Coin Swap II: Equal Winning Probability
Alice and Bob play a game with a biased coin that shows heads with probability $0 < p \leq 1$. Alice goes first. The rules are asymmetric:
- On Alice's turn, she flips the coin once. If she gets heads, she wins. If tails, she passes the coin to Bob.
- On Bob's turn, he flips the coin twice. If either flip is heads, he wins. If both are tails, he passes the coin back to Alice.
The game continues until someone flips heads.
Find the value of $p$ such that Alice and Bob are equally likely to win. Round to the nearest thousandth.
Hints
- Let $a = P(\text{Alice wins})$ and set up a recursion by conditioning on what happens in the first full cycle (Alice's flip and Bob's two flips).
- After one cycle where Alice fails (probability $1-p$) and Bob also fails (probability $(1-p)^2$), the game resets to the same state. This gives $a = p + (1-p)^3 a$.
- Solve for $a$ to get $a = 1/(p^2 - 3p + 3)$, then set $a = 1/2$ and solve the resulting quadratic $p^2 - 3p + 1 = 0$. Discard the root greater than 1.
Worked Solution
How to Think About It: This is a geometric-series-style recursion. Alice wins on her turn with probability $p$. Bob fails both his flips with probability $(1-p)^2$, returning the coin to Alice in the same state as the start. So the game has a recursive structure: after one full cycle (Alice fails, Bob fails), we are right back where we started.
Quick Estimate: At $p = 0.5$, Alice wins on her first flip with probability 0.5, which already exceeds half -- so Alice wins more than half the time when $p = 0.5$. We need $p$ lower so Bob gets more chances. At very small $p$, Alice rarely wins on turn one and Bob rarely wins on either flip, leading to many cycles. By symmetry, neither player has a strong advantage at very small $p$ -- but Alice's single flip vs. Bob's two flips suggests Bob gains advantage as $p$ grows from 0. The fair $p$ is somewhere below 0.5, maybe around 0.4.
Formal Solution:
Let $a = P(\text{Alice wins})$. Condition on the first round:
- With probability $p$: Alice flips heads on her first flip, she wins immediately.
- With probability $(1-p)$: Alice flips tails, coin goes to Bob.
- Bob flips twice. He wins (at least one heads) with probability $1-(1-p)^2$.
- Bob fails (both tails) with probability $(1-p)^2$, coin returns to Alice.
So: $$a = p + (1-p)(1-p)^2 \cdot a = p + (1-p)^3 a$$
Solving: $$a - (1-p)^3 a = p$$ $$a(1 - (1-p)^3) = p$$ $$a = \frac{p}{1-(1-p)^3}$$
Expand the denominator using the factorization $1 - q^3 = (1-q)(1+q+q^2)$ with $q = 1-p$: $$1-(1-p)^3 = p(1 + (1-p) + (1-p)^2) = p(3 - 3p + p^2)$$
So: $$a = \frac{p}{p(3-3p+p^2)} = \frac{1}{3-3p+p^2} = \frac{1}{p^2-3p+3}$$
For $a = 1/2$: $$\frac{1}{p^2-3p+3} = \frac{1}{2} \implies p^2 - 3p + 3 = 2 \implies p^2 - 3p + 1 = 0$$
By the quadratic formula: $$p = \frac{3 \pm \sqrt{5}}{2}$$
Since $p \leq 1$, we discard $\frac{3+\sqrt{5}}{2} \approx 2.618$.
$$\boxed{p = \frac{3-\sqrt{5}}{2} \approx 0.382}$$
Answer: $p = \dfrac{3-\sqrt{5}}{2} \approx 0.382$
Intuition
The recursion trick here -- conditioning on one full cycle and noting the game "resets" -- is one of the most powerful patterns in probability. It converts an infinite geometric series into a one-line equation. You see the same structure in gambler's ruin, the secretary problem, and virtually every optimal stopping problem.
The answer $p \approx 0.382$ is the reciprocal of the golden ratio $\phi = (1+\sqrt{5})/2$, since $3-\sqrt{5} = 2(\phi^{-1})$. This is a beautiful coincidence that shows up in several combinatorial probability problems. The practical takeaway: Alice's advantage from going first is exactly canceled by Bob's advantage of getting two flips per turn when $p \approx 38.2\%$. At lower $p$, Bob's two-flip advantage dominates (more cycling, more chances for Bob); at higher $p$, Alice's first-mover advantage dominates.