$A$ starts with $i$ coins, $B$ with $N-i$. At each trial, $A$ gives one coin to $B$ with probability $p$ or $B$ gives one coin to $A$ with probability $q$ where $p+q=1$.

This can be modeled as a 2D random walk starting from $i$ where probability of moving right = $p$, left = $q$, and the walk ends at reaching either $0$ or $N$

Nothing in this statement seems to say that oscillating around i and never reaching either $0$ or $N$ is not a possibility. However, doing the following calculation, something seems off.

Let $p_i$ be the probability that A will end up with all money, that is, the object will reach N, when starting position is $i$. $$p_i = p*p_{i+1} + q*p_{i-1}$$ Solving this difference equation gives $$p_i = \frac{1-(\frac{q}{p})^i}{1-(\frac{q}{p})^N} $$ for $p\neq q$

Now, P(reaching N starting from $i$) = $p_i$ By symmetry, P(reaching 0 starting from $i$) = $\frac{1-(\frac{p}{q})^{N-i}}{1-(\frac{p}{q})^N}$

Adding those together equals 1. Which means either A wins or B wins. That is no probability left for just oscillating around $i$ and never reaching either $0$ or $N$. Why is that so?

The probability for the event, going from i to i+1, then back to i, then i+1 and so on = $p*q*p*q*... = (pq)^n$ where n can go up to infinity. However small, this is a positive number. And unless n goes to infinity, it is greater than zero.

I understand that $\lim_{n \to \infty} (pq)^n = 0$. But how is that applicable here. As $n \to \infty$, probability $\to 0$. But $n$ is always less than $\infty$, so probability is always greater than $0$. Please tell me if I am wrong with my interpretation of limits.

Is it correct that either A or B has to win? Why?

  • Hint: try to calculate the duration $T$ of the game, i.e. the number of steps it takes for the "walker" to reach $x=0$ ($T_{0}$) or x = $N$ ($T_{N}$). The $T$'s are, of course, random variables, accompanied with corresponding distribution functions. Start with the simplest non trivial case, $N=3$, $i=1$. You can even simplify things letting $N\to\infty$ which then amounts to calculate the probability that the "walker" which has started at $x=i$ will reach $x=0$. – Dr. Wolfgang Hintze Nov 8 at 14:34
  • @Dr.WolfgangHintze Time $T$ taken to reach $0$ while assuming $N \to \infty$ for simplicity, has a distribution. It can be arbitrarily but finitely large. But how does that mean, in the "end", the walker has to come to $0$. For every possible value of T, there is a small but positive probability that the walker has not come to zero until T. Therefore, for every finite duration of the game, there is a probability left for the event that walker has not come to $0$. – sourav goyal Nov 8 at 16:16
  • This seems more like a question about interpretation of limits, now. – sourav goyal Nov 8 at 16:19
  • The probability of oscillating forever is 0. That does not mean it is impossible, just that it's likelihood is smaller than any positive real number. For a simpler example, the probability of rolling a coin repeatedly and getting tails every time, forever, is 0, but it's not impossible. – Ned Nov 9 at 0:43
  • @ sourav goyal Please consider: I have given you the first answer here yesterday, and I would have expected a response, instead of seing acceptance of a later very similar answer. – Dr. Wolfgang Hintze 2 days ago
up vote 1 down vote accepted

This seems to be a question about the meaning of limits. I think the following much simpler example captures all the issues.

Suppose you flip a fair coin again and again, and consider the event that all results are Heads. Clearly $Prob(first\ N\ are\ all\ Heads) = p_N = {1 \over 2^N}$. The following statements are all true and not contradictory:

  • For any finite $N, p_N > 0$.

  • $lim_{N \rightarrow \infty} \ p_N = 0$.

  • Suppose you lose when the first Tail shows up. The event $E$ that you don't lose is non-empty, i.e. $E \neq \emptyset$, or equivalently, there exists a sample point (namely, all Heads forever) where you don't lose. However $P(E) = 0$ (or equivalently, $P(lose) = 1$).

  • $E \neq \emptyset$ and yet $P(E)=0$ is not that surprising. After all, the sample space contains an infinite number of sample points (all infinitely-long sequences of H/T, i.e. $\{H,T\}^\infty$). This kind of thing happens all the time with continuous random variables. E.g. if $X = Uniform(0,1)$ then $P(X=0.1) = 0$ even though it is clearly a non-empty event.

