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I have an infinitely large pantry.

I put some number (say, 100) of special potatoes in there. Every day, each potato has an equally-likely probability of either:

  1. Dying
  2. Living
  3. Living plus producing a single identical clone of itself

After infinitely many days, how many potatoes will I have? My intuition tells me 0 because that's the only stable end-state, but I could see arguments for infinity or undefined/non-convergence, as well.

(Not sure if these should be asked as separate questions)

What if I instead start with countably infinite potatoes?

What if I instead start with uncountably infinite potatoes?

Sarov
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  • Seems like this problem might be related to the problem of a Gaussian random walk. – Robert Shore Sep 10 '21 at 23:14
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    Reason for downvote? – Sarov Sep 11 '21 at 00:18
  • For countably infinitely many potatoes, it should continue to be countably infinitely many potatoes with probability 1 (we can model as sequences of functions $f_n : \mathbb{N} \to \mathbb{N}$ with initial value $f_0 = 1$ where the number of potatoes is $\sum_if_n(i)$). For uncountably infinitely many potatoes, I don't even know how to model it in probability – Brian Moehring Sep 11 '21 at 02:14
  • @Sarov: A very fun question. Interpretations galore! – A rural reader Sep 11 '21 at 02:52
  • On Day 0 let’s say you have 300 live potatoes. Overnight you can count on 100 dying. Of the 200 survivors, 100 replicate and 100 don’t — you start the day with $100 + 2\times 100$ live potatoes. – A rural reader Sep 11 '21 at 03:08

2 Answers2

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I interpret your question as:

Given $100$ potatoes, what is the expected number of potatoes after $n$ days, as $n$ goes to infinity?

Of course the $100$ potatoes and their offspring are independent, so the expected number of potatoes is simply $100$ times the expected number of potatoes when starting with a single potato. Of course this is also true for every following day. Let $E(P_n)$ denote the expected number of potatoes after $n$ days. Then $E(P_0)=100$ and $$E(P_{n+1})=E(P_n)\cdot\left(\tfrac13\cdot0+\tfrac13\cdot1+\tfrac13\cdot2\right)=E(P_n).$$ This shows that the expected number of potatoes is constant as the number of days increases.

Alternatively, with a bit of imagination I can interpret your question as follows:

Given $100$ potatoes, what is the probability that they and their offspring eventually die out?

Consider the process for just a single potato. Let $P$ denote the probability that the potato and its offspring eventually die out. Then from the three options we see that $$P=\frac13\cdot1+\frac13\cdot P+\frac13\cdot P^2.$$ It follows that $(1-P)^2=0$ and hence that $P=1$. So regardless of the number of starting potatoes, they will all die out eventually, almost surely.

Servaes
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    The facts that the expected number is constant and that the potatoes die out with probability $1$ are not inconsistent. They just require that there be a possibility of a steadily rising number of potatoes with decreasing probability. – Ross Millikan Sep 11 '21 at 02:04
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With $d_n$ and $l_n$ denoting the number of dead and live spuds you have you have a process that looks like, on Day $n+1$, \begin{align} d_{n+1} &= d_{n} + \alpha l_{n} \\ l_{n+1} &= (1 - \beta + 2\beta)(1 - \alpha) l_{n} \\ &= (1 - \alpha)(1 + \beta)l_n \end{align} where $\alpha$ is the fraction of the living that die daily and $\beta$ is the fraction of the spuds that live to the next day and reproduce themselves. So $\alpha = \tfrac{1}{3}$ and $\beta = \tfrac{1}{2}$. The latter equation tells you that $l_n = (1-\alpha)^n(1+\beta)^n l_0$. Here $(1 - \alpha)(1 + \beta) = 1$, so it's a steady state as far as live potatoes goes, but the cellar will accumulate.

A rural reader
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