I think that the solution is
You are $36$ and your daughter is $2$ (she will turn $3$ at her next birthday).
Strategy
If the player picks your age, roll the dice.
If the player picks your daughter's age, choose the cards.
In both cases you are, at least, twice as likely to win as your opponent which means that your earnings will be non-negative in the long run (the expectation value for your earnings is non-negative).
Reasoning
Using the dice, your age occurs with probability $\frac{1}{36}$
and the daughter's age occurs with probability $\frac{1}{18}$ (in both cases).
Using the deck of cards the parent's age occurs with probability $4 \times \frac{1}{13} \times \frac{4}{51} + \frac{1}{13} \times \frac{3}{51} = \frac{19}{663}$.
while the daughter's age occurs with probability $\frac{2}{13} \times \frac{4}{51} = \frac{8}{663}$ (in both cases).
Why I chose these numbers
Using just the dice, we can list all the possible products. Given that the daughter's age must be the first in two consecutive entries in this list, this leaves the possibilities for the daughter's current age as being either $1,2,3,4,5,8,9,15$ or $24$.
For all of these values, at least one of the two consecutive ages will arise as the product of the two dice in at least two of the $36$ possible outcomes. This means, to enact a simple strategy where the player's choice directly informs the parent's choice, the parent's age must occur with probability $ \leq \frac{1}{36}$ or $\geq \frac{1}{9}$. This means the parent's age must be one of $1,6,9,12,16,25,36$.
When analysing the cards, this probability reasoning must flip. Heuristically, we can get an idea for what happens by analysing the number of products when choosing two numbers from the range $1$ to $13$ and multiplying (the cards probability is subtly different but this will still give a good idea of how to proceed).
In this instance, the number of products is essentially unchanged for the values $1,2,3,4,5,6,8,9,12,15,25$ while the number of products increases for the values $16, 24$ and $36$. Hence, one of the ages must be from this group of three.
If the parent's age is $16$, this needs to be half as likely as the daughters age in the dice case but, in the case of the cards the probability of $16$ is $2 \times \frac{1}{13} \times \frac{4}{51} + \frac{1}{13} \times \frac{3}{51} = \frac{11}{663}$ while all the possibilities for the daughter's ages contain at least one value with probability $\frac{8}{663}$ so we cannot guarantee a win here.
If the daughter's age is $24$, and the parent is older than the daughter then the daughter's age is more likely than the parent's age for both the dice and the cards and the player will always pick it.
Hence the parent's age must be $36$. The daughter's age must then be, at least, twice as likely as the parent's age for the dice, in both years. This restricts the possibilities for the daughter's current age to be $2,3,4$ or $5$.
Returning to the cards, the probability for getting $36$ is at least twice as likely as getting a $2, 3$ or a $5$ but less than twice as likely as getting a $4$ or a $6$. Hence, to make it work for both years, the daughter must be currently $2$ (turning $3$ at the next birthday).
