answer:This game can be analyzed by considering the number of heads and tails at each step. Since the game ends when all coins show the same face, we can focus on the difference between the number of heads and tails. Let's denote this difference as d. Initially, d is equal to the difference between the number of heads and tails in the random initial configuration, which is a random variable. However, since the number of coins is even, d must be an even number as well. At each step, the player flips a coin, which either increases or decreases d by 1, or leaves it unchanged if the flipped coin already shows the same face as the majority. Thus, d changes parity at each step. Since d starts as even and changes parity at each step, it will be even at every step made by P1 and odd at every step made by P2. Now, the game ends when d equals n (all heads) or -n (all tails). Since n is even, the game can only end during P1's turn if d equals n, and during P2's turn if d equals -n. This implies that P1 can win only if the game ends during their turn, which occurs with probability 1/2, as the game can end either during P1's or P2's turn with equal probability. Therefore, the probability that P1 wins is 1/2. This result arises from the symmetry of the game and the random behavior of the players, rather than the specific details of the sequence of moves.

answer:The fact that n is even played a crucial role in the solution by introducing a parity constraint on the difference d between the number of heads and tails. This constraint, combined with the fact that d changes parity at each step, led to the key observation that d will be even at every step made by P1 and odd at every step made by P2. If n were odd, this parity argument would not hold. In that case, d could be either even or odd initially, and it would not necessarily alternate between even and odd at each step in the same way. As a result, the game could end during either P1's or P2's turn with d equal to n, which is odd. In other words, when n is odd, the game is no longer symmetric with respect to the parity of d, and P1's probability of winning would not be 1/2. In fact, when n is odd, P1 has a slight advantage, but calculating the exact probability of winning would require a more complex analysis, possibly involving recurrence relations or Markov chains. The evenness of n simplifies the problem by introducing a parity constraint that leads to a symmetric situation, allowing us to conclude that P1's probability of winning is 1/2.

answer:Let's explore this further. When n is odd, the game is indeed asymmetric, and P1 has a slight advantage. To see why, consider the last step of the game. If the game ends with all heads, it must be P1's turn, since they started with the first move and the number of moves must be odd (as n is odd). On the other hand, if the game ends with all tails, it could be either P1's or P2's turn. Since P1 has the first move, they have a slight advantage in terms of controlling the final outcome. Specifically, when n is odd, P1 can win in one more situation than P2: when the game ends with all heads during P1's turn. This additional winning scenario gives P1 a slight edge over P2. However, calculating the exact probability of P1's win when n is odd is more involved and would require a more detailed analysis, as I mentioned earlier.

answer:Fractional reserve banking is the practice by which banks accept deposits from customers and only keep a small fraction of that money in reserve, lending out the rest to generate interest income. This system is based on the assumption that only a small percentage of depositors will request their money back at any given time, allowing the bank to lend out more money than it physically holds. This practice essentially creates new money due to the multiplier effect, as loans are spent and redeposited into the banking system, and is a key component of modern monetary systems.