Q.$A$ and $B$ throw a pair of dice alternately. $A$ wins the game if he gets a total of $6$ and $B$ wins if she gets a total of $7$. If $A$ starts the game, find the probability of winning the game by $A$ in third throw of the pair of dice.

A wins on her 3rd throw means A and B both fail on their first two turns, then A succeeds. With a pair of dice ($36$ outcomes):

So $P(A\text{ fails})=\dfrac{31}{36}$ and $P(B\text{ fails})=\dfrac{5}{6}$.

A starts, so the order is A, B, A, B, A. The required sequence (A fail, B fail, A fail, B fail, A win) has probability …

Single-throw probabilities

A pair of dice has $36$ equally likely outcomes.

• A wins by throwing a total of $6$: the pairs are $(1,5),(2,4),(3,3),(4,2),(5,1)$, so $P(A)=\dfrac{5}{36}$, and $P(A\text{ fails})=\dfrac{31}{36}$.
• B wins by throwing a total of $7$: the pairs are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$, so $P(B)=\dfrac{6}{36}=\dfrac{1}{6}$, and $P(B\text{ fails})=\dfrac{5}{6}$.

The required sequence

A starts, and the players alternate: A, B, A, B, A, … For A to win on her third throw, everyone before that must have failed (otherwise the game would already be over):

1. A fails (1st turn): $\dfrac{31}{36}$
2. B fails (1st turn): $\dfrac{5}{6}$
3. A fails (2nd turn): $\dfrac{31}{36}$
4. B fails (2nd turn): $\dfrac{5}{6}$ …

Method: Probability of a fixed sequence of independent trials

Use this for "player wins on the $k$-th turn" or "first success happens exactly here" problems in an alternating game: the outcome is one specific ordered chain of wins and losses.

Steps

Step 1: Find each single-turn success probability (and its complement)

Count the favourable outcomes over the sample space for one turn, and take $1-p$ for a failure on that turn. For a pair of dice there are $36$ outcomes, so count the ways to make each required total.

Step 2: Write the exact required sequence in order …

Common Mistakes

Mistake 1: Miscounting the winning totals

Why it's wrong: a total of $6$ arises in $5$ ways and a total of $7$ in $6$ ways, so a failure to make $6$ has probability $\tfrac{31}{36}$ (not $\tfrac{30}{36}$). Correct approach: enumerate the pairs for each total carefully.

Mistake 2: Not forcing all earlier turns to be losses

Why it's wrong: if any earlier turn were a win, the game would already be over, contradicting "wins on the third throw". Correct approach: every preceding turn must be a failure. …