Which game is more favourable? Two people $A$ and $B$ play a match consisting of a series of games. They have probability $p$ and $q$ respectively, where $p+q=1$. The match has two sets of rules that they can follow. I found that the probability of $A$ winning using one set of rules is $$A_1=\frac{p^2}{1-2pq}$$ and the probability of $A$ winning in the other set of rules is $$A_2=\frac{p^2(1+q)}{1-pq}$$ Let's say Person $A$ was not as good as Person $B$, so $p<q$, and has the option to choose the set of rules they'll both follow. How could I use the above probabilities to determine which set of rules is more favourable for Person $A$?

Since $p < \frac{1}{2}$, we have $2pq < q$. This leads to $1 - p - 2pq > 1 - p - q = 0$. And hence $q-pq-2pq^2 > 0$.

Hence $1+q-2pq-2pq^2>1-pq$ which shows $\frac{1+q}{1-pq} > \frac{1}{1-2pq}$ and the second set of rules is better.

Some additional notes: (you can ignore if it is not relevant any more)

The first formula is not the formula for the set of rules stating a player wins if he wins two consecutive games. (I read the original post as well.)

The logic seemed correct but one needs to consider the case where $A$ and $B$ alternates for not only $2$ games, but also $4$ games, $6$ games and so on.

So $A_1 = p^2 + (2pq+2p^2q^2+2p^3q^3+...)A_1$ actually and you need to use the geometric series formula to solve for it.

Edit again: In fact I was wrong, the formula should be $A_1 = p^2 + (2pq+2p^2q^2+2p^3q^3+...)p^2$ which actually coincide with your formula for $A_1$.