What is the probability of having all wealth concentrated into one single pocket after n trades? On Joe Rogan's podcast #1769, Dr. Jordan Peterson said something like this:

*

*Take a population of 10 people, each starting with 100 dollars.

*They will "trade" based on a coin flip. Whoever "loses" has to give the other person one dollar.

*In the long run, all wealth will be concentrated into one pocket.

At first, I couldn't believe this was true because of the equilibrium randomness would bring to this system (compared to an "informed trader" scenario). Each player would win as much as they would lose, resulting in a zero-sum endless game.
A player's expected wealth after $ n $ trades:
$$ E[w] = w_0 + \sum_{i=0}^n {1 \over 2} \times 1 + \sum_{i=0}^n {1 \over 2} \times -1$$
$$ E[w] = w_0 + {n \over 2} - {n \over 2} $$
$$ E[w] = w_0 $$
I ran a Python simulation, and in the very first run I got this:
It took 70431 individual trades for all wealth to be concentrated in one pocket.
Am I missing something here?

This is the code for the simulation:
https://gist.github.com/victorvalentee/1c04b2d16005f7372273b153fe9ece23

Here is an order-of-magnitude calculation of how long it takes to have only one player remaining.
For any player, let $p$ be the probability that they have exited the game after $n$ turns. So the probability that $8$ players have exited and $2$ remain is $45p^8(1-p)^2$. The probability that $9$ players have exited and $1$ remains is $10p^9(1-p)$. The most likely scenario is that there is only one player remaining when $10p>45(1-p)$, i.e. when $p>9/11$.
An exit occurs when a player goes to negative wealth. By the reflection principle, a player has a $9/11$ probability of negative wealth by the $n^{th}$ turn iff they have a $9/22$ probability of negative wealth at the $n^{th}$ turn. Since the standard deviation of their wealth is $\sqrt{n}$ after $n$ turns, this happens if $$\frac{9}{22}=.409=\Phi\left(\frac{-100}{\sqrt{n}}\right)$$
$$-0.23=\frac{-100}{\sqrt{n}}$$
$$n\sim 189000$$
where $\Phi$ is the cumulative normal probability.
So, by this estimate, the modal scenario is that all but one player has dropped out after 189000 turns, which is the same order of magnitude as the 70000 that you found in your experiment.
