A and B play a game with colored balls. A starts with 6 balls: 2 orange, 2 yellow and 2 green. B starts with 4 balls: 2 pinks ...

Question

CONTD.

$A$ and $B$ play a game with colored balls. $A$ starts with $6$ balls: $2$ orange, $2$ yellow and $2$ green. $B$ starts with $4$ balls: $2$ pinks and $2$ gray. Player "$A$" plays first, and both alternate turns. On each turn, the player places a ball at random on the table. The ball is placed there for rest of the game. Player wins and game ends when they have placed two balls of the same color on the table. Find the probability that $A$ wins the game.

My son tried forming a spiral case and got stuck at the conditional probability. In this case can we assume the total probability to be $1$. I am asking this because here cases are not the same $B$ has a clear advantage because of less balls and thus can win faster.

Kindly provide help for the above problem. (Please use basic probability terms and methods.) Suggest additional tags as well. Thank you.

Answer 1

On their first turn, neither player can win.

On the second turn, player A win with probability $\frac{1}{5}$, as one of the five balls left in their pile matches the colour already there.

On their second turn, player B wins with probability $\frac{4}{5} \times \frac{1}{3} = \frac{4}{15}$, as they require that player A did not win on their 2nd turn, and also that they pull out the one ball of their pile that matches the one already drawn.

Turn 3. Player A wins with probability $\frac{2}{4} \times \frac{4}{5} \times \frac{2}{3} = \frac{4}{15}$. That is, they must not have won on turn 2, Player B must not have won on turn 2, and 2 of the 4 balls left in their pile match those already on the table.

If the game is still in play, Player B wins, as each of their remaining two balls match the colour of one already drawn. The probability it gets this far is $\frac{2}{4} \times \frac{4}{5} \times \frac{2}{3} = \frac{4}{15}$.

Notice the total probabilities do indeed sum to one: $\frac{1}{5} + 3 \times \frac{4}{15} = 1$. This is a good check we haven't made a mistake.

The probability player A wins is therefore $P(A$ wins in turn 2$) + P(A$ wins in turn 3$) = \frac{1}{5} + \frac{4}{15} = \frac{3 + 4}{15} = \frac{7}{15}$.

Answer 2

• In the first round, each player places their first ball on the table. Neither player can win yet.
• On the second round, if A draws a ball of the same color as the first, A will win. This happens with probability $\frac{1}{5}$. (One of the five remaining balls matches the one already on the table.)

• On the third round, if A hasn't already won, A can win. The way this happens is if neither player draws a matching ball in the second round, then A draws a matching ball in the third round. This happens with probability $\frac{4}{5} \cdot \frac{2}{3} \cdot \frac{2}{4}$.

  ($\frac{4}{5}$ = probability that A draws a non-matching ball in round two; $\frac{2}{3}$ = probability that B draws a non-matching ball in round two; $\frac{2}{4}$ = probability that A draws a matching ball in round three. You can compute these probabilities as the number of remaining balls that match a color on the table, divided by the total number of remaining balls)

• If A doesn't win in the third round, A loses. This is because at this point B has drawn one pink ball and one grey ball, so the next ball B draws is guaranteed to match. Hence the game is over by round three and we've considered all possibilities.

The total probability is $$\frac{1}{5} + \frac{4}{5}\frac{2}{3}\frac{2}{4} = \frac{6}{30} + \frac{8}{30} = \frac{14}{30},$$ or slightly less than half the time.

Answer 3

After the first turn, nobody can win. Player $B$ automatically wins after the third turn. Hence, A can only win after his second or his third turn. Let $a_n$ be the event that A wins after $n$ turns, and let $b_n$ be the event that B wins after his $n$-th turn. So, we want the following probability: $$P(a_2)+P(a_3\cap\overline{a_2}\cap\overline{b_2}).$$ Because B can't win in the first turn (neither can A), we have $P(a_2)=\frac{1}{5}.$ (In the first turn, A draws some ball and has $5$ left, of which one is the right one.) We also have $P(b_2\vert\overline{a_2})=\frac{1}{3}.$ The second probability equals $$P(a_3\cap\overline{a_2}\cap\overline{b_2})=P(a_3\vert\overline{a_2}\cap\overline{b_2}) \cdot P(\overline{a_2}\cap\overline{b_2})=P(a_3\vert\overline{a_2}\cap\overline{b_2}) \cdot P(\overline{b_2}\vert\overline{a_2})\cdot P(\overline{a_2}).$$ Now, we have three probabilities to calculate: $$P(\overline{a_2})=1-P(a_2)=\frac{4}{5}.$$ $$P(\overline{b_2}\vert\overline{a_2})=1-P(b_2\vert\overline{a_2})=\frac{2}{3}$$ $$P(a_3\vert\overline{a_2}\cap\overline{b_2})=\frac{1}{2}.$$ Explanation for the last one: If $\overline{a_2}$, then A got two balls of different colors on the table. So, A has four balls left, of which two would lead to a win. Now we can calculate our wanted probability: $$P(a_2)+P(a_3\cap\overline{a_2}\cap\overline{b_2})=\frac{1}{5}+\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}=\frac{7}{15}.$$