Explanation:
Okay, so Alice and Bob are playing this game where they take turns flipping coins. Alice goes first. If she gets heads, she wins. If not, it's Bob's turn. Then Bob flips; if he gets tails, he wins. If not, it goes back to Alice. The question is, what's the probability that Alice wins the game?

Hmm, let me think. So, the game can end on the first turn if Alice flips heads. If she doesn't, then Bob gets a chance. If Bob flips tails, he wins. If he doesn't, then it's back to Alice. So, it's like a sequence of turns where each player has a chance to win on their turn, and if they don't, the other player gets a chance. So, the game could, in theory, go on forever, but the probability of that happening is zero, right? Because each time there's a chance to end the game.

So, maybe I can model this as an infinite series? Let me try to break it down.

First, the probability that Alice wins on her first turn is the probability she flips heads. Since it's a fair coin, that's 1/2.

If she doesn't get heads, which is probability 1/2, then it's Bob's turn. Now, Bob has a chance to win by flipping tails. The probability he flips tails is 1/2. If he does, then Bob wins, so Alice doesn't. If he doesn't flip tails, which is probability 1/2, then it goes back to Alice. So, the game is back to the starting situation, but now it's Alice's turn again. Wait, but the game is symmetric? Or is there a difference?

Wait, actually, the game is structured so that Alice starts, then Bob, then Alice, etc. So, if the game gets back to Alice's turn, it's the same as the original setup, right? So, maybe we can model this recursively.

Let me denote the probability that Alice wins as P. So, P is the probability that Alice wins starting from her turn.

So, when Alice starts, she has a 1/2 chance to win immediately. If she doesn't (probability 1/2), then Bob gets to flip. Now, Bob has a 1/2 chance to win, and 1/2 chance to pass the turn back to Alice. So, if Bob doesn't win, then the game is back to Alice's turn, and the probability of Alice winning from that point is again P. Wait, is that right?

Wait, if the game gets back to Alice's turn, then the situation is the same as the original, so the probability is P. Therefore, we can write an equation for P.

So, P = probability Alice wins on first turn + probability game continues to Bob's turn * probability Alice wins from there.

But when the game continues to Bob's turn, what is the probability that Alice wins? Let me think. If it's Bob's turn, then Bob can win with probability 1/2, and if he doesn't, then it's Alice's turn again. So, if we let Q be the probability that Alice wins when it's Bob's turn, then maybe?

Alternatively, maybe we can express P in terms of itself. Let me try again.

Starting from Alice's turn:

P = P(Alice wins on first turn) + P(Alice doesn't win on first turn) * P(Bob doesn't win on his turn) * P(Alice wins from there)

Wait, so breaking it down:

First, Alice flips. She can win with probability 1/2. If she doesn't (prob 1/2), then Bob flips. Bob can win with probability 1/2, or not (prob 1/2). If Bob doesn't win, then it's back to Alice's turn, and the probability of Alice winning from there is again P. Therefore, the total probability P is:

P = (1/2) + (1/2) * [ (1/2) * 0 + (1/2) * P ]

Wait, because if Alice doesn't win (prob 1/2), then Bob has a 1/2 chance to win (which would mean Alice loses, so 0 for Alice's winning probability) and 1/2 chance to pass it back to Alice, which would give Alice a P chance. Therefore:

P = 1/2 + (1/2) * [ (1/2)*0 + (1/2)*P ]

Simplify that:

P = 1/2 + (1/2)*(0 + (1/2)P) = 1/2 + (1/2)*(1/2 P) = 1/2 + (1/4)P

So, then we can solve for P:

P - (1/4)P = 1/2

(3/4)P = 1/2

Multiply both sides by 4/3:

P = (1/2) * (4/3) = 2/3

So, the probability that Alice wins is 2/3?

Wait, that seems straightforward, but let me check if I did that correctly.

Alternatively, maybe I can think of it as an infinite series. Let's see.

The probability that Alice wins can be broken down into the probability she wins on her first turn, plus the probability the game goes to Bob's turn and he doesn't win, then Alice wins on her second turn, and so on.

So, first term: Alice wins on first turn: 1/2.

Second term: Alice fails (1/2), Bob fails (1/2), then Alice wins on her second turn: (1/2)*(1/2)*(1/2) = 1/8.

Third term: Alice fails, Bob fails, Alice fails, Bob fails, Alice wins: (1/2)^4 * 1/2 = (1/2)^5 = 1/32?

Wait, let me check:

Wait, for Alice to win on her second turn, the game must have gone through two cycles: Alice fails, Bob fails, then Alice succeeds. So that's (1/2)^2 * 1/2 = (1/2)^3 = 1/8. Similarly, for her third turn, it would be (1/2)^4 * 1/2 = (1/2)^5, etc.

So, the total probability is the sum over n=1 to infinity of (1/2)^{2n - 1} ?

Wait, let me think again. For the first term, n=1: (1/2)^{2*1 - 1} = (1/2)^1 = 1/2. Then n=2: (1/2)^{3} = 1/8, n=3: (1/2)^5 = 1/32, etc. So the series is 1/2 + 1/8 + 1/32 + ... which is a geometric series with first term a = 1/2 and common ratio r = 1/4.

Because 1/8 divided by 1/2 is 1/4, and 1/32 divided by 1/8 is 1/4, etc. So the sum is a / (1 - r) = (1/2) / (1 - 1/4) = (1/2) / (3/4) = (1/2) * (4/3) = 2/3. So same result as before. So that seems to confirm that the probability is 2/3.

Alternatively, maybe I can think of it as states. Let me define P as the probability that Alice wins starting from her turn, and Q as the probability that Alice wins starting from Bob's turn.

