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# 数学代写|运筹学代写Operations Research代考|MATH2730 Absorbing Markov Chains

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## 数学代写|运筹学代写Operations Research代考|Absorbing Markov Chains

The transient probabilities $p_{i j}^{(n)}$ also come in handy when there are absorbing states. Recall that state $i$ is absorbing if $p_{i i}=1$. The Markov chain model with absorbing states is an extremely useful model with interesting applications. Many probability problems can be formulated in terms of an absorbing Markov chain.

Example 7.9 (Success runs). If we repeatedly toss a fair die, what is the probability distribution of the number of throws needed to toss three sixes in a row?
The trick for this type of problem is to define a Markov chain with an absorbing state. Choose a Markov chain with the four states $0,1,2$, and 3 , where state 3 is absorbing and corresponds to the situation where the last three throws are sixes. State 0 corresponds to the first toss of the die or to the situation where the last throw did not give a six. For $i=1,2$, we say that the process is in state $i$ if the last $i$ throws, but not the one before that, have been sixes. Moreover, after a toss that puts the process in state 3 , only fictitious throws are carried out, leaving the process in state 3 . If we define $X_n$ as the state after the $n$th throw, then $\left{X_n\right}$ is a Markov chain with state space $I={0,1,2,3}$. This Markov chain’s one-step transition probabilities are given by
$$p_{i 0}=\frac{5}{6}, \quad p_{i, i+1}=\frac{1}{6} \quad \text { for } i=0,1,2, \quad p_{33}=1, \quad \text { otherwise } p_{i j}=0$$

## 数学代写|运筹学代写Operations Research代考|Longest run in Monte Carlo casino

A reasoning similar to that used in Example 7.9 can be used to analyze the following situation. On August 18, 1913, a memorable event took place at the Monte Carlo casino: 26 times in a row, the roulette ball fell on black. Suppose that, up to that date, the roulette wheel had been spun about 8 million times since the casino’s opening in 1861. What is the probability for the European-style roulette that in $n$ spins with $n \approx 8 \times 10^6$, the ball falls on the same color red or black 26 or more times in a row? (A European roulette wheel contains the numbers 0 through 36 , where the 0 is green and half of the numbers 1 through 36 are red and half are black.) To calculate this probability, we can define a Markov chain with 27 states $0,1, \ldots, 26$. For $i=1, \ldots, 26$; the state $i$ corresponds to the situation where the ball has fallen on the same color in the last $i$ spins of the roulette wheel but not the time before that. State 0 corresponds to the situation where the ball has fallen on the zero (or the game begins). State 26 is taken to be absorbing (imagine that the roulette game is only played fictitiously once state 26 is reached). Now, define $X_n$ as the state after the $n$th spin of the roulette wheel; then $\left{X_n\right}$ is a Markov chain with state space $I={0,1, \ldots, 26}$. The one-step transition probabilities are given by
\begin{aligned} & p_{00}=\frac{1}{37}, \quad p_{01}=\frac{36}{37} \ & p_{i, i+1}=p_{i 1}=\frac{18}{37}, \quad p_{i 0}=\frac{1}{37} \text { for } i=1, \ldots, 25 \ & p_{26,26}=1, \quad p_{i j}=0 \text { otherwise. } \end{aligned}

## 数学代写|运筹学代写Operations Research代考|Absorbing Markov Chains

$$p_{i 0}=\frac{5}{6}, \quad p_{i, i+1}=\frac{1}{6} \quad \text { for } i=0,1,2, \quad p_{33}=1, \quad \text { otherwise } p_{i j}=0$$

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