Explanation:
Okay, so I need to figure out how many possible integer values of x there are such that x³ is a positive factor of 10!. Let me start by recalling what 10! is. 10 factorial is the product of all integers from 1 to 10, so 10! = 10 × 9 × 8 × ... × 1. Let me compute that value first, but maybe I don't need the exact number. Instead, I think I need to look at the prime factorization of 10! because factors are related to prime exponents.

Right, so if I can find the prime factors of 10! and their exponents, then I can figure out what exponents x³ can have. Since x³ divides 10!, then x must be such that when you cube it, all the exponents in its prime factorization are less than or equal to those in 10!.

Let me recall how to find the prime factorization of 10!. The primes less than or equal to 10 are 2, 3, 5, 7. Let me find the exponents of each prime in 10!.

Starting with 2: The exponent of 2 in 10! is given by the sum of floor(10/2) + floor(10/4) + floor(10/8) + ... until the division results in zero. Let me compute that:

floor(10/2) = 5

floor(10/4) = 2

floor(10/8) = 1

floor(10/16) = 0, so we stop here. So total exponent of 2 is 5 + 2 + 1 = 8.

Similarly for 3:

floor(10/3) = 3

floor(10/9) = 1

floor(10/27) = 0. So total exponent of 3 is 3 + 1 = 4.

For 5:

floor(10/5) = 2

floor(10/25) = 0. So exponent of 5 is 2.

For 7:

floor(10/7) = 1

floor(10/49) = 0. So exponent of 7 is 1.

So the prime factorization of 10! is 2⁸ × 3⁴ × 5² × 7¹.

Now, if x³ divides 10!, then x must be such that when we take its prime factorization, each exponent is less than or equal to the exponent in 10! divided by 3, right? Wait, let me think.

Suppose x has a prime factorization of 2^a × 3^b × 5^c × 7^d. Then x³ would be 2^(3a) × 3^(3b) × 5^(3c) × 7^(3d). For x³ to divide 10!, we need 3a ≤ 8, 3b ≤ 4, 3c ≤ 2, and 3d ≤ 1. Therefore, each exponent in x's prime factorization must satisfy:

For 2: 3a ≤ 8 ⇒ a ≤ 8/3 ≈ 2.666..., so since a must be an integer, a ≤ 2.

For 3: 3b ≤ 4 ⇒ b ≤ 4/3 ≈ 1.333..., so b ≤ 1.

For 5: 3c ≤ 2 ⇒ c ≤ 2/3 ≈ 0.666..., so c ≤ 0.

For 7: 3d ≤ 1 ⇒ d ≤ 1/3 ≈ 0.333..., so d ≤ 0.

Wait, so for primes 5 and 7, their exponents in x must be zero? Because c and d have to be integers, so c can only be 0, and d can only be 0. Therefore, x cannot have any factors of 5 or 7? Because if x had a factor of 5, then c would be at least 1, and 3*1=3 which is greater than 2, which is the exponent of 5 in 10!. Similarly for 7, 3*1=3 which is greater than 1. Therefore, x cannot have 5 or 7 as prime factors. Therefore, x must be a divisor of 2^8 × 3^4, but with exponents such that 3a ≤ 8 and 3b ≤ 4. Therefore, the exponents for 2 can be 0,1,2 and exponents for 3 can be 0,1. Wait, let me check:

For 2: 3a ≤ 8 ⇒ a ≤ floor(8/3) = 2. So a can be 0,1,2.

For 3: 3b ≤ 4 ⇒ b ≤ floor(4/3) = 1. So b can be 0,1.

For 5 and 7: c and d must be 0.

Therefore, the number of possible x is the number of combinations of exponents for 2 and 3. Since exponents for 2: 3 choices, exponents for 3: 2 choices. So total number of x is 3*2=6? Wait, but hold on. Wait, but x is an integer, so each combination of exponents gives a unique x. But wait, but maybe there are more possibilities? Wait, but according to the above, since 5 and 7 cannot be in x, so x is of the form 2^a * 3^b where a is 0,1,2 and b is 0,1. Therefore, total number of x is 3*2=6. But wait, but that seems low. Let me check again.

Wait, but maybe I made a mistake here. Let me think again. Let me check the exponents again.

Original exponents in 10!:

2^8, 3^4, 5^2, 7^1.

If x³ divides 10!, then for each prime in x's factorization, the exponent multiplied by 3 must be less than or equal to the exponent in 10!.

So for prime 2: exponent in x³ is 3a ≤ 8 ⇒ a ≤ 8/3 ≈ 2.666, so maximum integer a is 2. So a can be 0,1,2.

For prime 3: 3b ≤ 4 ⇒ b ≤ 1.333, so maximum integer b is 1. So b can be 0,1.

For prime 5: 3c ≤ 2 ⇒ c ≤ 0.666, so c=0.

For prime 7: 3d ≤ 1 ⇒ d ≤ 0.333, so d=0.

Therefore, x can only have primes 2 and 3, with exponents as above. Therefore, the number of possible x is (number of choices for a) × (number of choices for b) = 3 × 2 = 6. But wait, but 6 seems low. Let me check with examples.

For example, x can be 1 (which is 2^0 * 3^0), 2 (2^1 * 3^0), 4 (2^2 * 3^0), 3 (2^0 * 3^1), 6 (2^1 * 3^1), 12 (2^2 * 3^1). So that's 6 values. Let me check if their cubes divide 10!.

Check x=1: 1³=1 divides 10! yes.

x=2: 2³=8 divides 10! yes, since 10! has 2^8.

x=4: 4³=64. 64 is 2^6. Since 10! has 2^8, yes, 2^6 divides 2^8.

x=3: 3³=27. 10! has 3^4, so 27 divides 3^4, yes.

x=6: 6³=216. Let's factor 216: 2³ * 3³. 10! has 2^8 and 3^4, so yes, 2³ and 3³ are present.

x=12: 12³ = (2² * 3)^3 = 2^6 * 3^3. Again, 10! has 2^8 and 3^4, so yes, 2^6 and 3^3 are present. So all 6 values are valid.

But wait, are there more values? For example, what about x=2^0 * 3^0 =1, which is included. What about x=2^0 * 3^1=3, included. What about x=2^1 * 3^0=2, included. x=2^2 * 3^0=4, included. x=2^1 * 3^1=6, included. x=2^2 * 3^1=12, included. So that's 6. So is that all?

But wait, maybe there are other primes? Wait, but earlier we saw that 5 and 7 cannot be in x, because if x had 5, then x³ would have 5^3, but 10! only has 5^2. Similarly for 7. So x cannot have 5 or 7. Therefore, x can only have 2 and 3. Therefore, only 6 possibilities. But the answer seems small. Wait, but maybe I missed something?

Wait, but maybe x can have other primes? Wait, but if x has a prime factor not in 10!'s prime factors, then x³ would have that prime factor, which 10! doesn't have, so that would not divide. Therefore, x can only have primes 2,3,5,7. But as we saw, 5 and 7 can't be in x. Therefore, only 2 and 3. Therefore, only 6 possibilities. But maybe I need to check if there are more exponents?

Wait, for example, for prime 2, a can be 0,1,2. So exponents 0,1,2. For prime 3, b can be 0,1. So total combinations 3*2=6. So that's correct. Therefore, answer is 6? But let me check again.

Wait, but maybe I made a mistake in the exponents. Let me check again for prime 2. The exponent in 10! is 8. So 3a ≤ 8. So a can be 0,1,2 because 3*2=6 ≤8, 3*3=9>8. So yes, a can be 0,1,2. For prime 3, exponent in 10! is 4. So 3b ≤4. So b can be 0,1 because 3*1=3 ≤4, 3*2=6>4. So yes, b=0,1. So that's correct. Therefore, 3*2=6. So answer is 6? But I feel like maybe there's something wrong here. Let me check with another approach.

Alternatively, think of x³ divides 10! So x divides 10!^{1/3}. But since 10! is not a perfect cube, but we can think of the exponents. Wait, but maybe not helpful.

Alternatively, think of all divisors of 10! which are perfect cubes. Wait, but the question is not asking for x such that x is a cube divisor of 10! but rather x³ divides 10!.

Wait, but if x³ divides 10!, then x is a divisor of 10!^{1/3}, but since 10! isn't a perfect cube, but we can think of the exponents. Wait, but maybe not helpful.