So, when it's Alice's turn, she can win with probability 1/2, or pass to Bob's turn with probability 1/2. Therefore:

P = 1/2 + (1/2) * Q

Similarly, when it's Bob's turn, he can win with probability 1/2 (so Alice loses, so 0) or pass back to Alice's turn with probability 1/2. Therefore:

Q = (1/2)*0 + (1/2)*P = (1/2)P

So, substituting Q into the first equation:

P = 1/2 + (1/2)*(1/2 P) = 1/2 + (1/4)P

Which is the same equation as before. Then solving:

P - (1/4)P = 1/2 => (3/4)P = 1/2 => P = 2/3. Same answer.

So, seems like 2/3 is the answer. But let me check with another approach.

Alternatively, think of the game as a sequence of rounds. Each round consists of Alice's turn and Bob's turn. But actually, the game might end on Alice's turn or Bob's turn. So, maybe not exactly rounds. But let's see.

The probability that Alice wins on the first round (her first turn) is 1/2. The probability that the game continues to Bob's turn and he doesn't win is (1/2)*(1/2) = 1/4. Then, the probability that Alice wins on the second round (her second turn) is (1/4)*(1/2) = 1/8. Similarly, the probability that the game continues to the third round is (1/4)^2, and so on.

Therefore, the total probability is the sum over n=1 to infinity of (probability game reaches Alice's nth turn) * (probability she wins on that turn). The probability that the game reaches Alice's nth turn is (probability that both Alice and Bob failed in the previous n-1 rounds). Each round, the probability that both fail is (1/2)*(1/2) = 1/4. Therefore, the probability that the game reaches Alice's nth turn is (1/4)^{n-1}. Then, the probability she wins on that turn is 1/2. Therefore, the total probability is sum_{n=1}^\infty (1/4)^{n-1} * 1/2.

Which is again a geometric series with first term a = 1/2 and ratio r = 1/4. So sum is (1/2) / (1 - 1/4) = 2/3. Same answer again. So that seems to confirm.

Alternatively, maybe think of the game as a Markov chain with states: Alice's turn, Bob's turn, Alice wins, Bob wins. But that might be overcomplicating.

Alternatively, think recursively. Let me try again.

Let me denote P as the probability that Alice wins. So, when Alice starts, she has a 1/2 chance to win immediately. If she doesn't (prob 1/2), then Bob gets to flip. Now, Bob has a 1/2 chance to win, which would mean Alice loses, and 1/2 chance to not win, which brings us back to Alice's turn. Therefore, the probability that Alice wins is:

P = 1/2 + (1/2) * [ (1/2)*0 + (1/2)*P ]

Which is the same equation as before. So, solving gives P = 2/3.

Alternatively, maybe think of the game as Alice and Bob taking turns with probabilities. Since Alice goes first, she has a higher chance. Intuitively, since she has the first chance to win, her probability should be higher than 1/2, which 2/3 is. So that makes sense.

Alternatively, maybe think of the game as a sequence of trials where each trial consists of Alice and Bob each flipping once. But actually, the game might end before both flip. For example, if Alice flips heads on her first turn, the game ends. So, the trials aren't necessarily both flipping.

But maybe we can model it as such. Let me try.

The probability that Alice wins on the first trial (her first flip) is 1/2. The probability that the game continues to the second trial (both Alice and Bob failed their first flips) is (1/2)*(1/2) = 1/4. Then, in the second trial, Alice has another chance to flip, etc. So, the probability that Alice wins is the sum over all trials of the probability that the game reaches trial n and Alice wins on that trial.

In trial 1: probability Alice wins is 1/2.

In trial 2: probability game reaches trial 2 is 1/4, and then Alice wins with probability 1/2, so total 1/4 * 1/2 = 1/8.

In trial 3: probability game reaches trial 3 is (1/4)^2, and then Alice wins with 1/2, so (1/4)^2 * 1/2 = 1/32.

So, same as before, sum is 1/2 + 1/8 + 1/32 + ... = 2/3.

Alternatively, maybe think of the game as Alice having a 1/2 chance to win, and if not, Bob has a 1/2 chance to win, and if not, it's back to Alice. So, the probability that Alice wins is the probability that she wins on her first turn, plus the probability that both fail and then Alice wins from there. So:

P = 1/2 + (1/2 * 1/2) * P

Which is P = 1/2 + (1/4)P => P - (1/4)P = 1/2 => (3/4)P = 1/2 => P = 2/3. Same result.

So, all approaches lead to 2/3. Therefore, I think that is the correct answer.

But just to make sure, let me think of another way. Let me imagine that the game is played many times, and compute the expected number of times Alice wins or something. Wait, maybe not necessary. Alternatively, think of the probability that Alice wins as P, and Bob's probability as Q. Since the game must end eventually, P + Q = 1. Also, since the game is symmetric except for who starts, maybe we can find relations between P and Q.

But let me try. When Alice starts, she has a 1/2 chance to win, and 1/2 chance to pass to Bob. If it's Bob's turn, he has a 1/2 chance to win, and 1/2 chance to pass back to Alice. Therefore, the probability that Alice wins is:

P = 1/2 + (1/2) * [ (1/2)*0 + (1/2)*P ] = same as before.

Alternatively, if we think of Q as the probability that Bob wins when it's his turn. Wait, maybe not necessary. But since P + Q = 1, if we can find another equation.

Alternatively, when it's Bob's turn, the probability that Bob wins is Q. So, when it's Bob's turn, he can win with probability 1/2, or pass to Alice's turn with probability 1/2. Therefore, Q = 1/2 + (1/2)*P? Wait, no. Wait, if it's Bob's turn, then the probability that Bob wins is 1/2 (he wins immediately) plus the probability that he doesn't win (1/2) times the probability that Bob wins from Alice's turn. Wait, but if it goes back to Alice's turn, then the probability that Bob wins is the probability that Alice doesn't win, which is 1 - P. Therefore, Q = 1/2 + (1/2)*(1 - P). But since when it's Bob's turn, the probability that Bob wins is Q, and the probability that Alice wins is 1 - Q. Wait, maybe this is getting too convoluted.

