Q.$A$ and $B$ throw a die alternatively till one of them gets a '6' and wins the game. Find their respective probabilities of winning, if $A$ starts first.

Concept: Geometric Distribution — the game is a sequence of independent trials (die rolls), each with success probability $p = \frac{1}{6}$ (getting a 6). The player who gets the first success wins.

Step 1: Let $P(A)$ be the probability that $A$ wins. $A$ starts, so $A$ wins on the first roll with probability $\frac{1}{6}$.

Step 2: If $A$ fails (probability $\frac{5}{6}$), then $B$ gets a turn. $B$ wins immediately with probability $\frac{1}{6}$, but only if $A$ has already failed. So $B$'s chance from this branch is $\frac{5}{6} \cdot \frac{1}{6}$.

Step 3: If both fail in the first round (probability $\frac{5}{6} \cdot \frac{5}{6}$), the game resets exactly to the start — $A$ rolls again. This gives a recurrence: …

Why the Geometric Distribution?

When two players take turns rolling a fair die, and the first to roll a 6 wins, each roll is an independent trial with success probability $p = \frac{1}{6}$. The game stops at the first success. This is exactly the setting of the geometric distribution — the number of trials until the first success.

But here, the "trials" are not independent in the sense of who gets to roll — they alternate between A and B. So we need to think in terms of rounds: each round consists of A's turn followed by B's turn. However, the game can end in the middle of a round if A wins immediately.

The key insight: A wins if the first success occurs on an odd-numbered trial (1st, 3rd, 5th, ...), and B wins if it occurs on an even-numbered trial (2nd, 4th, 6th, ...).

Here $p = \frac{1}{6}$, so $q = 1-p = \frac{5}{6}$.

Step-by-step solution

1. Probability that A wins on her first turn (trial 1)

A rolls and gets a 6 immediately. This happens with probability:

2. Probability that A wins on her second turn (trial 3)

For this to happen, both A and B must fail on their first turns, then A succeeds on her second turn. That's two failures followed by a success:

3. Probability that A wins on her third turn (trial 5)

Now we need four failures (A fails twice, B fails twice) then A succeeds:

4. Pattern for A's total probability

A wins on trials 1, 3, 5, 7, ... — that is, on odd-numbered trials $2n-1$ for $n = 1, 2, 3, \dots$. Each such trial requires $2n-2$ failures before the success. So:

This is an infinite geometric series with first term $p$ and common ratio $q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$.

5. Computing A's probability

6. Probability that B wins …

Method: Winning on Alternating Turns (first-success via geometric series)

Use this for "two players take turns, first to achieve a success wins" games. You sum the probabilities of all the turns on which a chosen player can win; those probabilities form an infinite geometric series (or, equivalently, you set up a one-step recurrence).

Steps

Step 1: Fix the single-trial success and failure probabilities.

Let $p$ be the chance of success on one trial and $q=1-p$ the chance of failure (here, rolling a six, $p=\tfrac16$, $q=\tfrac56$).

Step 2: Identify which trials each player wins on.

The player who starts wins on trials $1,3,5,\dots$ (odd); the second player wins on $2,4,6,\dots$ (even). Winning on trial $k$ requires $k-1$ failures followed by a success, with probability $q^{\,k-1}p$.

Step 3: Sum the geometric series for each player.

For the starter,

$P(\text{starter})=p+q^{2}p+q^{4}p+\cdots=\frac{p}{1-q^{2}},$ …

Common Mistakes

Mistake 1: Assuming the starter's probability is exactly double the opponent's.

Why it's wrong: the ratio is $P(A):P(B)=6:5$, not $2:1$. Correct approach: sum the geometric series; the head-start is only the first roll, and after that the turns are symmetric.

Mistake 2: Summing only the first term or a finite number of turns.

Why it's wrong: a player can win on the 1st, 3rd, 5th, ... trial, so the win probability is an infinite series. Correct approach: use $\dfrac{p}{1-q^2}$ (or the recurrence $P(A)=p+q^2P(A)$). …