Explanation:
Okay, let me try to figure out the probability that Alice wins this game. Hmm, so Alice goes first. She flips a coin. If it's heads, she wins. If it's tails, then it's Bob's turn. Then Bob flips a coin. If it's tails, he wins. If it's heads, then it goes back to Alice. And this continues until someone wins. So we need to find the probability that Alice is the one who wins.

Let me think. Let me denote the probability that Alice wins as P. So, we need to find P.

Okay, let's break it down. On Alice's first turn, she has a 50% chance of getting heads and winning immediately. If she gets tails, which is also 50%, then it's Bob's turn. Now, Bob has his turn. He flips the coin. There's a 50% chance he gets tails and wins, which would mean Alice loses. If he gets heads, which is 50%, then it goes back to Alice, and the situation is the same as the beginning, right? So it's like the game resets, but it's Alice's turn again.

Wait, so maybe I can express this as a recursive probability? Let me try.

So, P is the probability that Alice wins. So, starting from the beginning, Alice has a 1/2 chance to win immediately. If she doesn't, which is 1/2 chance, then Bob gets to flip. Then Bob has a 1/2 chance to win, which would mean Alice loses. If Bob doesn't win, which is also 1/2 chance, then it's back to Alice again, and the probability that Alice wins from that point is again P. Wait, but is it exactly P? Because the game is now back to Alice's turn, same as the start. So maybe yes.

So, writing this as an equation. Let me try:

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

So, breaking down:

P = (1/2) * 1 + (1/2) * [probability that Bob doesn't win * probability Alice wins from the new state]

Wait, when Alice flips tails (prob 1/2), then Bob gets to flip. Now, Bob's chance to win is 1/2. If he wins, then Alice loses. If he doesn't, then it's back to Alice. So the probability that Alice wins after Alice's first flip is tails is equal to the probability that Bob doesn't win times P. Wait, because if Bob doesn't win, the game is back to Alice's turn, so the probability Alice wins is again P. So:

So, putting it all together:

P = (1/2)*1 + (1/2)*[ (probability Bob doesn't win) * P ]

But probability Bob doesn't win is 1/2. Therefore:

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

Simplify:

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

So then, subtract (1/4)P from both sides:

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

(3/4)P = 1/2

Multiply both sides by 4/3:

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

Wait, so the probability that Alice wins is 2/3?

That seems plausible? Let me check. Let me think of another way.

Alternatively, maybe think of all possible turns.

The game can end on the first flip (Alice wins with probability 1/2), or if both Alice and Bob flip tails, then it goes back to the start. Wait, no. Let me think again.

Wait, actually, the game can end on Alice's first turn, or on Bob's first turn, or on Alice's second turn, etc.

Alternatively, maybe think of the game as a sequence of rounds. Each round consists of Alice's turn and Bob's turn. But actually, the game can end on Alice's turn before Bob gets to flip. So maybe not exactly rounds.

Alternatively, think of the game as a sequence of trials where each trial is Alice's turn followed by Bob's turn, except that if Alice wins on her turn, the game ends immediately.

Alternatively, maybe model it as an infinite series. The probability that Alice wins on her first turn is 1/2. The probability that Alice wins on her second turn would be the probability that both Alice and Bob failed on their first turns, and then Alice succeeded on her second turn. Similarly for the third turn, etc.

So let's try that approach.

Probability Alice wins on first turn: 1/2.

Probability Alice wins on second turn: probability that Alice failed first (1/2), Bob failed (1/2), then Alice succeeds (1/2). So (1/2)^3 = 1/8.

Probability Alice wins on third turn: (1/2)^5 = 1/32? Wait:

Wait, for Alice to win on her second turn, it's (Alice tails, Bob tails, Alice heads). That's (1/2)^3.

For Alice to win on her third turn, it's (Alice tails, Bob tails, Alice tails, Bob tails, Alice heads). That's (1/2)^5.

Similarly, for the nth turn, it's (1/2)^{2n - 1} ?

Wait, let me check:

First win: 1 turn, probability 1/2 = (1/2)^1.

Second win: 3 flips, probability (1/2)^3.

Third win: 5 flips, probability (1/2)^5.

So the total probability that Alice wins is the sum over n=1 to infinity of (1/2)^{2n - 1}.

So that's a geometric series. Let me compute that.

