第 1 / 1 页
非代码面试题
显示 18 / 18 道匹配题目
答题状态：未尝试未正确已正确
ID题目领域难度题型进度权限
501Biased Gambler's Ruin ProbabilityA gambler starts with \3 and plays a sequence of independent rounds. Each round she wins \1 with probability p = 0.4 and loses \1 with probability q = 0.6. She stops when her fortune reaches \0 (ruin) or \8. What is the probability that she is ruined?概率简单数值题未尝试免费503Asymmetric Step-Size Gambler's RuinA gambler starts with \3. Each round she wins \1 with probability \tfrac 2 3 or loses \2 with probability \tfrac 1 3 . She stops when her fortune first hits \0 or below (ruin) or reaches \5 or above (victory). Note that if her fortune is \1 and she loses, she goes to -\1, which counts as ruin. What is the probability that she achieves victory?概率中等数值题未尝试免费504Three-Player Elimination GameThree players A, B, C hold tokens: A has 2, B has 1, C has 1 (total 4). Each round, two players are chosen uniformly at random (from the \binom 3 2 = 3 pairs) and they play a fair game: each of the two has probability \tfrac 1 2 of winning one token from the other. A player with 0 tokens is eliminated. Play continues until one player holds all 4 tokens. What is the probability that player A (who starts with 2 tokens) ends up with all 4?概率中等数值题未尝试免费505Gambler's Ruin with a Partially Reflecting BarrierConsider a Markov chain on \ 0, 1, 2, \ldots\ with absorbing state 0. From state k \ge 1, the chain moves to k+1 with probability p and to k-1 with probability q = 1 - p, where 0 < p < 1. However, there is a **reflecting barrier** at state N: from state N, the chain moves to N-1 with probability 1 (it is always pushed back). Starting from state k with 1 \le k \le N: **(a)** Find the probability r k of eventual absorption at 0. **(b)** For p = q = \tfrac 1 2 and N = 4, compute r 2 and the expected absorption time E[T \mid X 0 = 2].概率困难derivation未尝试免费506Martingale Approach to Biased Ruin ProbabilityA gambler starts with \4 and repeatedly bets \1 on a biased coin that lands heads with probability p = 0.55. She gains \1 on heads and loses \1 on tails. The game ends when her fortune reaches \0 or \10. Using the martingale (q/p) X n and the optional stopping theorem, find the exact probability that she reaches \10 before going broke.概率中等数值题未尝试免费507Gambler's Ruin with State-Dependent Win ProbabilityA gambler's fortune evolves on \ 0, 1, 2, 3, 4\ with absorbing barriers at 0 and 4. At state k (1 \le k \le 3), she wins \1 with probability p k = k 2k+1 and loses \1 with probability q k = k+1 2k+1 . Starting from state 2, find the probability of ruin (absorption at 0).概率中等数值题未尝试免费508Three-Outcome Random Walk to AbsorptionA particle moves on \ -5, -4, \ldots, 4, 5\ with absorbing barriers at -5 and 5. At each step, the particle moves +1 with probability p = 0.3, stays put with probability s = 0.2, and moves -1 with probability q = 0.5. Starting from position 1, what is the probability that the particle is absorbed at +5?概率中等数值题未尝试免费512Asymmetric Jump Ruin with Upward LeapA gambler on \ 0, 1, \ldots, 6\ starts at position 3. Each round, she jumps +2 with probability 1/3 and -1 with probability 2/3. Positions 0 and 6 are absorbing (but a jump from 5 that would land at 7 is truncated to 6, i.e., from state 5 the gambler moves to 6 with probability 1/3 and to 4 with probability 2/3). What is the probability that the gambler is absorbed at 0?概率中等数值题未尝试免费514Gambler's Ruin with Alternating BiasA gambler moves on \ 0, 1, 2, 3, 4, 5, 6\ with absorbing barriers at 0 and 6. The win probability depends on the current position: at odd states (k = 1, 3, 5), p = 0.6; at even states (k = 2, 4), p = 0.4. (Win means +1, loss means -1.) Starting from state 3, find the probability of reaching 6 before 0, and the expected number of steps until absorption.