SOS game with $n=14$ and $n=2000$ The board consists of a row of $n$ squares, initially empty. Players take turns selecting an empty square and writing either an S or O in it. The player who first succeeds in completing SOS in consecutive squares wins the game
a) Show that if $n=2000$ the second player can win the game
b) Who, if anyone wis the game if $n=14$?
Is there a simpler solution than the one presented in Ferguson? Because I don't quite understand it
 A: I'm going to write out my solution in hope it differs from one written in your book, so you might find some use of it. But you should have added the solution and pointed out which part you didn't understand as pointed out in comment by TonyK. Also, I'm assuming no SOS at end means draw.
Name first player = P, second = D.
P plays something. D plays S far away. P plays something. If D can win, he wins. If not, he puts S three squares apart from his previous S. That is, a board has S _ _ S, somewhere.
Now, if at any point P puts on one of the two blank positions, D can easily win on the other blank position.
D adopts following strategy. If he can win at any point, he does that. If he sees three blanks in a row ( _ _ _ ), he plays O in the middle. Obviously, P can't win after that move, so he must make a move. If D sees S _ S, S _ O, O _ S or O _ O, he plays O in the middle. After none of these moves, P can't win.
So the only remaining scenario is that there is no single blank and no three blanks in a row. Meaning only two blanks in a row on the table. But then, there is even number of moves made, so it's not D's turn.
Therefore, P must have filled one of the two blanks in S _ _ S at some point, so D won.
Now as for b) part. Obviously, by adopting the same strategy (without initial 4 moves), D can force at least a draw. For $n = 16$, D can win for the same argument like in $n = 2000$. Now to illustrate why P can prevent D from making smart opening when $n = 14$, and ultimately force a draw.
Positions are labeled $1 - 14$. Call configuration S _ _ S a trap. Also, P will play just O's (he would play S just if it led him to victory at that move, which D will not allow of course).
P plays O7.
If D puts S on 1, 2, 3, 9, 10 or 11, P can sabotage his trap by putting O two squares to the right of S. If D puts S on 4, 5, 12, 13 or 14, P can sabotage the trap by putting O two places to the left of S. If D puts S on 6 or 8, he looses. This was the key argument that differs $n = 14$ from $n = 16$, where D has nine blanks in a row and can make a trap same way as for $n = 2000$. I think this should convince you enough to believe that P can force a draw, but anyways, here is the rest of the proof.
Now P adopts following strategy. At any time if he can win, he does that. If D plays S, P responds with O that sabotages the trap that S can make as described above. Notice that this can't lead to S _ O S or S O _ S, since then, two moves before would look like _ _ _ S or S _ _ _, and P would have already sabotaged the corresponding traps by playing O two squares apart from S. Now, if the trap after D's move is already sabotaged or D plays O, then let's prove P can play O somewhere not letting D win after that. If he sees _ _ _, he plays O in the middle. If he sees X _ X, he fills the blank with O. And if only remaining are two blanks in a row, we know it's not a trap since there are no traps, so there is O at one end of the two blanks and he plays O next to it.