Alternatively, since when it's Bob's turn, the probability that Alice wins is Q, and the probability that Bob wins is 1 - Q. Wait, maybe not. Let me try again.

Let me define P as the probability that Alice wins starting from her turn, and Q as the probability that Alice wins starting from Bob's turn. Then, as before:

P = 1/2 + (1/2)*Q

Q = (1/2)*0 + (1/2)*P = (1/2)P

Therefore, substituting Q into the first equation:

P = 1/2 + (1/2)*(1/2 P) = 1/2 + (1/4)P => P = 2/3. Same result.

Alternatively, if I think of Q as the probability that Alice wins when it's Bob's turn, then Q = probability that Bob doesn't win * probability Alice wins from her turn. Which is (1/2)*P. So, same thing.

Alternatively, if I think of the game as a whole, since Alice has the first chance, her probability is higher. If the game was symmetric, like both players have the same chance to win on their turn, but Alice goes first, then her probability would be higher. For example, if both players had a 1/2 chance to win on their turn, then the probability that Alice wins would be 1/2 + 1/2 * 1/2 * 1/2 + ... which is same as the previous series. So, yeah, 2/3.

Alternatively, think of it as a geometric distribution. The game ends when either Alice gets heads or Bob gets tails. The probability that Alice wins is the probability that the first success (either Alice's heads or Bob's tails) is Alice's heads.

But how to model that? Let me think. Each "round" consists of Alice flipping and then Bob flipping. But the game could end in the middle of a round.

Alternatively, think of the game as a sequence of independent trials where each trial is Alice flipping, and if she fails, Bob flips. So, each trial has a probability of Alice winning, Bob winning, or the game continuing.

So, in each trial:

- Alice flips: probability 1/2 she wins, 1/2 continues.

- If continues, Bob flips: probability 1/2 he wins, 1/2 continues.

Therefore, the probability that Alice wins in the first trial is 1/2.

The probability that the game continues to the next trial is 1/2 * 1/2 = 1/4.

Then, the probability that Alice wins in the second trial is (1/4) * 1/2 = 1/8.

Similarly, third trial: (1/4)^2 * 1/2, etc.

So, same as before, sum is 1/2 + 1/8 + 1/32 + ... = 2/3.

Alternatively, think of the probability that Alice wins as the sum over all n ≥ 1 of the probability that the game lasts for n-1 full rounds and then Alice wins on the nth round.

Each full round (Alice and Bob both flipping) has probability 1/4 of continuing. So, the probability that the game lasts for n-1 rounds is (1/4)^{n-1}, and then Alice wins on the nth round with probability 1/2. Therefore, the total probability is sum_{n=1}^\infty (1/4)^{n-1} * 1/2 = 2/3.

Alternatively, think of the game as a sequence of independent events where each "cycle" consists of Alice and Bob each flipping once. The probability that Alice wins in a cycle is 1/2 (Alice gets heads). The probability that Bob wins in a cycle is 1/2 * 1/2 = 1/4 (Alice fails, Bob gets tails). The probability that the cycle continues is 1/4 (both fail). Therefore, the probability that Alice wins is the probability that she wins in the first cycle, plus the probability the cycle continues times the probability she wins in subsequent cycles. So:

P = 1/2 + 1/4 * P

Solving:

P - 1/4 P = 1/2 => 3/4 P = 1/2 => P = 2/3. Same answer again.

So, all these different approaches lead to the same answer, 2/3. Therefore, I think that is the correct probability.

But just to make sure, let me think of a concrete example with small probabilities.

Suppose the game is played, and we simulate the possibilities.

First, Alice flips:

- 50% chance she gets heads (wins).

- 50% chance she gets tails, then Bob flips:

   - 50% chance Bob gets tails (he wins).

   - 50% chance Bob gets heads, then it's Alice's turn again.

So, the probability that Alice wins is 1/2 + (1/2 * 1/2 * P) = 1/2 + 1/4 P. Which is the same equation as before. So, solving gives P = 2/3.

Alternatively, if I think of the game as a tree:

First level: Alice flips H (prob 1/2) -> Alice wins.

First level: Alice flips T (prob 1/2) -> Bob flips T (prob 1/2) -> Bob wins.

First level: Alice flips T (prob 1/2) -> Bob flips H (prob 1/2) -> back to Alice.

So, the probability that Alice wins is 1/2 + (1/2 * 1/2 * P) = 1/2 + 1/4 P. So, same equation.

Alternatively, if I think of the game as a recursive function. Let me define P as the probability Alice wins. Then:

P = probability Alice wins on first turn + probability game continues * P

But the probability game continues is the probability Alice fails and Bob fails, which is 1/2 * 1/2 = 1/4. Wait, but earlier we had P = 1/2 + 1/4 P. Wait, but that would be if the game continues with probability 1/4, but actually, when the game continues, it's not just the probability of continuing, but the probability that Alice wins from that point. Wait, but in this case, if the game continues, it's back to Alice's turn, so the probability is P again. Therefore, P = 1/2 + (probability game continues) * P. Wait, but the probability game continues is 1/4? Wait, no. Wait, the game continues only if Alice fails and Bob fails. But if Alice fails and Bob succeeds, the game ends with Bob winning. So, the probability that the game continues is 1/2 * 1/2 = 1/4. Therefore, P = 1/2 + (1/4) P. Wait, but earlier we had P = 1/2 + 1/4 P, which gives P = 2/3. Wait, but earlier equations had P = 1/2 + 1/4 P. Wait, but in the previous equations, we had P = 1/2 + (1/2)*(1/2)*P. Which is same as 1/2 + 1/4 P. So, same thing. Therefore, solving gives P = 2/3. So, that's consistent.

