# The game of balls

$6$ balls marked as $1,2,3,4,5$ and $6$ are kept in a box. Two players A and B start to take out $1$ ball at a time from the box one after another without replacing the ball till the game is over. The number marked on the ball is added each time to the previous sum to get the sum of numbers marked on the balls taken out. If this sum is even, then $1$ point is given to the player.

The first player to get $2$ points is declared winner. At the start of the game, the sum is $0$. If A starts to take out the ball, find the number of ways in which A can win the match.

MY APPROACH: In order to make cases, I prepared the following table and tried to list the ways in which A can win.

Number of ways for:

CASE $1$:$$3!=6$$

CASE $2$: $${3 \choose 1} \cdot {3 \choose 2} \cdot 2!=18$$

CASE $3$: $${3 \choose 3} \cdot {3 \choose 2} \cdot 2! \cdot 3! = 36$$

CASE $4$: $${3 \choose 3} \cdot {3 \choose 2} \cdot 2! \cdot 3! = 36$$

CASE $5$: $${3 \choose 3} \cdot {3 \choose 2} \cdot 2! \cdot 3!= 36$$

CASE $6$: $${3 \choose 3} \cdot {3 \choose 2} \cdot 2! \cdot 3!= 36$$

Total number of ways: $$\Rightarrow 36 \cdot 4 + 18 + 6 = 168$$

But my books says that the answer should be $96$ which is grossly different from my answer.

Where have I faltered in the whole approach?

• I think the book's answer is wrong and you are right unless we both missed something. – Arnaud Mortier Aug 8 '18 at 15:07
• @drhab In case $3$ the points are allowed as follows: $\color{red}1-0-0-1-\color{red}1$. $A$ reaches $2$ points first. – Arnaud Mortier Aug 8 '18 at 15:12
• Perhaps the sums are kept separate? So, maybe you have A's total and B's total as separate running totals. Winning scenarios would be: $$\begin{matrix}E & E & E & & \\ E & E & O & O & O \\ E & O & E & & \\ E & O & O & E & O \\ O & E & O & O & E \\ O & O & O & E & E\end{matrix}$$ But it looks like this gives 168 still. – InterstellarProbe Aug 8 '18 at 15:53

Player $A$ can either win on their second move, or on their third move.

$A$ wins on their second move iff the sequence starts with $E$ and then has an even number of $O$s, so

$$EEE\qquad EOO$$

For $A$ to win on their third move it is necessary that the first five throws contain exactly two $O$s (so that the last move by $A$ brings them a point). There are ${5\choose 2}=10$ such sequences, from which we subtract the two starting with $EEE$ and $EOO$, so only $8$ configurations to study:

I write the sequence of points earned and stop when either $A$ wins or it is clear that they won't.

\begin{align*}OOEEE &\qquad 0111& B\\ OEOEE &\qquad 00111& A\\ OEEOE &\qquad 000&not \ A \\ OEEEO &\qquad000&not \ A \\ EOEOE &\qquad 10011&A\\ EOEEO &\qquad 10001&A\\ EEOOE &\qquad 1101&B\\ EEOEO &\qquad 11001&A\\ \end{align*}

Therefore your list of configurations is complete, and your count is correct.