Let me write it as:

Sum_{n=1}^∞ (1/2)^{2n - 1} = (1/2) * Sum_{n=1}^∞ (1/4)^{n - 1} * (1/2)^{1} ? Wait, maybe better to factor out (1/2).

Wait, (1/2)^{2n - 1} = (1/2)^{-1} * (1/2)^{2n} = 2 * (1/4)^n. Wait, no:

Wait 2n -1 = 2(n - 1) + 1. Maybe not helpful. Let me think.

Alternatively, write the general term:

For n-th term, exponent is 2n -1. So Sum_{n=1}^∞ (1/2)^{2n - 1} = (1/2) * Sum_{n=1}^∞ (1/4)^{n - 1} } ?

Wait:

Let me note that (1/2)^{2n - 1} = (1/2) * (1/4)^{n - 1} ?

Wait, let me check with n=1: (1/2)^{2(1)-1} = (1/2)^1 = 1/2. On the other hand, (1/2) * (1/4)^{1 - 1} = (1/2)*1 = 1/2. Correct.

n=2: (1/2)^{3} = 1/8. On the other hand, (1/2) * (1/4)^{2 - 1} = (1/2)*(1/4) = 1/8. Correct.

Therefore, Sum_{n=1}^∞ (1/2)^{2n - 1} = (1/2) * Sum_{n=1}^∞ (1/4)^{n - 1} }.

The sum Sum_{n=1}^∞ (1/4)^{n - 1} is a geometric series with first term 1 and ratio 1/4. The sum is 1 / (1 - 1/4) ) = 1 / (3/4) ) = 4/3.

Therefore, total probability is (1/2) * (4/3) = 2/3. So same answer as before.

Therefore, the probability that Alice wins is 2/3. That matches with the previous method.

So that seems to confirm it. Alternatively, maybe another approach.

Let me think again about the recursive equation.

Let me think of P as the probability that Alice wins starting from her turn. Then, when she flips, she has 1/2 chance to win, else it's Bob's turn. Let me think of Q as the probability that Alice wins given that it's Bob's turn. Then, maybe we can set up equations for P and Q.

So, P = probability Alice wins when it's her turn.

Q = probability Alice wins when it's Bob's turn.

So, starting with P:

When it's Alice's turn, she flips. With probability 1/2, she gets heads and wins. With probability 1/2, she gets tails, and it becomes Bob's turn. Therefore:

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

Similarly, when it's Bob's turn, he flips. With probability 1/2, he gets tails and wins, meaning Alice loses, so probability Alice wins is 0. With probability 1/2, he gets heads, and it becomes Alice's turn again. So:

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

Therefore, we have:

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

and

Q = (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. Therefore, solving:

P - (1/4)P = 1/2 => (3/4)P = 1/2 => P = (1/2)*(4/3) = 2/3. Same result. So this seems consistent.

So, seems like the answer is 2/3. But just to make sure, let me think of another way.

Alternatively, think of the game as a Markov chain with states: Alice's turn, Bob's turn, Alice wins, Bob wins. We can model the probabilities.

But that's similar to previous methods.

Alternatively, think about expected number of turns or something else? Maybe not necessary.

Alternatively, think about odds. Since Alice goes first, she has an advantage. Since the game is symmetric except for who goes first, so maybe the probability is higher for Alice. 2/3 is higher than 1/2, so that makes sense.

Alternatively, think about the game as a sequence of rounds where each round consists of Alice and Bob each flipping once. Except that if Alice wins in her turn, the game ends before Bob gets to flip.

But maybe we can model it as such. The probability that Alice wins in the first round is 1/2 (she gets heads). The probability that Bob wins in the first round is 1/2 (he gets tails after Alice failed). The probability that the game continues to the next round is 1/2 * 1/2 = 1/4 (both flip heads). So in this case, the game is like a geometric distribution with probability of ending each round being 1/2 + 1/2 = 1? Wait, no. Wait, in each round, the game can end with Alice winning, Bob winning, or continuing.

Wait, the probability that the game ends in the first round is 1/2 (Alice wins) + 1/2 * 1/2 (Bob wins) = 1/2 + 1/4 = 3/4. Wait, no. Wait:

Wait, the first round: Alice flips. If heads (prob 1/2), Alice wins. If tails (prob 1/2), Bob flips. If Bob gets tails (prob 1/2), Bob wins. If Bob gets heads (prob 1/2), game continues. Therefore, the probability the game ends in the first round is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4. The probability the game continues is 1/2 * 1/2 = 1/4. So each round, there's a 3/4 chance the game ends and 1/4 chance it continues.

