Consecutive Wins Game
Alice and Bob play a points-based game. Suppose $0 < p < 1$. The game consists of up to $3$ points:
- On the first point, Alice wins the point with probability $p$ and Bob wins with probability $1 - p$.
- On each subsequent point, the winner of the previous point wins again with probability $p$, and the loser wins with probability $1 - p$.
A player wins the game as soon as they win two consecutive points. If all $3$ points are played and no player has won two in a row, the entire game is replayed from scratch under the same rules, repeating until someone wins.
Find the probability that Bob wins the overall game when $p = 1/3$.
Hints
- Think about what can happen in a single 3-point round. There are only a few distinct point sequences -- enumerate them and figure out which ones give Bob a win, which give Alice a win, and which result in a tie.
- Since a tie causes a full restart under identical conditions, you can set up a single equation $b = P(\text{Bob wins a round}) + P(\text{tie}) \cdot b$ and solve for $b$.
- Track the probabilities carefully: if Bob wins point 1 (probability $1-p$), he then wins point 2 with probability $p$ (the previous winner's advantage). The direct $BB$ path has probability $(1-p) \cdot p$, and the $ABB$ path has probability $p^2(1-p)$.
Worked Solution
How to Think About It: This is a momentum game -- whoever just won a point has probability $p$ of winning the next one, so winning streaks are self-reinforcing when $p > 1/2$ and fragile when $p < 1/2$. Since $p = 1/3$, the previous winner actually has a *disadvantage* on the next point. That means Bob, who starts as the underdog on point 1, benefits from the reversal mechanic. Intuitively his chance should be better than $1 - p = 2/3$ at winning the overall game -- maybe not by a lot, but meaningfully above $1/2$. The game can tie and restart, so we need a recursion.
Quick Estimate: Bob wins point 1 with probability $2/3$. If he does, he then has only a $1/3$ chance of winning point 2 (since he just won). So the direct $BB$ path has probability $(2/3)(1/3) = 2/9 \approx 0.22$. There is also the $ABB$ path where Alice wins point 1, Bob wins point 2 (prob $2/3$), then Bob wins point 3 (prob $1/3$): that is $(1/3)(2/3)(1/3) = 2/27 \approx 0.07$. So Bob wins in a single round with probability roughly $0.29$. The tie probability is $(1 - p)^2 = 4/9 \approx 0.44$, so about $44\%$ of the time the game restarts. Across restarts, Bob's win probability should be around $0.29 / (1 - 0.44) \approx 0.52$. So we expect something just above $1/2$.
Approach: Enumerate all outcomes of a 3-point round, set up a recursion for Bob's overall win probability $b$, and solve.
Formal Solution:
Let $b$ denote Bob's probability of winning the overall game. In a single 3-point round there are four possible point sequences:
- $BB$ (Bob, Bob): Probabilities $(1-p) \cdot p$. Bob wins two consecutive -- Bob wins the round.
- $ABB$ (Alice, Bob, Bob): Probabilities $p \cdot (1-p) \cdot p = p^2(1-p)$. Bob wins the round.
- $BAA$ (Bob, Alice, Alice): Probabilities $(1-p) \cdot (1-p) \cdot p = p(1-p)^2$. Alice wins the round.
- $AA$ (Alice, Alice): Probabilities $p \cdot p = p^2$. Alice wins the round.
- $BAB$ (Bob, Alice, Bob): Probabilities $(1-p)(1-p)(1-p) = (1-p)^3$. No consecutive winner -- tie.
- $ABA$ (Alice, Bob, Alice): Probabilities $p(1-p)(1-p) = p(1-p)^2$. No consecutive winner -- tie.
Probability Bob wins a round outright:
$$P(\text{Bob wins round}) = p(1-p) + p^2(1-p) = p(1-p)(1+p) = p(1-p^2)$$
Probability of a tie:
$$P(\text{tie}) = (1-p)^3 + p(1-p)^2 = (1-p)^2$$
By the law of total probability, conditioning on the first round:
$$b = p(1-p^2) + (1-p)^2 \cdot b$$
Solving for $b$:
$$b\bigl[1 - (1-p)^2\bigr] = p(1-p^2)$$
$$b = \frac{p(1-p^2)}{1 - (1-p)^2} = \frac{p(1-p)(1+p)}{p(2-p)} = \frac{(1-p)(1+p)}{2-p} = \frac{1-p^2}{2-p}$$
(Note we cancelled a factor of $p$, which is valid since $p > 0$.)
Plugging in $p = 1/3$:
$$b = \frac{1 - 1/9}{2 - 1/3} = \frac{8/9}{5/3} = \frac{8}{9} \cdot \frac{3}{5} = \frac{8}{15}$$
Answer: The probability that Bob wins is $\dfrac{8}{15} \approx 0.533$.
Intuition
The key structural insight is that the restart-on-tie mechanic turns this into a geometric trials problem. Each round is an independent attempt with three possible outcomes (Bob wins, Alice wins, tie), and the game just keeps going until one of the decisive outcomes occurs. So you can collapse the whole infinite-horizon game into a single ratio: Bob's win probability is just his per-round win probability divided by the probability that the round is decisive (i.e., not a tie). This pattern -- recursion that reduces to a ratio because restarts are memoryless -- shows up constantly in quant interviews and in practice whenever you model repeated strategic interactions.
The other subtlety worth noting is the "hot hand" mechanic where the previous winner has probability $p$ of winning the next point. When $p < 1/2$, this actually means momentum is anti-persistent: winning a point makes you *less* likely to win the next one. That is why Bob, despite being the underdog on the first point, ends up with a better-than-even chance overall. In real markets, mean-reverting dynamics produce a similar flavor -- the player (or strategy) that just "lost" is often better positioned for the next move.