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?
0.45652
To find Jack's long-term probability of winning, we need to determine the steady-state probabilities of the Markov chain. The steady-state probabilities represent the long-term proportion of time that Jack spends in each state (deck), given an infinite number of hands.
Let's denote the steady-state probability of being in deck #1 as P1 and the steady-state probability of being in deck #2 as P2.
To find the steady-state probabilities, we set up the following system of equations:
P1 = (7/15)P1 + (1/2)P2 [From deck #1, Jack can either win with probability 7/15 and stay in deck #1, or lose and move to deck #2 with probability 1/2]
P2 = (4/9)P2 + (1/2)P1 [From deck #2, Jack can either win with probability 4/9 and stay in deck #2, or lose and move to deck #1 with probability 1/2]
We can solve this system of equations to find the values of P1 and P2.
Rearranging the equations, we have:
(8/15)P1 - (1/2)P2 = 0
-(1/2)P1 + (5/9)P2 = 0
Multiplying the first equation by 4/5 and the second equation by 2/5, we get:
(32/75)P1 - (2/5)P2 = 0
-(1/5)P1 + (10/45)P2 = 0
Adding the two equations, we have:
(32/75)P1 - (1/5)P1 - (2/5)P2 + (10/45)P2 = 0
Combining like terms, we get:
(11/75)P1 - (4/45)P2 = 0
To satisfy this equation, we set P1 = (4/45) and P2 = (11/75).
Thus, Jack's long-term probability of winning is P1 + P2 = (4/45) + (11/75) = 67/675 ≈ 0.0993 or approximately 9.93%.
Therefore, Jack's long-term probability of winning is approximately 9.93%.
To find Jack's long-term probability of winning, we need to compute the steady-state probabilities of the Markov chain. The steady-state probabilities represent the long-term proportions of time Jack spends in each state.
To determine these probabilities, we can use the concept of detailed balance, which states that for any two states in the Markov chain, the product of the transition probabilities from the first state to the second state and the steady-state probability of the first state is equal to the product of the transition probabilities from the second state to the first state and the steady-state probability of the second state.
Let's denote the steady-state probabilities as P1 for deck #1 and P2 for deck #2. Using the detailed balance equation, we can set up the following equations:
(7/15) * P1 = (4/9) * P2 // probability of winning with deck #1
(4/9) * P2 = P2 // probability of winning with deck #2
Simplifying the equations, we get:
(105/135) * P1 = (60/135) * P2
(7/9) * P1 = (4/9) * P2
Next, we need to consider the fact that the steady-state probabilities P1 and P2 must sum up to 1. Therefore, we have another equation: P1 + P2 = 1.
We can solve this system of equations to find the values of P1 and P2. Let's substitute P2 = 1 - P1 into the second equation and solve for P1:
(7/9) * P1 = (4/9) * (1 - P1)
7P1 = 4 - 4P1
11P1 = 4
P1 = 4/11
Then we can calculate P2 by substituting the value of P1 into the equation P1 + P2 = 1:
4/11 + P2 = 1
P2 = 1 - 4/11
P2 = 7/11
Therefore, Jack's long-term probability of winning is 4/11 when deck #1 is used and 7/11 when deck #2 is used.
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.