Therefore, the probability that Alice wins can be thought of as the probability she wins in the first round plus the probability the game continues times the probability she wins in subsequent rounds.

So, P = (probability Alice wins in first round) + (probability game continues) * P

Therefore:

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

Which is same equation as before. Solving:

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

Same answer again. So that seems to confirm once again.

Alternatively, let me think of it as an infinite sum. Let me think of the probability that Alice wins on the first flip, plus the probability that the game continues and Alice wins on the second flip, etc.

Which we did earlier, and it summed up to 2/3. So all methods lead to the same answer. Therefore, I think 2/3 is the correct probability.

But just to be thorough, let me check with actual probabilities for first few terms and see if it approaches 2/3.

So, P = 1/2 + 1/8 + 1/32 + 1/128 + ...

Let me compute partial sums:

First term: 1/2 = 0.5

Second term: 0.5 + 1/8 = 0.5 + 0.125 = 0.625

Third term: 0.625 + 1/32 ≈ 0.625 + 0.03125 = 0.65625

Fourth term: +1/128 ≈ 0.65625 + 0.0078125 ≈ 0.6640625

Fifth term: +1/512 ≈ 0.6640625 + 0.001953125 ≈ 0.666015625

Sixth term: +1/2048 ≈ 0.666015625 + 0.00048828125 ≈ 0.66650390625

Continuing this, it approaches 2/3 ≈ 0.666666..., so that seems correct.

Therefore, I think the answer is 2/3. So, boxed answer is \boxed{\dfrac{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 need to analyze the sequence of turns and the conditions under which each player can win.

---

### **Understanding the Game Dynamics**

- **Alice's Turn:**
  - She flips a coin.
  - If it lands **Heads (H)**, she **wins immediately** (probability = 1/2).
  - If it lands **Tails (T)**, it becomes **Bob's turn** (probability = 1/2).

- **Bob's Turn:**
  - He flips a coin.
  - If it lands **Tails (T)**, he **wins**, so **Alice loses** (probability = 1/2).
  - If it lands **Heads (H)**, it becomes **Alice's turn again** (probability = 1/2).

This sequence continues until one of the players wins. So, the game is a **potentially infinite** sequence of moves, and we need to find the **overall probability** that Alice is the one who wins.

---

### **Modeling Using Recursion**

Let’s define:

- $ P $: the probability that **Alice wins** starting from **her turn**.

On her first turn, Alice has a:

- $ \frac{1}{2} $ chance of **winning immediately**.
- $ \frac{1}{2} $ chance of **not winning**, leading to Bob’s turn.

Let’s define:

- $ Q $: the probability that **Alice wins** starting from **Bob’s turn**.

On Bob’s turn:

- Bob has a $ \frac{1}{2} $ chance of winning (so Alice loses), probability = 0.
- Bob has a $ \frac{1}{2} $ chance of **not winning**, leading back to Alice’s turn.

So we can now write the following system of equations:

$$
P = \frac{1}{2} + \frac{1}{2}Q
$$
$$
Q = \frac{1}{2}(0) + \frac{1}{2}P = \frac{1}{2}P
$$

Substitute the second equation into the first:

$$
P = \frac{1}{2} + \frac{1}{2} \left( \frac{1}{2}P \right) = \frac{1}{2} + \frac{1}{4}P
$$

Solve for $ P $:

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

---

### **Alternative Confirmation via Infinite Series**

We can also sum the probabilities of Alice winning on her **first turn**, **second turn**, **third turn**, and so on:

- Alice wins on her **first** turn:
  $ \frac{1}{2} $

- Alice wins on her **second** turn:
  Both Alice and Bob must have failed first:
  $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $

- Alice wins on her **third** turn:
  $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{32} $

So the probability that Alice wins can be represented as:

$$
P = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{2n - 1} = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \dots
$$

This is a **geometric series**, where the first term is $ a = \frac{1}{2} $, and the common ratio is $ r = \frac{1}{4} $. The sum is:

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

---

### **Conclusion**

Using both recursive analysis and summation of a geometric series, we arrive at the same result. Therefore, the **probability that Alice wins the game** is:

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

[Assistant]
Explanation:
Okay, let me try to figure out the probability that Alice wins this game. Hmm, so Alice goes first. She flips a coin. If it's heads, she wins. If it's tails, then it's Bob's turn. Then Bob flips a coin. If it's tails, he wins. If it's heads, then it goes back to Alice. And this continues until someone wins. So we need to find the probability that Alice is the one who wins.

