Balancing the marble game (Red marbles are OP, please nerf)

Question

Ash and Bree are playing a simple game of chance:

They fill a bag with small marbles coloured red or blue. They take turns to draw a marble from the bag without looking, then:

• If Ash draws a red marble, she wins the game
• If Bree draws a blue marble, she wins
• If either player draws the opposite colour, the marble is returned to the bag, and the next player takes their turn.

Ash is set to take her turn first, but Bree complains that gives her a disadvantage. In order to balance out this disadvantage, they decide to add extra blue marbles to the bag until the chance of either player winning is equal.

What ratio of blue marbles to red should they use for optimal fairness?

Answer 1

How can Ash win? Looking at the game tree:

• Ash can win immediately with probability $r = \frac{\text{number of red marbles}}{\text{total number of marbles}}$.
• If Ash didn't win immediately (which happens with probability $1-r$), Bree would need to lose (with probability $r$) for Ash to have a chance at winning again.
• Once Bree loses, the game is back to how it was at the beginning.

Formalising the above, we get the following relationship for $p_A$, the probability that Ash wins:

$$p_A = r + (1-r) \cdot r \cdot p_A$$

Ash and Bree desire for $p_A = \frac{1}{2}$. Substituting and rearranging, we get:

$$r^2-3r+1 = 0$$

which can be solved

using standard quadratic equation solving techniques, to get \begin{align} r&=\frac{3\pm\sqrt{5}}{2}\\&\approx 2.62 \text{ or } 0.382 \end{align} Recall that $r$ is a probability bounded between $0$ and $1$, so we keep only the solution $r=\frac{3-\sqrt{5}}{2}\approx 0.382$.

Now, to finish up:

Notice that \begin{align}\text{blue marbles} &: \text{red marbles} \\= \text{chance of drawing a blue} &: \text{chance of drawing a red}\end{align} We had defined the chance of drawing a red as $r$, and we know that the chance of drawing a blue marble is $1 - r$. So, the answer we want is $$\text{blue marbles} : \text{red marbles} = 1 - r : r$$

But, there's a catch:

$r$ is not a rational number, so there is no solution involving integer numbers of balls. Sadly for Ash and Bree, they are going to have to settle with approximate fairness.
Ash and Bree can find "best approximations" using the technique of continued fractions. By applying this (or by noticing the relation between $r$ and the golden ratio), we see that the best approximations are actually $$\text{blue marbles} : \text{red marbles} = F_{n+1} : F_n$$ where $F_n$ represents the $n$-th Fibonacci number.

Answer 2

As we note, after a round of both losing, everything is the way it was initially. Let's proclaim chance to draw a red ball is r, chance to draw a blue ball is b = 1-r.

Probability A wins in a round is r = R/(R+B).

B gets a chance when A does not win, therefore in 1-r tries. Probability B wins in his try is b = B/(R+B) = (R+B)/(R+B) - R/(R+B) = 1-r. Probability B wins in a round is therefore b^2 (or (1-r)^2 if you prefer that form)

So, we want equal chances of winning: r = (1-r)^2, which is easily solved using ordinary boring math (and only solution between 0 and 1 makes sense).

However, alternative form of equation r = b^2 is more illustrative. You can easily rearrange that formula into something that has been seen many times before (divide both sides with b): r/b = b/1 = b/(r+b). Therefore, these two numbers are in golden ratio. This result also makes some sort of intuitive sense, obviously a perfectly fair game will have something to do with golden (perfect) ratio.