Believe it or not, it's the ties that make this problem really complicated to analyze. For that reason I'm going to ignore ties for now, and assume that all 52 cards have a unique order.
Suppose that the dealer has picked a card out of $n$, and we know (through card counting) that there are $l$ lower cards, and $n-1$ cards left. Then our winning chance is (assuming we pick our best choice of higher/lower):
$$p = \frac{\max(l, n-1-l)}{n-1}$$
Suppose we have seen the dealer's card and know our $p$ to win. Now there are two choices:
- See our card. We have an expected outcome of $pmb$, where $m$ is our multiplier and $b$ is our bet. We don't value the chance of going deeper into the rounds.
- Do not see our card. We have an expected outcome of $\frac{1}{2}b$.
Now as long as $m > 1$ we should always choose option one. That is because
$p$ can never be below $\frac{1}{2}$ (which is easy to see by symmetry - if one choice is < 50% chance to win, the other must be > 50%).
Our average win% in a round with $n$ cards left is:
$$\frac{1}{n}\sum_{l=0}^{n-1}\frac{\max(l, n-1-l)}{n-1} = \frac{1}{n(n-1)}\sum_{l=0}^{n-1}\max(l, n-1-l) = \frac{6n^2-4n-1+(-1)^n}{8n(n-1)}$$
Which for even $n$ simplifies to $\dfrac{3n - 2}{4n - 4}$.
Which gives the following winning percentages for the rounds:
r n p
01 52 0.7549019607843137
02 50 0.7551020408163265
03 48 0.7553191489361702
04 46 0.7555555555555555
05 44 0.7558139534883721
06 42 0.7560975609756098
07 40 0.7564102564102564
08 38 0.7567567567567568
09 36 0.7571428571428571
10 34 0.7575757575757576
We can see that even going into the deep rounds has a negligible effect on our chance to win. Also note that ties only improve on this, a tie is as if the round never happened while slightly improving our chances.
Now, when is it worth to play this game? If you only play a single round and then cash out it's worth it when
$$m > \frac{4\times 52 - 4}{3\times 52 - 2} = 1.324675$$
The most optimal play is to always play the full 9 or 10 rounds and then cash out.
For $9$ rounds your winning chances are $0.08057$, which makes it worth it if:
$$m^9 > \frac{1}{0.08057}$$
$$m > 1.322927$$
TL;DR: advanced strategizing isn't really applicable here. Wait for a day where the multiplier is $\geq 1.35$ to make a noticeable profit. In fact, assuming you can play the game as fast as you want, don't even bother card counting, just repeatedly play only round 1 as the advantage gained from card counting is so minimal, and faster throughput of games makes more money.
Best Answer
Without the option, this is a fair game.
If the option is priced on the cheap side of fair, we buy the option and the option favors the player. If the option is rich, we don't buy the option.
Assuming the option is cheap, we plan to exercise if our first card is in $[1,7]$ and we won't exercise if the first card is in $[8,13]$ (number from 1-13 is simpler than translating face-cards to numerals)
Your chance of getting any particular number after exercising the option is:
$P(x) = \begin{cases} \frac{7}{169} &x\in [1-7]\\\frac{20}{169}&x\in [8-13] \end{cases}$
Your chance of winning is for any given $x$ is $\frac {x-1}{13}$ your chance of loosing is $\frac{13-x}{13}$ and you have a $\frac {1}{13}$ chance of a push.
If the option is free. $E[X] = 10 (\sum_\limits{i=1}^7 \frac {7(i-1)}{13^3} + \sum_\limits{i=8}^{13} \frac {20(i-1)}{13^3}) - 10(\sum_\limits{i=1}^7 \frac {7(13-i)}{13^3} + \sum_\limits{i=8}^{13} \frac {20(13-i)}{13^3}) = 10(\frac {7\cdot21+20\cdot57 - 7\cdot 63 - 20\cdot 15}{13^3}) = \frac{546}{13^3}$
Paying for the option, this is a fixed cost regardless of how the game comes out.
You should be willing to pay up to, but no more than $\frac{546}{13^3}\approx \$2.485$
Note that we have ignored the push. If the option is close to fair value the push can be safely ignored. But if the option were free, our expected gain would be greater by a factor of $\frac{13}{12}$