[Combinatorics]
Problem: Amir and Brigitte play a card game. Amir starts with a hand of 6 cards: 2 red, 2 yellow and 2 green. Brigitte starts with a hand of 4 cards: 2 purple and 2 white. Amir plays first. Amir and Brigitte alternate turns. On each turn, the current player chooses one of their own cards at random and places it on the table. The cards remain on the table for the rest of the game. A player wins and the game ends when they have placed two cards of the same colour on the table. Determine the probability that Amir wins the game.

Solution: On her first two turns, Brigitte either chooses two cards of the same colour or two cards of different colours. If she chooses two cards of different colours, then on her third turn, she must choose a card that matches one of the cards that she already has.

Therefore, the game ends on or before Brigitte's third turn.

Thus, if Amir wins, he wins on his second turn or on his third turn. (He cannot win on his first turn.)

For Amir to win on his second turn, the second card he chooses must match the first card that he chooses.

On this second turn, there will be 5 cards in his hand, of which 1 matches the colour of the first card that he chose.

Therefore, the probability that Amir wins on his second turn is $\frac{1}{5}$.

Note that there is no restriction on the first card that he chooses or the first card that Brigitte chooses.

For Amir to win on his third turn, the following conditions must be true: (i) the colour of the second card that he chooses is different from the colour of the first card that he chooses, (ii) the colour of the second card that Brigitte chooses is different from the colour of the first card that she chooses, and (iii) the colour of the third card that Amir chooses matches the colour of one of the first two cards.

The probability of (i) is $\frac{4}{5}$, since he must choose any card other than the one that matches the first one.

The probability of (ii) is $\frac{2}{3}$, since Brigitte must choose either of the cards that does not match her first card.

The probability of (iii) is $\frac{2}{4}$, since Amir can choose either of the 2 cards that matches one of the first two cards that he chose.

Again, the cards that Amir and Brigitte choose on their first turns do not matter.

Thus, the probability that Amir wins on his third turn is $\frac{4}{5} \cdot \frac{2}{3} \cdot \frac{2}{4}$ which equals $\frac{4}{15}$.

Finally, the probabilty that Amir wins the game is thus $\frac{1}{5}+\frac{4}{15}$ which equals $\frac{7}{15}$.

Final answer: $\frac{7}{15}$