Correcting A Raffle Problem An interesting scenario that I'm trying to wrap my mind around.
Let's say you were running a giveaway, and have a probability of 10% for every participant to potentially win a prize. The way you would choose the participant would be using an RNG for each one (1,10) and if the output was 1, they win, otherwise, they don't. Technically not a 10% probability exactly, since the generation is random.
However, it turns out that by error, the raffle was actually conducted using a 7% chance instead.
If re-running the raffle was not an option, but you had to run a second one with the remaining supply of prizes so that you can help rectify the mistake, and restore balance to the force -- would you re-run it again with a 3% chance for each participant of winning (10% - 7%)?

My reading of your question is that the win event for each player is independent, so that more than one prize can be won.  Under this interpretation we can consider the case for a single player without further complication.  If my interpretation is not what you meant in your question then things could get more complicated.
Suppose the desired win-probability was $\pi$ but you accidentally used a lower value $\pi_* < \pi$.  The player wins the game if they either win on the first round (with the lower than intended win-probability) or if they lose on the first round but win on the second (your "make-up" round).  Consequently, you would need to set the new probability $p$ to attain the overall win-probability:
$$\pi = \pi_* + (1-\pi_*) p,$$
which gives:
$$p = \frac{\pi-\pi_*}{1-\pi_*}.$$
In the present case you stipulate that $\pi = 0.10$ and $\pi_* = 0.07$ so you would require the new win-probability to be $p = 0.03/0.90 = 1/30$.