概率困难数值题未尝试面试订阅515Ruin Probability Under a Martingale Betting StrategyA gambler uses a doubling strategy (Martingale system): she bets \1 on a fair coin, and after each loss she doubles her bet, resetting to \1 after any win. Her initial capital is \31 and the table maximum bet is \16 (so she can lose at most 4 consecutive times before hitting the cap, since 1 + 2 + 4 + 8 + 16 = 31). If the bet would exceed \16, she bets her remaining capital instead. Her target is \42 (a net gain of \11). What is the probability she reaches \42 before going broke?概率困难数值题未尝试面试订阅516Absorption in a Three-State Chain with Two TrapsA Markov chain has three states: A, B, and C. States A and B are absorbing. From state C, the chain moves to A with probability 1/5, to B with probability 1/5, and stays at C with probability 3/5. Starting from C, what is the probability of being absorbed at A? What is the expected number of steps until absorption?概率简单数值题未尝试免费517Two-Player Ruin with Unequal Bet SizesAlice and Bob play a sequence of fair-coin games. In each round, Alice wagers \1 and Bob wagers \2. If heads, Alice gains \1 from Bob (and Bob loses \1); if tails, Bob gains \2 from Alice (and Alice loses \2). Alice starts with \4 and Bob starts with \6. The game ends when either player is ruined (reaches \0). Note that the total wealth is not conserved — the net transfer is asymmetric. What is the probability that Alice is ruined?概率中等数值题未尝试免费518Gambler's Ruin with Killing (Absorption in the Interior)A particle moves on \ 0, 1, 2, 3, 4, 5\ . States 0 and 5 are absorbing barriers. At each step from a transient state k (1 \le k \le 4), the particle is "killed" (absorbed into a cemetery state \Delta) with probability c = 1/5. With the remaining probability 4/5, it moves +1 or -1 each with probability 1/2. Starting from state 2, find the probability that the particle reaches state 5 (without being killed or hitting 0).概率中等数值题未尝试免费519Ruin Probability via Probability Generating FunctionsConsider a random walk on \ 0, 1, 2, \ldots, N\ with absorbing barriers at 0 and N. At each step, the walker moves +1 with probability p and -1 with probability q = 1 - p. Let G k(s) = E[s \tau \mid X 0 = k] be the probability generating function of the absorption time \tau, for 0 < k < N. (a) Derive a recurrence relation for G k(s). (b) For N = 4, k = 2, and p = q = 1/2, find G 2(s) explicitly and use it to compute E[\tau] and Var (\tau).概率困难derivation未尝试面试订阅522Gambler's Ruin with Stochastic ResettingA gambler plays on \ 0, 1, 2, 3, 4, 5\ with absorbing barriers at 0 and 5. At each step from transient state k, with probability r = 1/4 the gambler is "reset" to state 3 (regardless of current position), and with probability 3/4 she takes a fair step (+1 or -1 each with probability 3/8). Starting from state 2, find the probability of reaching 5 before 0.概率中等数值题未尝试免费523Gambler's Ruin with Momentum (Correlated Steps)A gambler on \ 0, 1, \ldots, 8\ has absorbing barriers at 0 and 8. Her step probabilities depend on the previous outcome: after a win (+1), she wins again with probability 2/3 (momentum); after a loss (-1), she wins with probability 1/3. The first step is a win with probability 1/2. Starting from state 4, what is the probability of ruin (reaching 0)?概率困难数值题未尝试面试订阅524Absorption with a Long-Range JumpA particle moves on \ 0, 1, 2, 3, 4, 5\ with absorbing states 0 and 5. The transitions from transient states are: - From 1: to 0 w.p. 1/2, to 2 w.p. 1/2. - From 2: to 1 w.p. 1/3, to 3 w.p. 1/3, to 4 w.p. 1/3 (a long-range jump). - From 3: to 2 w.p. 1/2, to 4 w.p. 1/2. - From 4: to 3 w.p. 1/2, to 5 w.p. 1/2. Starting from state 2, what is the probability of being absorbed at 5?概率简单数值题未尝试免费525Expected Ruin Time via Martingale MethodA biased random walk on \ 0, 1, \ldots, 10\ with absorbing barriers at 0 and 10 moves +1 with probability p = 0.6 and -1 with probability q = 0.4. Starting from state 3, use the martingale M n = X n - n (where = p - q) and the optional stopping theorem to find the expected absorption time E[\tau].概率中等数值题未尝试免费