Game: $\frac{1}{3}$ winning probability vs. $\frac{2}{3}$ winning probability

Question

$A$ and $B$ play a round-based game. Each round $A$ wins with probability $\frac{1}{3}$ and $B$ with probability $\frac{2}{3}$. The loser of a round pays $1$ USD to the winner. The winner of the whole game is the one who wins all the USD from the other player. Now assume that $A$ starts with $n\geq 1$ USD and $B$ with $1$ USD. What is the winning probability of $B$? (Hint: Use recursion and the law of total probability)

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Let be $P(B)$ the probability that $B$ wins the game and $P(B|(k))$ the probability that $B$ wins the game if $B$ has $k\geq 1$ USD. So either one could understand the question in the sense that we must compute $P(B)$ or we must compute $P(B|(1))$. However, in both cases we must apply law of total probability which requires us to find a disjoint decomposition of $B$ or $(B\cap (1))$, respectively. As we don't know how the corresponding sets are defined we can't do this.

Maybe someone can help me to disentangle this problem or show me how to set up a proper recursion without knowing how the sets look like.

Answer 1

$\def\eqdef{\stackrel{\text{def}}{=}}$ Hint

If $\ p_k=P(B\,|\,(k)\,)\ $, then $\ p_k\ $ must satisfy the following recursion: $$ p_k=\frac{p_{k-1}}{3}+\frac{2p_{k+1}}{3}\ , $$ with boundary conditions $\ p_0=0, p_{n+1}=1\ $.

Reply to OP's query in comment below

There's a hidden assumption underlying the above equation, which was almost certainly intended, but not explicitly stated, by the setter of the problem you've described. From the way you framed your question, I had assumed you were aware of this. From your query, however, I'm now guessing that your awareness of this assumption is only partial. The giveaway is your referral to $\ (k)\ $ as a "set". However, there's no single event at which player $B$ can have $\ k\ $ dollars, since he or she could have that amount after any round beyond the $\ (k-1)^\text{th}\ $.

What you write as $\ P(B\,|\,(k)\,)\ $ should actually be $\ P\big(\mathcal{B}\,\big|\,K_r=k\,\big)\ $, where $\ \mathcal{B}\ $ is the event that player $B$ wins, and $\ K_r\ $ is the number of dollars $B$ has after the $\ r^\text{th}\ $ round. If you don't make the hidden assumption I refer to above, then, as the problem is currently given, $\ P\big(\mathcal{B}\,\big|\,K_r=k\,\big)\ $ is not necessarily independent of $\ r\ $. Once you make this assumption, however, it becomes intuitively obvious that $\ P\big(\mathcal{B}\,\big|\,K_r=k\,\big)\ $ is independent of $\ r\ $ (at least for $\ r\ge k-1\ $. If $\ r< k-1\ $, then $\ \big\{\,K_r=k\,\big\}=\varnothing\ $ is a null event, and $\ P\big(\mathcal{B}\,\big|\,K_r=k\,\big)\ $ is meaningless).

