Problem Statement
So to be clear the problem I'm about to solve is... Both have \$10 in and can win the full \$20. Loser gets nothing. One player is offered the option to be shown a card and if he would like to, switch to that card. The player does not get to look at his own card before switching. How much should the player offered the option be willing to pay for the option?
The wording of the question is ambiguous. This is the way I read it. The solution does not contradict the others solutions, but solves a different and similar game to the ones they solved. That being said, this may actually be the game OP wanted.
Edit
Thinking now that the two games are symmetric. The final pricing should be the same no matter if you (1) view the new card and not your original one and (2) view the original card and not the new one.
The Base Case
In the base case (no pay, fair game), it is trivial to show probability of winning a round is $\frac{6}{13}$, probability of losing a round is $\frac{6}{13}$, and probability of a tie is $\frac{1}{13}$. Generally probability of winning if we replay when we tie is...
P(win after any number of replays)$=P_W + P_W \cdot P_T + \cdots = \frac{P_W}{1-P_T} $
in the base game this evaluates to... $\frac{6/13}{1-1/13}=\frac{1}{2}$.
The Fun Case
Optimal Strategy
Consider that you have already paid to switch and are shown a card. Should you switch to it?
If you are shown a 2, the probability that the 2 is higher than your card is 0.
If you are shown a 3, the probability that the card is higher than your card is $\frac{1}{13}$.
If you are shown an $i$, the probability that the card is higher than your card is $\frac{i-2}{13}$ for $i \in \left\{2,3,\ldots,10,11=J,12=Q,13=K,14=A \right\}$.
That makes your probability of getting a higher card than your card greater than $\frac{1}{2}$ if the card shown to you is a 9, 10, Jack, Queen, King, or Ace.
The optimal strategy is to switch if the card shown to you is a 9, 10, Jack, Queen, King, or Ace.
Probability of Winning if you Pay and Play Correctly
You'll switch if you see a 9 or higher. I'll write games as (a,b,c) exmaple (9,2,10) if your opponent is dealt a 9, you are dealt a 2, and the card you are shown is a 10. In this case where you paid you'll switch to the 10 and win.
Probability of winning with optimal strategy...
The games that you will win given you play right and paid are...
$\left( 2,3,2 \right), \ldots ,\left(2,A,2 \right),$
$\left( 2,3,3 \right), \ldots ,\left(2,A,3 \right), \ldots$
$\left( 2,3,8 \right), \ldots ,\left(2,A,8 \right),$
Above are the games that you win when the opponent draws a 2 and you don't see a card greater than 8 (you don't switch). There are $12 \cdot 7$ possibilities above.
$\left(2,2,9 \right) ,\ldots ,\left(2,A,9 \right),$
$\left(2,2,10 \right) ,\ldots ,\left(2,A,10 \right),\ldots$
$\left(2,2,A \right), \ldots ,\left(2,A,A \right),\ldots$
The second block contains the games that you win when the opponent draws a 2 and you see a card greater than or equal to 9 and switch. There are $13 \cdot 6$ possibilities above.
There are clearly more cases but now we are on to the pattern.
Generally there are $\left(14-i\right) \cdot 7$ ways that you win (if the opponent draws card $i$ where $i$ is in $\left\{2,3,\ldots,10,11=J,12=Q,13=K,14=A \right\}$) and you don't see a card better than 8 so you don't switch.
If the opponent draws an $i\in \left\{2,3,\ldots,8 \right\}$ and you see a card better than 8 and switch there are $13\cdot 6$ ways you win for each $i$.
If the opponent draws an $i \in \left\{9,10,11=J,12=Q,13=K,14=A \right\}$ and you see a card better than 8 and switch there are $13 \cdot \left(14-i \right)$ ways that you win.
Overall there are...
$\sum_{i=2}^{14} \left[ \left(14-i \right) \cdot 7 \right] + \sum_{i=2}^{8} \left[ 13\cdot 6 \right] + \sum_{i=9}^{14} \left[ 13 \cdot \left(14-i \right) \right]$
$=546 + 546 + 195=1287$
ways to win the game if you pay and play right. In the above notation there are $13^3=2197$ games. Your probability of a win if you pay and play right on the first round is...
$P(Action \, win)=\frac{1287}{2197}=\frac{99}{169}$
Probability of a Tie if you Pay and Play Correctly
You tie these games...
$\left( 2,2,2 \right), \ldots ,\left(2,2,8\right),$
$\left(3,3,2 \right), \ldots ,\left(3,3,8 \right),\ldots$
$\left(8,8,2 \right), \ldots ,\left(8,8,8 \right),$
There are $7 \cdot 7$ no-switch games above.
$\left( 9,2,9 \right) , \ldots , \left(9,A,9 \right),$
$\left( 10,2,10 \right) , \ldots , \left(10,A,10 \right),\ldots$
$\left( A,2,A \right) , \ldots , \left(A,A,A \right)$
There are another $13 \cdot 6$ switch games above.
Overall the probability that you tie if you pay and play correctly is $\frac{127}{2197}$
Probability of Loss
By law of total probability the probability of a loss is $1-\frac{127}{2197} - \frac{1287}{2197}=\frac{783}{2197}$
Probability of Generally Winning with Advantage: Ad only Applied to First Round
$P=\frac{1287}{2197}+\frac{127}{2197} \cdot \frac{6}{13} + \frac{127}{2197}\cdot \frac{1}{13} \cdot \frac{6}{13} + \cdots = \frac{2701}{4394}$
Probability of Generally Winning: Pay once, Ad Applied Every Round
$P=\frac{1287}{2197}+\frac{127}{2197} \cdot \frac{1287}{2197} + \left( \frac{127}{2197} \right)^2 \frac{1287}{2197} + \cdots = \frac{143}{230}$
Pricing the Advantage
If the advantage only applies to the first round...
$E[W_{Ad}]-E[W_0]=\frac{2701}{4394} \cdot \$20 + \left( 1-\frac{2701}{4394} \right) \$0 - \frac{1}{2} \cdot \$20 - \frac{1}{2} \cdot \$0 = \frac{5040}{2197} \approx \$ 2.29$
If the advantage applied every round...
$E[W_{Ad}]-E[W_0]=\frac{143}{230} \cdot \$20 + \left( 1-\frac{143}{230} \right) \$0 - \frac{1}{2} \cdot \$20 - \frac{1}{2} \cdot \$0 = \frac{56}{23} \approx \$ 2.43$
Closing Comment
The \$2.29 number doesn't change round to round, so if the option is offered again the next round or any round after a tie it should still be taken at \$2.29. If the \$2.43 were to be paid, it would offer the advantage perpetually. These numbers could be reached if you had one and not the other with some geometric properties.