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
 A: 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.
A: 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
A: The answer by Aksakal already explains it enough why the wealth is ending up into one single pocket. In this answer we elaborate a bit further on the quantitative approach by Matt F.
One could compute the probability in terms of the probability of a specific player remaining in the game and consider the position of the players independent.
Then there are two ways

*

*The probability that only one player is in the game is equal to the probability that a player has reached the maximum amount of money. Let's call the probability for an independent player to win $p_{win}$.
The probability that none of the players have won is $$(1-p_{win})^n$$


*The probability that only one player is in the game is equal to the probability that no more than $n-2$ players have lost their money yet. Let's call this probability for an independent player to loose $p_{loose}$.
The probability that none of the players have won, is the probability that less than 9 have lost and is $$1-(p_{loose})^{n}-n(1-p_{loose})(p_{loose})^{n-1}$$
However, when I apply these formulas then I get small discrepancies. One way overestimates the number of trials before the game ends, and the other underestimates it.
The reason for the discrepancy is that the scores of the individual players are correlated. If one player has lost after $n$ games, then this reduces the probability for the other players to have lost after $n$ games.

In the image, we also plotted an inverse Gaussian distribution fitted with the method of moments, and it seems to agree reasonably well with the shape of the distribution determined from the simulations. So possibly an inverse gaussian could be used as a model for the distribution (but one would still need to figure out the mean and variance).
### function to simulate game
sim = function(k=100,n=5) {
  # x: is a variable that keeps track of values, 
  # the size of this variable will reduce 
  # once participants hit zero
  x = rep(k,n)    
  ### n_active: number of active participants
  ### these equals the lenght of 'x' but we do not 
  ### want to compute that length everytime
  n_active = n
  ### counts: keeping track of the number of games
  counts = 0
  while (n_active>1) {
    counts = counts + 1
    
    ### sample two players
    s = sample(1:n_active,2)
    ### add +1 and -1 to the wealth of the players
    x[s] = x[s] + c(1,-1)
    
    ### check for zero wealth and remove the participant from 'x' if neccesary
    if (x[s[1]]==0) {
      x = x[-s[1]]
      n_active = n_active -1 
    }
    if (x[s[2]]==0) {
      x = x[-s[2]]
      n_active = n_active -1 
    }
  }
  return(counts)
}

### simulations
set.seed(1)
y = replicate(10^3,sim())

### plot histogram
hist(y, breaks = seq(0,10^6,10^4), main = "histogram of simulations \n compared with approximations")


### add estimates based on an individual's probability to loose or win after n turns
n = seq(0,10^6,10^4)

p1 = pnorm(500,mean = 100,sd = sqrt(0.4*n))-pnorm(-500,mean = 100,sd = sqrt(0.4*n))
lines(n[-1],-diff(p1^5)*10^3, col = 2)

p2 = pnorm(0,mean = 100,sd = sqrt(0.4*n))*2
lines(n[-1],-diff(pbinom(1,5,p2))*10^3, col = 3)

### add inverse gaussian curve fitted with method of moments
lines(n,statmod::dinvgauss(n, mean(y), dispersion = var(y)/mean(y)^3)*10^7, col = 4)

legend(4*10^5, 80,
       c("estimate based on probability of winning", "estimate based on probability of loosing",
         "inverse Gauss distribution"),
       cex = 0.7, col = c(2,3,4), lty = 1)

