Question

For the gambler's ruin problem, let Mₖ denote the expected number of bets that will bemade if the player initially has $k, and stops at 0 or n. Let p be the probability of winning each bet,and q = 1 - p Show that Mₒ = Mₙ = 0 and

Mₖ =1+pMₖ₊₁ +qMₖ₋₁
for 0 < k < n (Hint: Compute the expectation of the number of bets X by condi**outcome of the first game. If A is the event that the player wins the first bet,
EX=E[X|A]P(A)+E[X|Aᶜ ] P(Aᶜ)

Answer 1

The expected number of bets in the gambler's ruin problem can be found using conditional expectations. By considering the events of winning or losing the first bet, we can derive the equation Mₖ = 1 + pMₖ₊₁ + qMₖ₋₁, where Mₖ is the expected number of bets with initial amount k. Additionally, it is shown that Mₒ and Mₙ are both equal to 0.

Step-by-step explanation:

In the gambler's ruin problem, we want to find the expected number of bets needed to reach either 0 or n dollars, given that the player initially has k dollars. Let Mₖ denote the expected number bets needed if the player initially has k dollars, and stops at 0 or n. We are asked to show that Mₒ = Mₙ = 0, and for 0 < k < n, Mₖ = 1 + pMₖ₊₁ + qMₖ₋₁, where p is the probability of winning each bet and q = 1 - p.

We can show this using a conditional expectation argument. Let A be the event that the player wins the first bet, and Aᶜ be the event that the player loses the first bet. The expectation of the number of bets X can be written as E[X] = E[X|A]P(A) + E[X|Aᶜ]P(Aᶜ).

Applying this approach to Mₖ, we can write Mₖ = E[Mₖ|A]P(A) + E[Mₖ|Aᶜ]P(Aᶜ). Since the player wins the first bet with probability p, we have P(A) = p and P(Aᶜ) = q. We can then substitute these values into the above expression and rearrange to obtain Mₖ = 1 + pMₖ₊₁ + qMₖ₋₁, which is the desired result.