Let $$ W_r=\cases{\hspace{0.8em}1&if $B$ wins round $\ r\ $\\ -1&if $B$ loses round $\ r\ $.} $$ Then $\ K_r=K_{r-1}+W_r\ $, provided $\ 2\le K_{r-1}\le n\ $. The hidden assumption referred to above is that $$ W_1,W_2,\dots,W_r,\dots $$ are independent. Then if the sequence $\ k_ 1,k_2,\dots, k_r\ $ satisfies the conditions $\ k_1=2, 1\le k_i\le n,\ $ and $\ \big|k_i-k_{i-1}\big|=1\ $ for $\ i=2,3,\dots,r\ $, then \begin{align} P\left(K_r=k_r\,\left|\,\bigcap_{i=1}^{r-1}\big\{K_i=k_i\big\}\right.\right)&=\frac{P\left(\bigcap_\limits{i=1}^r\big\{K_i=k_i\big\}\right)}{P\left(\bigcap_\limits{i=1}^{r-1}\big\{K_i=k_i\big\}\right)}\\ &=\frac{P\left(\big\{W_1=1\big\}\cap\bigcap_\limits{i=2}^r\big\{W_i=k_i-k_{i-1}\big\}\right)}{P\left(\big\{W_1=1\big\}\cap\bigcap_\limits{i=2}^{r-1}\big\{W_i=k_i-k_{i-1}\big\}\right)}\\ &=\frac{\prod_\limits{i=2}^rP\big(\big\{W_i=k_i-k_{i-1}\big\}\big)}{\prod_\limits{i=2}^{r-1}P\big(\big\{W_i=k_i-k_{i-1}\big\}\big)}\\ &=P\big(\,W_r=k_r-k_{r-1}\,\big)\\ &=P\big(\,K_r=k_r\,\big|\,K_{r-1}=k_{r-1}\,\big) \end{align} Also, \begin{align} P\big(\,K_r=k\,\big|\,K_{r-1}=0\,\big)&=\cases{1&if $\ k=0$\\ 0&otherwise}\\ P\big(\,K_r=k\,\big|\,K_{r-1}=n+1\,\big)&=\cases{1&if $\ k=n+1$\\ 0&otherwise} \end{align} Thus, under the above assumption, $\ K_r\ $ is a time-homogeneous Markov chain, with initial state, $\ K_0=1\ $, and $\ (n+2)\times(n+2)\ $ transition matrix, $\ P\ $, given by \begin{align} P_{ij}&\eqdef P\big(\,K_{r+1}=j\,\big|\,K_r=i\,\big)\\ &=\cases{\frac{2}{3}&if $\ 1\le j=i+1\le n+1$\\ \frac{1}{3}&if $\ 0\le j=i-1\le n-1$\\ 1&if $\ j=i\in\{0,n+1\}$\\ 0&otherwise,} \end{align} or $$ P=\pmatrix{1&0&0&0&\dots&\dots&0&0&0&0\\ \frac{1}{3}&0&\frac{2}{3}&0&\dots&\dots&0&0&0&0\\ 0&\frac{1}{3}&0&\frac{2}{3}&\dots&\dots&0&0&0&0\\ \vdots&&\ddots&\ddots&\ddots&&\vdots&\vdots&\vdots&\vdots\\ \vdots&\vdots&&\ddots&\ddots&\ddots&&\vdots&\vdots&\vdots\\ \vdots&\vdots&\vdots&&\ddots&\ddots&\ddots&&\vdots&\vdots\\ \vdots&\vdots&\vdots&\vdots&&\ddots&\ddots&\ddots&&\vdots\\ 0&0&0&0&\dots&\dots&\frac{1}{3}&0&\frac{2}{3}&0\\ 0&0&0&0&\dots&\dots&0&\frac{1}{3}&0&\frac{2}{3}\\ 0&0&0&0&\dots&\dots&0&0&0&1}\ . $$ This Markov chain has two absorbing states, at $\ k=0\ $ and $\ k=n+1\ $, and both of these are reachable from any of the other states, all of which are transient. The theory of such chains tells us $\ P^n\ $ converges to a limit $\ P_\infty\ $ as $\ n\rightarrow\infty\ $, and \begin{align} P_\infty&=\pmatrix{1&0&\dots&0&0\\ 1-p_1&0&\dots&0&p_1\\ 1-p_2&0&\dots&0&p_2\\ \vdots&\vdots&&\vdots&\vdots\\ 1-p_n&0&\dots&0&p_n\\ 0&0&\dots&0&1}\\ &=(\mathbf{1}-p)\,\varepsilon_0^T+p\,\varepsilon_{n+1}^T\ , \end{align} where $\ \varepsilon_j\ $ is the $\ (n+2)\times1\ $ column vector whose $\ j^{\,\text{th}}\ $ entry is $\ 1\ $ and all of whose other entries are $\ 0\ $, $\ \mathbf{1}\ $ the $\ (n+2)\times1\ $ column vector all of whose entries are $\ 1\ $, and $\ p\ $ the $\ (n+2)\times1\ $ column vector whose $\ j^{\,\text{th}}\ $ entry is $\ p_j\eqdef P\big(\mathcal{B}\,\big|\,K_r=j\,\big)\ $. You can now obtain the above recursion from the fact that for $\ 1\le k\le n+1\ $ \begin{align} p_k&=\varepsilon_k^Tp\\ &=\varepsilon_k^TP_\infty\varepsilon_{n+1}\\ &=\varepsilon_k^TPP_\infty\varepsilon_{n+1}\\ &=\left(\frac{1}{3}\varepsilon_{k-1}+\frac{2}{3}\varepsilon_{k+1}\right)P_\infty\varepsilon_{n+1}\\ &=\frac{p_{k-1}}{3}+\frac{2p_{k+1}}{3}\ . \end{align} Once you become familiar with the properties of time-homogeneous Markov chains, all the above rigmarole becomes completely redundant. You can simply state that the independence of $\ W_r\ $ implies that $\ K_r\ $ is a time-homogeneous Markov chain and write down the recursion without further ado.

