The only Bitcoin mining profitability calculator you’ll ever need

The first question potential new miners have to ask is “Will I make a profit?” When asked on a forum, Reddit or some other discussion platform, this question is often answered by links to complicated looking calculators that attempt to estimate the mining profit using a wide variety of parameters. The result of the calculations typically strongly depends on the choice of these parameters, in particular the difficulty. Most calculators use a preloaded value for the difficulty increase that is extremely conservative. There are further matters that complicated the calculations such as power consumption and associated costs. In the end, however, it is possible to provide an easy and very reasonable estimate for the upper limit of the amount of Bitcoins mined by a certain amount of hashpower over its lifetime. If this upper limit is already below the purchase price of the device, and it often is, then there is no point in performing the more advanced calculations.

Mining brings both income and expenses. The income is obvious, and the expenses consist of the purchase price of the mining hardware (or contract) and operational costs (mainly power, but one might factor in time spent maintaining the setup as well). Lets focus solely on the income earned from mining and make an estimate of that. Since the income of a setup scales inversely with the difficulty, we have seen mining income go down steadily since the inception of Bitcoin as the network grows and mining hardware is increased both in quantity and in efficiency. One month from now, a miner will earn less than he does today and a year from the earnings will probably be a tiny fraction of what he earns today.

To predict how the earnings will change, one needs to predict how the network hashrate and consequently the difficulty of mining will grow. Over the last 6-9 months the difficulty has grown exponentially, that is, with a (relatively) constant percentage increase per unit of time. Exponential growth is unsustainable on the long term, but on shorter timescales, there is little indication that the growth is slowing down as both Bitcoins popularity as well as the hashpower of new mining hardware increases.

So without further ado, lets get to the calculations. The next section will contain some math for the derivation of the income estimate. If you want, you can skip right past the math to the conclusion (but why would you want to?).

Warning: Math ahead!

Lets start with the amount of coins mined per day right now and call it X and take the expected difficulty increase per adjustment and call it D1.

The difficulty is adjusted every 2016 blocks. If blocks are found every 10 minutes on average, as is the goal, then it takes exactly 14 days to find 2016 blocks. However, if the network hashrate is increasing, blocks are found more rapidly and the difficulty is increased to compensate. More specifically, if blocks were found 10% too fast over the last 2016 blocks, the difficulty will be increased by 10%. So instead of 14 days, the time between difficulty adjustments becomes
\frac{14}{1 + D} days.
For example, if the difficulty is increasing by 25%, there are only 11.2 days between adjustments.

The amount mined during the first period is
X \frac{14}{1 + D}

During the second period, the difficulty is up by an amount D, so the mining profit has decreased by factor 1 + D and in the second period, total earnings are
\frac{X}{1 + D} \frac{14}{1 + D}

Similarly, total earnings in the third period are
\frac{X}{(1 + D)^2} \frac{14}{1 + D}

Adding up the total earnings for all these periods gives the following series
X \frac{14}{1 + D} + \frac{X}{1 + D} \frac{14}{1 + D} + \frac{X}{(1 + D)^2} \frac{14}{1 + D} + ...
which can be rewritten as
X \frac{14}{1 + D} (1 + \frac{1}{1 + D} + \frac{1}{(1 + D)^2} + \frac{1}{(1 + D)^3} + ...)

We can use the sum notation for an infinite series to write this expression more eloquently:
X \frac{14}{1 + D} \sum_{n=0}^{\infty} (\frac{1}{1 + D})^n

What we have here is a geometric series, that is, a series in which the next term is obtained by multiplying the previous one with a fixed value. In this case, this fixed value is
\frac{1}{1 + D}

The nice thing about geometric series is that if the multiplication factor is smaller than one, then the series adds up to finite number. For a multiplication factor r (with r less than 1) , the sum value can be computed as
\sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}

Using this wisdom, we can express the summation part of our intermediate result as
\sum_{n=0}^{\infty} (\frac{1}{1 + D})^n = \frac{1}{1 - \frac{1}{1 + D}} = \frac{1}{\frac{D}{1 + D}} = \frac{1 + D}{D}

and we can plug this result into our original calculation to obtain
X \frac{14}{1 + D} \frac{1 + D}{D}

The fractions cancel and we obtain …

–MATH ENDS HERE– The result

\frac{14 X}{D}

That doesn’t look too bad. Ok, so how does it work? Lets take the popular cloud mining service as an example. Currently, you pay 11.4 mBTC (I will express values in mBTC for clarity) for 1 GHash/s of mining power on that website. We first need to calculate the daily income from that at the current difficulty. You can compute that with one of the many calculators available, just put all the things like power costs to zero and only focus on the amount of income per day. Anyway, the daily income of 1 GHash/s is 0.082 mBTC.

Next, we need to make an estimate of the rate of difficulty increase. A quick glance at recent difficulty increases (You can use this site for that) shows that 20% is a good ballpark estimate.

We plug in the numbers and find:
\frac{14 \times 0.082}{0.2} = 5.74

So this 1 GHash/s will not yield more than 5.74 mBTC in total if the difficulty keeps increasing by 20% every adjustment. That’s pretty much half the current purchase price. If you assume that the difficulty will only increase by 10% on average, you end up earning about as much as you paid.

Limitations of this model

Of course the model described above is incredibly simple and has some limitations. It is intended as a simple way to show that mining typically isn’t at all profitable. Consequently, costs like electricity, man-hours and repairs/replacements are not included. Typically only electricity is factored in by other calculators, but right now power costs of modern mining hardware are only a small fraction of their current income.

The main limitation of the model is that the difficulty won’t be going up exponentially forever. As many people have rightly pointed out on various forums, an exponential increase in difficulty means that in a few years we’d be using all of our planets energy production for mining. Clearly not realistic. Apart from that, technological limitations will also play a role in slowing down the growth of the difficulty.

But does that matter for the general conclusion? Probably not. Because while difficulty growth will slow down eventually, for now there are no signs of that happening any time soon. If the difficulty keeps going up by 20% for 6 months, then after those 6 months the income generated by a miner is only 3.5% of what it would be today. And at those levels, operational costs (primarily power) will start to play a significant role and the net profit will be close to or below zero, whereas operational costs are now negligible for up-to-date hardware.

So after these 6 months it hardly matters if the difficulty will keep going up as fast, if the growth slows down or even halts completely. The profits have gone down enough to make continued operation rather uneconomical.


Using basic arithmetic and some knowledge about geometric series, we have derived a nice expression that allows you to make a good first estimate of the upper limit of mining profits:
\frac{14 X}{D}

While the model is very simple, it is good enough to show that almost any mining operation (hardware or cloud mining service) is not profitable for the miner. Anyone considering the purchase of mining hardware or a mining contract should apply this formula and only if the calculated profits are considerably above the cost of the device or contract is it worthwhile to look at more advanced calculations that factor in finer details such as operational costs.

But I bet that that won’t be necessary.

  1. D is the fractional difficulty increase, so a 10% difficulty increase corresponds to a value of 0.1 for D.

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