Ratio of turns for fair game between two Bernoulli players This is a self-written problem, which may or may not have a known solution, and may or may not be well-posed.
Setup: Consider a two-player game with discrete turns, where the game can be handicapped by varying the number of turns a player receives. Player one has a probability of $p_1$ of scoring 1 point on a turn.  Player two has a probability of $p_2$ of scoring 1 point on their turn.
Question: What is the ratio of turns each player should get, such that the game is fair (the expected difference in scores is zero)? To standardize the answers, use the convention $p_1 \geq p_2$.
Notes: Consider the player's total scores to be binomially distributed, e.g., given $n_1$ turns, player one's score is $X_1 \sim B(n_1, p_1).$

Attempted solution:

 I've looked at the probability of a win, loss, and a tie from player one's perspective (here $f(k, n, p)$ is the probability mass function of the Bernoulli distribution for k successes, over n trails, with a success probability of p)
 
$$\begin{align}P(\textrm{win}) &= \sum_{k_2=0}^{n_2} \left[f(k_2, n_2, p_2) \left(1-\sum_{k_1=0}^{\min(k_2,n_1)} f(k_1, n_1, p_1)\right)\right],\\&= \sum_{k_2=0}^{n_2} \left[f(k_2, n_2, p_2) \left(\sum_{k_1=k_2+1}^{n_1} f(k_1, n_1, p_1)\right)\right].\end{align}$$
$$\begin{align}P(\textrm{loss}) &= \sum_{k_2=0}^{n_2} \left[f(k_2, n_2, p_2) \left(\sum_{k_1=0}^{\min(k_2-1,n_1)} f(k_1, n_1, p_1)\right)\right],\\&= \sum_{k_2=0}^{n_2} \left[f(k_2, n_2, p_2) \left(1-\sum_{k_1=k_2}^{n_1} f(k_1, n_1, p_1)\right)\right].\end{align}$$
$$P(\textrm{tie}) = \sum_{k=0}^{\min(n_1,n_2)} f(k, n_2, p_2)\, f(k, n_1, p_1).$$
 These are just summing over all possible outcomes for player two, multiplied by the probability that player one has a higher, lower, or equal score respectively. Note that since $p_1 \geq p_2$ then it must be that $n_1 \leq n_2$ to achieve a fair game. Thus, on the second sums we must be careful to only sum the minimum of $k_2$ (or $k_2-1$) and $n_1$. Similarly, by convention, on the other sums, when the lower bound exceeds the upper bounds we consider the sum to be zero, i.e., when $k_2$ (or $k_2+1$) is greater than $n_1$.
 
The constraint we want is $P(\textrm{win}) = P(\textrm{loss})$, such that I tried setting these two probabilities equal to each other, but made no substantial progress.
 
 I also considered creating a new random variable $Z = X_1-X_2$, where $X_1$ and $X_2$ are the player's scores. One could then tune the ratio such that $\mathbb{E}(Z) = 0$. This seems much more promising and natural, but I did not find much useful in the way of subtracting binomial distributions when $p_1 \neq p_2$.

 A: The expected score for Player $i \in \{1, 2\}$ is simply $\operatorname{E}[X_i] = n_i p_i$, where $n_i$ is the number of trials for player $i$, and $p_i$ is the probability of receiving a point for any given trial.  So the requirement for the game to be fair in the sense that their expected number of points are equal, is $$n_1 p_1 = n_2 p_2,$$ or $$\frac{n_1}{n_2} = \frac{p_2}{p_1}.$$  In theory, this may not be possible; e.g., if $p_2/p_1 \ne \mathbb Q$, then no choice of integers $n_1, n_2$ will yield the desired ratio; however, we can make the game arbitrarily close to being fair as we please, given enough trials.
A: There may be a simplifying approach, depending on the scoring system used.
Assume that groups of turns are taken in rounds, where in round 1, first player 1 takes $T_1$ turns, with the chance of success on each turn equal to $p_1$.
Assume that in round 1, following the $T_1$ turns taken by player 1, player 2 takes $T_2$ turns, with the chance of success on each turn equal to $p_2$.
By Linearity of Expectation the expected scores of player 1 and player 2 at the end of round 1 are $(p_1 \times T_1)$ and $(p_2 \times T_2)$, respectively.
So, let $\frac{T_2}{T_1} = \frac{p_1}{p_2}.$
So, after $n$ rounds, the expected scoring of players 1 and 2 is $(n \times p_1 \times T_1)$ and $(n \times p_2 \times T_2)$, respectively.

However, that isn't the end of the analysis.
If the scoring system is simply total number of points scored, so that at the end of $n$ rounds, player 2 gives $1\$$ to  player 1 for each point that player 1 scored, and then player 1 gives $1\$$ to player 2 for each point that player 2 scored, then the game is fair.
However, suppose instead that only 1 round is played, and the rule is that the player that scored the most points in that round gets $1\$$ from the other player, for that round.
This would be similar to changing the rules of golf from stroke play to match play, where the person who is expected to have the fewest strokes is not necessarily the person who is expected to win the most holes.
Then unfortunately, as the OP's work indicated, you would have to look at the binomial distribution of points scored, for each player, in the round.
That is, you would have to examine $(T_1 + 1) \times (T_2 + 1)$ events, where the number of wins for player 1 was some element $x \in \{0,1,\cdots, T_1\}$ and the number of wins for player 2 was some element $y \in \{0,1,\cdots, T_2\}$.
Under this scoring system, the number of rounds played is irrelevant, because it is necessary that the chance that $x$ is greater than $y$ for a specific round must equal the chance that $y$ is greater than $x$ for that specific round.

So, under this more complicated scoring system, with $p_1, p_2$ fixed, what is the formula for deriving $T_1, T_2$ so that the (match play oriented) game is fair?
With $x \in \{0,1,\cdots, T_1\}$, the probability that player 1 scored exactly $x$ points is
$$A_x = \binom{T_1}{x} \times (p_1)^x (1 - p_1)^{T_1 - x}.\tag1 $$
Similarly, with $y \in \{0,1,\cdots, T_2\}$, the probability that player 2 scored exactly $y$ points is
$$B_y = \binom{T_2}{y} \times (p_2)^y (1 - p_2)^{T_2 - y}.\tag2 $$
For ease of syntax, assume that

*

*$A_x = 0 ~: ~x > T_1.$

*$B_y = 0 ~: ~y > T_2.$

*$M = \max(T_1, T_2)$.

Then, the chance of an overall win in the round by player 1 is
$$\sum_{x = 0}^{M} \left[A_x \times \sum_{0 \leq y < x} B_y \right]. \tag3 $$
The chance of a tie in the round is
$$\sum_{x = 0}^{M} \left[A_x \times B_x\right]. \tag4 $$
Similarly, the chance of player 2 winning the round is
$$\sum_{x = 0}^{M} \left[A_x \times \sum_{x < y \leq M} B_y \right]. \tag5 $$
Assume that $p_1 > p_2$.  This implies that $T_2$ must be $\geq T_1$.  Therefore, you can set $M = T_2$.
Then, in order to make the game as fair as possible, the expression in (4) above should be ignored, and $T_1, T_2$ must be derived, with $M = T_2 \geq T_1$ so that the expression in (3) above is approximately equal to the expression in (5) above.