Answer 2

Notation

We make the following notation local to this solution. This notation is required only if users need a completely rigorous proof, as requested by OP.

For the situation when $B$ has $1$ USD and $A$ has $n$ USD, let us construct the corresponding sample space $\Omega_n$. We define $$ \Omega_{n} = \{(R_1,R_2,\ldots,R_m) : m \geq 1, R_i \in \{A,B\} \text{ and the game is complete after $m$ steps}\} $$

Here, $R_1,R_2,\ldots,R_m$ are letters that either $A$ or $B$, and if $R_i = A$ then $A$ won in the $i$th round. Otherwise, if $R_i = B$ then $B$ won in the $i$th round. In addition, we stop the tuple at round $m$ when there is some winner of the game at the end of that round.

For example, if $n=2$ then I can write down some elements of $\Omega_2$ as follows :

• $(A)$ : So $A$ won the first round and $B$ has $0$ USD left, hence $A$ wins.

• $(B,A,A) $ : $B$ won the first round , but $A$ won the next two. Now, $B$ has $0$ USD left, so $A$ wins the game.

• $(B,B)$ : $B$ won all three rounds, and $A$ now has $0$ USD left, hence $B$ wins the game.

• $(B,A,B,B)$ : $B$ won the first round, $A$ won the next, and then $B$ won the next two so $A$ has $0$ USD left and $B$ wins the game.

An example of a tuple not in $\Omega_2$, for example, is $(B)$ because although $B$ won the first round, there is no winner at the end of this round. Similarly, $(B,A,B,A) \notin \Omega_2$. I hope that the description of elements of $\Omega_n$ are now clear to everybody.

To make $\Omega_n$ into a probability space, we need a probability assignment $p_n : \Omega_n \to [0,1]$. This is easy : we know that $B$ wins each round with probability $\frac 23$ and $A$ with probability $\frac 13$. Hence, if $(R_1,\ldots,R_m) \in \Omega_n$ and $M$ is the number of $A$s in the tuple $(R_1,\ldots,R_m)$, then $$ p_n((R_1,\ldots,R_m)) = \left(\frac{1}{3}\right)^M \left(\frac 23\right)^{m-M}. $$ Now, $(\Omega_n,p_n)$ is a proper probability space for each $n \geq 1$ because we have a sample space along with a probability assignment on it. We will refer to the elements of $\Omega_n$ as "games" because that's what each element represents.

By definition of $\Omega_n$, every sample element is a game that either has $B$ as a winner or $A$ as a winner at the end. Let $P_n$ be the probability of the event that, at the end of a game in $\Omega_n$, $B$ is the winner.

We will find a way to relate $P_n$ with $P_{n-1}$, which is the intention of the author here.

(There are ways to solve this question without changing the value of $n$ and using recursion. This approach strictly uses recursion by relating $P_n$ to $P_{n-1}$). For a proof that doesn't change the value of $n$, see the other answer to this question.

We will require the analysis of another "new" game in order to create the recursion. Here's the description of the "new" game : consider the old game where $A$ has probability $\frac 13$ of winning a round, $B$ has probability $\frac 23$ of winning a round, the winner gives $1$ USD to the loser, but the key difference is that $A$ starts with $1$ USD, while $B$ starts with $n$ USD.

This "new" game has its own sample space $\Omega'_n$, which is given by $$ \Omega'_{n} = \{(R_1,R_2,\ldots,R_m) : m \geq 1, R_i \in \{A,B\} \text{ and the "new" game is complete after $m$ steps}\} $$

So for example, $(B) \in \Omega'_{2}$ because $B$ winning one round means that $A$ has $0$ USD at the end of this round so $B$ is the winner of the "new" game. Similarly, $(A,A),(A,B,B) \in \Omega'_{2}$ as well.

The same kind of probability assignment $p'_n$ can given be easily specified for $\Omega'_2$ : $$ p'_n((R_1,\ldots,R_m)) = \left(\frac{1}{3}\right)^M \left(\frac 23\right)^{m-M}. $$ where $M$ is the number of $A$s in $R_1,\ldots,R_m$. Thus, we have another bunch of sample spaces $(\Omega'_n,p'_n), n \geq 1$ corresponding to the "new" game where $A$ starts with $1$ USD and $B$ starts with $n$ USD. We refer to elements of $\Omega'_n$ as "new" games.