Let me think. Let me denote the probability that Alice wins as P. So, we need to find P.

Okay, let's break it down. On Alice's first turn, she has a 50% chance of getting heads and winning immediately. If she gets tails, which is also 50%, then it's Bob's turn. Now, Bob has his turn. He flips the coin. There's a 50% chance he gets tails and wins, which would mean Alice loses. If he gets heads, which is 50%, then it goes back to Alice, and the situation is the same as the beginning, right? So it's like the game resets, but it's Alice's turn again.

Wait, so maybe I can express this as a recursive probability? Let me try.

So, P is the probability that Alice wins. So, starting from the beginning, Alice has a 1/2 chance to win immediately. If she doesn't, which is 1/2 chance, then Bob gets to flip. Then Bob has a 1/2 chance to win, which would mean Alice loses. If Bob doesn't win, which is also 1/2 chance, then it's back to Alice again, and the probability that Alice wins from that point is again P. Wait, but is it exactly P? Because the game is now back to Alice's turn, same as the start. So maybe yes.

So, writing this as an equation. Let me try:

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

So, breaking down:

P = (1/2) * 1 + (1/2) * [probability that Bob doesn't win * probability Alice wins from the new state]

Wait, when Alice flips tails (prob 1/2), then Bob gets to flip. Now, Bob's chance to win is 1/2. If he wins, then Alice loses. If he doesn't, then it's back to Alice. So the probability that Alice wins after Alice's first flip is tails is equal to the probability that Bob doesn't win times P. Wait, because if Bob doesn't win, the game is back to Alice's turn, so the probability Alice wins is again P. So:

So, putting it all together:

P = (1/2)*1 + (1/2)*[ (probability Bob doesn't win) * P ]

But probability Bob doesn't win is 1/2. Therefore:

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

Simplify:

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

So then, subtract (1/4)P from both sides:

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

(3/4)P = 1/2

Multiply both sides by 4/3:

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

Wait, so the probability that Alice wins is 2/3?

That seems plausible? Let me check. Let me think of another way.

Alternatively, maybe think of all possible turns.

The game can end on the first flip (Alice wins with probability 1/2), or if both Alice and Bob flip tails, then it goes back to the start. Wait, no. Let me think again.

Wait, actually, the game can end on Alice's first turn, or on Bob's first turn, or on Alice's second turn, etc.

Alternatively, maybe think of the game as a sequence of rounds. Each round consists of Alice's turn and Bob's turn. But actually, the game can end on Alice's turn before Bob gets to flip. So maybe not exactly rounds.

Alternatively, think of the game as a sequence of trials where each trial is Alice's turn followed by Bob's turn, except that if Alice wins on her turn, the game ends immediately.

Alternatively, maybe model it as an infinite series. The probability that Alice wins on her first turn is 1/2. The probability that Alice wins on her second turn would be the probability that both Alice and Bob failed on their first turns, and then Alice succeeded on her second turn. Similarly for the third turn, etc.

So let's try that approach.

Probability Alice wins on first turn: 1/2.

Probability Alice wins on second turn: probability that Alice failed first (1/2), Bob failed (1/2), then Alice succeeds (1/2). So (1/2)^3 = 1/8.

Probability Alice wins on third turn: (1/2)^5 = 1/32? Wait:

Wait, for Alice to win on her second turn, it's (Alice tails, Bob tails, Alice heads). That's (1/2)^3.

For Alice to win on her third turn, it's (Alice tails, Bob tails, Alice tails, Bob tails, Alice heads). That's (1/2)^5.

Similarly, for the nth turn, it's (1/2)^{2n - 1} ?

Wait, let me check:

First win: 1 turn, probability 1/2 = (1/2)^1.

Second win: 3 flips, probability (1/2)^3.

Third win: 5 flips, probability (1/2)^5.

So the total probability that Alice wins is the sum over n=1 to infinity of (1/2)^{2n - 1}.

So that's a geometric series. Let me compute that.

