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Source: http://www.decisionsciencenews.com/2017/06/19/counterintuitive-problem-everyone-room-keeps-giving-dollars-random-others-youll-never-guess-happens-next/

"Imagine a room full of 100 people with 100 dollars each. With every tick of the clock, every person with money gives a dollar to one randomly chosen other person. After some time progresses, how will the money be distributed?"

I've seen various blog posts on this that use code to simulate the distribution. My question is: what is the reason that the distribution becomes skewed rather than stay uniform, what math do I need to understand this problem?

I initially tried reasoning with expected value, that each person has at each moment an expected value of $1. So I thought it would be uniform. Clearly this can change as soon as one person goes broke, then the expected value drops.

Thanks

Yashmnash
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  • the links referenced by the article you link to are quite helpful...did you look at them? – lulu Jul 19 '17 at 17:43
  • I read the article referenced in it that talks about random walks and lattice points, and admittedly I didn't really understand it. – Yashmnash Jul 19 '17 at 17:52
  • @Yashmnash I looked at them as well, and didn't find a good explanation either. Admittedly I didn't study it in detail, and even if I would, I still would not necessarily understand with my limited math skills .... I am wondering something: does it have something to do that everytime you randomly pick just 2 people where 1 person gives a dollar to the other? That is, I am wondering if, when every 'turn' we have all 100 people give away 1 dollar to a random person, do we still get the skewed distribution? Or what if, every turn, every person gives a dollar to some random other person? – Bram28 Jul 19 '17 at 17:58
  • @Yashmnash if you don't mind, you can look at my answer. Thanks! – Fei Cao May 18 '21 at 01:26

4 Answers4

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A perfectly even distribution is unlikely because it is too regular. There are just so many irregular states that their probability is much higher. On the first move, the chance that the distribution stays exactly $100$ per person is the chance you select a derangement, of which there are about $\frac {100!}e,$ out of the $100^{100}$ choices for paying dollars. This is tiny, about $3 \cdot 10^{-43}$.

What you will really have is a(fast approach to a) stationary distribution. Each assignment of dollars to people, including one person having all $10,000,$ will have a probability to be what you see at any point in time. The nice regular distributions will have low probability, as will each specific irregular distribution. The irregular distributions as a whole will have very high probability. It is true that each person's expected value at the end of each payment is $100$, but that doesn't mean it is very likely. It is far more unlikely that they all have exactly $100$.

Added: there is only one arrangement with each person having exactly $100$. There are almost $10,000$ with one person having $101$, one person having $99$, and the rest having $100$. If you lump all the ones in the second group together, that factor $10,000$ will swamp the small amount that the specific distribution is less frequent. If everybody has different amounts of money, there are $100!\approx 10^{157}$ ways to assign the numbers to people.

Ross Millikan
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'Expected value' is not the same as what actually happens. A die thrown one time has an expected value of 3.5, but that is obviously not going to be an actual outcome.

And even in case where the 'expected value' is an actual outcome, it may in fact be the very least likely actual outcome (see, for example Polya's Urn)

So, this doesn't tell you why you get the distribution you do, but it does tell you why your reasoning using the 'expected value' doesn't work.

p.s. When someone goes broke they are not out of the game; they can still receive dollars.

Bram28
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Your question is equivalent to asking why in our world the Second Law of Thermodynamics holds.

Consider a box $W=1 \times L=100 \times H=10,000$ and fill it evenly with 10,000 unit cubes, shake it somewhat, and the result is the answer to your question: the even distribution is very unlike to be attained out of "chaos" (it has the lowest possible enthropy), and that is the "thermodynamic" sense of Ross answer.

G Cab
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I am not sure whether the OP is still interested in this question. This phenomenon is usually referred to as the "propagation of chaos": In the large population/limit, any finitely many players are unlikely to change money with each other, and each agent is feeling the influence of a so-called "mean-field". The 2019 paper "Cutoff for the mean-field zero-range process" (link: https://arxiv.org/pdf/1804.04608.pdf) published at the Annals of Probability contains the best result (in some sense) about this model: The authors established a profound result called "cutoff phenomenon" (this phenomenon is somehow ubiquitous in the study of various Markov chains). Me and my Ph.D advisor will soon submit another paper regarding this model (and including closely related models), I will post the arXiv link once it's available online.

Edit: The arXiv link of our paper is https://arxiv.org/pdf/2105.07341.pdf, you can found closely related models as well. There are also other models in econophysics, for instance the uniform reshuffling model investigated in https://arxiv.org/pdf/2104.01302.pdf.

Fei Cao
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