Three players, A, B and C, toss a coin cyclically in that order (that is A, B, C, A, B, C, A, B, …)

Q: Wait, why is the common ratio $q^3$ and not just $q$?

That is a brilliant observation! The common ratio represents the probability of the game *returning* to the same state (i.e., it is A's turn again). Since A, B, and C toss in order, it takes 3 tosses to get back to A. Each toss has a failure probability of $q$. Thus, the probability of the game cycling back to A without anyone winning is $q \times q \times q = q^3$. We are looking at the probability of the *cycle* repeating, not just a single toss.

Q: Is it possible for the game to go on forever?

Mathematically, the series is infinite, which might make it feel like the game never ends. However, the sum of the probabilities is 1, which means the probability of the game continuing indefinitely is 0. The 'Head' will eventually show up with probability 1. The infinite series is just a tool to capture the sum of all possible winning scenarios.

Q: Why can't we just say $\alpha = \beta = \gamma = 1/3$?

That would only be true if the game were perfectly fair and symmetric in terms of turn order. But here, A goes first. A has the 'first-mover advantage'. The math shows that $\alpha > \beta > \gamma$, which makes perfect physical sense! A gets the first shot, B gets the second, and C gets the third. The further down the order you are, the more likely it is that someone else has already won.

Question

Three players, $A, B$ and $C$, toss a coin cyclically in that order (that is $A, B, C, A, B, C, A, B, \dots$) till a head shows. Let $p$ be the probability that the coin shows a head. Let $\alpha, \beta$ and $\gamma$ be, respectively, the probabilities that $A, B$ and $C$ gets the first head. Prove that $\beta = (1 - p) \alpha$. Determine $\alpha, \beta$ and $\gamma$ (in terms of $p$).

Sigma Percentile · Accepted Answer

### The Cyclic Dance of Probability

Imagine three friends, $A$, $B$, and $C$, sitting in a circle. They are playing a game of chance, tossing a coin in a fixed, repeating order: $A$, then $B$, then $C$, and back to $A$.

The game ends the moment a 'Head' appears. This is not just a game; it is a beautiful, infinite dance of probabilities. As a student of JEE Advanced, you must learn to see the elegance in this repetition.

### Phase 1: The Logic of the Cycle

Before we touch the algebra, we must define our universe. Let the probability of getting a Head be $p$. This is our success condition.

If a player gets a Tail, they fail, and the coin passes to the next person. The probability of a Tail is $q = 1 - p$.

Every time the coin moves to the next player, a 'failure' has occurred. This is the fundamental building block of our solution.

### Phase 2: The Infinite Series of Player A

Focus your attention on Player $A$. $A$ can win on the very first toss. If that fails, $A$ must wait for $B$ and $C$ to both get Tails, so $A$ gets another chance on the fourth toss, then the seventh, and so on.

The probability of winning on the first toss is simply $p$. For the fourth toss, $A$, $B$, and $C$ must have rolled Tails first, which is $q 	imes q 	imes q = q^3$, followed by $A$ rolling a Head, which is $p$.

So the second term is $q^3 p$. This pattern continues: $p + q^3 p + q^6 p + \dots$.

This is an infinite Geometric Progression (GP) where the first term $a = p$ and the common ratio $r = q^3$. Using the sum formula $S_{\infty} = \frac{a}{1-r}$, we find:

$$\alpha = \frac{p}{1 - q^3}$$

### Phase 3: The Symmetry of Failure

Now, let us look at Player $B$. $B$ can only win if $A$ fails on the first toss. $B$'s winning turns are the second, fifth, eighth, and so on.

The probability for the second toss is $q 	imes p$. For the fifth toss, we need four Tails followed by a Head, giving $q^4 p$.

This creates the series $\beta = qp + q^4 p + q^7 p + \dots$. If we factor out a single $q$, we get $\beta = q(p + q^3 p + q^6 p + \dots)$.

Look closely at the bracket—it is exactly the series for $\alpha$! Thus, we have the elegant relationship:

$$\beta = q\alpha = (1-p)\alpha$$

Similarly, for Player $C$, who only gets a chance on the third, sixth, and ninth tosses, the probability is $\gamma = q^2 p + q^5 p + q^8 p + \dots$. Factoring out $q^2$, we see that $\gamma = q^2 \alpha$.

This is the beauty of cyclic games: the order of play creates a 'handicap' of factor $q$ for each subsequent player.

### Phase 4: The Grand Unification

Finally, we express these in terms of $p$. Substituting $q = 1-p$ into our expressions, we get:

$$\alpha = \frac{p}{1 - (1-p)^3}$$

$$\beta = \frac{p(1-p)}{1 - (1-p)^3}$$

$$\gamma = \frac{p(1-p)^2}{1 - (1-p)^3}$$

As a final sanity check, if you sum $\alpha + \beta + \gamma$, you get $\alpha(1 + q + q^2)$. Substituting the formula for $\alpha$, this simplifies perfectly to **1**.

This confirms that someone, eventually, must win. You have just mastered the cyclic probability game—a classic JEE concept that rewards those who look for patterns rather than brute force.

Visualized Solution

The Cyclic Game Setup

Defining Success and Failure

Player A's Winning Turns

Summing the GP for $α$

Player B's Winning Turns

Proving $β = (1 - p) α$

Player C's Winning Turns

Relating $γ$ to $α$

Final Expressions for $α, β, γ$

Key Takeaways and Summary

The Sigma Insight: Addition and Multiplication Theorems

The Cyclic Dance of Probability

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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List-I

List-II

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