Solution: Let us explore the structure of the problem. If we have 3 coins before the round, with probability 1/8 we will still have 3 coins after one round; with probability 3/8 - two; with probability 3/8 - one; with probability 1/8 - zero. If we have 2 coins, with probability 1/4 we will still have 2 coins after one round; with probability 1/2 - one; with probability 1/4 - zero. If we have 1 coin, with probability 1/2 we will still have 1 coin after one round; with probability 1/2 - zero. If we have 0 coins, we will have 0 coins with probability 1. Markov Chain is the model for this structure - for this question it has four states (0, 1, 2, 3), corresponding to the number of coins. In these terms, the question can be formulated as: what is the expected number of transitions needed to move from state '3' to state '0'? Denoting by t_i (for i = 0...3) the expected number of transitions needed to move from state i to state 0, one gets: t_0 = 0 t_1 = 1/2 * (t_1 + 1) + 1/2 * (t_0 + 1) t_2 = 1/4 * (t_2 +1) + 1/2 * (t_1 + 1) + 1/4 * (t_0 + 1) t_3 = 1/8 * (t_3 + 1) + 3/8 * (t_2 + 1) + 3/8 * (t_1 + 1) + 1/8 * (t_0 + 1) Solving these equations top-to-bottom, one gets: t_1 = 2; t_2 = 8/3; t_3 = 22/7 So the answer is 22/7

Nice problem! The answer is 22/7 ≈ 3.14 rounds. let E(n) = expected rounds from n coins. Each round, removed coins follow Binomial(n, ½). Solving the recurrence bottom-up: E(1) = 2 → E(2) = 8/3 → E(3) = 22/7 The diminishing returns make intuitive sense as coins drop out, the "last survivor" becomes increasingly stubborn to eliminate.