But here, I thought of the game continuing with probability 1/4, but actually, the game can end in two ways: Alice wins, Bob wins, or continues. So, the probability that the game continues is 1/4, and the probability that Alice wins is 1/2, and the probability that Bob wins is 1/4. Therefore, the total probability is 1/2 + 1/4 + 1/4 = 1, which checks out.

Therefore, the probability that Alice wins is 1/2 + (probability game continues) * P. Which is 1/2 + 1/4 * P. Therefore, solving gives P = 2/3.

So, all these different ways of thinking about it lead to the same answer. Therefore, I think the answer is 2/3.

But just to make sure, let me think of another angle. Suppose we consider the game as a sequence of independent trials where each trial is Alice flipping, and if she fails, Bob flips. The game ends when either Alice gets heads or Bob gets tails. So, the probability that Alice wins is the probability that the first success in this sequence is Alice's heads.

In probability theory, if we have a sequence of independent trials, each with probability p of success for Alice and q of success for Bob, and the game continues with probability r, then the probability that Alice wins is p / (p + q). Wait, is that a thing?

Wait, in this case, each trial is: Alice flips. If she gets heads (prob 1/2), Alice wins. If not (prob 1/2), then Bob flips. If he gets tails (prob 1/2), Bob wins. If not (prob 1/2), the game continues. So, in each "trial", the probability that Alice wins is 1/2, Bob wins is 1/2 * 1/2 = 1/4, and the game continues with probability 1/4. Therefore, the probability that Alice wins is (probability Alice wins in a trial) / (probability Alice wins or Bob wins in a trial). Wait, that would be (1/2) / (1/2 + 1/4) = (1/2) / (3/4) = 2/3. So, that's another way to think about it. So, the probability that Alice wins is the ratio of her winning probability to the total probability of the game ending in that trial. Since the game is memoryless, the probability is the same as the first trial. Therefore, 2/3.

Yes, that's a nice way to think about it. So, in each cycle, the probability that the game ends with Alice winning is 1/2, Bob winning is 1/4, and continuing is 1/4. Therefore, the probability that Alice wins is (1/2) / (1/2 + 1/4) = 2/3. So, that's another way to get the same answer.

Therefore, after all these different approaches and checks, I feel confident that the probability that Alice wins the game is 2/3.

**Final Answer**
The probability that Alice wins the game is \boxed{\dfrac{2}{3}}.

To determine the probability that Alice wins the game, we analyze the structure of the game and model it using recursive reasoning and probability theory.

---

### **Game Structure**

- **Alice starts first.**
- On her turn, she flips a fair coin:
  - **Heads (1/2 chance):** Alice wins immediately.
  - **Tails (1/2 chance):** It becomes Bob's turn.
- On Bob's turn, he flips a fair coin:
  - **Tails (1/2 chance):** Bob wins.
  - **Heads (1/2 chance):** It becomes Alice's turn again.

This process continues until one of the players wins.

---

### **Recursive Probability Model**

Let $ P $ be the probability that **Alice wins** starting from her turn.

- On her first turn, Alice has a $ \frac{1}{2} $ chance to win immediately.
- If she flips tails (probability $ \frac{1}{2} $), it becomes Bob's turn.
  - Bob has a $ \frac{1}{2} $ chance to win (tails), in which case Alice loses.
  - Bob has a $ \frac{1}{2} $ chance to flip heads, and the game returns to Alice's turn.
    - In this case, the probability that Alice wins is again $ P $.

So, we can write the recursive equation:

$$
P = \frac{1}{2} + \left( \frac{1}{2} \times \frac{1}{2} \times P \right)
$$

Simplifying:

$$
P = \frac{1}{2} + \frac{1}{4}P
$$

Solving for $ P $:

$$
P - \frac{1}{4}P = \frac{1}{2} \Rightarrow \frac{3}{4}P = \frac{1}{2} \Rightarrow P = \frac{2}{3}
$$

---

### **Alternative Approach: Infinite Series**

We can also model the game as an infinite series of rounds where Alice gets a chance to win on each of her turns.

- **First turn:** Alice wins with probability $ \frac{1}{2} $.
- **Second turn:** Alice only gets a second chance if both she and Bob failed on their first turns. That happens with probability $ \left( \frac{1}{2} \right)^2 = \frac{1}{4} $, and then Alice wins with probability $ \frac{1}{2} $, so total: $ \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} $.
- **Third turn:** Probability of reaching this point is $ \left( \frac{1}{4} \right)^2 = \frac{1}{16} $, and Alice wins with probability $ \frac{1}{2} $, so total: $ \frac{1}{32} $, and so on.

So the total probability that Alice wins is:

$$
P = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots
$$

This is a geometric series with first term $ a = \frac{1}{2} $ and common ratio $ r = \frac{1}{4} $:

$$
P = \frac{a}{1 - r} = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}
$$

---

### **Conclusion**

All approaches—recursive reasoning, infinite series, and probability modeling—lead to the same result. Therefore, the probability that Alice wins the game is:

$$
\boxed{\dfrac{2}{3}}
$$

[Assistant]
Explanation:
Okay, so Alice and Bob are playing this game where they take turns flipping coins. Alice goes first. If she gets heads, she wins. If not, it's Bob's turn. Then Bob flips; if he gets tails, he wins. If not, it goes back to Alice. The question is, what's the probability that Alice wins the game?

Hmm, let me think. So, the game can end on the first turn if Alice flips heads. If she doesn't, then Bob gets a chance. If Bob flips tails, he wins. If he doesn't, then it's back to Alice. So, it's like a sequence of turns where each player has a chance to win on their turn, and if they don't, the other player gets a chance. So, the game could, in theory, go on forever, but the probability of that happening is zero, right? Because each time there's a chance to end the game.