Let me write it as:

Sum_{n=1}^∞ (1/2)^{2n - 1} = (1/2) * Sum_{n=1}^∞ (1/4)^{n - 1} * (1/2)^{1} ? Wait, maybe better to factor out (1/2).

Wait, (1/2)^{2n - 1} = (1/2)^{-1} * (1/2)^{2n} = 2 * (1/4)^n. Wait, no:

Wait 2n -1 = 2(n - 1) + 1. Maybe not helpful. Let me think.

Alternatively, write the general term:

For n-th term, exponent is 2n -1. So Sum_{n=1}^∞ (1/2)^{2n - 1} = (1/2) * Sum_{n=1}^∞ (1/4)^{n - 1} } ?

Wait:

Let me note that (1/2)^{2n - 1} = (1/2) * (1/4)^{n - 1} ?

Wait, let me check with n=1: (1/2)^{2(1)-1} = (1/2)^1 = 1/2. On the other hand, (1/2) * (1/4)^{1 - 1} = (1/2)*1 = 1/2. Correct.

n=2: (1/2)^{3} = 1/8. On the other hand, (1/2) * (1/4)^{2 - 1} = (1/2)*(1/4) = 1/8. Correct.

Therefore, Sum_{n=1}^∞ (1/2)^{2n - 1} = (1/2) * Sum_{n=1}^∞ (1/4)^{n - 1} }.

The sum Sum_{n=1}^∞ (1/4)^{n - 1} is a geometric series with first term 1 and ratio 1/4. The sum is 1 / (1 - 1/4) ) = 1 / (3/4) ) = 4/3.

Therefore, total probability is (1/2) * (4/3) = 2/3. So same answer as before.

Therefore, the probability that Alice wins is 2/3. That matches with the previous method.

So that seems to confirm it. Alternatively, maybe another approach.

Let me think again about the recursive equation.

Let me think of P as the probability that Alice wins starting from her turn. Then, when she flips, she has 1/2 chance to win, else it's Bob's turn. Let me think of Q as the probability that Alice wins given that it's Bob's turn. Then, maybe we can set up equations for P and Q.

So, P = probability Alice wins when it's her turn.

Q = probability Alice wins when it's Bob's turn.

So, starting with P:

When it's Alice's turn, she flips. With probability 1/2, she gets heads and wins. With probability 1/2, she gets tails, and it becomes Bob's turn. Therefore:

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

Similarly, when it's Bob's turn, he flips. With probability 1/2, he gets tails and wins, meaning Alice loses, so probability Alice wins is 0. With probability 1/2, he gets heads, and it becomes Alice's turn again. So:

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

Therefore, we have:

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

and

Q = (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. Therefore, solving:

P - (1/4)P = 1/2 => (3/4)P = 1/2 => P = (1/2)*(4/3) = 2/3. Same result. So this seems consistent.

So, seems like the answer is 2/3. But just to make sure, let me think of another way.

Alternatively, think of the game as a Markov chain with states: Alice's turn, Bob's turn, Alice wins, Bob wins. We can model the probabilities.

But that's similar to previous methods.

Alternatively, think about expected number of turns or something else? Maybe not necessary.

Alternatively, think about odds. Since Alice goes first, she has an advantage. Since the game is symmetric except for who goes first, so maybe the probability is higher for Alice. 2/3 is higher than 1/2, so that makes sense.

Alternatively, think about the game as a sequence of rounds where each round consists of Alice and Bob each flipping once. Except that if Alice wins in her turn, the game ends before Bob gets to flip.

But maybe we can model it as such. The probability that Alice wins in the first round is 1/2 (she gets heads). The probability that Bob wins in the first round is 1/2 (he gets tails after Alice failed). The probability that the game continues to the next round is 1/2 * 1/2 = 1/4 (both flip heads). So in this case, the game is like a geometric distribution with probability of ending each round being 1/2 + 1/2 = 1? Wait, no. Wait, in each round, the game can end with Alice winning, Bob winning, or continuing.

Wait, the probability that the game ends in the first round is 1/2 (Alice wins) + 1/2 * 1/2 (Bob wins) = 1/2 + 1/4 = 3/4. Wait, no. Wait:

Wait, the first round: Alice flips. If heads (prob 1/2), Alice wins. If tails (prob 1/2), Bob flips. If Bob gets tails (prob 1/2), Bob wins. If Bob gets heads (prob 1/2), game continues. Therefore, the probability the game ends in the first round is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4. The probability the game continues is 1/2 * 1/2 = 1/4. So each round, there's a 3/4 chance the game ends and 1/4 chance it continues.

