Prove the gambler can expect to play forever in the gambler’s ruin scenario

Question

In mcs.pdf, Problem 21.2 says:

In a gambler’s ruin scenario, the gambler makes independent $1$ bets, where the probability of winning a bet is p and of losing is $q ::= 1-p$. The gambler keeps betting until he goes broke or reaches a target of T dollars.

Suppose $T = \infty$, that is, the gambler keeps playing until he goes broke. Let r be the probability that starting with n > 0 dollars, the gambler’s stake ever gets reduced to $n-1$ dollars.

(a) Explain why $$ r = q + pr^2. $$

(b) Conclude that if $p \le 1/2$, then r = 1.

(c) Prove that even in a fair game, the gambler is sure to get ruined no matter how much money he starts with!

(d) Let t be the expected time for the gambler’s stake to go down by 1 dollar. Verify that $$ t = q + p(1 + 2t). $$ Conclude that starting with a 1 dollar stake in a fair game, the gambler can expect to play forever!

Here the book views the above process as one 1-D random walk.

Walking one step to the right corresponds to winning a \$1 bet and thereby increasing his capital by \$1. Similarly, walking one step to the left corresponds to losing a \$1 bet.

I can understand (a,b,c). For (d), it is similar to (a) where $t$ is independent from $n$. Then we either directly have $n-1$ or $n\to n+1\to n-1$. Then in one fair game $p=q=\frac{1}{2}$ so we have $t=t+1$. But that is inconsistent.

Since the problem asks to prove infinity, then we can interpret $t$ as $\aleph_0$ due to "$A$ is infinite iff there is a bijection between $A$ and $A \cup \{b\}$". Is it fine to use this interpretation? If not, then how to solve (d)?

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P.s. Here (d) says "expect" which is different from probability. The probability to continue indefinitely is zero. The situation is possible because the time can be arbitrarily long. Then when multiplying the very great time with the very small probability to calculate the expectation, we may get one very great expectation, i.e. infinity.

Answer 1

The expected value has a very specific meaning in probability theory - it is more-or-less the weighted average of the different outcomes (in this problem's case - the number of steps until the person goes broke), weighted by the probability that this outcome happens. In your case, the outcomes are discrete - there is probabilty $p_n$ (which you can calculate) that first time that the gambler goes down to $n-1$ happens at time $n$, so the expected value is defined as: $E = \sum_n{np_n}$ - again, this is a weighted (by $p_n$) average of the numbers $n$.

So it's very possible that the expected value is $\infty$ - it just means that the probabilities $p_n$, although they sum up to 1 as probabilities must do ($\sum_n{p_n}=1$), don't go down fast enough that $\sum_n{np_n}$ remains finite.

So with that definition of the expected value in mind, let's solve your problem without doing suspicious computation with infinities:

Let's assume that the formula that your book derives, $t=q+p(1+2t)$, is correct. From this, we know that in your case of $p=q={1\over 2}$ the expected value cannot be any finite number, because no finite $t$ can fullfill this equation: for $p=q={1\over 2}$, the equation simplifies to $t=1+t$ which obviously has no finite solution. So if the expected value cannot be finite, there remain two options what the expected value might be:

1. The expected value may be infinite
2. The expected value may not be defined at all

But option 2 is not possible, as I noted the expected value is just an infinite sum $\sum_n{n p_n}$ , an an infinite sum of positive numbers is always defined (but can be infinite). We are left just with the possibility that it's infinite.