Suppose we introduce the following: a token S with an initial supply of a 1000. At $t_0$, a 1000 different parties $P_1,\dots,P_{1000}$ each buy a single token for $1$\$.

Whoever has the most tokens at time $t_N$ (suppose $N$ discrete time steps), receives $950$\$. After $t_N$ the tokens become useless and nothing happens.

If we denote by $\mathbf{P}$ the vector $(P_1,\dots,P_{1000})$, and by $n\in \{1,\dots,N\}$ the current time step, is there a way to make sense of the value $V(\mathbf{P}, n, i)$ of a token for party $i$ given $\mathbf{P}$ and $n$?

I guess that by arbitrage $V(\mathbf{P},1,i)=1$ for all $i$.

However, I imagine that as $n$ approaches $N$, the value of a token will depend on the party's current holdings compared to other parties, but I am not sure how to make precise what the exact value should be.

Edit with more details: initially, before $t_0$ the tokens are sold by some party $B$. Every $P_1,\dots,P_{1000}$ buys one and only one token.

Afterwards the tokens are sold on a "standard" exchange. Order books are public, as are past orders, and it is publicly visible who is selling and who is buying tokens.

In case of a tie at $t_N$, the $950$\$ is split evenly among all tied parties.

  • $\begingroup$ So far we only know that the surest way for a party to receive the $950\$$ is to try to get hold of 501 tokens before $t_N$. Trying to do less than that seems pointless. The guess that by arbitrage a token should always be worth $1\$$ I cannot follow. What is clear is that for the 501 tokens one should not pay more than $950\$$. $\endgroup$
    – Kurt G.
    May 19, 2022 at 20:42
  • $\begingroup$ @KurtG. Imagine a situation where at time $N-1$ there are three parties $a,b,c$ with $333$ tokens, and one party $d$ with $1$ token. Surely at that point the single token is worth more than $1$\$ to parties $a,b$ and $c$. In fact each of them would be crazy not to want to pay $954$\$ for it. $\endgroup$ May 19, 2022 at 20:46
  • $\begingroup$ I think they would be crazy unless they got their 333 tokens for free. $\endgroup$
    – Kurt G.
    May 19, 2022 at 20:48
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    $\begingroup$ @KurtG. And I disagree that it matter how much they payed up until this point. All that matters at time $N-1$ is that they are offered the choice of ending up with nothing, or ending up with $1$\$. Regardless of how much they have spend up until that point, you should take the dollar, right? $\endgroup$ May 19, 2022 at 20:59
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    $\begingroup$ Can we assume all players are smart and rational? How is the trading structured, a series of 1 on 1 transactions or are more complex constructions also allowed, i.e. I bid 1.1 for 500 coins but the trade is conditional on at least 500 being offered? $\endgroup$
    – Bob Jansen
    May 20, 2022 at 4:47

1 Answer 1


If you considered:

2 counterparties and 1 round, payout 95%

In this case, each counterparty has one token and could either buy the token from the other, or sell their token, or do nothing. All the scenarios are equivalent if the price bid/offered is 97.5 cents. There is no advantage to playing this game. The guaranteed loss is 2.5 cents every time. The rational order book at the exchange would be the minimal price differential from 97.5 cents, e.g. both counterparties bidding 97c and both offering 98c and no trading occurring.

Since there are only two players on party's entire stake, loss is not dependent upon the actions of another counterparty (since it maintains its own control of trading). There is no prisoner's dilemma here.

3 counterparties and 1 round, payout 95%

In this case one counterparty stands to lose their entire stake based on the actions of other counterparties. I.e if two counterparties trade (or collude) that third counterparty will fail to receive any payout at all, and lose the full \$1 stake.

Therefore it is in the interest of a counterparty to salvage some value. Either, they will sell their token at the lowest price increment, say 1c, to recover some value, or they will pay up to 284c for an additional token (since the payout is 285c either case is equivalent for a loss of 99c).

However, all parties are in the same situation. No-one can offer at 1c and bid at 284c, simultaneously on an exchange. The linear price asserts a mid-market price of 142.5c, where one should simultaneously bid 143c and offer at 142c, but again this is backwardation. If you can trade at 142.5cents mid price with another counterparty then the non-trading counterparty loses their 1$ stake and the remaining profit is shared between the trading counterparties, i.e the seller nets 42.5c and the buyer nets 42.5c.

Basically, there would be a mad scramble at exchange open where each counterparty enters a random buy or sell order (you are not allowed on an exchange to enter both simultaneously) at 142.5c and hopes to be filled.

Trading is not volatile or strategic in this case it is simply a random lottery.

This is a form of prisoner's dilemma, since no-one trading at all ensures no-one loses their full stake, but there is something to be gained by two counterparties making a successful trade; therefore everyone must attempt it.

4 counterparties and 1 round, payout 95% (\$3.80)

If a party sells a token for 0c they lose 100c. If the same party buys a token for 190c they will lose 100c (assuming other counterparties rationally settle so that another party also acquires 2 tokens and the pot is shared) If the same party buys 2 tokens for 380c they will lose 100c (acquiring the full pot) The mid price in this case is 95c.

But this gets more interesting once a counterparty has acquired 2 tokens and there are still two other counterparties with 1 token each.

I don't have time to keep thinking this through, but happy for someone to edit or take over this. I think it's quite an interesting game-theoretical exercise.

Collusion Risk

This game is highly susceptible to collusion. If you can arrange off exchange crosses, two counterparties can easily avoid the random lottery aspect at the expense of all other counterparties by combining their tokens to give a greater share. No rational player should ever buy one of these tokens, due to expected loss at outset, and if trading is not restricted to exchange only, then a rational player should expect the whole system is fraudulent.

  • $\begingroup$ Thanks for this. However, I don’t think N=1 is comparable/gives insight into the N>>1 case. In the N=1 case things are much more random and luck based. For large N there is time to change strategies. For example, you may start out with the intention to win, but half way through see that it’ll be impossible because someone else has large lead, and then switch to trying to offload your tokens to recover as much as you can (maybe even making a profit if there is fierce competition between a few players that want to win). $\endgroup$ May 25, 2022 at 16:19
  • $\begingroup$ Similarly, someone may swoop in at the last moment to profit of of players that played to win, but realize towards to end of the game that they’re losing. These player will all try to dump their tokens to recover at least some of their input, so someone else can come in an try to steal the win. $\endgroup$ May 25, 2022 at 16:20
  • $\begingroup$ With regards to collusion, that’s why trading is only possible on a public exchange. $\endgroup$ May 25, 2022 at 16:21
  • $\begingroup$ Ps, the fact that even for 4 players with N=1 it’s not a trivial matter to decide a strategy in my mind means that although the EV of playing is negative, it’s a very interesting and potentially fun game. And of course that’s the point. $\endgroup$ May 25, 2022 at 16:26
  • $\begingroup$ By induction it is easy to show for 2 or 3 players that for N >> 1 the same analysis as above leads to the case that everything occurs (random lottery based in the first round) and every successive round experiences no trading. This hints at similar action for larger numbers of players. That strategies are quickly determined an later rounds are entirely redundant. $\endgroup$
    – Attack68
    May 25, 2022 at 23:00

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