g Two players each put one dollar into a pot. They decide to throw a pair of dice alternately. The first one who throws the two dice so that the sum of the two faces is five (5) wins the pot. How much should the player who starts add to the pot to make the game a ‘fair game’? Give your answer as a fraction in reduced form (i.e. A/B where A and B do not share any common factors). Note: A ‘fair game’ would be one where the expected payment to each player is equal

The probability that the first player wins on the first turn is equivalent to the probability of getting 5 in the dices. Out of the 36 possible outcomes for a dice, only 4 are favourable in the event of obtaining 5 in the sum:

- first dice 1, second dice 4

- first dice 2, second dice 3

- first dice 3, second dice 2

- first dice 4, second dice 1

Thus, the probability that the first player wins on the first turn is 4/36 = 1/9.

Note that if the first player doesnt win in the first turn, then the second player will be at the same position the first player was at the start of the game. If we call P the probability that the first player wins, then, the second player will have a probability of P of winning after 'surviving' on the first turn.

As a result, the probability that the second player wins is P multiplied by 8/9 (the probability that the first player doenst win in the first turn). In short

Probability that the first player wins = P

Probability that the second player wins = 8/9 * P

Since the sum should be 1 (because both events are complementary from each other), then P + P*8/9 = 1, thus

17/9 P = 1

P = 9/17

Lets call E the extra amount of money player A should include into the pot (in dollard). We have that both players should have expected gain equal to 0, and we have that

- First player gains 1 if he wins (with probability 9/17)

- First player gains -1-E if the loses (with probability 8/17)

Therefore

9/17*1+ 8/17 (-1-E) = 0

9/17 = 8/17 + E 8/17

E = (1/17) / (8/17) = 1/8

The first player should add 1/8 more dollar to the pot.