Shooting sequence in shuffled Russian Roulette to remove advantage of going first? Alice and Bob are playing shuffled Russian Roulette. Starting with Alice, each player takes a turn to throw a dice. The first person to land a 1 (with probability $p=\frac{1}{6}$) is the winner.
But since Alice goes first, she has an advantage. It's not hard to prove her probability of winning is $\frac{6}{11}\approx 0.55$.
To combat this (and try to make it fair 50-50 odds), we adjust the sequence of turns.
Originally, it was ABABABA$\cdots$ (alternating turns).
We might think of changing it to ABBABABABA$\cdots$ (Bob gets 2 turns initially, then it goes back to alternating), but this turns out to give odds in favor of Bob.
My question is: is there a sequence of turns (with Alice's turn first) such that Bob and Alice will have equal odds of winning? Preferably, your answer should be for a generic $p$, not just $1/6$.
 A: I've made a simple python program to calculate a sequence of turns. The idea is that $a$ and $b$ are the probabilities of Alice or Bob having won so far. $a$ is initialized to $p$ and $b$ is initialized to 0. At each step, whoever has the lower probability is given the turn. If $c$ is given the turn, their probability is updated as $c = c + p\cdot(1 - a - b)$.
    from fractions import Fraction as F # use exact Fraction class because 32-bit floats are too imprecise
    def move(a, b, p):
      if a < b:
        a += p*(1-(a+b))
        return "A", (a, b)
      else:
        b += p*(1-(a+b))
        return "B", (a, b)
    
    n = 500 # number of turns to compute
    p = F(1, 6) # 1/6
    turns = ["A"]
    a = p; b = 0
    for _ in range(n):
      t, (a, b) = move(a, b, p)
      turns.append(t)
    
    print("".join(turns))

At the end is the turn order for the first 500 turns. Any insight about this turn order? I've found a way to compute it, but I don't really have a sense of if there's some deeper pattern emerging here.
Turn order for $p=1/6$ (first 500 turns):
ABBABAABBAABBAABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABBAABBAABABBABAABABBAABBAABBAABBABAABABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABABBAABBAABBABAABABBAABBABAABABBABAABABBAABBAABBAABBABAABBAABABBAABBABAABBAABABBAABBABAABABBABAABABBABAABABBABAABBAABABBABAABABBABAABBAABABBABAABABBABAABBAABABBAABBABAABABBABAABBAABABBABAABBAABBAABABBABAABBAABBAABBAABABBAABBAABBAABBAABABBAABBAABBAABBABAABBAABABBAABBAABBAABBAABABBABAABBAABBAABABBABAABABBAABBAABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABBAABABBABAA
