3 students play head and toss game. The one who gets head first is the winner. If we assume that the game is indefinitely continued, the probability of winning the game of 3 students is

To determine the probability of each student winning the game, we assume that the students take turns flipping a coin, and the first student to get a head wins. Since the game can be repeated indefinitely, we'll analyze the situation step by step.

### Step 1: Probability of Winning on the First Round
- **Student A:** The probability that Student A wins on the first round is \(\frac{1}{2}\).
- **Student B:** For Student B to win, Student A must first lose (which happens with probability \(\frac{1}{2}\)), and then Student B must win. The probability of this sequence is \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).
- **Student C:** Similarly, for Student C to win, both Student A and Student B must lose first (each with probability \(\frac{1}{2}\)), and then Student C must win. The probability of this sequence is \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\).

### Step 2: Probability of Winning on Subsequent Rounds
If all students lose in the first round, the situation resets, and the probabilities remain the same as the initial setup. Therefore, each student's chance of winning after a round where no one wins can be thought of as another iteration of the same scenario.

### Step 3: Calculate the Total Probability
The total probability that each student wins is a sum of the probabilities of them winning in the first round, the second round (after the first loop where no one wins), and so on.

- **Student A:** 
  - Wins on the first round: \(\frac{1}{2}\)
  - Wins after one full loop: \(\frac{1}{8} \times \frac{1}{2}\)
  - Wins after two full loops: \(\frac{1}{64} \times \frac{1}{2}\)
  - This forms an infinite geometric series: \[
  P_A = \frac{1}{2} + \frac{1}{2} \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{16} + \dots
  \]
  - Sum of this geometric series: \[
  P_A = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{2}{3}
  \]

- **Student B:**
  - The same method can be applied. The sequence would be \(\frac{1}{4} + \frac{1}{4} \times \frac{1}{4} + \dots\), giving:
  \[
  P_B = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}
  \]

- **Student C:**
  - Similarly, the probability sequence is \(\frac{1}{8} + \frac{1}{8} \times \frac{1}{4} + \dots\), giving:
  \[
  P_C = \frac{\frac{1}{8}}{1 - \frac{1}{4}} = \frac{1}{6}
  \]

### Conclusion:
The probabilities for each student are:
- **Student A:** \(\frac{2}{3}\)
- **Student B:** \(\frac{1}{3}\)
- **Student C:** \(\frac{1}{6}\)

However, there's a mistake in the original evaluation above as the probabilities should sum to 1. Let's correct it:

- **Student A:** \(P_A = \frac{4}{7}\)
- **Student B:** \(P_B = \frac{2}{7}\)
- **Student C:** \(P_C = \frac{1}{7}\)

So, the correct probabilities are:
- **Student A:** \(\frac{4}{7}\)
- **Student B:** \(\frac{2}{7}\)
- **Student C:** \(\frac{1}{7}\)

Would you like further details or have any questions?

### Related Questions:
1. What is the probability if there are 4 students playing?
2. How would the probabilities change if a biased coin was used?
3. What if the game stops after a certain number of rounds?
4. Can you generalize the probabilities for \(n\) students?
5. What is the expected number of rounds before someone wins?

### Tip:
For problems involving probabilities of repeated independent events, setting up the correct geometric series can simplify finding the solution.

Explore the probability of winning in a coin toss game involving 3 students. Learn how to calculate the probabilities using geometric and infinite series methods. Suitable for advanced high school students and beyond.

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