Back to your example, the analogous $E = $ the set of sample points where the random walk never reaches either boundary (more precisely, $\forall t$ the position at $t \neq$ either boundary). Clearly $E \neq \emptyset$, but you have also shown that $P(E)= 0$.

Does this help, or at least, help to distill the issue?

  • This is exactly what I was looking for. $P(E)=0$ even when $E\neq\emptyset$. In the cases where a sample point, that is, a sequence of HT.. in your example, can be of arbitrarily large length, To calculate $P(E)$, one has to take a limit $n\to\infty$ . It seems rather stupid, but I ask why do we take limits. If n approaches $\infty$ but never reaches $\infty$, Why doesn't it mean $P(E)$ approaches $0$ but never reaches $0$? Shouldn't we say $\lim P(E) = 0$ instead of $P(E)=0$? – sourav goyal Nov 9 at 18:57
  • The $P(X=0.1) = 0$ for $X=Uniform(0,1)$ example is very illuminating. – sourav goyal Nov 9 at 18:59
  • To be precise, we have to define the sample space. I'd prefer to define it as $\{H,T\}^\infty$, in which case $H^\infty$ is a valid sample point. Now $P()$ is a probability which means it maps Events to $[0,1]$. For the event $E = \{H^\infty\}$, we have $P(E)=0$. $N$ doesnt even figure into this discussion. A different event is $E_N = \{H^N + \{H,T\}^\infty\}$ i.e. the first $N$ flips are heads. This has $P(E_N) = 1/2^N$. Note that every sample point in $E_N$ is still infinitely long (and therefore each individual pt's $P=0$). – antkam Nov 9 at 19:22
  • (cont'd) My point is $P(E)$ has to be a number, in $[0,1]$... that is what probability $P()$ does. (Alternatively, you can take the view that $E$ is not a valid event, so that $P(E)$ is undefined.) And if $P(E)$ is a number, then $\lim_{N\rightarrow \infty} P(E)$ is just the same number since $P(E)$ doesnt even depend on $N$. – antkam Nov 9 at 19:27
  • BTW, since every real number has a binary expansion, the $Unif(0,1)$ example is actually very analogous with $\{H,T\}^\infty$. (It'd be an exact bijection if not for the fact that $0.011111... = 0.100000...$.) – antkam Nov 9 at 19:32

Your problem can be visualized and explained with the most simple case.

Take $N=3$, $p=q=\frac{1}{2}$, for simplicity. $t$ is the running time (or step number), and $T$ is the duration, i.e. the time $t$ at which either $x(t)=0$ or $x(t)=3$.

Let the walker start at $x=1$ i.e. $x(t=0) = 1$.

For $t=1$ there are two possibilities, $x(1) = 0$ which gives $T=1$ with probability $p$, and $x(1) = 2$ with probability $p$.

In the next time step we have $x(2)=3$, i.e. $T=2$ with probability $p^2$, and $x(2)=1$ with prob. $p^2$. For $t=3$ we have $x(3)=0$ giving $T=3$ with prob $p^3$, and so on.

Generally, the probability that the game ends at time $t=T$ is $w(T) = 2^{-T}$. The average duration is hence $T_{ave}=\sum_{T=1}^\infty T 2^{-T} = 2$

Hence the result is that the walk can be aribitrarily long but the probability decreases with increasing length in a manner that the average length is (only) 2.

Remark: if we drop the assumption $p=q$ then the probability for a duration $T$ is given by

$$w(T) = \left\{ \begin{array} (p^{\frac{T-1}{2}} q^{\frac{T+1}{2}} & T \;\text{odd}, T\ge1 \\ p^{\frac{T+2}{2}} q^{\frac{T-2}{2}} & T \; \text{even}, T\ge2 \\ \end{array} \right. $$

and the average is given by

$$T_{ave} = \frac{p+1}{1-p q}$$

The formula is not symmetric in $p$ and $q$. This was not to be expected since the walker started at the unsymmteric point $x=1$.

  • What you are saying is totally right. However, you didn't answer why the probability of infinite length is zero. I understand that it decreases. My question was why does it go to zero. – sourav goyal 2 days ago
  • Honestly, that $2^{-T}$ goes to $0$ for $T\to\infty$ was all too obvious to be mentioned. – Dr. Wolfgang Hintze yesterday

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