Alternatively, think of the exponents for each prime in x. For each prime, the exponent in x is floor(exponent in 10! / 3). Wait, no. Wait, if x³ divides 10!, then for each prime p, exponent of p in x³ is ≤ exponent of p in 10!. Therefore, exponent of p in x is ≤ floor(exponent of p in 10! / 3). Wait, no. Wait, exponent of p in x is a, then exponent in x³ is 3a. So 3a ≤ exponent in 10! So a ≤ floor(exponent in 10! / 3). Wait, but floor(exponent in 10! / 3) is not necessarily the same as floor(exponent in 10! / 3). Wait, for example, for prime 2, exponent in 10! is 8. So 8 / 3 is 2.666..., so floor(8/3)=2. So a can be 0,1,2. Similarly for prime 3, exponent is 4, 4/3=1.333..., floor is 1. So b can be 0,1. For prime 5, exponent is 2, 2/3≈0.666, floor is 0. So c=0. For prime 7, exponent is 1, 1/3≈0.333, floor is 0. So d=0. Therefore, the number of possible x is (floor(8/3)+1) * (floor(4/3)+1) * (floor(2/3)+1) * (floor(1/3)+1). Wait, but floor(8/3) is 2, so 2+1=3 choices for a. Similarly floor(4/3)=1, so 1+1=2 choices for b. For c: floor(2/3)=0, so 0+1=1 choice. For d: floor(1/3)=0, so 1 choice. Therefore total number is 3*2*1*1=6. So same answer. Therefore, 6 possible x. Therefore, answer is 6? But let me check again with examples. Wait, but maybe there are more numbers?

Wait, for example, suppose x is 1, 2, 3, 4, 6, 12. That's 6 numbers. Are there any others? Let me think. What about x= 2^0 * 3^0 =1, yes. 2^1 * 3^0=2, yes. 2^2 * 3^0=4, yes. 2^0 * 3^1=3, yes. 2^1 * 3^1=6, yes. 2^2 * 3^1=12, yes. What about 2^0 * 3^2? Wait, but 3^2 would mean b=2. But earlier we said that for prime 3, 3b ≤4. If b=2, then 3*2=6>4. Therefore, that's not allowed. Therefore, 3^2 is not allowed. Similarly, 2^3 would be a=3, but 3*3=9>8. So not allowed. Therefore, no. Therefore, only those 6. Therefore, answer is 6? But I feel like maybe I need to check if there are other primes? Wait, but 10! has primes 2,3,5,7. If x has any other prime factors, say 11, but 11 is not in 10! so x can't have 11. So no. Therefore, only 2,3,5,7. But 5 and 7 can't be in x. Therefore, only 2 and 3. Therefore, 6 possibilities. Therefore, answer is 6? But I have a doubt because sometimes problems like this can have more answers. Wait, let me check if x can be 1, 2, 3, 4, 6, 12. Let me check if there are other numbers. For example, 2^2 * 3^0 =4, which is included. What about 2^1 * 3^1=6. What about 2^0 * 3^1=3. What about 2^2 * 3^1=12. What about 2^0 * 3^0=1. What about 2^1 * 3^0=2. So that's all. So 6 numbers. Therefore, answer is 6? But maybe I need to check if there are numbers with exponents beyond that? For example, if x is 2^2 * 3^1=12, which is allowed. What about x= 2^2 * 3^1 * something? But we can't have anything else. So no. Therefore, answer is 6. Wait, but maybe I made a mistake in the exponents? Let me check again.

Wait, for example, take x= 2^2 * 3^1=12. Then x³= (2^2 * 3^1)^3= 2^6 * 3^3. Now, 10! has 2^8 and 3^4. So 2^6 divides 2^8 and 3^3 divides 3^4. So yes, that's okay. What about x= 2^2 * 3^1 * 5^0 * 7^0. So that's 12. What about x= 2^2 * 3^1 * 5^0 * 7^0. So that's same as 12. So no other primes. So yes, only 6 numbers. Therefore, answer is 6? But I have a feeling that maybe the answer is different. Wait, let me check with another approach.

Alternatively, think of all possible x such that x³ divides 10!. So x must be a divisor of 10!^{1/3}, but since 10! is not a perfect cube, but we can think of the exponents. Wait, but maybe not helpful. Alternatively, think of the exponents for each prime in x. For each prime, the exponent in x is floor(exponent in 10! / 3). Wait, but no, because for example, for prime 2, exponent in 10! is 8. So floor(8/3)=2, so exponent in x can be up to 2. Similarly for others. Therefore, the number of possible exponents for each prime is floor(exponent / 3) + 1? Wait, no. Wait, if the exponent in 10! is e, then the exponent in x can be from 0 to floor(e / 3). Therefore, the number of choices is floor(e / 3) + 1. Wait, for example, for prime 2, e=8, floor(8/3)=2, so number of choices is 2 + 1=3. For prime 3, e=4, floor(4/3)=1, so 1 +1=2. For prime 5, e=2, floor(2/3)=0, so 0 +1=1. For prime 7, e=1, floor(1/3)=0, so 0 +1=1. Therefore, total number of x is 3*2*1*1=6. Therefore, same answer. Therefore, answer is 6. Therefore, I think that is correct.

But let me check with another example. Suppose I take x= 2^2 * 3^1=12. Then x³= 2^6 * 3^3. 10! has 2^8 and 3^4. So yes, 2^6 divides 2^8 and 3^3 divides 3^4. So that's okay. What if I take x= 2^2 * 3^1 * 5^0 * 7^0. So same as 12. What if I take x= 2^2 * 3^1 * 5^0 * 7^0. So same. So no other possibilities. Therefore, answer is 6. Therefore, I think the answer is 6. But maybe I need to check if there are other numbers. Wait, for example, x= 2^0 * 3^0=1. x=2^1 * 3^0=2. x=2^2 * 3^0=4. x=2^0 * 3^1=3. x=2^1 * 3^1=6. x=2^2 * 3^1=12. So that's 6 numbers. So yes, 6. Therefore, answer is 6. Therefore, I think the answer is 6. Wait, but maybe I need to check if there are numbers with exponents beyond that? For example, if x= 2^2 * 3^1 * something? But no, since 5 and 7 can't be included. Therefore, no. Therefore, answer is 6. Therefore, I think the answer is 6. Wait, but maybe I made a mistake in the prime factorization of 10!? Let me check again.

Compute 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me compute the exponents again for primes 2,3,5,7.

For prime 2:

Number of multiples of 2 in 1-10: 5 (2,4,6,8,10)

Number of multiples of 4: 2 (4,8)

Number of multiples of 8: 1 (8)

Number of multiples of 16: 0. So total exponent of 2 is 5 + 2 + 1 = 8. Correct.

For prime 3:

Multiples of 3: 3,6,9. That's 3 numbers.

Multiples of 9: 9. That's 1 number.

Multiples of 27: 0. So total exponent of 3 is 3 + 1 = 4. Correct.

For prime 5:

Multiples of 5: 5,10. That's 2 numbers.

Multiples of 25: 0. So exponent of 5 is 2. Correct.

For prime 7:

Multiples of 7: 7. That's 1 number.

Multiples of 49: 0. So exponent of 7 is 1. Correct.

Therefore, prime factorization is correct. Therefore, exponents are correct. Therefore, answer is 6. Therefore, I think the answer is 6. Therefore, the number of possible integer values of x is 6. Therefore, I think the answer is \boxed{6}.

Wait, but hold on. Wait, I just thought of something. What if x is negative? The problem says "positive factor", so x³ must be a positive factor. Since x³ is positive, x must be positive. Therefore, x is a positive integer. Therefore, negative numbers are excluded. Therefore, no problem. Therefore, answer is 6. Therefore, I think that is correct.

But let me check once more. Let me list all possible x:

1. 1
2. 2
3. 3
4. 4
5. 6
6. 12

Yes, that's 6 numbers. Let me check if there are more numbers. For example, is 8 a possible x? Let's check. If x=8, then x³=512. Let's see if 512 divides 10!.

But 10! is 3628800. Let me divide 3628800 by 512. 3628800 / 512. Let me compute 512 × 7000 = 3,584,000. 3,628,800 - 3,584,000 = 44,800. 512 × 87 = 44,544. 44,800 - 44,544 = 256. 256 is 512 / 2. So total is 7000 + 87 + 0.5 = 7087.5. Therefore, 512 does not divide 10! because it's not an integer. Therefore, x=8 is invalid. Therefore, 8 is not allowed. Therefore, x=8 is not in the list. Similarly, x=9: x³=729. Let me check if 729 divides 10!.

10! is 3628800. 3628800 / 729. Let me compute 729 × 4980 = 729 × 5000 - 729 × 20 = 3,645,000 - 14,580 = 3,630,420. Which is larger than 3,628,800. So 729 × 4980 is too big. Let me try 729 × 4960 = 729 × (5000 - 40) = 3,645,000 - 29,160 = 3,615,840. Then 3,628,800 - 3,615,840 = 12,960. 12,960 / 729 = 17.8... So not an integer. Therefore, 729 does not divide 10!. Therefore, x=9 is invalid. Therefore, x=9 is not allowed. Similarly, x=5: x³=125. 10! has 5², so 5³ would need 5^3, but 10! only has 5². Therefore, 125 does not divide 10!. Therefore, x=5 is invalid. Similarly, x=7: 7³=343. 10! has only one 7, so 343 does not divide 10!. Therefore, invalid. Therefore, x=7 invalid. Therefore, numbers like 5,7,8,9, etc., are invalid. Therefore, only the 6 numbers listed before. Therefore, answer is 6. Therefore, I think the answer is 6. Therefore, I think the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To determine how many possible integer values of $ x $ satisfy the condition that $ x^3 $ is a **positive factor** of $ 10! $, we begin by analyzing the prime factorization of $ 10! $.

---

### Step 1: Prime Factorization of $ 10! $

We compute the prime factorization of $ 10! $ by counting the powers of each prime number less than or equal to 10:

- **Prime 2**:
  $$
  \left\lfloor \frac{10}{2} \right\rfloor + \left\lfloor \frac{10}{4} \right\rfloor + \left\lfloor \frac{10}{8} \right\rfloor = 5 + 2 + 1 = 8
  $$
- **Prime 3**:
  $$
  \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor = 3 + 1 = 4
  $$
- **Prime 5**:
  $$
  \left\lfloor \frac{10}{5} \right\rfloor = 2
  $$
- **Prime 7**:
  $$
  \left\lfloor \frac{10}{7} \right\rfloor = 1
  $$

So, the prime factorization of $ 10! $ is:
$$
10! = 2^8 \times 3^4 \times 5^2 \times 7^1
$$

---

### Step 2: Understanding the Condition $ x^3 \mid 10! $

We are told that $ x^3 $ must divide $ 10! $. That means the prime factorization of $ x^3 $ must be a subset of the prime factorization of $ 10! $.

Let $ x = 2^a \times 3^b \times 5^c \times 7^d $, where $ a, b, c, d $ are non-negative integers. Then:
$$
x^3 = 2^{3a} \times 3^{3b} \times 5^{3c} \times 7^{3d}
$$

For $ x^3 $ to divide $ 10! $, we must have:
- $ 3a \leq 8 $ → $ a \leq \left\lfloor \frac{8}{3} \right\rfloor = 2 $
- $ 3b \leq 4 $ → $ b \leq \left\lfloor \frac{4}{3} \right\rfloor = 1 $
- $ 3c \leq 2 $ → $ c \leq \left\lfloor \frac{2}{3} \right\rfloor = 0 $
- $ 3d \leq 1 $ → $ d \leq \left\lfloor \frac{1}{3} \right\rfloor = 0 $

So, the possible values for the exponents are:
- $ a \in \{0, 1, 2\} $ → 3 choices
- $ b \in \{0, 1\} $ → 2 choices
- $ c = 0 $ → 1 choice
- $ d = 0 $ → 1 choice

---

### Step 3: Counting the Number of Valid $ x $

Each combination of $ a $ and $ b $ gives a unique value of $ x $, since $ c = d = 0 $, and the other primes (5 and 7) cannot be included in $ x $.

So the total number of valid $ x $ is:
$$
3 \times 2 \times 1 \times 1 = 6
$$

---

### Step 4: Listing the Valid $ x $

The valid values of $ x $ are:
- $ 2^0 \times 3^0 = 1 $
- $ 2^1 \times 3^0 = 2 $
- $ 2^2 \times 3^0 = 4 $
- $ 2^0 \times 3^1 = 3 $
- $ 2^1 \times 3^1 = 6 $
- $ 2^2 \times 3^1 = 12 $

Each of these values satisfies the condition that $ x^3 $ divides $ 10! $, and no other values are possible due to the constraints on the exponents.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many possible integer values of x there are such that x³ is a positive factor of 10!. Let me start by recalling what 10! is. 10 factorial is the product of all integers from 1 to 10, so 10! = 10 × 9 × 8 × ... × 1. Let me compute that value first, but maybe I don't need the exact number. Instead, I think I need to look at the prime factorization of 10! because factors are related to prime exponents.

Right, so if I can find the prime factors of 10! and their exponents, then I can figure out what exponents x³ can have. Since x³ divides 10!, then x must be such that when you cube it, all the exponents in its prime factorization are less than or equal to those in 10!.

Let me recall how to find the prime factorization of 10!. The primes less than or equal to 10 are 2, 3, 5, 7. Let me find the exponents of each prime in 10!.

Starting with 2: The exponent of 2 in 10! is given by the sum of floor(10/2) + floor(10/4) + floor(10/8) + ... until the division results in zero. Let me compute that:

floor(10/2) = 5

floor(10/4) = 2

floor(10/8) = 1

floor(10/16) = 0, so we stop here. So total exponent of 2 is 5 + 2 + 1 = 8.

Similarly for 3:

floor(10/3) = 3

floor(10/9) = 1

floor(10/27) = 0. So total exponent of 3 is 3 + 1 = 4.

For 5:

floor(10/5) = 2

floor(10/25) = 0. So exponent of 5 is 2.

For 7:

floor(10/7) = 1

floor(10/49) = 0. So exponent of 7 is 1.

So the prime factorization of 10! is 2⁸ × 3⁴ × 5² × 7¹.

Now, if x³ divides 10!, then x must be such that when we take its prime factorization, each exponent is less than or equal to the exponent in 10! divided by 3, right? Wait, let me think.

Suppose x has a prime factorization of 2^a × 3^b × 5^c × 7^d. Then x³ would be 2^(3a) × 3^(3b) × 5^(3c) × 7^(3d). For x³ to divide 10!, we need 3a ≤ 8, 3b ≤ 4, 3c ≤ 2, and 3d ≤ 1. Therefore, each exponent in x's prime factorization must satisfy:

For 2: 3a ≤ 8 ⇒ a ≤ 8/3 ≈ 2.666..., so since a must be an integer, a ≤ 2.

For 3: 3b ≤ 4 ⇒ b ≤ 4/3 ≈ 1.333..., so b ≤ 1.

For 5: 3c ≤ 2 ⇒ c ≤ 2/3 ≈ 0.666..., so c ≤ 0.

For 7: 3d ≤ 1 ⇒ d ≤ 1/3 ≈ 0.333..., so d ≤ 0.

Wait, so for primes 5 and 7, their exponents in x must be zero? Because c and d have to be integers, so c can only be 0, and d can only be 0. Therefore, x cannot have any factors of 5 or 7? Because if x had a factor of 5, then c would be at least 1, and 3*1=3 which is greater than 2, which is the exponent of 5 in 10!. Similarly for 7, 3*1=3 which is greater than 1. Therefore, x cannot have 5 or 7 as prime factors. Therefore, x must be a divisor of 2^8 × 3^4, but with exponents such that 3a ≤ 8 and 3b ≤ 4. Therefore, the exponents for 2 can be 0,1,2 and exponents for 3 can be 0,1. Wait, let me check:

For 2: 3a ≤ 8 ⇒ a ≤ floor(8/3) = 2. So a can be 0,1,2.

For 3: 3b ≤ 4 ⇒ b ≤ floor(4/3) = 1. So b can be 0,1.

For 5 and 7: c and d must be 0.

Therefore, the number of possible x is the number of combinations of exponents for 2 and 3. Since exponents for 2: 3 choices, exponents for 3: 2 choices. So total number of x is 3*2=6? Wait, but hold on. Wait, but x is an integer, so each combination of exponents gives a unique x. But wait, but maybe there are more possibilities? Wait, but according to the above, since 5 and 7 cannot be in x, so x is of the form 2^a * 3^b where a is 0,1,2 and b is 0,1. Therefore, total number of x is 3*2=6. But wait, but that seems low. Let me check again.

Wait, but maybe I made a mistake here. Let me think again. Let me check the exponents again.

Original exponents in 10!:

2^8, 3^4, 5^2, 7^1.

If x³ divides 10!, then for each prime in x's factorization, the exponent multiplied by 3 must be less than or equal to the exponent in 10!.

So for prime 2: exponent in x³ is 3a ≤ 8 ⇒ a ≤ 8/3 ≈ 2.666, so maximum integer a is 2. So a can be 0,1,2.

For prime 3: 3b ≤ 4 ⇒ b ≤ 1.333, so maximum integer b is 1. So b can be 0,1.

For prime 5: 3c ≤ 2 ⇒ c ≤ 0.666, so c=0.

For prime 7: 3d ≤ 1 ⇒ d ≤ 0.333, so d=0.

Therefore, x can only have primes 2 and 3, with exponents as above. Therefore, the number of possible x is (number of choices for a) × (number of choices for b) = 3 × 2 = 6. But wait, but 6 seems low. Let me check with examples.

For example, x can be 1 (which is 2^0 * 3^0), 2 (2^1 * 3^0), 4 (2^2 * 3^0), 3 (2^0 * 3^1), 6 (2^1 * 3^1), 12 (2^2 * 3^1). So that's 6 values. Let me check if their cubes divide 10!.

Check x=1: 1³=1 divides 10! yes.

x=2: 2³=8 divides 10! yes, since 10! has 2^8.

x=4: 4³=64. 64 is 2^6. Since 10! has 2^8, yes, 2^6 divides 2^8.

x=3: 3³=27. 10! has 3^4, so 27 divides 3^4, yes.

x=6: 6³=216. Let's factor 216: 2³ * 3³. 10! has 2^8 and 3^4, so yes, 2³ and 3³ are present.

x=12: 12³ = (2² * 3)^3 = 2^6 * 3^3. Again, 10! has 2^8 and 3^4, so yes, 2^6 and 3^3 are present. So all 6 values are valid.

But wait, are there more values? For example, what about x=2^0 * 3^0 =1, which is included. What about x=2^0 * 3^1=3, included. What about x=2^1 * 3^0=2, included. x=2^2 * 3^0=4, included. x=2^1 * 3^1=6, included. x=2^2 * 3^1=12, included. So that's 6. So is that all?

But wait, maybe there are other primes? Wait, but earlier we saw that 5 and 7 cannot be in x, because if x had 5, then x³ would have 5^3, but 10! only has 5^2. Similarly for 7. So x cannot have 5 or 7. Therefore, x can only have 2 and 3. Therefore, only 6 possibilities. But the answer seems small. Wait, but maybe I missed something?

Wait, but maybe x can have other primes? Wait, but if x has a prime factor not in 10!'s prime factors, then x³ would have that prime factor, which 10! doesn't have, so that would not divide. Therefore, x can only have primes 2,3,5,7. But as we saw, 5 and 7 can't be in x. Therefore, only 2 and 3. Therefore, only 6 possibilities. But maybe I need to check if there are more exponents?

Wait, for example, for prime 2, a can be 0,1,2. So exponents 0,1,2. For prime 3, b can be 0,1. So total combinations 3*2=6. So that's correct. Therefore, answer is 6? But let me check again.

Wait, but maybe I made a mistake in the exponents. Let me check again for prime 2. The exponent in 10! is 8. So 3a ≤ 8. So a can be 0,1,2 because 3*2=6 ≤8, 3*3=9>8. So yes, a can be 0,1,2. For prime 3, exponent in 10! is 4. So 3b ≤4. So b can be 0,1 because 3*1=3 ≤4, 3*2=6>4. So yes, b=0,1. So that's correct. Therefore, 3*2=6. So answer is 6? But I feel like maybe there's something wrong here. Let me check with another approach.

Alternatively, think of x³ divides 10! So x divides 10!^{1/3}. But since 10! is not a perfect cube, but we can think of the exponents. Wait, but maybe not helpful.

Alternatively, think of all divisors of 10! which are perfect cubes. Wait, but the question is not asking for x such that x is a cube divisor of 10! but rather x³ divides 10!.

Wait, but if x³ divides 10!, then x is a divisor of 10!^{1/3}, but since 10! isn't a perfect cube, but we can think of the exponents. Wait, but maybe not helpful.

Alternatively, think of the exponents for each prime in x. For each prime, the exponent in x is floor(exponent in 10! / 3). Wait, no. Wait, if x³ divides 10!, then for each prime p, exponent of p in x³ is ≤ exponent of p in 10!. Therefore, exponent of p in x is ≤ floor(exponent of p in 10! / 3). Wait, no. Wait, exponent of p in x is a, then exponent in x³ is 3a. So 3a ≤ exponent in 10! So a ≤ floor(exponent in 10! / 3). Wait, but floor(exponent in 10! / 3) is not necessarily the same as floor(exponent in 10! / 3). Wait, for example, for prime 2, exponent in 10! is 8. So 8 / 3 is 2.666..., so floor(8/3)=2. So a can be 0,1,2. Similarly for prime 3, exponent is 4, 4/3=1.333..., floor is 1. So b can be 0,1. For prime 5, exponent is 2, 2/3≈0.666, floor is 0. So c=0. For prime 7, exponent is 1, 1/3≈0.333, floor is 0. So d=0. Therefore, the number of possible x is (floor(8/3)+1) * (floor(4/3)+1) * (floor(2/3)+1) * (floor(1/3)+1). Wait, but floor(8/3) is 2, so 2+1=3 choices for a. Similarly floor(4/3)=1, so 1+1=2 choices for b. For c: floor(2/3)=0, so 0+1=1 choice. For d: floor(1/3)=0, so 1 choice. Therefore total number is 3*2*1*1=6. So same answer. Therefore, 6 possible x. Therefore, answer is 6? But let me check again with examples. Wait, but maybe there are more numbers?

Wait, for example, suppose x is 1, 2, 3, 4, 6, 12. That's 6 numbers. Are there any others? Let me think. What about x= 2^0 * 3^0 =1, yes. 2^1 * 3^0=2, yes. 2^2 * 3^0=4, yes. 2^0 * 3^1=3, yes. 2^1 * 3^1=6, yes. 2^2 * 3^1=12, yes. What about 2^0 * 3^2? Wait, but 3^2 would mean b=2. But earlier we said that for prime 3, 3b ≤4. If b=2, then 3*2=6>4. Therefore, that's not allowed. Therefore, 3^2 is not allowed. Similarly, 2^3 would be a=3, but 3*3=9>8. So not allowed. Therefore, no. Therefore, only those 6. Therefore, answer is 6? But I feel like maybe I need to check if there are other primes? Wait, but 10! has primes 2,3,5,7. If x has any other prime factors, say 11, but 11 is not in 10! so x can't have 11. So no. Therefore, only 2,3,5,7. But 5 and 7 can't be in x. Therefore, only 2 and 3. Therefore, 6 possibilities. Therefore, answer is 6? But I have a doubt because sometimes problems like this can have more answers. Wait, let me check if x can be 1, 2, 3, 4, 6, 12. Let me check if there are other numbers. For example, 2^2 * 3^0 =4, which is included. What about 2^1 * 3^1=6. What about 2^0 * 3^1=3. What about 2^2 * 3^1=12. What about 2^0 * 3^0=1. What about 2^1 * 3^0=2. So that's all. So 6 numbers. Therefore, answer is 6? But maybe I need to check if there are numbers with exponents beyond that? For example, if x is 2^2 * 3^1=12, which is allowed. What about x= 2^2 * 3^1 * something? But we can't have anything else. So no. Therefore, answer is 6. Wait, but maybe I made a mistake in the exponents? Let me check again.

Wait, for example, take x= 2^2 * 3^1=12. Then x³= (2^2 * 3^1)^3= 2^6 * 3^3. Now, 10! has 2^8 and 3^4. So 2^6 divides 2^8 and 3^3 divides 3^4. So yes, that's okay. What about x= 2^2 * 3^1 * 5^0 * 7^0. So that's 12. What about x= 2^2 * 3^1 * 5^0 * 7^0. So that's same as 12. So no other primes. So yes, only 6 numbers. Therefore, answer is 6? But I have a feeling that maybe the answer is different. Wait, let me check with another approach.

Alternatively, think of all possible x such that x³ divides 10!. So x must be a divisor of 10!^{1/3}, but since 10! is not a perfect cube, but we can think of the exponents. Wait, but maybe not helpful. Alternatively, think of the exponents for each prime in x. For each prime, the exponent in x is floor(exponent in 10! / 3). Wait, but no, because for example, for prime 2, exponent in 10! is 8. So floor(8/3)=2, so exponent in x can be up to 2. Similarly for others. Therefore, the number of possible exponents for each prime is floor(exponent / 3) + 1? Wait, no. Wait, if the exponent in 10! is e, then the exponent in x can be from 0 to floor(e / 3). Therefore, the number of choices is floor(e / 3) + 1. Wait, for example, for prime 2, e=8, floor(8/3)=2, so number of choices is 2 + 1=3. For prime 3, e=4, floor(4/3)=1, so 1 +1=2. For prime 5, e=2, floor(2/3)=0, so 0 +1=1. For prime 7, e=1, floor(1/3)=0, so 0 +1=1. Therefore, total number of x is 3*2*1*1=6. Therefore, same answer. Therefore, answer is 6. Therefore, I think that is correct.

But let me check with another example. Suppose I take x= 2^2 * 3^1=12. Then x³= 2^6 * 3^3. 10! has 2^8 and 3^4. So yes, 2^6 divides 2^8 and 3^3 divides 3^4. So that's okay. What if I take x= 2^2 * 3^1 * 5^0 * 7^0. So same as 12. What if I take x= 2^2 * 3^1 * 5^0 * 7^0. So same. So no other possibilities. Therefore, answer is 6. Therefore, I think the answer is 6. But maybe I need to check if there are other numbers. Wait, for example, x= 2^0 * 3^0=1. x=2^1 * 3^0=2. x=2^2 * 3^0=4. x=2^0 * 3^1=3. x=2^1 * 3^1=6. x=2^2 * 3^1=12. So that's 6 numbers. So yes, 6. Therefore, answer is 6. Therefore, I think the answer is 6. Wait, but maybe I need to check if there are numbers with exponents beyond that? For example, if x= 2^2 * 3^1 * something? But no, since 5 and 7 can't be included. Therefore, no. Therefore, answer is 6. Therefore, I think the answer is 6. Wait, but maybe I made a mistake in the prime factorization of 10!? Let me check again.

Compute 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me compute the exponents again for primes 2,3,5,7.

For prime 2:

Number of multiples of 2 in 1-10: 5 (2,4,6,8,10)

Number of multiples of 4: 2 (4,8)

Number of multiples of 8: 1 (8)

Number of multiples of 16: 0. So total exponent of 2 is 5 + 2 + 1 = 8. Correct.

For prime 3:

Multiples of 3: 3,6,9. That's 3 numbers.

Multiples of 9: 9. That's 1 number.

Multiples of 27: 0. So total exponent of 3 is 3 + 1 = 4. Correct.

For prime 5:

Multiples of 5: 5,10. That's 2 numbers.

Multiples of 25: 0. So exponent of 5 is 2. Correct.

For prime 7:

Multiples of 7: 7. That's 1 number.

Multiples of 49: 0. So exponent of 7 is 1. Correct.

Therefore, prime factorization is correct. Therefore, exponents are correct. Therefore, answer is 6. Therefore, I think the answer is 6. Therefore, the number of possible integer values of x is 6. Therefore, I think the answer is \boxed{6}.

Wait, but hold on. Wait, I just thought of something. What if x is negative? The problem says "positive factor", so x³ must be a positive factor. Since x³ is positive, x must be positive. Therefore, x is a positive integer. Therefore, negative numbers are excluded. Therefore, no problem. Therefore, answer is 6. Therefore, I think that is correct.

But let me check once more. Let me list all possible x:

1. 1
2. 2
3. 3
4. 4
5. 6
6. 12

Yes, that's 6 numbers. Let me check if there are more numbers. For example, is 8 a possible x? Let's check. If x=8, then x³=512. Let's see if 512 divides 10!.

But 10! is 3628800. Let me divide 3628800 by 512. 3628800 / 512. Let me compute 512 × 7000 = 3,584,000. 3,628,800 - 3,584,000 = 44,800. 512 × 87 = 44,544. 44,800 - 44,544 = 256. 256 is 512 / 2. So total is 7000 + 87 + 0.5 = 7087.5. Therefore, 512 does not divide 10! because it's not an integer. Therefore, x=8 is invalid. Therefore, 8 is not allowed. Therefore, x=8 is not in the list. Similarly, x=9: x³=729. Let me check if 729 divides 10!.

10! is 3628800. 3628800 / 729. Let me compute 729 × 4980 = 729 × 5000 - 729 × 20 = 3,645,000 - 14,580 = 3,630,420. Which is larger than 3,628,800. So 729 × 4980 is too big. Let me try 729 × 4960 = 729 × (5000 - 40) = 3,645,000 - 29,160 = 3,615,840. Then 3,628,800 - 3,615,840 = 12,960. 12,960 / 729 = 17.8... So not an integer. Therefore, 729 does not divide 10!. Therefore, x=9 is invalid. Therefore, x=9 is not allowed. Similarly, x=5: x³=125. 10! has 5², so 5³ would need 5^3, but 10! only has 5². Therefore, 125 does not divide 10!. Therefore, x=5 is invalid. Similarly, x=7: 7³=343. 10! has only one 7, so 343 does not divide 10!. Therefore, invalid. Therefore, x=7 invalid. Therefore, numbers like 5,7,8,9, etc., are invalid. Therefore, only the 6 numbers listed before. Therefore, answer is 6. Therefore, I think the answer is 6. Therefore, I think the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To determine how many possible integer values of $ x $ satisfy the condition that $ x^3 $ is a **positive factor** of $ 10! $, we begin by analyzing the prime factorization of $ 10! $.

---

### Step 1: Prime Factorization of $ 10! $

We compute the prime factorization of $ 10! $ by counting the powers of each prime number less than or equal to 10:

- **Prime 2**:
  $$
  \left\lfloor \frac{10}{2} \right\rfloor + \left\lfloor \frac{10}{4} \right\rfloor + \left\lfloor \frac{10}{8} \right\rfloor = 5 + 2 + 1 = 8
  $$
- **Prime 3**:
  $$
  \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor = 3 + 1 = 4
  $$
- **Prime 5**:
  $$
  \left\lfloor \frac{10}{5} \right\rfloor = 2
  $$
- **Prime 7**:
  $$
  \left\lfloor \frac{10}{7} \right\rfloor = 1
  $$

So, the prime factorization of $ 10! $ is:
$$
10! = 2^8 \times 3^4 \times 5^2 \times 7^1
$$

---

### Step 2: Understanding the Condition $ x^3 \mid 10! $

We are told that $ x^3 $ must divide $ 10! $. That means the prime factorization of $ x^3 $ must be a subset of the prime factorization of $ 10! $.

Let $ x = 2^a \times 3^b \times 5^c \times 7^d $, where $ a, b, c, d $ are non-negative integers. Then:
$$
x^3 = 2^{3a} \times 3^{3b} \times 5^{3c} \times 7^{3d}
$$

For $ x^3 $ to divide $ 10! $, we must have:
- $ 3a \leq 8 $ → $ a \leq \left\lfloor \frac{8}{3} \right\rfloor = 2 $
- $ 3b \leq 4 $ → $ b \leq \left\lfloor \frac{4}{3} \right\rfloor = 1 $
- $ 3c \leq 2 $ → $ c \leq \left\lfloor \frac{2}{3} \right\rfloor = 0 $
- $ 3d \leq 1 $ → $ d \leq \left\lfloor \frac{1}{3} \right\rfloor = 0 $

So, the possible values for the exponents are:
- $ a \in \{0, 1, 2\} $ → 3 choices
- $ b \in \{0, 1\} $ → 2 choices
- $ c = 0 $ → 1 choice
- $ d = 0 $ → 1 choice

---

### Step 3: Counting the Number of Valid $ x $

Each combination of $ a $ and $ b $ gives a unique value of $ x $, since $ c = d = 0 $, and the other primes (5 and 7) cannot be included in $ x $.

So the total number of valid $ x $ is:
$$
3 \times 2 \times 1 \times 1 = 6
$$

---

### Step 4: Listing the Valid $ x $

The valid values of $ x $ are:
- $ 2^0 \times 3^0 = 1 $
- $ 2^1 \times 3^0 = 2 $
- $ 2^2 \times 3^0 = 4 $
- $ 2^0 \times 3^1 = 3 $
- $ 2^1 \times 3^1 = 6 $
- $ 2^2 \times 3^1 = 12 $

Each of these values satisfies the condition that $ x^3 $ divides $ 10! $, and no other values are possible due to the constraints on the exponents.

---

### Final Answer

$$
\boxed{6}
$$

Solution:
First, we factor $10!:$
\begin{align*} 10!&=10\cdot 9\cdot 8\cdot 7 \cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\\ &=2^8\cdot 3^4 \cdot 5^2 \cdot 7.\end{align*}
Hence, $x$ can be $1, 2^1, 2^2, 3, 2^1\cdot 3,\text{ or }2^2\cdot 3$ for a total of $\boxed{6}$ possible values of $x.$


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 asking for the probability that Alice wins the game. Hmm, let me think about how to approach this.

First, let me try to understand the structure of the game. It's a turn-based game with alternating turns between Alice and Bob. Each 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 indefinitely if both keep flipping tails, but the probability of that happening gets smaller each time. But since we're dealing with probabilities, we can model this as an infinite series or use recursion.

Let me try to break it down step by step. Let's denote the probability that Alice wins the game as P. So, we need to find P.

On Alice's first turn, she flips a coin. There's a 50% chance she gets heads, which would mean she immediately wins. If she gets tails (which is also 50% chance), then it's Bob's turn.

Now, when it's Bob's turn, he flips a coin. There's a 50% chance he gets tails and wins, which would mean Alice loses. If he gets heads (another 50% chance), then it goes back to Alice's turn. So, the game essentially resets, but with both players having already had a turn without winning.

So, if we think about this, the probability that Alice wins can be broken down into two scenarios:

1. Alice wins on her first turn: probability 1/2.
2. Alice doesn't win on her first turn (probability 1/2), Bob doesn't win on his turn (probability 1/2), and then the game is back to Alice's turn, so the probability of Alice winning from that point is the same as the original probability P.

Wait, so if both Alice and Bob fail on their first turns, the game is back to the starting situation, but with both players having had a turn. Therefore, the probability of Alice winning in this case would be P again. Therefore, we can write an equation for P.

Let me formalize this:

P = Probability Alice wins on first turn + Probability game continues * Probability Alice wins from the new state.

So, P = (1/2) + (Probability Alice doesn't win first turn) * (Probability Bob doesn't win his turn) * P.

Wait, let me check that. If Alice doesn't win on her first turn (prob 1/2), then Bob gets to flip. If Bob doesn't win on his turn (prob 1/2), then it's back to Alice's turn, and the probability of Alice winning from there is again P. Therefore, the equation is:

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

Simplifying that:

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

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 is the probability that Alice wins 2/3? That seems plausible, but let me check if my reasoning is correct.

Alternatively, maybe I need to consider more turns? Let me think again.

Another way to approach this is to model the game as an infinite series. The probability that Alice wins can be considered as the sum of the probabilities that she wins on her first turn, or on her second turn, or her third turn, etc.

So, let's think about the different ways Alice can win:

- She can win on the first turn: probability 1/2.

- If she doesn't win on the first turn (prob 1/2), Bob doesn't win on his turn (prob 1/2), then Alice gets another chance. So, the probability that Alice wins on her second turn is (1/2)*(1/2)*(1/2) = 1/8.

Wait, hold on. Let me think again. For Alice to win on her second turn, the game must have gone through one full cycle of Alice and Bob both failing. So, the probability of that is (1/2)*(1/2) = 1/4, and then Alice has another chance to win, which is 1/2. So, the probability of Alice winning on her second turn is (1/4)*(1/2) = 1/8.

Similarly, the probability that Alice wins on her third turn would be the probability that both Alice and Bob failed twice in a row, and then Alice succeeds. That would be (1/4)^2 * (1/2) = 1/32.

Wait, so in general, the probability that Alice wins on her nth turn is (1/4)^{n-1} * (1/2). Therefore, the total probability P is the sum from n=1 to infinity of (1/4)^{n-1} * (1/2).

This is a geometric series with first term a = 1/2 and common ratio r = 1/4.

The sum of an infinite geometric series is a / (1 - r). Therefore:

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

So that's the same result as before. Therefore, the probability that Alice wins is 2/3. That seems consistent.

But let me check again with another approach to make sure.

Another way is to use recursion. Let me define P as the probability that Alice wins starting from her turn. Similarly, let me define Q as the probability that Alice wins starting from Bob's turn. Wait, maybe that's helpful.

So, when it's Alice's turn, she has a 1/2 chance to win immediately. If she fails (prob 1/2), then it becomes Bob's turn. So, the probability that Alice wins from her turn is P = 1/2 + 1/2 * Q, where Q is the probability that Alice wins when it's Bob's turn.

Now, when it's Bob's turn, he has a 1/2 chance to win (so Alice loses) and 1/2 chance to pass the turn back to Alice. Therefore, Q = probability that Alice wins from Bob's turn is equal to the probability that Bob doesn't win (1/2) times the probability that Alice wins from her turn, which is P. Therefore, Q = 1/2 * P.

So, substituting back into the equation for P:

P = 1/2 + 1/2 * Q = 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 result. So that seems to confirm it again.

Alternatively, maybe I can think in terms of expected number of turns or something else, but I think the above methods are sufficient.

But let me think if there's another angle. For example, what is the probability that Bob wins? Let me denote that as Q. Then, since the game must end eventually, P + Q = 1. If I can compute Q, then I can find P.

But let's try that. Let me think about Q, the probability that Bob wins.

Similarly, Bob can win on his first turn, which is when Alice fails on her first turn (prob 1/2) and then Bob gets heads (prob 1/2). So, the probability that Bob wins on his first turn is (1/2)*(1/2) = 1/4.

Alternatively, if both Alice and Bob fail on their first turns, then it's back to Alice's turn, and the probability that Bob wins from there is the same as the original probability Q? Wait, no. Wait, if the game resets, then the probability that Bob wins from that point is the same as the original probability that Bob wins starting from Alice's turn. Wait, maybe I need to define Q as the probability that Bob wins starting from his turn. Wait, this might get confusing.

Alternatively, since P + Q = 1, and if I can find another equation relating P and Q, then I can solve for P.

But maybe this is more complicated. Let me try.

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, and Q = 1/2 * P. Then solving gives P = 2/3. Alternatively, if I define Q as the probability that Bob wins starting from his turn, then maybe?

Alternatively, maybe not. Let me try again.

Alternatively, think of the game as a sequence of rounds. Each round consists of Alice flipping and then Bob flipping. However, the game can end in the middle of a round if Alice gets heads on her flip, or Bob gets tails on his flip.

But maybe considering rounds isn't the easiest way. However, let's try.

In each round, first Alice flips. If she gets heads (prob 1/2), she wins. If not (prob 1/2), then Bob flips. If he gets tails (prob 1/2), he wins. If not (prob 1/2), the round ends and the game continues to the next round.

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

Therefore, the probability that Alice wins can be considered 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

Probability Alice wins in the first round is 1/2. Probability game continues is 1/4. Therefore:

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

Which is the same equation as before. Solving:

P - (1/4)P = 1/2 => (3/4)P = 1/2 => P = 2/3. Same result again. So that seems to confirm it again.

Therefore, regardless of the approach—whether using recursion, infinite series, or considering rounds—the probability that Alice wins is 2/3.

But just to be thorough, let me think about the first few terms of the infinite series and see if they add up to 2/3.

The probability that Alice wins on her first turn: 1/2.

Probability she wins on her second turn: (1/2)^2 * (1/2) = 1/8? Wait, no. Wait, for her second turn, the game must have gone through one full cycle of Alice and Bob both failing. So, the probability is (probability both fail once) * probability Alice succeeds. So, (1/2 * 1/2) * 1/2 = 1/8. Then, the probability she wins on her third turn is (probability both fail twice) * probability Alice succeeds: (1/4)^2 * 1/2 = 1/32. Then, fourth turn: (1/4)^3 * 1/2 = 1/128, etc.

So the total probability is 1/2 + 1/8 + 1/32 + 1/128 + ...

Let me compute this sum. Let's factor out 1/2:

1/2 [1 + 1/4 + 1/16 + 1/64 + ...]

The series inside the brackets is a geometric series with first term 1 and ratio 1/4. The sum is 1 / (1 - 1/4) = 4/3. Therefore, total P = (1/2)*(4/3) = 2/3. Yep, same result. So that's consistent.

Alternatively, if I compute partial sums:

First term: 1/2 = 0.5

Second term: 0.5 + 0.125 = 0.625

Third term: 0.625 + 0.03125 = 0.65625

Fourth term: 0.65625 + 0.0078125 = 0.6640625

Fifth term: + 0.001953125 = 0.666015625

And so on, approaching 2/3 ≈ 0.666666..., which is consistent. So that seems to check out.

Another way to think about it: since Alice has the first chance to win, she has an advantage. The probability is 2/3, which is more than 1/2, which makes sense. If both players had equal chances, it would be 1/2, but since Alice goes first, she has a higher chance. So 2/3 seems reasonable.

Alternatively, if we think about the game as a sequence of trials where each trial consists of Alice and Bob each flipping once. But the game can end before the full trial if Alice wins on her flip. However, the probability that Alice wins in the first trial is 1/2. The probability that Bob wins in the first trial is 1/2 * 1/2 = 1/4. The probability that the trial is a tie (both flip heads) is 1/2 * 1/2 = 1/4. Then, if it's a tie, the game restarts. Therefore, the probability that Alice wins is the probability she wins in the first trial plus the probability of a tie times the probability she wins in subsequent trials. So:

P = 1/2 + 1/4 * P

Which again gives P = 2/3. So same result.

Therefore, all methods lead to the same answer. Therefore, I think the probability that Alice wins is 2/3.

