A person starts with 64 rupees and makes six bets in a game, where the probability of a win is equal to the probability of a loss. In each bet, the person stakes half of the money currently held. For each win, the stake is gained in addition to the remaining money; for each loss, the stake is lost. The person wins exactly three bets and loses exactly three bets, in random order. What will be the final net result relative to the starting amount?

Introduction / Context:
This question tests understanding of percentage change and multiplicative growth rather than straightforward probability. The order of wins and losses might look important, but when each bet is for half the current capital, the final result depends only on how many wins and how many losses occur, not on their sequence. Recognising this simplifies the problem dramatically and avoids tedious case-by-case listing of different orders of wins and losses.

Given Data / Assumptions:

Initial capital is Rs. 64.

The person makes 6 bets.

Each bet is for half the money currently held.

Three bets are wins and three bets are losses.

Probability of a win equals probability of a loss, but we are already told the result pattern (3 wins, 3 losses), so we focus on the final amount.

Concept / Approach:
When the person bets half of the current capital, two different multiplicative factors apply. On a win, the person keeps the half that was not staked plus wins an equal amount to the stake. So the money is multiplied by 1.5. On a loss, the person loses the staked half and keeps only half of the capital, so the money is multiplied by 0.5. Since there are exactly three wins and three losses, the final capital is the initial capital multiplied by 1.5^3 and 0.5^3, in any order.

Step-by-Step Solution:
Step 1: After a win, factor = 1.5 (because capital becomes capital + half of capital = 1.5 times).Step 2: After a loss, factor = 0.5 (because capital is reduced to half).Step 3: The order of these factors does not matter, because multiplication is commutative.Step 4: Total multiplicative factor for three wins and three losses = 1.5^3 * 0.5^3.Step 5: Compute 1.5^3 = (3/2)^3 = 27/8.Step 6: Compute 0.5^3 = (1/2)^3 = 1/8.Step 7: Combined factor = (27/8) * (1/8) = 27/64.Step 8: Final capital = initial capital * combined factor = 64 * (27/64) = 27 rupees.Step 9: Net result compared to starting 64 rupees = final 27 rupees, so net loss = 64 - 27 = 37 rupees.

Verification / Alternative check:
To check, you can take a specific order such as win, win, win, loss, loss, loss and track the rupees step by step. After each win, multiply by 1.5, after each loss multiply by 0.5. Any order you choose will always give 27 rupees at the end, confirming that the order does not matter and that the multiplicative approach is correct.

Why Other Options Are Wrong:
A gain of Rs. 27 or Rs. 37 assumes the final amount is greater than the starting 64 rupees, which contradicts the fact that multiplying by 1.5 three times and by 0.5 three times yields an overall factor less than 1. A loss of Rs. 27 would correspond to a final amount of 37 rupees, which does not match the product 64 * 27 / 64. The option of no gain and no loss would mean the overall factor is 1, which is clearly not the case here.

Common Pitfalls:
Many learners mistakenly average the gains and losses and think that three wins and three losses cancel out. They forget that successive percentage changes do not simply cancel but combine multiplicatively. Another common error is to treat each win as plus a fixed amount, which is incorrect because each stake is half of the remaining money, not half of the original 64 rupees. Always convert percentage-style problems into multiplicative factors on the current amount and then multiply those factors in sequence.

Final Answer:
The person ends with Rs. 27, so the net outcome is a loss of Rs. 37 compared with the starting 64 rupees.