It is not mentioned that if this game is going to be played like forever or once, but by wording I believe this is one time game, so you can only play this game only once. So the rest will be according to this assumption:
Let's say
Instead of $ \left\{ 1,\ldots ,30\right\}$, let's make it $ \left\{ 1,2,3 ,4\right\}$, so the first player optimal option will be obviously $3$. Because second player is forced to choose between $2$ and $4$. If he chooses $2$, he will win if the outcome is $1$ or $2$ (or 50% chance to win), if he chooses $4$, he can only win if outcome is $4$ (which is 25% chance to win). So the optimal choose will be $2$ if you want to increase his chance to win and but if he wants to maximize the expected value, he needs to go for $4$.
So, let's say the first players want to increase to chance to win instead of money;
The first player becomes advantageous with the higher expected value, $3\times0.25+4\times0.25=1.75$ whereas for second player it is $1\times0.25+2\times0.25=0.75$ and their chances to win the game is equal!
similarly, if the first players maximizes the expected value, he needs to go for
$4$ and the expected value of second player becomes $1$, whereas the first player's expectec value is $1.5$, which is $0.25$ is lower but the chance to win of first player was $50\%$ before, but now it is $75\%$!
So whatever the second player thinks to play, the first player will be
advantageous!
Let's make it
$\left\{ 1,2,3,4,5,6\right\}$ now, the player one will go for $5$ if he wants to optimize the expected value at the end but his chance to win will drop drastically compared to $ \left\{ 1,2,3 ,4\right\}$. He will naturally force to second player to go to $4$. and the $E(P1)=11$ points and $E(P2)=10$. But the chance to win the game for first player is only 33%, whereas second player has chance to win 66% which is as twice as first player. But statistically, for player 1, going for $5$ was the optimal by means of Expected Value, but what if he wants to win? he would go for $4$, so players two would go to $5$ and the chances to win would be reversed than before. Player 1 would have chance to win 66% whereas player 2 would have 33% chance to win with a little higher expected value than player 1.
Similarly, nothing would change much with $ \left\{ 1,\ldots ,30\right\}$, It actually depends on that
The question is, you want to win or you want to optimize/maximize your profit with highest expected value?
Think about it,
if this question was asked for 1 million dollars to 30 million dollars, you would think of winning the game or maximizing your expected value?
So,
I believe there are two answers to this question depending on you would like to win or you would like to maximize your expected value, because you may even be okay with 1 million dollar to win? so it depends on the value of $30\$$ in your mind.
As a result,
Being first player is advantageous in general, because he/she can think of the expected value, chance to win, and the value of the winning the game for both sides and force the second player to play whatever he/she wants. For me, 30 dollars is not that much, so I would go for $22$ previously answered to maximize my expected value, but some other might think of going for $16$ to maximize his chance to win against the player who would think the same way.