To find Jack's long-term probability of winning, we can calculate the steady-state probabilities of the Markov chain. The steady-state probabilities represent the long-term probabilities of being in each state.
Let's denote the steady-state probabilities as \(P_1\) and \(P_2\) for being in deck #1 and deck #2, respectively. Since the dealer switches decks based on Jack's win or loss, the transition probabilities between the decks are as follows:
\(P(\text{deck #2} \to \text{deck #2}) = 1\) (if Jack wins)
\(P(\text{deck #1} \to \text{deck #2}) = 1\) (if Jack wins)
\(P(\text{deck #2} \to \text{deck #1}) = \frac{4}{9}\) (if Jack loses)
\(P(\text{deck #1} \to \text{deck #1}) = \frac{8}{15}\) (if Jack loses)
To find the steady-state probabilities, we need to solve the following system of equations:
\(P_1 = \frac{8}{15}P_1\) (since \(P(\text{deck #1} \to \text{deck #1}) = \frac{8}{15}\))
\(P_2 = 1 + \frac{4}{9}P_1\) (since \(P(\text{deck #2} \to \text{deck #1}) = \frac{4}{9}\))
Simplifying the equations, we have:
\(\frac{7}{15}P_1 = 0\) (after subtracting \(P_1\) from both sides of the first equation)
\(P_2 = 1 + \frac{4}{9}\times 0\) (substituting \(0\) for \(P_1\) in the second equation)
The first equation tells us that \(P_1 = 0\), which means Jack has zero probability of being in deck #1 in the long-term.
The second equation tells us that \(P_2 = 1\), which means Jack has a 100% probability of being in deck #2 in the long-term.
Therefore, Jack's long-term probability of winning is zero, since he has zero probability of being in deck #1, where he has a higher probability of winning.
Exercise: Frequency interpretations 0.0/2.0 points (ungraded) Jack is a gambler who pays for his MIT tuition by spending weekends in Las Vegas. His latest game of choice is blackjack, which he plays using a fixed strategy. However, at this special blackjack table, the dealer uses one of 2 decks of cards for each hand. Using his fixed strategy, Jack wins with probability 7/15 when deck #1 is used and with probability 4/9 when deck #2 is used, whenever deck #1 is used, ifjack wins, the dealer switches to deck #2 for the next hand, and if Jack loses, the dealer keeps using deck #1 for the next hand. Whenever deck #2 is used fjack wins. the dealer keeps using deck #2 for the next hand, and ifjack loses, the dealer switches to deck #1 for the next hand. Jack's wins and losses can be modeled as the transitions of the following Markov chain, whose states correspond to the particular deck being used (win) 2 ) ) (win) (loss) What is Jack's long-term probability of winning?
2 answers
0.45652