So, maybe I can model this as an infinite series? Let me try to break it down.

First, the probability that Alice wins on her first turn is the probability she flips heads. Since it's a fair coin, that's 1/2.

If she doesn't get heads, which is probability 1/2, then it's Bob's turn. Now, Bob has a chance to win by flipping tails. The probability he flips tails is 1/2. If he does, then Bob wins, so Alice doesn't. If he doesn't flip tails, which is probability 1/2, then it goes back to Alice. So, the game is back to the starting situation, but now it's Alice's turn again. Wait, but the game is symmetric? Or is there a difference?

Wait, actually, the game is structured so that Alice starts, then Bob, then Alice, etc. So, if the game gets back to Alice's turn, it's the same as the original setup, right? So, maybe we can model this recursively.

Let me denote the probability that Alice wins as P. So, P is the probability that Alice wins starting from her turn.

So, when Alice starts, she has a 1/2 chance to win immediately. If she doesn't (probability 1/2), then Bob gets to flip. Now, Bob has a 1/2 chance to win, and 1/2 chance to pass the turn back to Alice. So, if Bob doesn't win, then the game is back to Alice's turn, and the probability of Alice winning from that point is again P. Wait, is that right?

Wait, if the game gets back to Alice's turn, then the situation is the same as the original, so the probability is P. Therefore, we can write an equation for P.

So, P = probability Alice wins on first turn + probability game continues to Bob's turn * probability Alice wins from there.

But when the game continues to Bob's turn, what is the probability that Alice wins? Let me think. If it's Bob's turn, then Bob can win with probability 1/2, and if he doesn't, then it's Alice's turn again. So, if we let Q be the probability that Alice wins when it's Bob's turn, then maybe?

Alternatively, maybe we can express P in terms of itself. Let me try again.

Starting from Alice's turn:

P = P(Alice wins on first turn) + P(Alice doesn't win on first turn) * P(Bob doesn't win on his turn) * P(Alice wins from there)

Wait, so breaking it down:

First, Alice flips. She can win with probability 1/2. If she doesn't (prob 1/2), then Bob flips. Bob can win with probability 1/2, or not (prob 1/2). If Bob doesn't win, then it's back to Alice's turn, and the probability of Alice winning from there is again P. Therefore, the total probability P is:

P = (1/2) + (1/2) * [ (1/2) * 0 + (1/2) * P ]

Wait, because if Alice doesn't win (prob 1/2), then Bob has a 1/2 chance to win (which would mean Alice loses, so 0 for Alice's winning probability) and 1/2 chance to pass it back to Alice, which would give Alice a P chance. Therefore:

P = 1/2 + (1/2) * [ (1/2)*0 + (1/2)*P ]

Simplify that:

P = 1/2 + (1/2)*(0 + (1/2)P) = 1/2 + (1/2)*(1/2 P) = 1/2 + (1/4)P

So, then we can solve for P:

P - (1/4)P = 1/2

(3/4)P = 1/2

Multiply both sides by 4/3:

P = (1/2) * (4/3) = 2/3

So, the probability that Alice wins is 2/3?

Wait, that seems straightforward, but let me check if I did that correctly.

Alternatively, maybe I can think of it as an infinite series. Let's see.

The probability that Alice wins can be broken down into the probability she wins on her first turn, plus the probability the game goes to Bob's turn and he doesn't win, then Alice wins on her second turn, and so on.

So, first term: Alice wins on first turn: 1/2.

Second term: Alice fails (1/2), Bob fails (1/2), then Alice wins on her second turn: (1/2)*(1/2)*(1/2) = 1/8.

Third term: Alice fails, Bob fails, Alice fails, Bob fails, Alice wins: (1/2)^4 * 1/2 = (1/2)^5 = 1/32?

Wait, let me check:

Wait, for Alice to win on her second turn, the game must have gone through two cycles: Alice fails, Bob fails, then Alice succeeds. So that's (1/2)^2 * 1/2 = (1/2)^3 = 1/8. Similarly, for her third turn, it would be (1/2)^4 * 1/2 = (1/2)^5, etc.

So, the total probability is the sum over n=1 to infinity of (1/2)^{2n - 1} ?

Wait, let me think again. For the first term, n=1: (1/2)^{2*1 - 1} = (1/2)^1 = 1/2. Then n=2: (1/2)^{3} = 1/8, n=3: (1/2)^5 = 1/32, etc. So the series is 1/2 + 1/8 + 1/32 + ... which is a geometric series with first term a = 1/2 and common ratio r = 1/4.

Because 1/8 divided by 1/2 is 1/4, and 1/32 divided by 1/8 is 1/4, etc. So the sum is a / (1 - r) = (1/2) / (1 - 1/4) = (1/2) / (3/4) = (1/2) * (4/3) = 2/3. So same result as before. So that seems to confirm that the probability is 2/3.

Alternatively, maybe I can think of it as states. Let me define P as the probability that Alice wins starting from her turn, and Q as the probability that Alice wins starting from Bob's turn.

So, when it's Alice's turn, she can win with probability 1/2, or pass to Bob's turn with probability 1/2. Therefore:

P = 1/2 + (1/2) * Q

Similarly, when it's Bob's turn, he can win with probability 1/2 (so Alice loses, so 0) or pass back to Alice's turn with probability 1/2. Therefore:

Q = (1/2)*0 + (1/2)*P = (1/2)P

So, substituting Q into the first equation:

P = 1/2 + (1/2)*(1/2 P) = 1/2 + (1/4)P

Which is the same equation as before. Then solving:

P - (1/4)P = 1/2 => (3/4)P = 1/2 => P = 2/3. Same answer.

So, seems like 2/3 is the answer. But let me check with another approach.