Let $Q_n$ be the probability(in $\Omega'_n$) that $A$ is the winner at the end of a "new" game. For example, $A$ wins at the end of the "new" game $(A,A,A)$.

We will find ways to calculate both $P_n$ and $Q_n$ using recursive formulas dependent on each other.

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Let $E_n \subset \Omega_n$ be the event that $B$ wins the usual game and $F_n \subset \Omega'_n$ be the event that $A$ wins the "new" game. Observe that $P_n$ is the probability(in $\Omega_n$) that $E_n$ occurs, similarly $Q_n$ is the probability $F_n$ occurs.

Let $(R_1,\ldots,R_m) \in E_n$. Then, $B$ wins at the end of the game, which means that $A$ has $0$ USD left. Now, during every such game, there must be a first point during which $A$ has exactly $1$ USD left and $B$ has $n$ USD. Define $f_n: E_n \to \mathbb N$ as $f_n((R_1,\ldots,R_m)) = i$ where $i$ is the smallest round, after whose completion $A$ has $1$ USD left.

For example, consider the game $$ (B,A,B,B,A,B,B) \in \Omega_3. $$ In this game, $B$ wins at the end, but there are two occasions when $A$ has exactly $1$ USD left : at the end of the fourth round and at the end of the sixth round. In this case, $f((B,A,B,B,A,B,B)) = 4$.

Similarly, $$ (B,B) \in \Omega_2 \implies f((B,B)) = 1. $$

Now, consider the following "split" functions $G_{1,n}$ and $G_{2,n}$ defined on $E_n$ : if $l \in E_n$ is a game, then $G_{1,n}(l)$ is the tuple formed by taking all the rounds in a game until the end of the $f(l)$th round. Whatever comes after that is defined by $G_{2,n}(l)$.

For example, $$ f_3((B,A,B,B,A,B,B)) = 4\\ \implies G_{1,3}(B,A,B,B,A,B,B) = (B,A,B,B),G_{2,3}((B,A,B,B,A,B,B)) = (A,B,B)\\ f_2((B,B)) = 1 \\ \implies G_{1,2}((B,B)) = (B), G_{2,2}((B,B)) = (B) $$

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What follows is the first central claim of this entire argument, along with a proof (that, if self-evident, can be skipped).

The "split" map $l \to (G_{1,n}(l), G_{2,n}(l))$ is a bijection between $E_{n}$ and $E_{n-1} \times (\Omega'_n \setminus F_n)$.

Here are examples of this bijection : take $l = (B,A,B,B,A,B,B)$. THen, if you look just at $G_{1,3}(l) = (B,A,B,B)$, then just by itself, $(B,A,B,B)$ represents an entire game in which $A$ had $2$ USD, $B$ had $1$ USD, and $B$ won. That is, $(B,A,B,B) \in E_{n-1}$. On the other hand, $(A,B,B)$ is a "new" game in which $A$ had $1$ USD, $B$ has $2$ USD, and $A$ lost : which is why $(A,B,B) \in (\Omega'_n \setminus F_n)$.

Proof : Let $(R_1,\ldots,R_m)\in E_n$. By definition , $$G_{1,n}((R_1,\ldots,R_m))=(R_1,\ldots,R_{f_n((R_1,\ldots,R_m))}).$$

Now, imagine that we were playing the game in which $A$ has $n$ USD to begin with. If the sequence of winners is given by $R_1,\ldots,R_{f_n((R_1,\ldots,R_m))}$ then at the end of round $f_n((R_1,\ldots,R_m))$ , $A$ has $1$ USD and $B$ has $n$ USD.

However, if $A$ instead had $n-1$ USD to begin with instead of $n$, then the same sequence of rounds $(R_1,\ldots,R_{f_n((R_1,\ldots,R_m))})$ played in that order clearly result in $A$ having $0$ USD at the end of round $f_n((R_1,\ldots,R_m))$ (because $A$ has one less dollar). It follows that $G_{1,n}((R_1,\ldots,R_m)) \in E_{n-1}$.

