Type-I vs. type-II error in statistical hypotheses testing Let us consider standard statistical hypotheses testing:
$$\alpha=P\{\text{type}-I  \text{ error}\}=P\{\text{Rejecting } H_0 \text{  when }H_0\text{ is true}\}$$
and
$$\beta=P\{\text{type}-II  \text{ error}\}=P\{\text{Accepting } H_0 \text{  when }H_1\text{ is true}\}.$$
My question is as follows: could you give an example of $\alpha$ and $\beta$
when tossing 10 coins so that I can see that it does not hold this equation:
$$\alpha=1-\beta.$$
More specifically, compose an outcome so that inequality between $\alpha$ and $1-\beta$ is seen.
 A: Suppose you have two boxes of dice, one is a box of fair dice in which all faces are equally likely. The other has loaded dice for which the probability of getting a six is is 1/3. The labels are missing so you will roll a sample of 50 dice from each box to try to identify which box has the loaded dice.
Let $H_0: \text{FAIR},$ so that $p_0(6) = 1/6$ and let
$H_a: \text{LOADED},$ so that $p_a(6) = 1/3.$
In the figure below, blue bars represent the null
distribution under which the number of 6's seen in $n = 50$ trials is $\mathsf{Binom}(n = 50,\, p = 1/6).$ And
let the brown bars represent the alternative
distribution under which the number of 6's seen is
$\mathsf{Binom}(n = 50,\, p = 1/3).$

You choose critical value $c = 10.5$ (dotted line).
Thus $$\alpha = P(S \ge  11 \,|\, p=1/6) = .2014,$$
and $$\beta = P(S \le 10 \,|\, p = 1/3) = .0284.$$
Thus the 'power' of the test is $$1 - \beta =
P(\text{Rej } H_0 | H_a \text{ True}) = P(S \ge 11 | p=1/3) \\
= 1-.0284 = .9716.$$
sum(dbinom(11:50, 50, 1/6))
[1] 0.2013702
sum(dbinom(0:10, 50, 1/3))
[1] 0.02844031

Note: In many practical applications it seems reasonable to design an experiment so that the significance level $\alpha$ is approximately the same as the power $1 - \beta.$ 
Here, perhaps you chose to make them different because you think it would be more serious to sell a loaded die to a customer
who wants a fair one, than to sell a fair die to someone in the market for a loaded one.
[If you wanted significance level and power to be more nearly equal, you could pick the critical value $c$ to be near the middle of the region where the two distributions 'overlap'.]
A: Let’s say we’re testing a coin with probability $p$ of heads and our null hypothesis is $p=0$ (that it never comes up heads. We flip 10 times and our procedure is that we reject if we see a head, otherwise we accept. 
If the null hypothesis is true we will definitely not see a head, so $\alpha=0.$ However if not and we have $p>0,$ then the probability of 10 tails in a row is $(1-p)^n,$ so we have $\beta =(1-p)^n\ne 1-\alpha.$
There is no reason whatsoever to expect $\alpha=1-\beta.$ Usually, since the null hypothesis is a simple hypothesis, $\alpha$ is just a number, whereas since the alternative hypothesis is compound, $\beta$ depends on the effect size. So there’s usually a particular effect size for which they’re equal (in this silly example it’s $p=0$), but there is no real significance to this value. (One would hope they were both small for effect sizes of interest. Traditionally, the rule of thumb is to fix 5% for $\alpha$ and then set sample size so that $\beta$ is less than 20%.)
