Conditional Probability With Independence

An alternating-turns game with a fixed per-trial success probability produces a geometric series for each player's winning chance, since the first player wins on turn 1, 3, 5, and so on. This infinite-series probability model is a challenging JEE Advanced favourite. The probability of rolling a six on any throw is 1/6 and of not rolling a six is 5/6. The first player wins immediately with probability 1/6, or if both miss once (probability (5/6)² = 25/36) and then the first player wins from the identical restarted situation. So if p is the first player's win probability, p = 1/6 + (25/36)p. Solving, p(1 − 25/36) = 1/6, so p(11/36) = 1/6, giving p = (1/6)(36/11) = 6/11. Option 5/11 is the second player's complementary probability. Option 1/2 wrongly assumes symmetry between players. Option 1/6 gives only the first-throw win. The method sums the geometric series Σ(5/6)^{2k}(1/6) = (1/6)/(1 − 25/36) = 6/11, which agrees exactly with the recursive self-similarity argument and demonstrates that the two standard techniques, infinite-series summation and one-step recursion, are fully equivalent here. The game terminates with probability one because the chance of both players missing forever is zero. Plausibility check: 6/11 exceeds 1/2, correctly reflecting the advantage held by whoever rolls first, and 6/11 + 5/11 = 1, confirming that the two win probabilities partition certainty with no leftover probability for an unending game.