A psychologically meaningful model can guide us.
Derivation of a useful test
Any variation in the observations can be attributed to variations among the subjects. We might imagine that each subject, at some level, comes up with a numeric value for the result of method 1 and a numeric value for the result of method 2. They then compare these results. If the two are sufficiently different, the subject makes a definite choice, but otherwise the subject declares a tie. (This relates ties to the existence of a threshold of discrimination.)
The variation among the subject causes variation in the experimental observations. There will be a certain chance $\pi_1$ of favoring method 1, a certain chance $\pi_2$ of favoring method 2, and a certain chance $\pi_0$ of a tie.
It is fair to assume the subject respond independently of one another. Accordingly, the likelihood of observing $n_1$ subjects favoring method 1, $n_2$ subjects favoring method 2, and $n_0$ subjects giving ties, is multinomial. Apart from an (irrelevant) normalizing constant, the logarithm of the likelihood equals
$$n_1 \log(\pi_1) + n_2 \log(\pi_2) + n_0 \log(\pi_0).$$
Given that $\pi_0 + \pi_1 + \pi_2=0$, this is maximized when $\pi_i = n_i/n$ where $n = n_0+n_1+n_2$ is the number of subjects.
To test the null hypothesis that the two methods are considered equally good, we maximize the likelihood subject to the restriction implied by this hypothesis. Bearing in mind the psychological model and its invocation of a hypothetical threshold, we will have to live with the possibility that $\pi_0$ (the chance of ties) is nonzero. The only way to detect a tendency to favor one model over the other lies in how $\pi_1$ and $\pi_2$ are affected: if model 1 is favored, then $\pi_1$ should increase and $\pi_2$ decrease, and vice versa. Assuming the variation is symmetric, the no-preference situation occurs when $\pi_1=\pi_2$. (The size of $\pi_0$ will tell us something about the threshold--about discriminatory ability--but otherwise gives no information about preferences.)
When there is no favored model, the maximum likelihood occurs when $\pi_1=\pi_2 = \frac{n_1+n_2}{2}/n$ and, once again, $\pi_0 = n_0/n$. Plugging in the two previous solutions, we compute the change in maximum likelihoods, $G$:
$$\eqalign{
G &=\left(n_1\log\frac{n_1}{n} + n_2\log\frac{n_2}{n} + n_0\log\frac{n_0}{n}\right) \\
&-\left(n_1\log\frac{(n_1+n_2)/2}{n} + n_2\log\frac{(n_1+n_2)/2}{n} + n_0\log\frac{n_0}{n}\right) \\
&=n_1 \log\frac{2n_1}{n_1+n_2} + n_2 \log\frac{2n_2}{n_1+n_2}.
}$$
The size of this value--which cannot be negative--tells us how credible the null hypothesis is: when $G$ is small, the data are "explained" almost as well with the (restrictive) null hypothesis as they are in general; when the value is large, the null hypothesis is less credible.
The (asymptotic) maximum likelihood estimation theory says that a reasonable threshold for this change is one-half the $1-\alpha$ quantile of a chi-square distribution with one degree of freedom (due to the single restriction $\pi_1=\pi_2$ imposed by the null hypothesis). As usual, $\alpha$ is the size of this test, often taken to be 5% ($0.05$) or 1% ($0.01$). The corresponding quantiles are $3.841459$ and $6.634897$.
Example
Suppose that out of $n=20$ subjects, $n_1=3$ favor method 1 and $n_2=9$ favor method 2. That implies there are $n_0 = 20 - 3 - 9 = 8$ ties. The likelihood is maximized, then, for $\pi_1 = 3/20 = 0.15$ and $\pi_2 = 9/20 = 0.45$, where it has a value of $-20.208\ldots$. Under the null hypothesis the likelihood is instead maximized for $\pi_1 = \pi_2 = 6/20 = 0.30$, where its value is only $-21.778$. The difference of $G = -20.208 - (-21.778) = 1.57$ is less than one-half the $\alpha = $5% threshold of $3.84$. We therefore do not reject the null hypothesis.
About ties and alternative tests
Looking back at the formula for $G$, notice that the number of ties ($n_0$) does not appear. In the example, if we had instead observed $n=100$ subjects and among them $3$ favored method 1, $9$ favored method 2, and the remaining $100 - 3 - 9 = 88$ were tied, the result would be the same.
Splitting the ties and assigning half to method 1 and half to method 2 is intuitively reasonable, but it results in a less powerful test. For instance, let $n_1=5$ and $n_2=15$. Consider two cases:
$n=20$ subjects, so there were $n_0=0$ ties. The maximum likelihood test would reject the null for any value of $\alpha$ greater than $0.02217$. Another test frequently used in this situation (because there are no ties) is a binomial test; it would reject the null for any value of $\alpha$ greater than $0.02660$. The two tests therefore would typically give the same results, because these critical values are fairly close.
$n=100$ subjects, so there were $n_0=80$ ties. The maximum likelihood test would still reject the null for any value of $\alpha$ greater than $0.02217$. The binomial test would reject the null only for any value of $\alpha$ greater than $0.3197$. The two tests give entirely different results. In particular, the $80$ ties have weakened the ability of the binomial test to distinguish a difference that the maximum likelihood theory suggests is real.
Finally, let's consider the $3 \times 1$ contingency table approach suggested in another answer. Consider $n=20$ subjects with $n_1=3$ favoring method 1, $n_2=10$ favoring method 2, and $n_0=7$ with ties. The "table" is just the vector $(n_0,n_1,n_2)=(7,3,10)$. Its chi-squared statistic is $3.7$ with two degrees of freedom. The p-value is $0.1572$, which would cause most people to conclude there is no difference between the methods. The maximum likelihood result instead gives a p-value of $0.04614$, which would reject this conclusion at the $\alpha=$5% level.
With $n=100$ subjects suppose that only $1$ favored method 1, only $2$ favored method 2, and there were $97$ ties. Intuitively there is very little evidence that one of these methods tends to be favored. But this time the chi-squared statistic of $182.42$ clearly, incontrovertibly, (but quite wrongly) shows there is a difference (the p value is less than $10^{-15}$).
In both situations the chi-squared approach gets the answer entirely wrong: in the first case it lacks power to detect a substantial difference while in the second case (with lots of ties) it is extremely overconfident about an inconsequential difference. The problem is not that the chi-squared test is bad; the problem is that it tests a different hypothesis: namely, whether $\pi_1=\pi_2=\pi_0$. According to our conceptual model, this hypothesis is psychological nonsense, because it confuses information about preferences (namely, $\pi_1$ and $\pi_2$) with information about thresholds of discrimination (namely, $\pi_0$). This is a nice demonstration of the need to use a research context and subject matter knowledge (however simplified) in selecting a statistical test.
