Let’s do a bit of statistics. You need to know a few things:
- variance is the square of standard deviation
- the variance of a Bernoulli trial with probability p is p(1-p)
- the variance of the sume of independent events is the sum of the variance
So let’s look at Emmy+B and Emmy+C for a baker with a fraction p of all the rolls.
We’ll look at the expected reward for one block and the variance. We assume in both cases that 32 endorsements are received and that the block is baked at priority 0.
Emmy+B expect block reward
p \times 16 + 32 \times p \times 2 = 80 \times p ꜩ
Emmy+C expected block reward
p \times 32 \times \frac{40}{32} + 32 \times p \times \frac{40}{32} = 80 \times p ꜩ
Emmy+B block reward variance
p (1-p) \times 16^2 + 32 \times p(1-p) \times 2^2 = 384 \times p(1-p) ꜩ^2
Emmy+C block reward variance
p (1-p) \times 40^2 + 32 \times p(1-p) \times \frac{40}{32}^2 = 1650 \times p(1-p) ꜩ^2
The standard deviation is found by taking the square root:
Emmy+B block reward standard deviation:
19.60_\ldots \times \sqrt{p(1-p)} ꜩ
Emmy+C block reward standard deviation:
40.62_\ldots \times \sqrt{p(1-p)} ꜩ
The standard deviation is thus multiplied by about 2.07. This does not depend on the length of time (whether per block or per year) and it does not depend on the number of rolls.
A good metric is the ratio of yearly reward by yearly standard deviation. Taking a year to be 128 cycles:
Yearly ratio for Emmy+B
2956._\ldots \times \sqrt{\frac{p}{1-p}}
Yearly ratio for Emmy+C
1426._\ldots \times \sqrt{\frac{p}{1-p}}
There are about 80,000 rolls now. So if one owns a single roll, p = 1 / 80,000 and the Emmy+B and Emmy+C ratio are, respectively, about 10.45 and 5.04.
How to understand that 5.04? Given an annual reward of about 6.55\% (excluding tx feees), a baker will one roll would receive 6.55\% \pm \frac{6.55\%}{5.04} \simeq 6.55\% \pm 1.3\%. With Emmy+B this would be 6.55\% \pm 0.63\%.
For a baker with 100 rolls, Emmy+C gives 6.55\% \pm 0.013\% where Emmy+B gave 6.55\% \pm 0.0063\%.
Hope this helps.