Therefore, the probability that Alice wins can be thought of as the probability she wins in the first round plus the probability the game continues times the probability she wins in subsequent rounds.

So, P = (probability Alice wins in first round) + (probability game continues) * P

Therefore:

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

Which is same equation as before. Solving:

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

Same answer again. So that seems to confirm once again.

Alternatively, let me think of it as an infinite sum. Let me think of the probability that Alice wins on the first flip, plus the probability that the game continues and Alice wins on the second flip, etc.

Which we did earlier, and it summed up to 2/3. So all methods lead to the same answer. Therefore, I think 2/3 is the correct probability.

But just to be thorough, let me check with actual probabilities for first few terms and see if it approaches 2/3.

So, P = 1/2 + 1/8 + 1/32 + 1/128 + ...

Let me compute partial sums:

First term: 1/2 = 0.5

Second term: 0.5 + 1/8 = 0.5 + 0.125 = 0.625

Third term: 0.625 + 1/32 ≈ 0.625 + 0.03125 = 0.65625

Fourth term: +1/128 ≈ 0.65625 + 0.0078125 ≈ 0.6640625

Fifth term: +1/512 ≈ 0.6640625 + 0.001953125 ≈ 0.666015625

Sixth term: +1/2048 ≈ 0.666015625 + 0.00048828125 ≈ 0.66650390625

Continuing this, it approaches 2/3 ≈ 0.666666..., so that seems correct.

Therefore, I think the answer is 2/3. So, boxed answer is \boxed{\dfrac{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 need to analyze the sequence of turns and the conditions under which each player can win.

---

### **Understanding the Game Dynamics**

- **Alice's Turn:**
  - She flips a coin.
  - If it lands **Heads (H)**, she **wins immediately** (probability = 1/2).
  - If it lands **Tails (T)**, it becomes **Bob's turn** (probability = 1/2).

- **Bob's Turn:**
  - He flips a coin.
  - If it lands **Tails (T)**, he **wins**, so **Alice loses** (probability = 1/2).
  - If it lands **Heads (H)**, it becomes **Alice's turn again** (probability = 1/2).

This sequence continues until one of the players wins. So, the game is a **potentially infinite** sequence of moves, and we need to find the **overall probability** that Alice is the one who wins.

---

### **Modeling Using Recursion**

Let’s define:

- $ P $: the probability that **Alice wins** starting from **her turn**.

On her first turn, Alice has a:

- $ \frac{1}{2} $ chance of **winning immediately**.
- $ \frac{1}{2} $ chance of **not winning**, leading to Bob’s turn.

Let’s define:

- $ Q $: the probability that **Alice wins** starting from **Bob’s turn**.

On Bob’s turn:

- Bob has a $ \frac{1}{2} $ chance of winning (so Alice loses), probability = 0.
- Bob has a $ \frac{1}{2} $ chance of **not winning**, leading back to Alice’s turn.

So we can now write the following system of equations:

$$
P = \frac{1}{2} + \frac{1}{2}Q
$$
$$
Q = \frac{1}{2}(0) + \frac{1}{2}P = \frac{1}{2}P
$$

Substitute the second equation into the first:

$$
P = \frac{1}{2} + \frac{1}{2} \left( \frac{1}{2}P \right) = \frac{1}{2} + \frac{1}{4}P
$$

Solve for $ P $:

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

---

### **Alternative Confirmation via Infinite Series**

We can also sum the probabilities of Alice winning on her **first turn**, **second turn**, **third turn**, and so on:

- Alice wins on her **first** turn:
  $ \frac{1}{2} $

- Alice wins on her **second** turn:
  Both Alice and Bob must have failed first:
  $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $

- Alice wins on her **third** turn:
  $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{32} $

So the probability that Alice wins can be represented as:

$$
P = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{2n - 1} = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \dots
$$

This is a **geometric series**, where the first term is $ a = \frac{1}{2} $, and the common ratio is $ r = \frac{1}{4} $. The sum is:

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

---

### **Conclusion**

Using both recursive analysis and summation of a geometric series, we arrive at 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.