Now consider, by definition, $$ G_{2,n}((R_1,\ldots,R_m))=(R_{f_n((R_1,\ldots,R_m))+1},\ldots,R_m). $$ In the usual game, this is the phase of play which begins with $A$ having $1$ USD and $B$ having $n$ USD, and finishes with $A$ losing. However, we can treat this phase as a "new" game because $A$ is effectively starting from $1$ USD and loses from that point on. Hence, it is obvious that $$ G_{2,n}((R_1,\ldots,R_m)) \in \Omega'_n \setminus F_{n} $$ because $F_{n}$ consists of those "new" games in which $A$ wins, but we want $B$ to win instead, so we take the complement.

Now, there is an obvious map the other way, the "concatenation" map : given $(R_1,\ldots,R_m) \in E_{n-1}$ and $(S_1,\ldots,S_k) \in \Omega'_n \setminus F_{n}$, just play the "new" game after the old game. That is, consider the element $$ (R_1,\ldots,R_m,S_1,\ldots,S_k) $$ we claim that this element is in $E_n$. But that's obvious : if $B$ has $1$ USD and $A$ has $n$ USD, then because $(R_1,\ldots,R_m) \in E_{n-1}$, we see that at the end of round $R_m$, $B$ has $n$ USD and $A$ has $1$ USD left. Now, from that point on, $(S_1,\ldots,S_k)$ is chosen so that $A$ loses from this point on. All in all, the entire tuple represents a game in which $B$ has $1$ USD, $A$ has $n$ USD, and yet $B$ wins. Hence, the game belongs in $E_{n-1}$.

It is not difficult to see that the "split" and "concatenation" maps that have been described between $E_n$ and $E_{n-1} \times (\Omega'_n \setminus F_n)$ are inverses of each other. We are done with the proof of the main result. $\blacksquare$

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Now, with a proof just like the above, we make a similar but equally important claim. Let us define the analogous notation.

Let $(R_1,\ldots,R_m) \in F_n$. Then, $A$ wins at the end of the game, which means that $B$ has $0$ USD left. During every such game, there must be a first point during which $B$ has exactly $1$ USD left and $A$ has $n$ USD. Define $f'_n: E_n \to \mathbb N$ as $f'_n((R_1,\ldots,R_m)) = i$ where $i$ is the smallest round, after whose completion $B$ has $1$ USD left.

Consider the following "split" functions $G'_{1,n}$ and $G'_{2,n}$ defined on $F_n$ : if $l \in F_n$ is a game, then $G'_{1,n}(l)$ is the tuple formed by taking all the rounds in a game until the end of the $f'(l)$th round. Whatever comes after that is defined by $G'_{2,n}(l)$.

The second central claim follows, with a proof entirely analogous to that of the first claim.

The "split" map $l \to (G'_{1,n}(l), G'_{2,n}(l))$ is a bijection between $F_{n}$ and $F_{n-1} \times (\Omega_n \setminus E_n)$.

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As a result of these bijection, we derive the following equations.

We have $P_1= \frac 23, Q_1 = \frac 13$. Furthermore, for every $n \geq 2$, $$ P_n = P_{n-1}(1-Q_{n}) \\ Q_n = Q_{n-1}(1-P_n) $$

Proof : We observe the following "probability-preserving property" of the "split" map we defined in our first claim. Let $(R_1,\ldots,R_m) \in E_n$ and let $M$ be the number of $A$s in $(R_1,\ldots,R_m)$, with $M_1,M_2$ being the number of $A$s in $G_{1,n}((R_1,\ldots,R_m))$ and $G_{2,n}((R_1,\ldots,R_m))$ respectively.

Because the "split" map literally splits the tuple $(R_1,\ldots,R_m)$ into two parts, it is obvious that $M = M_1+M_2$. Therefore, $$ p_n((R_1,\ldots,R_m)) = \left(\frac 13\right)^M \left(\frac 23\right)^{m-M} \\ = \left[\left(\frac 13\right)^{M_1} \left(\frac 23\right)^{f_n((R_1,\ldots,R_m))-M_1}\right]\left[\left(\frac 13\right)^{M_2} \left(\frac 23\right)^{m-f_n((R_1,\ldots,R_m))-M_2}\right] \\ = p_{n-1}(G_{1,n}((R_1,\ldots,R_m)) \times p'_{n}(G_{2,n}(R_1,\ldots,R_m)) $$

