# What is the distribution of the time-to-ruin for a gambler's ruin problem that allows "pauper bets"?

In another question on this site I have derived the distribution for the time-to-ruin in the gambler's ruin problem where the wealth of the gambler follows a discrete-time random walk. In this standard problem, we assume that the gambler starts with an amount of wealth that is a positive integer multiple of the betting amount, such that there is no "residual wealth" lower than the betting amount when the gambler gets down near to zero wealth.

In the present question I would like to examine a variant of this standard problem. Here we suppose that (until ruin) the gambler has some positive wealth $$w>0$$ and bets in unit increments as normal. However, if the wealth of the gambler gets into the range $$0 (i.e., less than the standard betting amount but more than nothing) then the gambler is allowed a "pauper bet" where the gambler bets his remaining wealth (even though it is less than one unit). The pauper bet is treated the same as the standard bets, in the sense that a loss loses the bet amount and a win returns the bet amount as profit. Thus, when the gambler makes a "pauper bet" his wealth either goes to zero or to $$2w$$ (with equal probability). Note that the gambler is only allowed to make a pauper bet whenever his wealth is less than one unit.

The stochastic process for this variant of the gambler's ruin can be described formally as follows. Let $$w_0 > 0$$ be the gambler's starting wealth. Unlike in the standard problem, this value can be any positive real value; not just an integer. Let $$\{ Q_t | t \in \mathbb{N} \}$$ denote the sign values for the bet outcomes. The time-series for the gambler's wealth is described recursively by:

$$w_{t+1} = w_t + Q_i \cdot \min(1, w_t) \quad \quad \quad \mathbb{P}(Q_i = -1) = \mathbb{P}(Q_i = 1) = \frac{1}{2}.$$

The time-to-ruin is defined as:

$$T \equiv \min \{ t \in \mathbb{N} | w_t = 0 \}.$$

I would like to derive the distribution of the time-to-ruin for this variant of the gambler's ruin problem. The remainder amounts in the "pauper bets" complicate the situation substantially, so I am looking for techniques to deal with this variation in the derivation of the distribution. Can anyone show how to derive the distribution of the time-to-ruin in this case?

As a smaller—but still interesting—question, are the time-to-ruin probabilities even continuous with respect to the starting wealth $$w_0$$? Intuitively it seems certain that they would be continuous at non-integer values, but it might be possible that there would be a "jump" at the integer values. I think they would be continuous, but a proof of this would be nice.

• "are the time-to-ruin probabilities even continuous with respect to the starting wealth" I suspect that this is not the case. If the fractional part of starting wealth is $>0.5$ then the first time that a pauper bet is neccesary, will only need a single successful bet to get the the gambler out of the dangerous zone of being one step away from the pauper bet. If the wealth is $<0.5$ then they will need two successive successful bets in order to be out of the danger zone. So around 0.5 there is a discontinuity in the way that the bets work out. Feb 22, 2023 at 13:03

Some preliminary attempt.

### Waiting time from pauper bet to pauper bet

Say we just got out of the pauper bet zone. Then the distribution of waiting time untill reaching the pauper betting zone again is for odd $$t=2n+1$$ values:

$$\begin{array}{} P(T_\text{next pauper zone} = t=2n+1)& =& \frac{P_\text{binom}(n,2n,0.5)-P_\text{binom}(n-1,2n,0.5)}{2}\\ & = &{2n\choose n}0.5^{2n}\frac{\left(1-\frac{n}{n+1}\right)}{2}\\ &= &\frac{(2n)!}{(n+1)!(n)!} 0.5^{2n+1}\end{array}$$

### Inside the pauper zone

Each time that we reach the 'pauper zone' (when $$0) we have a $$\frac{1}{2^k}$$ probability of getting out of it, where $$k$$ depends on the value when we enter the zone. For some values we need multiple successes before we get out of the zone.

### Connecting all together

Say that we start with wealth $$1, then the time to ruin will be a mixture of a sum of variables.

Let denote $$n(k)$$ be the number of neccesary success to get out of pauper zone $$k$$ times. Let $$K$$ be the number of times that we get out of the pauper zone, where $$P(K > k) = 0.5^{n(k)}$$ and $$n(0) = 0$$, the first betting sequence we get for free, because we start outsode the pauper zone.

Let $$T_i$$ be independently drawn variables for times untill hitting the pauper zone.

Then we can consider the following equivalent process for getting the hitting time:

$$T = \sum_{i=0}^K T_i$$

That is, the time untill 'ruin' is a sum of independent waiting times, where the number of terms in the sum depends on the number of times $$K$$ that we 'get out of the pauper zone'.

If we define $$Z_k = \sum_{i=0}^k T_i$$ a variable whose mass distribution function is still unclear (It is a convolution $$T_i$$, this I still have to work out), but let's say that it is $$p_{Z_k}(t)$$, then the probability mass function for the time untill ruin is

$$P(T_{\text{ruin}} = t) = \sum_{k=0}^\infty P(K = k) p_{Z_k}(t) = \sum_{k=0}^\infty \left(0.5^{n(k)} - 0.5^{n(k+1)}\right) p_{Z_k}(t)$$

Todo-list:

• Create an expression for $$p_{Z_k}(t)$$ which is a convolution of variables distributed according to $$\frac{(2n)!}{(n+1)!(n)!} 0.5^{2n+1}$$.
• Create an expression for $$n(k)$$. This I believe is very difficult. For some values it is easy. For example, when we start with a value like $$w_0 = 1 + \frac{1}{2^{x}-1}$$ then we will have a constant $$n(k) = x$$. E.g. when we have $$w_0 = 1+\frac{1}{3}$$, then every time we will get into the pauper zone with a wealth of $$1/3$$ and exit again, after two successes, with a wealth of $$4/3$$.
• The waiting time untill hitting the barrier at a distance $x$ multiple times is similar to the waiting time of hitting a barrier at a distance $kx$ a single time. Feb 22, 2023 at 16:59