But just to make sure, let me think if there's any possible mistake in my reasoning. For example, is there a case where the game could end in a different way? Let me think.

Suppose Alice flips tails, Bob flips tails, Alice flips tails, Bob flips tails, etc. The game could go on forever, but the probability of that is zero, since it's the limit of (1/4)^n as n approaches infinity, which is zero. Therefore, the game will end with probability 1, so the total probability is well-defined.

Another possible mistake could be in the setup of the equations. For example, in the first approach, I considered P = 1/2 + 1/2 * 1/2 * P. But is that correct?

Yes, because if Alice fails (prob 1/2), then Bob has a chance. If Bob fails (prob 1/2), then it's back to Alice's turn, so the probability of Alice winning from there is P. Therefore, the combined probability is 1/2 * 1/2 * P. So that term is correct.

Alternatively, if I thought that after both fail, the probability is P, but maybe I need to consider something else? No, because the game is symmetric in terms of the structure. So, after both fail, it's the same as the original situation, just with both players having had a turn. Therefore, the probability is the same.

Alternatively, maybe someone might think that after both fail, the probability is different? But no, because the game is memoryless. The probability of Alice winning from that point is the same as the original probability.

Therefore, I think the reasoning is sound.

Another way to think about it is to consider the odds. Since Alice has the first chance, she has a higher probability. Let me think in terms of odds. Let me suppose that the probability Alice wins is P, and Bob is 1 - P.

When Alice flips, she has a 1/2 chance to win immediately. If she doesn't, then Bob has a 1/2 chance to win. If he doesn't, then it's back to Alice. So, the game is like a competition where each "cycle" has a chance for Alice or Bob to win, or to continue.

Alternatively, think of it as a Markov chain with states: Alice's turn, Bob's turn, and absorbing states (Alice wins, Bob wins). But that might be overcomplicating.

Alternatively, think of the game as a sequence of independent trials where each trial consists of Alice flipping and then Bob flipping. However, the game can end in the middle of a trial. But as we saw earlier, the probability that Alice wins in the first trial is 1/2, Bob wins is 1/4, and the probability of continuing is 1/4. Therefore, the probability that Alice wins is 1/2 + 1/4 * P, leading to P = 2/3.

Alternatively, if I think of the game as a competition between Alice and Bob, where each has a chance to win on their turn, and the game continues otherwise. Since Alice goes first, her probability is higher.

Alternatively, maybe using conditional probability. Let me define P as the probability that Alice wins. Then, on her first turn, she can win with probability 1/2. If she doesn't, then Bob gets to play. The probability that Bob wins on his turn is 1/2, and if he doesn't, then it's back to Alice. Therefore, the probability that Alice wins is:

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

Which is:

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

Which again gives P = 2/3.

Therefore, all these different ways of thinking lead to the same answer. Therefore, 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 probability and recursion.

---

### **Game Structure**

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

The game continues in this alternating fashion until one of the players wins.

---

### **Defining the Probability**

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

We can break this down into two mutually exclusive cases:

1. **Alice wins on her first turn:** This happens with probability $ \frac{1}{2} $.
2. **Alice does not win on her first turn (probability $ \frac{1}{2} $), and Bob does not win on his turn (probability $ \frac{1}{2} $):** The game resets to Alice's turn, and the probability of Alice winning from this point is again $ P $.

So, the total probability that Alice wins is:

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

---

### **Solving the Equation**

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

Subtract $ \frac{1}{4}P $ from both sides:

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

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

$$
P = \frac{1}{2} \div \frac{3}{4} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}
$$

---

### **Alternative Verification via Infinite Series**

We can also model the game as an infinite series of rounds where Alice has a chance to win on her turn. The probability that Alice wins on her $ n $-th turn is:

$$
\left( \frac{1}{2} \times \frac{1}{2} \right)^{n-1} \times \frac{1}{2} = \left( \frac{1}{4} \right)^{n-1} \times \frac{1}{2}
$$

So the total probability that Alice wins is:

$$
P = \sum_{n=1}^{\infty} \left( \frac{1}{4} \right)^{n-1} \times \frac{1}{2}
$$

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 conditional probability—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 asking for the probability that Alice wins the game. Hmm, let me think about how to approach this.

First, let me try to understand the structure of the game. It's a turn-based game with alternating turns between Alice and Bob. Each 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 indefinitely if both keep flipping tails, but the probability of that happening gets smaller each time. But since we're dealing with probabilities, we can model this as an infinite series or use recursion.

Let me try to break it down step by step. Let's denote the probability that Alice wins the game as P. So, we need to find P.

On Alice's first turn, she flips a coin. There's a 50% chance she gets heads, which would mean she immediately wins. If she gets tails (which is also 50% chance), then it's Bob's turn.

Now, when it's Bob's turn, he flips a coin. There's a 50% chance he gets tails and wins, which would mean Alice loses. If he gets heads (another 50% chance), then it goes back to Alice's turn. So, the game essentially resets, but with both players having already had a turn without winning.

So, if we think about this, the probability that Alice wins can be broken down into two scenarios:

1. Alice wins on her first turn: probability 1/2.
2. Alice doesn't win on her first turn (probability 1/2), Bob doesn't win on his turn (probability 1/2), and then the game is back to Alice's turn, so the probability of Alice winning from that point is the same as the original probability P.

Wait, so if both Alice and Bob fail on their first turns, the game is back to the starting situation, but with both players having had a turn. Therefore, the probability of Alice winning in this case would be P again. Therefore, we can write an equation for P.

Let me formalize this:

P = Probability Alice wins on first turn + Probability game continues * Probability Alice wins from the new state.

So, P = (1/2) + (Probability Alice doesn't win first turn) * (Probability Bob doesn't win his turn) * P.

Wait, let me check that. If Alice doesn't win on her first turn (prob 1/2), then Bob gets to flip. If Bob doesn't win on his turn (prob 1/2), then it's back to Alice's turn, and the probability of Alice winning from there is again P. Therefore, the equation is:

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

Simplifying that:

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

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 is the probability that Alice wins 2/3? That seems plausible, but let me check if my reasoning is correct.

Alternatively, maybe I need to consider more turns? Let me think again.

Another way to approach this is to model the game as an infinite series. The probability that Alice wins can be considered as the sum of the probabilities that she wins on her first turn, or on her second turn, or her third turn, etc.

So, let's think about the different ways Alice can win:

- She can win on the first turn: probability 1/2.

- If she doesn't win on the first turn (prob 1/2), Bob doesn't win on his turn (prob 1/2), then Alice gets another chance. So, the probability that Alice wins on her second turn is (1/2)*(1/2)*(1/2) = 1/8.

Wait, hold on. Let me think again. For Alice to win on her second turn, the game must have gone through one full cycle of Alice and Bob both failing. So, the probability of that is (1/2)*(1/2) = 1/4, and then Alice has another chance to win, which is 1/2. So, the probability of Alice winning on her second turn is (1/4)*(1/2) = 1/8.

Similarly, the probability that Alice wins on her third turn would be the probability that both Alice and Bob failed twice in a row, and then Alice succeeds. That would be (1/4)^2 * (1/2) = 1/32.

Wait, so in general, the probability that Alice wins on her nth turn is (1/4)^{n-1} * (1/2). Therefore, the total probability P is the sum from n=1 to infinity of (1/4)^{n-1} * (1/2).

This is a geometric series with first term a = 1/2 and common ratio r = 1/4.

The sum of an infinite geometric series is a / (1 - r). Therefore:

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

So that's the same result as before. Therefore, the probability that Alice wins is 2/3. That seems consistent.

But let me check again with another approach to make sure.

Another way is to use recursion. Let me define P as the probability that Alice wins starting from her turn. Similarly, let me define Q as the probability that Alice wins starting from Bob's turn. Wait, maybe that's helpful.

So, when it's Alice's turn, she has a 1/2 chance to win immediately. If she fails (prob 1/2), then it becomes Bob's turn. So, the probability that Alice wins from her turn is P = 1/2 + 1/2 * Q, where Q is the probability that Alice wins when it's Bob's turn.

Now, when it's Bob's turn, he has a 1/2 chance to win (so Alice loses) and 1/2 chance to pass the turn back to Alice. Therefore, Q = probability that Alice wins from Bob's turn is equal to the probability that Bob doesn't win (1/2) times the probability that Alice wins from her turn, which is P. Therefore, Q = 1/2 * P.

So, substituting back into the equation for P:

P = 1/2 + 1/2 * Q = 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 result. So that seems to confirm it again.

Alternatively, maybe I can think in terms of expected number of turns or something else, but I think the above methods are sufficient.

