How to find probability of a given sample generated by a given distribution This is a homework problem I have.

Let a random variable $x$ follow a Gaussian distribution with mean $\mu = 10$ and variance $1$.
Which of the following samples of size 4 has the largest probability
of being generated from this distribution?
i.  9, 9.1, 10, 11
ii.  1, 2, 0, 11
iii.  9.9, 10.1, 9.8, 10.3
iv.  11, 9, 12, 8
v.  10, 10, 10, 1

My friend argues that iii. is the right answer because $\Pi_i \, p(x_i;\mu=10,\sigma^2 =1)$ would be higher for it. But I don't think that is true. If it were, a sample size of a million with all values equal to 10 would become most likely to be generated.
In my opinion, we need to take sample mean and variance, and find their likelihood from the distribution of sample mean and variance.
Please help clear my conceptual confusion.
 A: Find $S=\frac14\sum_{i=1}^4(X_i-10)^2$. Select the sample for which the corresponding $S$ is closest to $1$.
A: Find the product of the probability densities of each of the four points in the sample, proportional to:
$$\prod_{i=1}^4 P(x_i| {\cal N}(\mu, \sigma))$$
The values are:
i) $.00621522$
ii) $9.66 \times 10^{-56}$
iii) $0.0235$
iv) $0.000170674$
v) $6.527 \times 10^{-20}$
Hence the answer is iii.
A: The stated 'probability' criterion does not make sense. So one is left to imagine what might have been intended. One approach for parts (i)-(iv) is to do a Kolmogorov-Smirnov goodness-of-fit test of each sample to $\mathsf{Norm}(\mu=10, \sigma=1).$ 
Roughly speaking this test compares the 'distance' of the empirical CDF
of the sample from the CDF of the candidate distribution. This distance
has a known distribution when the candidate distribution is correct (the
null hypothesis is true). 
The P-value of the test is the probability that
a sample of the given size would be 'farther' from the candidate distribution
then the one tested (under the null distribution). Thus large P-values might be taken as an indication
of a relatively good fit and small ones as an indication of a relatively bad fit.
In R statistical software the test for (iv) gives the following output:
x4
##  11  9 12  8

ks.test(x4, pnorm, 10, 1)

        One-sample Kolmogorov-Smirnov test

data:  x4 
D = 0.3413, p-value = 0.6347
alternative hypothesis: two-sided 


Of the four samples tested, (iv) has the largest P-value. P-values
  for samples (i)-(iii) are 0.1875, 0.0058, and 0.3737, respectively. So,
  of the four samples tested, sample (iv) has the best fit to $\mathsf{Norm}(10,1)$ according to the K-S criterion.

The K-S test is for continuous distributions, and so does not work with
samples such as (v) which have tied values. I believe that sample (iv) can
be eliminated because if samples of size 4 are taken from
$\mathsf{Norm}(10,1)$ and rounded to two places, the probability
of seeing only two unique values is about 0.00006. (And about 0.006, if rounded
to one place.) Moreover, the 'unrounded' sample $(9.99, 10.00, 10.01, 1.00)$ has K-S P-value 0.1941.

Note: Another criterion might be to try matching sample means and SDs with
$\mu = 10$ and $\sigma = 1.$ Samples (iii) and (iv) are the only plausible choices:
mean(x3); sd(x3)
## 10.025
## 0.2217356
mean(x4); sd(x4)
## 10
## 1.825742

Again here, I'd choose sample (iv) because it has no ties, the smallest difference
in means, and the most favorable ratio of SDs. (See @Henry's Comment on variances.)
