Question

Consider the following gambling problem. suppose you go to a casino to play an unusual gambling game, where the probability of winning is 0.45. you start with $3.00 and you want to double your stake. if you ever win enough to reach $6.00, you quit and walk out. similarly, if you lose your original stake and have no money left, you quit also. (a) the first betting scheme is simple: you bet 1.00 on every bet until you either go broke or you reach $6.00. draw a markov chain, with transition probabilities, corresponding to this betting scheme. you should have 7 states, corresponding to the amount of money you have at any time. In the gambling problem described, if you bet $1.00 on every bet until you either go broke or reach $6.00, what kind of Markov chain with transition probabilities corresponds to this betting scheme?

A. Absorbing Markov Chain with 7 states
B. Non-absorbing Markov Chain with 7 states
C. Markov Chain with 2 states
D. Continuous Markov Chain

Answer 1

The betting scheme corresponds to an Absorbing Markov Chain with 7 states. The states represent the amount of money the player has and the transition probabilities represent the chances of winning or losing a bet.

Step-by-step explanation:

The betting scheme described in the gambling problem corresponds to an Absorbing Markov Chain with 7 states. The states represent the amount of money the player has at any given time, ranging from $0.00 to $6.00. The transition probabilities represent the chances of winning or losing a bet.

To visualize this as a Markov chain, we can represent the states as nodes and the transitions as edges between the nodes. The probabilities of transitioning from one state to another can be depicted as the weights on the edges. In this case, the transition probabilities depend on the outcome of each bet, which has a 0.45 probability of winning and a 0.55 probability of losing.