But let me think if there's another angle. For example, what is the probability that Bob wins? Let me denote that as Q. Then, since the game must end eventually, P + Q = 1. If I can compute Q, then I can find P.

But let's try that. Let me think about Q, the probability that Bob wins.

Similarly, Bob can win on his first turn, which is when Alice fails on her first turn (prob 1/2) and then Bob gets heads (prob 1/2). So, the probability that Bob wins on his first turn is (1/2)*(1/2) = 1/4.

Alternatively, if both Alice and Bob fail on their first turns, then it's back to Alice's turn, and the probability that Bob wins from there is the same as the original probability Q? Wait, no. Wait, if the game resets, then the probability that Bob wins from that point is the same as the original probability that Bob wins starting from Alice's turn. Wait, maybe I need to define Q as the probability that Bob wins starting from his turn. Wait, this might get confusing.

Alternatively, since P + Q = 1, and if I can find another equation relating P and Q, then I can solve for P.

But maybe this is more complicated. Let me try.

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, and Q = 1/2 * P. Then solving gives P = 2/3. Alternatively, if I define Q as the probability that Bob wins starting from his turn, then maybe?

Alternatively, maybe not. Let me try again.

Alternatively, think of the game as a sequence of rounds. Each round consists of Alice flipping and then Bob flipping. However, the game can end in the middle of a round if Alice gets heads on her flip, or Bob gets tails on his flip.

But maybe considering rounds isn't the easiest way. However, let's try.

In each round, first Alice flips. If she gets heads (prob 1/2), she wins. If not (prob 1/2), then Bob flips. If he gets tails (prob 1/2), he wins. If not (prob 1/2), the round ends and the game continues to the next round.

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

Therefore, the probability that Alice wins can be considered 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

Probability Alice wins in the first round is 1/2. Probability game continues is 1/4. Therefore:

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

Which is the same equation as before. Solving:

P - (1/4)P = 1/2 => (3/4)P = 1/2 => P = 2/3. Same result again. So that seems to confirm it again.

Therefore, regardless of the approach—whether using recursion, infinite series, or considering rounds—the probability that Alice wins is 2/3.

But just to be thorough, let me think about the first few terms of the infinite series and see if they add up to 2/3.

The probability that Alice wins on her first turn: 1/2.

Probability she wins on her second turn: (1/2)^2 * (1/2) = 1/8? Wait, no. Wait, for her second turn, the game must have gone through one full cycle of Alice and Bob both failing. So, the probability is (probability both fail once) * probability Alice succeeds. So, (1/2 * 1/2) * 1/2 = 1/8. Then, the probability she wins on her third turn is (probability both fail twice) * probability Alice succeeds: (1/4)^2 * 1/2 = 1/32. Then, fourth turn: (1/4)^3 * 1/2 = 1/128, etc.

So the total probability is 1/2 + 1/8 + 1/32 + 1/128 + ...

Let me compute this sum. Let's factor out 1/2:

1/2 [1 + 1/4 + 1/16 + 1/64 + ...]

The series inside the brackets is a geometric series with first term 1 and ratio 1/4. The sum is 1 / (1 - 1/4) = 4/3. Therefore, total P = (1/2)*(4/3) = 2/3. Yep, same result. So that's consistent.

Alternatively, if I compute partial sums:

First term: 1/2 = 0.5

Second term: 0.5 + 0.125 = 0.625

Third term: 0.625 + 0.03125 = 0.65625

Fourth term: 0.65625 + 0.0078125 = 0.6640625

Fifth term: + 0.001953125 = 0.666015625

And so on, approaching 2/3 ≈ 0.666666..., which is consistent. So that seems to check out.

Another way to think about it: since Alice has the first chance to win, she has an advantage. The probability is 2/3, which is more than 1/2, which makes sense. If both players had equal chances, it would be 1/2, but since Alice goes first, she has a higher chance. So 2/3 seems reasonable.

Alternatively, if we think about the game as a sequence of trials where each trial consists of Alice and Bob each flipping once. But the game can end before the full trial if Alice wins on her flip. However, the probability that Alice wins in the first trial is 1/2. The probability that Bob wins in the first trial is 1/2 * 1/2 = 1/4. The probability that the trial is a tie (both flip heads) is 1/2 * 1/2 = 1/4. Then, if it's a tie, the game restarts. Therefore, the probability that Alice wins is the probability she wins in the first trial plus the probability of a tie times the probability she wins in subsequent trials. So:

P = 1/2 + 1/4 * P

Which again gives P = 2/3. So same result.

Therefore, all methods lead to the same answer. Therefore, I think the probability that Alice wins is 2/3.

But just to make sure, let me think if there's any possible mistake in my reasoning. For example, is there a case where the game could end in a different way? Let me think.

Suppose Alice flips tails, Bob flips tails, Alice flips tails, Bob flips tails, etc. The game could go on forever, but the probability of that is zero, since it's the limit of (1/4)^n as n approaches infinity, which is zero. Therefore, the game will end with probability 1, so the total probability is well-defined.

Another possible mistake could be in the setup of the equations. For example, in the first approach, I considered P = 1/2 + 1/2 * 1/2 * P. But is that correct?

Yes, because if Alice fails (prob 1/2), then Bob has a chance. If Bob fails (prob 1/2), then it's back to Alice's turn, so the probability of Alice winning from there is P. Therefore, the combined probability is 1/2 * 1/2 * P. So that term is correct.

Alternatively, if I thought that after both fail, the probability is P, but maybe I need to consider something else? No, because the game is symmetric in terms of the structure. So, after both fail, it's the same as the original situation, just with both players having had a turn. Therefore, the probability is the same.

Alternatively, maybe someone might think that after both fail, the probability is different? But no, because the game is memoryless. The probability of Alice winning from that point is the same as the original probability.

Therefore, I think the reasoning is sound.

Another way to think about it is to consider the odds. Since Alice has the first chance, she has a higher probability. Let me think in terms of odds. Let me suppose that the probability Alice wins is P, and Bob is 1 - P.

When Alice flips, she has a 1/2 chance to win immediately. If she doesn't, then Bob has a 1/2 chance to win. If he doesn't, then it's back to Alice. So, the game is like a competition where each "cycle" has a chance for Alice or Bob to win, or to continue.

Alternatively, think of it as a Markov chain with states: Alice's turn, Bob's turn, and absorbing states (Alice wins, Bob wins). But that might be overcomplicating.

Alternatively, think of the game as a sequence of independent trials where each trial consists of Alice flipping and then Bob flipping. However, the game can end in the middle of a trial. But as we saw earlier, the probability that Alice wins in the first trial is 1/2, Bob wins is 1/4, and the probability of continuing is 1/4. Therefore, the probability that Alice wins is 1/2 + 1/4 * P, leading to P = 2/3.

Alternatively, if I think of the game as a competition between Alice and Bob, where each has a chance to win on their turn, and the game continues otherwise. Since Alice goes first, her probability is higher.

Alternatively, maybe using conditional probability. Let me define P as the probability that Alice wins. Then, on her first turn, she can win with probability 1/2. If she doesn't, then Bob gets to play. The probability that Bob wins on his turn is 1/2, and if he doesn't, then it's back to Alice. Therefore, the probability that Alice wins is:

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

Which is:

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

Which again gives P = 2/3.

Therefore, all these different ways of thinking lead to the same answer. Therefore, 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 probability and recursion.

---

### **Game Structure**

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

The game continues in this alternating fashion until one of the players wins.

---

### **Defining the Probability**

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

We can break this down into two mutually exclusive cases:

1. **Alice wins on her first turn:** This happens with probability $ \frac{1}{2} $.
2. **Alice does not win on her first turn (probability $ \frac{1}{2} $), and Bob does not win on his turn (probability $ \frac{1}{2} $):** The game resets to Alice's turn, and the probability of Alice winning from this point is again $ P $.

So, the total probability that Alice wins is:

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

---

### **Solving the Equation**

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

Subtract $ \frac{1}{4}P $ from both sides:

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

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

$$
P = \frac{1}{2} \div \frac{3}{4} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}
$$

---

### **Alternative Verification via Infinite Series**

We can also model the game as an infinite series of rounds where Alice has a chance to win on her turn. The probability that Alice wins on her $ n $-th turn is:

$$
\left( \frac{1}{2} \times \frac{1}{2} \right)^{n-1} \times \frac{1}{2} = \left( \frac{1}{4} \right)^{n-1} \times \frac{1}{2}
$$

So the total probability that Alice wins is:

$$
P = \sum_{n=1}^{\infty} \left( \frac{1}{4} \right)^{n-1} \times \frac{1}{2}
$$

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 conditional probability—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.