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Question: In a game, a man wins ₹ 1000 if he gets an even number greater than or equal to 4 on a fair dice and...

In a game, a man wins ₹ 1000 if he gets an even number greater than or equal to 4 on a fair dice and loses ₹ 200 for getting any other number on the dice. If he decides to throw the dice until he wins or maximum of three times, then his expected gain/loss in (₹) is --

A

22009loss\frac{2200}{9} \text{loss}

B

38009loss\frac{3800}{9} \text{loss}

C

22009gain\frac{2200}{9} \text{gain}

D

38009gain\frac{3800}{9} \text{gain}

Answer

38009 gain\frac{3800}{9} \text{ gain}

Explanation

Solution

To calculate the expected gain/loss, we need to consider all possible scenarios for the game and their respective probabilities and net gains/losses.

1. Determine probabilities for a single throw:

A fair die has 6 outcomes: {1, 2, 3, 4, 5, 6}.

Winning condition: An even number greater than or equal to 4. These numbers are {4, 6}.

Number of winning outcomes = 2.

Probability of winning on a single throw (PWP_W) = 2/6 = 1/3.

Losing condition: Any other number. These numbers are {1, 2, 3, 5}.

Number of losing outcomes = 4.

Probability of losing on a single throw (PLP_L) = 4/6 = 2/3.

(Note: PW+PLP_W + P_L = 1/3 + 2/3 = 1)

2. Define gain/loss amounts:

  • If he wins, he gains ₹ 1000.
  • If he loses, he loses ₹ 200.

3. Analyze scenarios based on the maximum of three throws:

  • Scenario 1: Wins on the 1st throw (W1)

    • Probability: P(W1) = PWP_W = 1/3
    • Net Gain: ₹ 1000
    • Contribution to Expected Value: (1/3) * 1000 = 1000/3

  • Scenario 2: Loses on the 1st throw, Wins on the 2nd throw (L1W2)

    • Probability: P(L1W2) = PL∗PWP_L * P_W = (2/3) * (1/3) = 2/9
    • Net Gain: -₹ 200 (from 1st throw) + ₹ 1000 (from 2nd throw) = ₹ 800
    • Contribution to Expected Value: (2/9) * 800 = 1600/9

  • Scenario 3: Loses on the 1st, Loses on the 2nd, Wins on the 3rd throw (L1L2W3)

    • Probability: P(L1L2W3) = PL∗PL∗PWP_L * P_L * P_W = (2/3) * (2/3) * (1/3) = 4/27
    • Net Gain: -₹ 200 (from 1st) - ₹ 200 (from 2nd) + ₹ 1000 (from 3rd) = ₹ 600
    • Contribution to Expected Value: (4/27) * 600 = 2400/27

  • Scenario 4: Loses on the 1st, Loses on the 2nd, Loses on the 3rd throw (L1L2L3)

    • This is the case where he reaches the maximum three throws without winning.
    • Probability: P(L1L2L3) = PL∗PL∗PLP_L * P_L * P_L = (2/3) * (2/3) * (2/3) = 8/27
    • Net Gain: -₹ 200 (from 1st) - ₹ 200 (from 2nd) - ₹ 200 (from 3rd) = -₹ 600 (a loss of ₹ 600)
    • Contribution to Expected Value: (8/27) * (-600) = -4800/27

4. Calculate the Total Expected Gain/Loss:

Expected Value (E) = Sum of contributions from all scenarios

E=10003+16009+240027+(−480027)E = \frac{1000}{3} + \frac{1600}{9} + \frac{2400}{27} + \left(-\frac{4800}{27}\right)

To sum these fractions, find a common denominator, which is 27:

E=1000×927+1600×327+240027−480027E = \frac{1000 \times 9}{27} + \frac{1600 \times 3}{27} + \frac{2400}{27} - \frac{4800}{27}

E=900027+480027+240027−480027E = \frac{9000}{27} + \frac{4800}{27} + \frac{2400}{27} - \frac{4800}{27}

E=9000+4800+2400−480027E = \frac{9000 + 4800 + 2400 - 4800}{27}

E=9000+240027E = \frac{9000 + 2400}{27}

E=1140027E = \frac{11400}{27}

Simplify the fraction by dividing both numerator and denominator by 3:

E=11400÷327÷3E = \frac{11400 \div 3}{27 \div 3}

E=38009E = \frac{3800}{9}

Since the expected value is positive, it represents a gain.