Alternatively, think of the game as a sequence of rounds. Each round consists of Alice's turn and Bob's turn. But actually, the game might end on Alice's turn or Bob's turn. So, maybe not exactly rounds. But let's see.

The probability that Alice wins on the first round (her first turn) is 1/2. The probability that the game continues to Bob's turn and he doesn't win is (1/2)*(1/2) = 1/4. Then, the probability that Alice wins on the second round (her second turn) is (1/4)*(1/2) = 1/8. Similarly, the probability that the game continues to the third round is (1/4)^2, and so on.

Therefore, the total probability is the sum over n=1 to infinity of (probability game reaches Alice's nth turn) * (probability she wins on that turn). The probability that the game reaches Alice's nth turn is (probability that both Alice and Bob failed in the previous n-1 rounds). Each round, the probability that both fail is (1/2)*(1/2) = 1/4. Therefore, the probability that the game reaches Alice's nth turn is (1/4)^{n-1}. Then, the probability she wins on that turn is 1/2. Therefore, the total probability is sum_{n=1}^\infty (1/4)^{n-1} * 1/2.

Which is again a geometric series with first term a = 1/2 and ratio r = 1/4. So sum is (1/2) / (1 - 1/4) = 2/3. Same answer again. So that seems to confirm.

Alternatively, maybe think of the game as a Markov chain with states: Alice's turn, Bob's turn, Alice wins, Bob wins. But that might be overcomplicating.

Alternatively, think recursively. Let me try again.

Let me denote P as the probability that Alice wins. So, when Alice starts, she has a 1/2 chance to win immediately. If she doesn't (prob 1/2), then Bob gets to flip. Now, Bob has a 1/2 chance to win, which would mean Alice loses, and 1/2 chance to not win, which brings us back to Alice's turn. Therefore, the probability that Alice wins is:

P = 1/2 + (1/2) * [ (1/2)*0 + (1/2)*P ]

Which is the same equation as before. So, solving gives P = 2/3.

Alternatively, maybe think of the game as Alice and Bob taking turns with probabilities. Since Alice goes first, she has a higher chance. Intuitively, since she has the first chance to win, her probability should be higher than 1/2, which 2/3 is. So that makes sense.

Alternatively, maybe think of the game as a sequence of trials where each trial consists of Alice and Bob each flipping once. But actually, the game might end before both flip. For example, if Alice flips heads on her first turn, the game ends. So, the trials aren't necessarily both flipping.

But maybe we can model it as such. Let me try.

The probability that Alice wins on the first trial (her first flip) is 1/2. The probability that the game continues to the second trial (both Alice and Bob failed their first flips) is (1/2)*(1/2) = 1/4. Then, in the second trial, Alice has another chance to flip, etc. So, the probability that Alice wins is the sum over all trials of the probability that the game reaches trial n and Alice wins on that trial.

In trial 1: probability Alice wins is 1/2.

In trial 2: probability game reaches trial 2 is 1/4, and then Alice wins with probability 1/2, so total 1/4 * 1/2 = 1/8.

In trial 3: probability game reaches trial 3 is (1/4)^2, and then Alice wins with 1/2, so (1/4)^2 * 1/2 = 1/32.

So, same as before, sum is 1/2 + 1/8 + 1/32 + ... = 2/3.

Alternatively, maybe think of the game as Alice having a 1/2 chance to win, and if not, Bob has a 1/2 chance to win, and if not, it's back to Alice. So, the probability that Alice wins is the probability that she wins on her first turn, plus the probability that both fail and then Alice wins from there. So:

P = 1/2 + (1/2 * 1/2) * P

Which is P = 1/2 + (1/4)P => P - (1/4)P = 1/2 => (3/4)P = 1/2 => P = 2/3. Same result.

So, all approaches lead to 2/3. Therefore, I think that is the correct answer.

But just to make sure, let me think of another way. Let me imagine that the game is played many times, and compute the expected number of times Alice wins or something. Wait, maybe not necessary. Alternatively, think of the probability that Alice wins as P, and Bob's probability as Q. Since the game must end eventually, P + Q = 1. Also, since the game is symmetric except for who starts, maybe we can find relations between P and Q.

But let me try. When Alice starts, she has a 1/2 chance to win, and 1/2 chance to pass to Bob. If it's Bob's turn, he has a 1/2 chance to win, and 1/2 chance to pass back to Alice. Therefore, the probability that Alice wins is:

P = 1/2 + (1/2) * [ (1/2)*0 + (1/2)*P ] = same as before.

Alternatively, if we think of Q as the probability that Bob wins when it's his turn. Wait, maybe not necessary. But since P + Q = 1, if we can find another equation.

Alternatively, when it's Bob's turn, the probability that Bob wins is Q. So, when it's Bob's turn, he can win with probability 1/2, or pass to Alice's turn with probability 1/2. Therefore, Q = 1/2 + (1/2)*P? Wait, no. Wait, if it's Bob's turn, then the probability that Bob wins is 1/2 (he wins immediately) plus the probability that he doesn't win (1/2) times the probability that Bob wins from Alice's turn. Wait, but if it goes back to Alice's turn, then the probability that Bob wins is the probability that Alice doesn't win, which is 1 - P. Therefore, Q = 1/2 + (1/2)*(1 - P). But since when it's Bob's turn, the probability that Bob wins is Q, and the probability that Alice wins is 1 - Q. Wait, maybe this is getting too convoluted.

Alternatively, since when it's Bob's turn, the probability that Alice wins is Q, and the probability that Bob wins is 1 - Q. Wait, maybe not. Let me try again.

Let me define P as the probability that Alice wins starting from her turn, and Q as the probability that Alice wins starting from Bob's turn. Then, as before:

P = 1/2 + (1/2)*Q

Q = (1/2)*0 + (1/2)*P = (1/2)P

Therefore, substituting Q into the first equation:

P = 1/2 + (1/2)*(1/2 P) = 1/2 + (1/4)P => P = 2/3. Same result.

