High Low Card Game probability of winning There is an infinite deck of regular playing cards (2-10, JQKA, four suits).  The dealer deals face down one card to you and one card to me.  The player with the higher card dealt wins.
We each put in $10 as an initial bet.
Now before you look at your cards, you can pay $x to have the option of replacing your card with a new (random card) if you wish, after you view it.  
How much would you pay for this option?
What is the new probability that you win?
Edit; in the event you tie, the dealer throws out both of your cards and deals both of you a new card
Edit2 : all suits are the same.
 A: 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 $\frac{13}{12}$
A: If you are allowed one switch, your winning probability is $p(52) = \frac{5}{8}$, so you should be willing to pay up to 2 dollars for the right to switch.
$p(N) = \sum_{i=1}^{N} \frac{1}{N} \max(\frac{i-1}{N} + \frac{1}{2N}, \frac{1}{2}).$
so:
$p(2n) = \frac{5}{8}$
$p(2n+1) = \frac{5}{8} - \frac{1}{(2n+1)^2}$
(I assumed 10 dollars payoff and 52 values, I see that it might be 20 and 13, but the principle and formula should be the same)
For N=13 and a total pot of 20 you should be willing to bet at most $10 ( 2 - \frac{1}{p(13)}) = \frac{840}{211} = 3.981...$.
