Final answer:
The student's question is about drawing a Markov chain for a gambling game with a winning probability of 0.45. The chain will have 7 states representing amounts from $0.00 to $6.00, including transition probabilities. States at $0.00 and $6.00 are absorbing, ending the game once reached.
Step-by-step explanation:
A Markov Chain for a Gambling Game
The student's question involves creating a Markov chain for a gambling game with distinct states corresponding to different monetary values. In this game, a player bets $1.00 on each turn with a winning probability of 0.45, aiming to double their initial stake of $3.00 to $6.00. The player stops either upon losing all money or reaching the goal. We will represent each monetary amount as a state in our chain, resulting in seven states from $0.00 to $6.00. The transition probabilities will be 0.45 for winning a bet and moving to the next higher state and 0.55 for losing a bet and moving to a lower state. The states at $0.00 and $6.00 are absorbing states, meaning once they are reached, the player stops playing.
To visualize the transitions, consider state $3.00: there is a 0.45 probability to move up to $4.00 if the player wins, and a 0.55 probability to move down to $2.00 if the player loses. This logic applies to intermediate states ($1.00 to $5.00), except for the endpoints which do not have outgoing transitions because they signify the end of the game.