What is tied data in the context of a rank correlation coefficient? I am not in statistics field. 
I have seen the word "tied data" while reading about Rank Correlation Coefficients.


*

*What is tied data?

*What is an example of tied data?

 A: It means data that have the same value; for instance if you have 1,2,3,3,4 as the dataset then the two 3's are tied data. If you have 1,2,3,4,5,5,5,6,7,7 as the dataset then the 5's and the 7's are tied data.
A: It's simply two identical data values, such as observing 7 twice in the same data set.
This comes up in the context of statistical methods that assume data has a continuous and so identical measurements are impossible (or technically, the probability identical values is zero). Practical complications arise when these methods are applied to data that are rounded or clipped so that identical measurements are not only possible but fairly common.
A: The question is of fundamental importance: 
What is a tied observation/data/pair ?
Altough often mentioned only in nonparametric methods, this notion is independent of nonparametric methods. It is mentioned in nonparametric methods because this situation will cause calculation complication in obtaining the statistics used in nonparametric methods, like Wilcoxon Signed Ranked statistics $T^+$.
(So I don't think @Ming-Chih Kao's answer is proper by introducing nonparametric tests first. But since the title is 'What is tied data in the context of a rank correlation coefficient?', I will buy it.)
To illustrate, I think the best way is to work with the simplest example of Wilcoxon Signed Ranked Test:
Let us have a sample of paired data of size 10:
Define the difference random variable $Z_{i}=X_{i}-Y_{i}$
$(X_{i},Y_{i})$: (1,-1) (1,2) (1,2) (1,-1) (2,1) (2,1) (2,3) (2,3) (3,2) (3,0)
$Z_{i}$: 2 -1 -1 2 1 1 -1 -1 1 3
Take the absolute value of these $Z_{i}$'s in order to have a rank.
$|Z_{i}|$: 2 1 1 2 1 1 1 1 1 3
Now the problem arise, with so many identical 1's and 2's, how can we make a ranking? We give them the term "tied" to show this case. And by the term "tied group"(which is an equivalent relation), we simply group those tied observations into groups by their values. In this example, we have 3 tied groups(Think why):$\{(1,-1) (1,-1)\},\{ (1,2) (1,2) (2,1) (2,1) (2,3) (2,3) (3,2) \},\{(3,0)\}$ Attention that the bracket does not mean a set but just a notation.
Let us try the very easy way of doing this, we rank from left to right and give:
$R_{i}$: 8 1 2 9 3 4 5 6 7 10
But here again we should ask why so other ranking is not suitable since there is no difference between those identical $|Z_{i}|$'s, like:
$R_{i}$: 8 7 6 9 5 4 3 2 1 10
Therefore we may just take the mean of those identical $|Z_{i}|$'s and assign again:
$R_{i}$: 8 7 6 9 5 4 3 2 1 10
The bold represents the first tied group consists of those $|Z_{i}|=1$ observations; the italic represents the second tied group consists of those $|Z_{i}|=2$ observations.
We assign to each of the observation in the first group the rank$\frac{1+\cdots+7}{7}=4$;we assign to each of the observation in the second group the rank$\frac{8+9}{2}=8.5$. Therefore we have:
$R_{i}$: 8.5 4 4 8.5 4 4 4 4 4 10
This modified the rankings and make each of the tied observation has the same influence in calculating the ranked statistics, thus in the rank test.
What are the solutions to tied observation/data/pair ?
(1)Assign the average rank. This is just what we did above. By assigning the same rank to the tied data in the same group, we make their influence in the ranked test just the same and therefore eliminate the possible inaccuracy caused by tied observations.
(2)Assign the random rank. Just assign ranks randomly to each of the tied group element. The only restriction is that $MaxRank_{first group}<MinRank_{second group}$ since if $MaxRank_{first group}>MinRank_{second group}$, that breaks the ranking law; if $MaxRank_{first group}=MinRank_{second group}$, then we have to merge two tied groups into one.
(3)Perturbation of data. This requires very careful consideration about the nature of the data. This works only if the data is not categorical(discrete). In the above example, we can just make a 
This will put different weights manually to each of the elements in the tied group. For a continuous distribution, for example, it makes little difference if you perturb it in $\epsilon$ manner.
(@John D. Cook 's answer is a bit misleading in this way. A better way of saying this point is that when the distribution is continuous, $P{X=x}=0$. However, we shall observe ties since our measurement is of limited accuracy, i.e. any sample space in reality is actually finite.)
(@quarkdown27 's answer is simple but correct in each word.)
A: "Tied data" comes up in the context of rank-based non-parametric statistical tests.
Non-parametric tests: testing that does not assume a particular probability distribution, eg it does not assume a bell-shaped curve. 
rank-based: a large class of non-parametric tests start by converting the numbers (eg "3 days", "5 days", and "4 days") into ranks (eg "shortest duration (3rd)", "longest duration (1st)", "second longest duration (2nd)"). A traditional parametric testing method is then applied to these ranks. 
Tied data is an issue since numbers that are identical now need to be converted into rank. Sometimes ranks are randomly assigned, sometimes an average rank is used. Most importantly, a protocol for breaking tied ranks needs to be described for reproducibility of the result. 
