Pricing a Winner-Take-Most Token Contest with Trading

# How would the following be priced?


# How would the following be priced?












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.

## Answer by Attack68 (score 1)

https://quant.stackexchange.com/a/71020

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.