Question

A and B throw one dice for a stake of Rs.11, which is to be won by the player who first throws a six. The game ends when stake is won by A or B. If A has the first throw, what are their respective expectations?

• A. 5 and 6
• B. 6 and 5
• C. 11 and 0
• D. 10 and 1

Answer 1

The Correct Option is B

Let's analyze the expectations of A and B in this dice game. First, let's establish the probability of throwing a 6. When a dice is rolled, the probability of getting a six is \( \frac{1}{6} \), and the probability of not getting a six is \( \frac{5}{6} \).

Player A has the first throw. He wins immediately if he throws a 6. If he does not, the chance moves to Player B. The game continues until one throws a 6.

Let's denote:

• \( E_A \) as A's expectation.
• \( E_B \) as B's expectation.

The key to solving the problem is setting up equations to express these expectations considering the rules of the game.

1. A's Throw:

• If A rolls a 6 on his first throw, he wins the stake of Rs.11.
• If A does not roll a 6, the probability of this is \( \frac{5}{6} \), B gets to throw, and the expectation updates to the combination of B's expectation.

The equation representing A's expectation is:
\( E_A = \frac{1}{6} \times 11 + \frac{5}{6} \times E_B \)

2. B's Throw:

• If B rolls a 6, he wins the stake of Rs.11.
• If B does not roll a 6, the probability of this is \( \frac{5}{6} \), it again becomes A's turn, and the expectation updates to A's cycle.

The equation for B's expectation is:
\( E_B = \frac{1}{6} \times 11 + \frac{5}{6} \times E_A \)

Now, solve these simultaneous equations:

1. From A's equation: \( E_A = \frac{1}{6} \times 11 + \frac{5}{6} \times E_B \)
2. From B's equation: \( E_B = \frac{1}{6} \times 11 + \frac{5}{6} \times E_A \)

Substitute equation 2 into equation 1:

• \( E_A = \frac{11}{6} + \frac{5}{6} \left(\frac{11}{6} + \frac{5}{6}E_A \right) \)
• \( E_A = \frac{11}{6} + \frac{55}{36} + \frac{25}{36}E_A \)
• \( E_A - \frac{25}{36}E_A = \frac{11}{6} + \frac{55}{36} \)

Solve for \( E_A \):

• \( \frac{11}{6} + \frac{55}{36} = \frac{161}{36} \)
• \( \frac{11}{36}E_A = \frac{161}{36} \)
• \( E_A = \frac{161}{11} \)
• \( E_A = 6 \)

Plug \( E_A = 6 \) back into one of the original equations to find \( E_B \):

• \( E_B = \frac{11}{6} + \frac{5}{6} \times 6 \)
• \( E_B = \frac{11}{6} + \frac{30}{6} = \frac{41}{6} \)
• \( E_B = 5 \)

Therefore, the expectations are:

Player	Expectation (Rs.)
A	6
B	5

Thus, the respective expectations of A and B are 6 and 5.