Now, because of the "concatenation" side of the bijection established in the first claim and this probability preserving property, \begin{align} &P_n \\&= \sum_{(R_1,\ldots,R_m) \in E_n} p_n(R_1,\ldots,R_m) \\&= \sum_{(S_1,\ldots,S_k) \in E_{n-1},(T_1,\ldots,T_l) \in \Omega'_n \setminus F_n} p_n(S_1,\ldots,S_k,T_1,\ldots,T_l) \\ &= \sum_{(S_1,\ldots,S_k) \in E_{n-1},(T_1,\ldots,T_l) \in \Omega'_n \setminus F_n}p_{n-1}((S_1,\ldots,S_k)) \times p'_{n}((T_1,\ldots,T_l)) \\ &= \left[\sum_{(S_1,\ldots,S_k) \in E_{n-1}} p_{n-1}((S_1,\ldots,S_k)) \right]\left[\sum_{(T_1,\ldots,T_l) \in \Omega'_n \setminus F_n} p'_{n}((T_1,\ldots,T_l))\right] \\&= P_{n-1}(1-Q_{n}) \end{align}

The proof of the other result is similar and follows from the second key central claim.

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Thus, we have the equations $$ P_{n}=P_{n-1}(1-Q_n) \\ Q_n = Q_{n-1}(1-P_n) $$ Solving these equations, $$ P_{n} = P_{n-1}(1-Q_{n-1}(1-P_n)) = P_{n-1} - P_{n-1}Q_{n-1} + P_{n-1}Q_{n-1}P_n \\ \implies P_{n} =\frac{P_{n-1}(1-Q_{n-1})}{1-P_{n-1}Q_{n-1}} \\ Q_n = \frac{Q_{n-1}(1-P_{n-1})}{1-P_{n-1}Q_{n-1}} $$

This along with the initial values $$ P_1 = \frac 23, Q_1 = \frac 13 $$ furnishes the answer (the initial values are obvious). For example, $$ P_2 = \frac{4}{7} , Q_2 = \frac 17 \\ P_3 = \frac{8}{15}, Q_3 = \frac{1}{15} $$

But now, we begin to see a pattern here. Indeed, it is not difficult to observe that $$ P_i = \frac{2^i}{2^{i+1}-1}, Q_i = \frac{1}{2^{i+1}-1} $$

These can easily be proved by induction, providing the answer to the question.

Answer 3

Answer:† $\frac{2^{n}}{2^{n+1}-1}$.

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Fix $n\ge 1$.

Suppose we're at some point of the game when $A$ has $m$ dollars and $B$ has $\left(n+1-m\right)$ dollars ($m\in\left\{ 0,1,2,\dots,n,n+1\right\} $).

Let $p_{m}$ be the probability $B$ wins the game at this point of the game. (So, $p_{n}$ will be our answer.)

Then

• $p_{n+1}=0$.
• $p_{0}=1$.
• For every $m\in\left\{ 1,2,\dots,n\right\} $, $p_{m}=\frac{2}{3}p_{m-1}+\frac{1}{3}p_{m+1}$ or $p_{m-1}\overset{\star}{=}\frac{3}{2}\left(p_{m}-\frac{1}{3}p_{m+1}\right)$.

So,

• $p_{n-1}\overset{\star}{=}\frac{3}{2}\left(p_{n}-\frac{1}{3}p_{n+1}\right)=\frac{3}{2}p_{n}$.
• $p_{n-2}\overset{\star}{=}\frac{3}{2}\left(p_{n-1}-\frac{1}{3}p_{n}\right)=\frac{3}{2}\left(\frac{3}{2}p_{n}-\frac{1}{3}p_{n}\right)=\frac{7}{4}p_{n}$.
• $p_{n-3}\overset{\star}{=}\frac{3}{2}\left(p_{n-2}-\frac{1}{3}p_{n-1}\right)=\frac{3}{2}\left(\frac{7}{4}p_{n}-\frac{1}{2}p_{n}\right)=\frac{15}{8}p_{n}$.

We can show by induction (omitted) that for any integer $k\leq n$, $p_{n-k}=\frac{2^{k+1}-1}{2^{k}}p_{n}$.

So, $p_{n}=\frac{2^{n}}{2^{n+1}-1}p_{n-n}=\frac{2^{n}}{2^{n+1}-1}p_{0}=\frac{2^{n}}{2^{n+1}-1}$.

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†I assume $n$ is an integer but if $n$ can be a non-integer, then change the answer to $\frac{2^{\lceil n \rceil}}{2^{\lceil n\rceil+1}-1}$.