Your strategy of betting on single numbers so as to double your initial stake would work for the first $24$ bets, and has a probability of $1-\left(1-\frac{1}{37}\right)^{24} \approx 0.481893977$ of succeeding in at least one of these.
The $25$th round hits the problem that you cannot reach double as you have less than $\$\frac{3000}{36}\approx \$83.33$ left. So you might say bet everything that remains on a single number, which you will probably lose, ending the game. This has an overall probability about $\left(1-\frac{1}{37}\right)^{25} \approx 0.50410316$.
But there is a probability of $\frac{1}{37}\left(1-\frac{1}{37}\right)^{24} \approx 0.014002865$ still to deal with if you are still playing after $25$ rounds. What happens to this depends on exactly how much you have left at this stage, and that depends on the casino betting rules.
If you must bet whole numbers of dollars, so rounding up your earlier bets, you will have $\$34$ left after losing the first $24$ rounds and $\$1224$ left after winning the $25$th round. Following the same strategy as before will lead to an overall probability of winning of about $0.48745373$ and of losing of about $0.51254627$
If you can bet amounts including arbitrary fractions of dollars, so not rounding up your earlier bets, you will have about $\$50.70501$ left after losing the first $24$ rounds and $\$1825.38036$ left after winning the $25$th round. Following the same strategy as before will lead to an overall probability of winning of about $0.49027046$ and of losing of about $0.50972954$
First round
In the first round the player must beat the dealer. The dealer's card is excluded, leaving $311$ possible cards for the player. On average
$23$ cards tie (there are $24$ cards in each rank, but the dealer has one of them.)
$144$ cards win
$144$ cards lose.
Second round
The second round takes place if the first round was a tie, so the probability it happens is $23/311$.
In the second round the player wins the bet if the card values relative to the dealer are either a win or a tie. There are now $310\cdot309=95790$ possible deals.
As before, non-ties are split equally on average.
For $12$ of the ranks there are $24\cdot23=552$ ways to tie.
For the rank that occurred in the first round tie there are $22\cdot21=462$ ways to tie.
So in the second round there are
$7086$ ways to tie $(12\cdot552+1\cdot462)$
$44352$ ways for the player's card to be higher.
$44352$ ways for the player's card to be lower.
Conclusion
The overall probability of a win is$$\frac{144}{311}+\frac{23}{311}\cdot\frac{7086+44352}{95790}\approx0.502375$$
Incidentally, the house advantage comes from the way the betting works, and happens in a way that is not relevant to the question asked here.
Best Answer
Suppose that we win nothing if we do not call. Let $p$ be our probability of winning. Then when we call the expectation of our gain is $$(30p-10(1-p))\$=(40p-10)\$ .$$ So when $p>1/4$ the expectation is a positive number of dollars.