user1286: We do not know, given two perfect players White and Black, whether the game would necessarily end in a draw, or necessarily end in a win (for either Black or White).
However, can we prove that if there is a winning strategy, then it is for White? In other words, can we prove that Black must either lose or draw?

user885: In theory, chess can be "solved", since it is a "finite" game with "perfect information". More precisely, there exists a strategy such that one player has a guaranteed win, or both players have a guaranteed draw given perfect play. Here's a technical article on the basic (well, basic for those familiar with economics/mathematics) concepts of Game Theory for those interested in the specifics. Essentially, every game which has "perfect information", i.e. each player can see all the pieces, and is aware of all the legal moves of said pieces at all points during the game (a counterexample of a perfect information game would be a card game, where you aren't able to see your opponent's hand), ** a finite number of players and a finite number of legal moves**, i.e. the game doesn't go on indefinitely, then it has a guaranteed winning or drawing strategy for one of the players.
In practice, we have neither the technology nor the intelligence (ok, maybe if all the best chess minds of today collaborated on finding the strategy, we may have sufficient intelligence required. MAYBE.) and time to do it manually.
To answer your question: Yes, there exists a winning (or drawing strategy). No, we don't know whether it is for white or for black.
Yes, chess is doomed to get solved someday. But we won't have the technology (in my opinion the only means of doing so) for it for many, many decades (hopefully even centuries) to come.

user1430: It can theoretically be proven, but not with current technology. 
If you take a brute force approach, there is some difficulty due to the number of positions.
In analysis of the Shannon Number, it is suggested that the game tree complexity is at least 10^123 for games of max length 80 moves. Let's assume that it is 10^123 for the purposes of this discussion. 

10 ^ 81 = Estimated number of atoms in the universe
10 ^ 12 = Operations per second of a terahertz processor core (your
  processor probably runs at about 1/300th of this speed.)
10 ^ 7 = Rounded-off seconds per year
10 ^ 12 = 1 trillion years

Let's also assume that our processors can evaluate a chess position in only 1 processor cycle.
So, let's make every atom in the universe operate as a terahertz processor core for 1 trillion years.
Can we evaluate each position for 80-max length games?
No.

10^81 x 10^12 x 10^7 x 10^12 = 10^112

We fall short to the tune of being only 0.0000000001% complete with the calculation.
With advanced pruning (throwing out bad lines and their descendants), better technology, and some crafty programming... maybe we'll see 40-max games solved in our lifetime! We can also prune out positions that we've seen before (we can arrive there via transposition), but keep in mind it will take at least a CPU cycle to determine that we've evaluated the position before!
However, this should help you see why it's so far out of reach at the moment.
References

Shannon Number
Number of Atoms in the Universe


user291: If there is such a proof, no one has found it, and I doubt very much that such a proof exists (it is hard to imagine a mathematically provable "guaranteed-draw" strategy as White). One would certainly expect White to have an advantage if anybody did, but there are some downsides to going first as well (you have to disclose information before your opponent), so it is theoretically possible that the downsides outweigh the upsides. That said, the probability of that being the case seems to be infinitesimal.

1286: <p>We do not know, given two perfect players White and Black, whether the game would necessarily end in a draw, or necessarily end in a win (for either Black or White).</p>

<p>However, can we prove that if there is a winning strategy, then it is for White? In other words, can we prove that Black must either lose or draw?</p>
 885: <p>In theory, chess can be "solved", since it is a "finite" game with "perfect information". More precisely, there exists a strategy such that one player has a guaranteed win, or both players have a guaranteed draw given perfect play. Here's a technical article on the basic (well, basic for those familiar with economics/mathematics) concepts of <a href="http://www.econ.ucsd.edu/~jsobel/200Cs09/Kartik_Notes.pdf" rel="nofollow noreferrer">Game Theory</a> for those interested in the specifics. <strong>Essentially, every game which has "perfect information",</strong> i.e. each player can see all the pieces, and is aware of all the legal moves of said pieces at all points during the game (a counterexample of a perfect information game would be a card game, where you aren't able to see your opponent's hand), ** a finite number of players and a finite number of legal moves**, i.e. the game doesn't go on indefinitely, <strong>then it has a guaranteed winning or drawing strategy for one of the players.</strong></p>

<p>In practice, we have neither the technology nor the intelligence (ok, maybe if all the best chess minds of today collaborated on finding the strategy, we may have sufficient intelligence required. MAYBE.) and time to do it manually.</p>

<p>To answer your question: Yes, there exists a winning (or drawing strategy). No, we don't know whether it is for white or for black.</p>

<p>Yes, chess is doomed to get solved someday. But we won't have the technology (in my opinion the only means of doing so) for it for many, many decades (hopefully even centuries) to come.</p>
 1430: <p>It can theoretically be proven, but not with current technology. </p>

<p>If you take a brute force approach, there is some difficulty due to the number of positions.</p>

<p>In analysis of the <a href="http://en.wikipedia.org/wiki/Shannon_number" rel="nofollow noreferrer">Shannon Number</a>, it is suggested that the game tree complexity is at least 10^123 for games of max length 80 moves. Let's assume that it is 10^123 for the purposes of this discussion. </p>

<blockquote>
  <p>10 ^ 81 = Estimated number of atoms in the universe</p>
  
  <p>10 ^ 12 = Operations per second of a terahertz processor core (your
  processor probably runs at about 1/300th of this speed.)</p>
  
  <p>10 ^ 7 = Rounded-off seconds per year</p>
  
  <p>10 ^ 12 = 1 trillion years</p>
</blockquote>

<p>Let's also assume that our processors can evaluate a chess position in only 1 processor cycle.</p>

<p><strong>So, let's make every atom in the universe operate as a terahertz processor core for 1 trillion years.</strong></p>

<p><strong>Can we evaluate each position for 80-max length games?</strong></p>

<p><strong>No.</strong></p>

<blockquote>
  <p>10^81 x 10^12 x 10^7 x 10^12 = 10^112</p>
</blockquote>

<p>We fall short to the tune of being only 0.0000000001% complete with the calculation.</p>

<p>With advanced pruning (throwing out bad lines and their descendants), better technology, and some crafty programming... maybe we'll see 40-max games solved in our lifetime! We can also prune out positions that we've seen before (we can arrive there via transposition), but keep in mind it will take at least a CPU cycle to determine that we've evaluated the position before!</p>

<p>However, this should help you see why it's so far out of reach at the moment.</p>

<p><strong>References</strong></p>

<ul>
<li><a href="http://en.wikipedia.org/wiki/Shannon_number" rel="nofollow noreferrer">Shannon Number</a></li>
<li><a href="http://en.wikipedia.org/wiki/Observable_universe#Matter_content" rel="nofollow noreferrer">Number of Atoms in the Universe</a></li>
</ul>
 291: <p>If there is such a proof, no one has found it, and I doubt very much that such a proof exists (it is hard to imagine a mathematically provable "guaranteed-draw" strategy as White). One would certainly expect White to have an advantage if anybody did, but there are some downsides to going first as well (you have to disclose information before your opponent), so it is theoretically possible that the downsides outweigh the upsides. That said, the probability of that being the case seems to be infinitesimal.</p>
