expected value - Gambler's ruin / how long does it take to win or …
Dec 12, 2022 · As shown here, Expected time for winning in biased Gambler's Ruin we can derive the expected value of $\tau_i$, $\sigma_i$, and $\tau_i+\sigma_i$ by some computations with the two above recurrences. Such computations can be avoided by the use of martingales. See this question: Gambler's ruin and martingale
probability - Gambler's ruin model - Mathematics Stack Exchange
Nov 22, 2017 · The Question is, For the gambler's ruin model mentioned above, let βn = P(min(τ0,τc)> n) β n = P (m i n (τ 0, τ c)> n) be the probability that the player's fortue has not hit 0 0 or c c by time n n. (a) Find any explicit, simple expression γn γ n such that βn <γn β n <γ n for all n ∈ N n ∈ N, and such that limn→∞γn = 0 lim ...
Gambler's ruin and martingale - Mathematics Stack Exchange
Jan 30, 2017 · The solutions says this: As a consequence: (1 − p p)a = P(ST = b)(1 − p p)b + (1 − P(ST = b))(1 − p p)0. I don't understand this equation. I understand that E(MT) = E(M0) = (1 − p p)a and it also makes sense that ST is either b or 0. But I don't understand why this MT is Bernoulli distributed with these parameters.
Gambler's Ruin (Infinite Capital) - Mathematics Stack Exchange
Consider a standard Gambler's Ruin game in which A starts with $1, B has infinite capital, and A wins w.p. p. Let q = 1 − p From standard results listed on Wikipedia, the probability that A is ever ruined is. lim n2 → ∞q p − (q p)1 + n2 1 − (q p)1 + n2 = q p. This all works out fine, since we're given that p ∈ (0.5, 1), so q <p.
probability - Expected game duration - gambler's ruin
Oct 19, 2021 · A gambler with initial capacity k units plays against an opponent with initial capital a-k units. At each game the gambler either wins one unit or loses one unit with probability 1/2. Whenever the opponent loses the game, the gambler returns one unit so that the game may continue. Show that the expected duration of the game is k (2a - 1 - k).
Gambler’s Ruin in finite time - solve by Markov chains?
Jan 2, 2019 · 2. Consider the Gambler's Ruin Problem: an infinite Markov process of steps with binary outcomes of plus or minus, each of equal probability. At each step of the process there may be a play of the game. At each play of the game starting with step 1, the gambler's initial fortune, B, increases by one unit with probability 1/2 or decreases by one ...
probability - Random Walk: Gambler's Ruin - Infinite Fortune ...
Aug 15, 2020 · 1. The basic idea is that if you win less than half the time your expected value goes to negative infinity linearly and your variance increases sublinearly (O(n−−√) O (n)), so you almost surely are negative (go broke) eventually. One approach is to estimate the property of every going one lower than where you started. Let's call this z z.
Gambler's ruin stopping time - Mathematics Stack Exchange
Nov 17, 2015 · I'm trying to show that the expected stopping time of the Gambler's Ruin game is $x(n-x)$, where the gambler starts with \$$x$ and the game stops at \$0 or \$$n$.
linear algebra - recurrence solution to gambler's ruin
Gambler's Ruin - Probability of Losing in t Steps. 2. Justifying a recurrence relationship. 2. Staircase ...
Problem with Gambler's ruin - Mathematics Stack Exchange
Oct 12, 2019 · Recurrence relation in the Gambler's Ruin problem. 1. More aggressive variation of gambler's ruin. 0.