Example of determined games When do we say a game is determined? I searched a lot and found that: 
A (class of) game(s) is determined if for all instances of the game there is a winning strategy for one of the players (not necessarily the same player for each instance)
I don't know if it's true. I couldn't find any example of determined games and how to prove that a game is determined. The following image is from here. "Is this game determined for all finite graphs?" [Asked in the source].

 A: To complement DukeZhou's answer, let me say a bit about the more abstract side of things. It turns out that when we restrict attention to two-player, perfect information games pretty much every game you can think of is determined, even if we don't know who has the winning strategy or what it is. Note that the existence of a winning strategy is in no way equivalent to the existence of a simple, or known, winning strategy.

Let's start with the definition of a finite-length (two-person, perfect information) game. Abstractly, we have a set of moves, $X$, which might be infinite, and two players $1$ and $2$; the game proceeds by having players alternate elements of $X$. We have some set $L$ of finite strings of elements of $X$, which we think of as the set of "legal plays;" the first player who causes the play to leave $L$ loses.
For example, a very silly case is the game My dad makes more money than yours. We take $X=\mathbb{N}$ and $L$ is the set of all strings of naturals of length at most $2$ where the second bit of the string is greater than the first bit of the string (if the string has length $2$). Basically, player $2$ just has to play a bigger number than player $1$. In particular, a winning strategy for player $2$ is "play whatever $1$ just did, plus one." To see how this goes:


*

*Player $1$ plays $17$. The string $\langle 17\rangle$ is in $L$, so they haven't lost yet.

*Player $2$ plays $18$. The string $\langle 17,18\rangle$ is in $L$, so they haven't lost yet.

*Now it's player $1$'s turn again, and they're screwed. No matter what they play, the result will be a string of length $3$ - which can't be in $L$! So whatever they do now, they lose.
A less silly example is Tic-tac-toe. One way to interpret this in the formalism above is:


*

*$X$ is the $18$-element set $\{1,2,3,4,5,6,7,8,9\}\times\{$circle, cross$\}$ - here the numbers $1$-$9$ represent the squares on the board, and the circle or cross represents what symbol is drawn there. E.g. the move "$(3, $ circle$)$"  should be thought of as "draw a circle in the top-right square."

*Now $L$ takes into account the rules of the game. The three rules of tic-tac-toe are: 


*

*(1) $1$ plays crosses, $2$ plays circles; 

*(2) you can't play in a square that already has a symbol; and 

*(3) if someone gets three-in-a-row, they win.


*We'll define $L$ accordingly. $L$ is the set of finite strings of elements of $X$ such that $\sigma\in L$ iff:


*

*(1) Every odd bit of $\sigma$ has second coordinate "cross," and every even bit has second coordinate "circle."

*(2) We never have two bits of $\sigma$ with the same left coordinate.

*(3) No proper substring of $\sigma$ has a three-in-a-row (this is tedious, but not difficult, to write down formally). That is: once one player gets three-in-a-row, the other player has no legal moves and immediately loses.
At this point it's good to stop and do an exercise:

Show that chess, with the rule that if the game goes on for five billion moves or if it's a draw/stalemate at some point then player $2$ wins, can be expressed using the above formalism.


Now here's the cool bit:

Theorem (Zermelo): Every finite-length (two-player, perfect-information) game is determined.

And as a consequence, the answer to the question in the source you link to is "yes."
For example, chess - with the modification above - is determined; phrasing this in the language of normal chess, either white has a winning strategy, or black has a strategy which will never lose (but might allow the game to go on forever). We don't know who has the winning strategy - the proof is highly nonconstructive - but we know one exists!
So the question of whether a finite (two-person, perfect information) game is determined is completely answered. Is there anything else to do?
It turns out that the answer is yes! We can talk about infinite-length games - the idea being that players $1$ and $2$ play forever, producing an infinite sequence of elements of $X$, and player $1$ wins iff the infinite sequence so produced is an element of the "payoff set." Here our formalism is a bit different: rather than a set of moves $X$ and a set $L$ of finite strings of elements of $X$, we have a set of moves $X$ and a set $P$ of infinite strings of elements of $X$ (the "payoff set"), with the rule that player $1$ wins iff the infinite string produced over the course of the game is in $P$.
The simplest type of infinite game is where player $2$ wins if they "survive forever:" we have some set $D$ of finite strings of elements of $X$ corresponding to moments when player $2$ has lost, and our payoff set is the set of all infinite strings of elements of $X$ extending $D$ ($1$ wins if $2$ loses); these are called "closed games" (for topological reasons - ask me in the comments if you're interested and I'll explain more), and Gale and Stewart showed that every closed game is determined. An example of a closed game which is not (equivalent to) a finite game is chess with the rule that player $2$ wins if the game goes on forever.
However, it doesn't stop there. We can even talk about infinite-length games where the rule for who wins an infinitely long play is really complicated - maybe sometimes $2$ wins if the game goes on forever, but other times $1$ wins if the game goes on forever, depending on how it goes on forever. For example, consider the game with $X=\{0, 1\}$, $L$ is the set of all finite binary strings, and player $1$ wins iff the infinite binary string produced is the decimal expansion of a rational number; it's not hard to show that $2$ has a winning strategy here. It turns out that still many games of this type are determined; in fact, every reasonably-definable game is determined (and this fact is referred to in the second-to-last slide of the linked article).

Now, the axiom of choice proves the existence of a nondetermined game (see the Fact, and following parenthetical remark, on the second-to-last slide of the linked article). But this proof is highly nonconstructive, and the game so produced is not definable in any nice way. There is a surprisingly deep question at this point:

How are definability and determinacy related? That is, for what notions of "nicely definable" does nicely definable imply determined?

This is a major topic in set theory and descriptive set theory, and has surprising connections to large cardinals. A survey of determinacy in set theory can be found here.
A: I like to reduce everything to the most simple form.  My understanding is that all trivial games are determined because triviality in relation to games means the gametree is tractable to at least one competitor.
It's wiggly because this is subjective: Tic-Tac-Toe is considered trivial, but is distinctly non-trivial to the average 5 year old.  (See: Bounded Rationality.)
Finite, non-trivial deterministic games of perfect information have an interesting quality in that, at some point, they collapse into a state of tractability.  (This seems similar to a wave function collapse, but I'm not aware of any mathematical literature on the subject.)  This condition is what allowed Berlekamp to do the initial Go endgame analyses.
This is related the economic game theory concept of a strictly determined game and a formal use of determinacy in set theory in relation to games (see Noah Schweber's answer).  
