Suspect unfair die If I claim to have a fair die that rolls 1-6 uniformly but my die actually only rolls 1-5 uniformly (and never produces a 6) how many rolls would you need to see before you had over 50% confidence that I was messing with you?
 A: You just need to compute the number of rolls that gives a $50\%$ chance of getting at least one $6$ on the assumption that it is a standard die.  That corresponds to less than $50\%$ chance of having no $6$s.  Can you do that?
A: Power Computation for a Chi-squared Goodness-of-Fit Test: Detecting an Unfair Die
Perhaps you are using the chi-squared goodness-of-fit (GOF) test, with
test statistic $$Q = \sum_{i=1}^6 \frac{(X_i - E_i)^2}{E_i},$$
where $X_i$ is the number of occurrences of face $i$ in $n$ tosses
of the die and $E_i = n/6$ is the expected number of occurrences of each
face for a fair die. Notice that $Q = 0$ in the (unlikely) event that
each face appears exactly 1/6 th of the time. The worse the agreement,
the larger $Q$ becomes. (Although it is called a 'goodness-of-fit statistic',
it is large for 'bad' fits.)
Then for a fair die, $Q \stackrel{aprx}{\sim} Chisq(df = \nu)$ provided 
$n$ is large enough that the expected counts exceed 5 (so you need
$n/6 > 5$ and $n > 30).$ You have $k = 6$ categories and $\nu - k-1 = 6-1 = 5$
degrees of freedom. 
It is typical to test at the 5% level of significance. You will reject the
null hypothesis that the die is fair at the 5% level if $Q > q^*,$ where
the 'critical value'
$q^*$ is chosen to cut 5% from the upper tail of $Chisq(5).$ From tables
or from R statistical software, you have $q^* = 11.07.$
 q.crit = qchisq(.95, 5);  q.crit
 ## 11.0705

However, for the dishonest die you describe, values of $Q$ are inflated
and do not have the distribution $Chisq(5)$. The result is that you will
reject the null hypothesis more often than 5% of the time, if you judge
your dishonest die using the test described above. 
In particular, you have asked to have a rejection rate of 50% instead of 5%
when your dishonest die is used. To find the $n$ that will accomplish that,
you need to know the distribution of $Q$ for your dishonest die. 
Then $Q$ has a noncentral chi-squared distribution $Chisq(\nu = 5, \lambda)$, with noncentrality
parameter $\lambda.$ Here is how to
find $\lambda$ for your dishonest die. The null hypothesis assumes probabilities
$p_0 = (1/6, 1/6, 1/6, 1/6, 1/6, 1/6)$ for the six faces. Under the alternative
hypothesis that the die is unfair the probabilities are
$p_1 = (1/5, 1/5, 1/5, 1/5, 1/5, 0).$ Then 
$$\lambda = nS = n\sum_{i=1}^6 \frac{(p_{1i} - p_{0i})^2}{p_{0i}}.$$
For your unfair die, $S = 0.2,$ so you have $\lambda = ns = 0.2n.$
 p.0 = rep(1/6, 6)
 p.1 = c(rep(1/5,5),0)
 s = sum((p.1 - p.0)^2/p.0);  s
 ## 0.2

Now you want to know $n$ such that 
$$P(Q > 11.07 | \nu = 5, \lambda = 0.2n) = .5.$$
This probability is called the power of the test at the 5% level,
against the alternative face probabilities $p_1.$
We can find $n$ by computing the power for many values of $n$, and
picking the smallest one that makes the power exceed 0.5. The answer is
$n = 41$ (when $\lambda = ns = 8.2$)  If you were fussier and wanted a 95% chance of detecting
your die is dishonest, then you would want $n = 106.$
 n = 30:1000
 pwr = 1 - pchisq(11.0705, 4, .2*n)
 min(n[pwr > .5])
 ## 41
 min(n[pwr > .95])
 ## 106

The figure below shows the PDF of the central chi-squared distribution $Chisq(\nu = 5)$ for a fair die (green), and the noncentral chi-squared distribution $Chisq(\nu =5, \lambda=8.2)$ for your dishonest die (blue).
You can see that about half the area under the latter curve is to the
right of the critical value $q^* = 11.07.$

Reference: The power of the GOF test is not routinely explained in
basic statistics courses because it takes specialized software such as R
to find probabilities for the noncentral chi-squared distribution. However,
you can find an formal explanation in a paper by Guenther (1977) in The
American Statistician. (The preview page you can get in Google has the
formula for $\lambda.$ He uses detecting an unfair die as one of his
examples. If interested, you can probably find a university
library with access to the article.)
Note: In your specific case where one of the faces on your unfair die
is impossible, you could use the method proposed by @RossMillikan, and
get a smaller answer. For $X \sim Binom(4, 1/6),$ one has $P(X = 0) \approx .48.$ [Even quicker, you could likely just look at the die.
A typical unfair die might have faces 1, 2, 3 (with a common corner) slightly less likely than 1/6, and
faces 4, 5, 6 slightly more likely. This might be done by embedding a lead
weight in the plastic just beneath the 123-corner.] 