Alternatively, if I think of Q as the probability that Alice wins when it's Bob's turn, then Q = probability that Bob doesn't win * probability Alice wins from her turn. Which is (1/2)*P. So, same thing.

Alternatively, if I think of the game as a whole, since Alice has the first chance, her probability is higher. If the game was symmetric, like both players have the same chance to win on their turn, but Alice goes first, then her probability would be higher. For example, if both players had a 1/2 chance to win on their turn, then the probability that Alice wins would be 1/2 + 1/2 * 1/2 * 1/2 + ... which is same as the previous series. So, yeah, 2/3.

Alternatively, think of it as a geometric distribution. The game ends when either Alice gets heads or Bob gets tails. The probability that Alice wins is the probability that the first success (either Alice's heads or Bob's tails) is Alice's heads.

But how to model that? Let me think. Each "round" consists of Alice flipping and then Bob flipping. But the game could end in the middle of a round.

Alternatively, think of the game as a sequence of independent trials where each trial is Alice flipping, and if she fails, Bob flips. So, each trial has a probability of Alice winning, Bob winning, or the game continuing.

So, in each trial:

- Alice flips: probability 1/2 she wins, 1/2 continues.

- If continues, Bob flips: probability 1/2 he wins, 1/2 continues.

Therefore, the probability that Alice wins in the first trial is 1/2.

The probability that the game continues to the next trial is 1/2 * 1/2 = 1/4.

Then, the probability that Alice wins in the second trial is (1/4) * 1/2 = 1/8.

Similarly, third trial: (1/4)^2 * 1/2, etc.

So, same as before, sum is 1/2 + 1/8 + 1/32 + ... = 2/3.

Alternatively, think of the probability that Alice wins as the sum over all n ≥ 1 of the probability that the game lasts for n-1 full rounds and then Alice wins on the nth round.

Each full round (Alice and Bob both flipping) has probability 1/4 of continuing. So, the probability that the game lasts for n-1 rounds is (1/4)^{n-1}, and then Alice wins on the nth round with probability 1/2. Therefore, the total probability is sum_{n=1}^\infty (1/4)^{n-1} * 1/2 = 2/3.

Alternatively, think of the game as a sequence of independent events where each "cycle" consists of Alice and Bob each flipping once. The probability that Alice wins in a cycle is 1/2 (Alice gets heads). The probability that Bob wins in a cycle is 1/2 * 1/2 = 1/4 (Alice fails, Bob gets tails). The probability that the cycle continues is 1/4 (both fail). Therefore, the probability that Alice wins is the probability that she wins in the first cycle, plus the probability the cycle continues times the probability she wins in subsequent cycles. So:

P = 1/2 + 1/4 * P

Solving:

P - 1/4 P = 1/2 => 3/4 P = 1/2 => P = 2/3. Same answer again.

So, all these different approaches lead to the same answer, 2/3. Therefore, I think that is the correct probability.

But just to make sure, let me think of a concrete example with small probabilities.

Suppose the game is played, and we simulate the possibilities.

First, Alice flips:

- 50% chance she gets heads (wins).

- 50% chance she gets tails, then Bob flips:

   - 50% chance Bob gets tails (he wins).

   - 50% chance Bob gets heads, then it's Alice's turn again.

So, the probability that Alice wins is 1/2 + (1/2 * 1/2 * P) = 1/2 + 1/4 P. Which is the same equation as before. So, solving gives P = 2/3.

Alternatively, if I think of the game as a tree:

First level: Alice flips H (prob 1/2) -> Alice wins.

First level: Alice flips T (prob 1/2) -> Bob flips T (prob 1/2) -> Bob wins.

First level: Alice flips T (prob 1/2) -> Bob flips H (prob 1/2) -> back to Alice.

So, the probability that Alice wins is 1/2 + (1/2 * 1/2 * P) = 1/2 + 1/4 P. So, same equation.

Alternatively, if I think of the game as a recursive function. Let me define P as the probability Alice wins. Then:

P = probability Alice wins on first turn + probability game continues * P

But the probability game continues is the probability Alice fails and Bob fails, which is 1/2 * 1/2 = 1/4. Wait, but earlier we had P = 1/2 + 1/4 P. Wait, but that would be if the game continues with probability 1/4, but actually, when the game continues, it's not just the probability of continuing, but the probability that Alice wins from that point. Wait, but in this case, if the game continues, it's back to Alice's turn, so the probability is P again. Therefore, P = 1/2 + (probability game continues) * P. Wait, but the probability game continues is 1/4? Wait, no. Wait, the game continues only if Alice fails and Bob fails. But if Alice fails and Bob succeeds, the game ends with Bob winning. So, the probability that the game continues is 1/2 * 1/2 = 1/4. Therefore, P = 1/2 + (1/4) P. Wait, but earlier we had P = 1/2 + 1/4 P, which gives P = 2/3. Wait, but earlier equations had P = 1/2 + 1/4 P. Wait, but in the previous equations, we had P = 1/2 + (1/2)*(1/2)*P. Which is same as 1/2 + 1/4 P. So, same thing. Therefore, solving gives P = 2/3. So, that's consistent.

But here, I thought of the game continuing with probability 1/4, but actually, the game can end in two ways: Alice wins, Bob wins, or continues. So, the probability that the game continues is 1/4, and the probability that Alice wins is 1/2, and the probability that Bob wins is 1/4. Therefore, the total probability is 1/2 + 1/4 + 1/4 = 1, which checks out.

Therefore, the probability that Alice wins is 1/2 + (probability game continues) * P. Which is 1/2 + 1/4 * P. Therefore, solving gives P = 2/3.

