from 50 to 25 BTC in just a few hours. Looking at http://blockchain.info/charts/transaction-fees the total transaction fees are roughly 25 BTC per day. With roughly 144 blocks per day, that is 0.17 BTC per block, making the total inccome for a block approximately 25.17 BTC.

With a difficulty of 3438908.96015914 that gives us an average value per share of 25.17/3438908.96015914 = 0.0000073191818369147015 BTC.

Assuming a pool that also gives you namecoins through merged mining, a share would have an additional value of approx. 50/1119016.08618347 = 0.000044682110129918345 NMC. With namecoins currently trading at 225 NMC to 1 BTC that puts the total value of a share at 0.000007517768993047672 BTC.

0.000007517768993047672 BTC per share, as opposed to 0.000007269747553550557 BTC which you get with the simplified formula (minted coins divided by difficulty). This 3.4% difference can be worth noting as many would have you believe that the simpler formula shows you the full value of a share.

Disclaimer: I run a pool that pays out transaction fees and namecoins.

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Thanks for the complete answer! –  ConstableJoe Nov 27 '12 at 19:04
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