I have been playing a lot of dice based games and re-rolling has been a major mechanic in a lot of them. knowing when to re-roll and just keep what is currently face up seemed like a basic stats problem.
Lets say I had a $6$ sided die each with a unique numeric side. There would then be a $1$ in $6$ chance I roll any one side. Lets also say I wanted to get the highest/best side possible. In the first roll I would have a $1/6$ chance with $5$ undesirable outcomes, and $1$ desirable.
If I re-rolled would the number of total undesirable outcome be as simple as $5^2$ and the total number of desirable out comes be $2$? Thus making my odds of succeeding ($1$ out of $6$ odds) + ($2$ out of $25$ odds)?
No. There are two ways to calculate this.
One way is to consider the events sequentially. In the first roll, you have a 1/6 chance of success (at which point you stop rolling) and a 5/6 chance of failure (in which case you roll again). The second roll has the same 1/6 chance of success. Letting $S_1$ and $S_2$ be the events of the first and second roll succeeding respectively, and $S$ be overall success. Then we have $$ P(S) = P(S_1) + P(\neg S_1)P(S_2) = \frac{1}{6} + \frac{5}{6}\frac{1}{6} = \frac{11}{36}. $$ For further rerolls, you could continue this process and get a geometric series.
The other is to consider the die rolling event as a whole. Let's say it's a standard d6 and 6 is the successful face. The setup appears to have one or two die rolls depending on whether the first roll is a 6, but it is equivalent for you to roll the die again if the first roll is a 6 and ignore the result. Therefore, this situation is equivalent to rolling the die twice and seeing if either roll is a 6. Of the 36 equally likely pairs of die rolls, 11 have at least one 6 in them. So the success probability is again 11/36.
11/36 is 83% bigger than 1/6, meaning you nearly double your odds by having a reroll. This has diminishing returns, though, and no matter how many rerolls you get, there's always a chance you could fail all of them.