So, all these different ways of thinking about it lead to the same answer. Therefore, I think the answer is 2/3.

But just to make sure, let me think of another angle. Suppose we consider the game as a sequence of independent trials where each trial is Alice flipping, and if she fails, Bob flips. The game ends when either Alice gets heads or Bob gets tails. So, the probability that Alice wins is the probability that the first success in this sequence is Alice's heads.

In probability theory, if we have a sequence of independent trials, each with probability p of success for Alice and q of success for Bob, and the game continues with probability r, then the probability that Alice wins is p / (p + q). Wait, is that a thing?

Wait, in this case, each trial is: Alice flips. If she gets heads (prob 1/2), Alice wins. If not (prob 1/2), then Bob flips. If he gets tails (prob 1/2), Bob wins. If not (prob 1/2), the game continues. So, in each "trial", the probability that Alice wins is 1/2, Bob wins is 1/2 * 1/2 = 1/4, and the game continues with probability 1/4. Therefore, the probability that Alice wins is (probability Alice wins in a trial) / (probability Alice wins or Bob wins in a trial). Wait, that would be (1/2) / (1/2 + 1/4) = (1/2) / (3/4) = 2/3. So, that's another way to think about it. So, the probability that Alice wins is the ratio of her winning probability to the total probability of the game ending in that trial. Since the game is memoryless, the probability is the same as the first trial. Therefore, 2/3.

Yes, that's a nice way to think about it. So, in each cycle, the probability that the game ends with Alice winning is 1/2, Bob winning is 1/4, and continuing is 1/4. Therefore, the probability that Alice wins is (1/2) / (1/2 + 1/4) = 2/3. So, that's another way to get the same answer.

Therefore, after all these different approaches and checks, I feel confident that the probability that Alice wins the game is 2/3.

**Final Answer**
The probability that Alice wins the game is \boxed{\dfrac{2}{3}}.

To determine the probability that Alice wins the game, we analyze the structure of the game and model it using recursive reasoning and probability theory.

---

### **Game Structure**

- **Alice starts first.**
- On her turn, she flips a fair coin:
  - **Heads (1/2 chance):** Alice wins immediately.
  - **Tails (1/2 chance):** It becomes Bob's turn.
- On Bob's turn, he flips a fair coin:
  - **Tails (1/2 chance):** Bob wins.
  - **Heads (1/2 chance):** It becomes Alice's turn again.

This process continues until one of the players wins.

---

### **Recursive Probability Model**

Let $ P $ be the probability that **Alice wins** starting from her turn.

- On her first turn, Alice has a $ \frac{1}{2} $ chance to win immediately.
- If she flips tails (probability $ \frac{1}{2} $), it becomes Bob's turn.
  - Bob has a $ \frac{1}{2} $ chance to win (tails), in which case Alice loses.
  - Bob has a $ \frac{1}{2} $ chance to flip heads, and the game returns to Alice's turn.
    - In this case, the probability that Alice wins is again $ P $.

So, we can write the recursive equation:

$$
P = \frac{1}{2} + \left( \frac{1}{2} \times \frac{1}{2} \times P \right)
$$

Simplifying:

$$
P = \frac{1}{2} + \frac{1}{4}P
$$

Solving for $ P $:

$$
P - \frac{1}{4}P = \frac{1}{2} \Rightarrow \frac{3}{4}P = \frac{1}{2} \Rightarrow P = \frac{2}{3}
$$

---

### **Alternative Approach: Infinite Series**

We can also model the game as an infinite series of rounds where Alice gets a chance to win on each of her turns.

- **First turn:** Alice wins with probability $ \frac{1}{2} $.
- **Second turn:** Alice only gets a second chance if both she and Bob failed on their first turns. That happens with probability $ \left( \frac{1}{2} \right)^2 = \frac{1}{4} $, and then Alice wins with probability $ \frac{1}{2} $, so total: $ \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} $.
- **Third turn:** Probability of reaching this point is $ \left( \frac{1}{4} \right)^2 = \frac{1}{16} $, and Alice wins with probability $ \frac{1}{2} $, so total: $ \frac{1}{32} $, and so on.

So the total probability that Alice wins is:

$$
P = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots
$$

This is a geometric series with first term $ a = \frac{1}{2} $ and common ratio $ r = \frac{1}{4} $:

$$
P = \frac{a}{1 - r} = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}
$$

---

### **Conclusion**

All approaches—recursive reasoning, infinite series, and probability modeling—lead to the same result. Therefore, the probability that Alice wins the game is:

$$
\boxed{\dfrac{2}{3}}
$$

Solution:
Alice has a $1/2$ chance of winning the game on her first turn. If she doesn't, then the probability that she wins the game on her second turn is $1/8,$ since she must not win on her first flip ($1/2$ chance), Bob must not win on his first flip ($1/2$ chance), and then Alice must win on her second flip ($1/2$ chance). The probability that she wins the game on her third turn is $1/32,$ and in general, the probability that she wins the game on her $k^\text{th}$ turn is $(1/2)^{2k-1}.$ Thus, the probability that Alice wins is an infinite geometric series with first term $1/2$ and common ratio $1/4.$ So, the probability that Alice wins the game is $$\frac{\frac12}{1-\frac14} = \boxed{\frac{2}{3}}.$$OR

Note that the only difference between the odds of Alice or Bob winning is who goes first. Because Bob goes second, the odds of him winning on his $k^\text{th}$ flip is half of the odds that Alice wins on her $k^\text{th}$ flip, since Alice must first get a tails before Bob gets a chance to win. Thus, if $a$ is Alice's chance of winning, and $b$ is Bob's chance of winning, then $a = 2b.$ Also, since someone must win, $a + b = 1.$ It follows that $a = 2/3$ and $b = 1/3,$ so Alice has a $\boxed{\frac{2}{3}}$ chance of winning the game.