A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin.

Q: Wait, why do we set $P(X) = \frac{1}{2}$? The problem doesn't explicitly say the probability is $0.5$.

That is a brilliant observation! The problem states that the probability of winning is equal for both players. Since the game must eventually end (someone must win), the sum of their probabilities must be $P(X) + P(Y) = 1$. If $P(X) = P(Y)$, then $2P(X) = 1$, which forces $P(X) = \frac{1}{2}$. It is a beautiful consequence of the symmetry of the game.

Q: Why is the common ratio $r = \frac{1-p}{2}$? I am getting confused with the terms.

Think about the transition from one winning turn to the next. For X to win on turn 3 instead of turn 1, X must fail (probability $1-p$) AND Y must fail (probability $\frac{1}{2}$). So, to get from the probability of winning on turn 1 to the probability of winning on turn 3, we multiply by $(1-p) \cdot \frac{1}{2}$. This multiplier is constant for every subsequent turn, making it our common ratio $r$.

Q: Can we solve this using a recursive relation instead of an infinite series?

Absolutely! You can define $P(X)$ as the probability of X winning. If X fails on the first turn (probability $1-p$) and Y fails on the second turn (probability $\frac{1}{2}$), the game effectively 'resets' to the starting position. Thus, $P(X) = p + (1-p)(\frac{1}{2})P(X)$. Solving this linear equation will yield the exact same result. Both methods are elegant; the G.P. method is just more visual!

Question

A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p' is :-

Options:

(A) 1/3

(B) 2/5

(C) 1/4

(D) 1/5

Sigma Percentile · Accepted Answer

### The Dance of Probability: Unraveling the Infinite Game

My dear student, welcome to one of the most elegant problems in probability theory. When we look at a game involving alternating turns and infinite possibilities, it is easy to feel overwhelmed.

But today, we are going to strip away the complexity and look at the beautiful, rhythmic structure hidden beneath the surface. Imagine you are standing in a room with Player X and Player Y. They are tossing coins, and the tension is palpable. Let us break this down, step by step.

### Phase 1: Defining the Players and the Timeline

First, let us define our variables. Player X is using a biased coin where the probability of heads is $p$. This means the probability of tails is $1-p$.

Player Y, on the other hand, is using a fair coin, so their probability of heads is $\frac{1}{2}$ and tails is $\frac{1}{2}$.

Now, visualize the timeline. The game is a sequence of turns: $1, 2, 3, 4, 5, \dots$. Player X goes on $1, 3, 5, \dots$, and Player Y goes on $2, 4, 6, \dots$.

The game ends the moment someone throws a head. This means X can only win on odd-numbered turns.

### Phase 2: The Anatomy of a Win

Let us calculate the probability of X winning on specific turns.

1. **Turn 1:** X wins immediately. The probability is simply $P_1 = p$.

2. **Turn 3:** For X to win here, X must fail on turn 1, Y must fail on turn 2, and then X must succeed on turn 3. The probability is $P_3 = (1-p) \cdot \frac{1}{2} \cdot p$.

3. **Turn 5:** X must fail, Y must fail, X must fail again, Y must fail again, and finally X succeeds. The probability is $P_5 = (1-p)^2 \cdot (\frac{1}{2})^2 \cdot p$.

Do you see the pattern emerging? Each time we move to the next possible winning turn for X, we are multiplying by the probability of both players failing their respective turns: $(1-p) \cdot \frac{1}{2}$.

### Phase 3: The Infinite Geometric Progression

The total probability of X winning, $P(X)$, is the sum of all these mutually exclusive cases: $P(X) = p + p(\frac{1-p}{2}) + p(\frac{1-p}{2})^2 + \dots$. This is a classic infinite Geometric Progression (G.P.) where the first term $a = p$ and the common ratio $r = \frac{1-p}{2}$.

Using the sum formula for an infinite G.P., $S_{\infty} = \frac{a}{1-r}$, we get:

$$P(X) = \frac{p}{1 - \frac{1-p}{2}}$$

Let us simplify the denominator with care. $1 - \frac{1-p}{2} = \frac{2 - (1-p)}{2} = \frac{1+p}{2}$. Thus, our expression becomes:

$$P(X) = \frac{p}{\frac{1+p}{2}} = \frac{2p}{1+p}$$

### Phase 4: The Final Symmetry

We are told the game is fair, meaning $P(X) = P(Y)$. Since the game must end, $P(X) + P(Y) = 1$, which implies $P(X) = \frac{1}{2}$.

Now, we simply equate our derived expression to this value:

$$\frac{2p}{1+p} = \frac{1}{2}$$

Cross-multiplying gives us $4p = 1 + p$. Subtracting $p$ from both sides, we find $3p = 1$, or **$p = \frac{1}{3}$**.

There it is! The math is clean, logical, and deeply satisfying. You have successfully navigated an infinite series to find a single, precise value.

Remember, in JEE Advanced, the complexity is often just a mask for a beautiful, simple pattern. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process.

Select Answer:

Visualized Solution

$P l a y er X^{'} s P r o babi l i t i es$

$P l a y er Y^{'} s P r o babi l i t i es$

$T h e G am e S e q u e n ce$

$X W in so n T u r n 1$

$X W in so n T u r n 3$

$X W in so n T u r n 5$

$T o t a l P r o babi l i t y f or X$

$I d e n t i f y in g t h e G eo m e t r i c P r o g r ess i o n$

$S u m o f I n f ini t e G . P .$

$S im pl i f y in g P (X)$

$U s in g t h e G i v e n C o n d i t i o n$

$S o l v in g f or p$

The Sigma Insight: Addition and Multiplication Theorems

The Dance of Probability

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PlayerX′sProbabilities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PlayerY′sProbabilities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheGameSequence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XWinsonTurn1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XWinsonTurn3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XWinsonTurn5

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TotalProbabilityforX

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IdentifyingtheGeometricProgression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SumofInfiniteG.P.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingP(X)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } UsingtheGivenCondition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solvingforp