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

it's a rigged game :) you have an absorbing state: one guy got all money. if somehow at any given moment one guy got all the money, then nobody has anything left to trade, and the game stops. it's pretty much certainty that if you play long enough this is bound to happen eventually, i.e. at some point you'll observe the sequence of trades that will lead to all money in one pocket. this is not about the equilibrium, this is related to a problem known as first-hitting-time in a random walk.
there's actually another absorbing state if we disallow short selling and margin trading: a player hits zero wealth and stops trading, while others continue trading. that's why estimating probabilities in the game with more than two traders is not straightforward, you must account for players dropping off over time
