Question: suppose now that two players (Aand B),each having $2$,agree to keep playing the game and betting $1$ at a time until one player is broke. The probability of Awinning a single bet is $\frac{1}{3}$, so Bwins the bet with probability $\frac{2}{3}$. The number of dollars that player A has before each bet $(0,1,2,3,\text{or }4)$ provides the states of a Markov chain with transition matrix can you explain easy to understand and calculate?
1. **Problem Statement:**
Two players, A and B, each start with $2$ dollars. They play a game betting $1$ dollar each round until one player is broke (has $0$ dollars). Player A wins each bet with probability $\frac{1}{3}$, and player B wins with probability $\frac{2}{3}$. The state of the game is the number of dollars player A has before each bet, which can be $0,1,2,3,$ or $4$.
2. **Markov Chain States:**
The states represent player A's current amount of money:
- $0$: A is broke (absorbing state)
- $1, 2, 3$: intermediate states
- $4$: B is broke (absorbing state)
3. **Transition Probabilities:**
- From state $i$ (where $1 \leq i \leq 3$), the next state is:
- $i+1$ with probability $\frac{1}{3}$ (A wins $1$ dollar)
- $i-1$ with probability $\frac{2}{3}$ (A loses $1$ dollar)
- States $0$ and $4$ are absorbing, so:
- $P(0 \to 0) = 1$
- $P(4 \to 4) = 1$
4. **Transition Matrix $P$: (states ordered $0,1,2,3,4$)**
$$
P = \begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
\frac{2}{3} & 0 & \frac{1}{3} & 0 & 0 \\
0 & \frac{2}{3} & 0 & \frac{1}{3} & 0 \\
0 & 0 & \frac{2}{3} & 0 & \frac{1}{3} \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
$$
5. **Goal:**
Calculate the probability that player A eventually wins (reaches state $4$) starting from state $2$ (both players have $2$ dollars).
6. **Define $u_i$ as the probability that starting from state $i$, player A eventually wins (reaches $4$):**
- $u_0 = 0$ (A already broke)
- $u_4 = 1$ (A already won)
For $i=1,2,3$, the probabilities satisfy:
$$
u_i = \frac{1}{3} u_{i+1} + \frac{2}{3} u_{i-1}
$$
7. **Write the system of equations:**
$$
\begin{cases}
u_1 = \frac{1}{3} u_2 + \frac{2}{3} u_0 = \frac{1}{3} u_2 + 0 \\
u_2 = \frac{1}{3} u_3 + \frac{2}{3} u_1 \\
u_3 = \frac{1}{3} u_4 + \frac{2}{3} u_2 = \frac{1}{3} \times 1 + \frac{2}{3} u_2
\end{cases}
$$
8. **Substitute $u_0=0$ and $u_4=1$:**
$$
\begin{cases}
u_1 = \frac{1}{3} u_2 \\
u_2 = \frac{1}{3} u_3 + \frac{2}{3} u_1 \\
u_3 = \frac{1}{3} + \frac{2}{3} u_2
\end{cases}
$$
9. **Express $u_1$ in terms of $u_2$: $u_1 = \frac{1}{3} u_2$**
10. **Substitute $u_1$ into the second equation:**
$$
u_2 = \frac{1}{3} u_3 + \frac{2}{3} \times \frac{1}{3} u_2 = \frac{1}{3} u_3 + \frac{2}{9} u_2$$
Multiply both sides by $9$ to clear denominators:
$$9 u_2 = 3 u_3 + 2 u_2$$
Simplify:
$$9 u_2 - 2 u_2 = 3 u_3 \implies 7 u_2 = 3 u_3 \implies u_3 = \frac{7}{3} u_2$$
11. **Use the third equation:**
$$u_3 = \frac{1}{3} + \frac{2}{3} u_2$$
Substitute $u_3 = \frac{7}{3} u_2$:
$$\frac{7}{3} u_2 = \frac{1}{3} + \frac{2}{3} u_2$$
Multiply both sides by $3$:
$$7 u_2 = 1 + 2 u_2$$
Simplify:
$$7 u_2 - 2 u_2 = 1 \implies 5 u_2 = 1 \implies u_2 = \frac{1}{5}$$
12. **Calculate $u_1$ and $u_3$:**
$$u_1 = \frac{1}{3} u_2 = \frac{1}{3} \times \frac{1}{5} = \frac{1}{15}$$
$$u_3 = \frac{7}{3} u_2 = \frac{7}{3} \times \frac{1}{5} = \frac{7}{15}$$
13. **Final answer:**
The probability that player A eventually wins starting from $2$ dollars is:
$$\boxed{\frac{1}{5} = 0.2}$$
This means player A has a 20% chance to win the game starting with $2$ dollars when the probability of winning each bet is $\frac{1}{3}$.