Years ago while on a Wikipedia browsing binge, I read a maths article about how two (or more) mistrusting parties can reach an transactional equilibrium, but I've wracked my brain and I can't remember what it was!
So here's a scenario:
- Two traders exist and want to trade mutually desirable assets.
- No third party exists that can enforce honest trade between the parties, or penalize false trade.
- Each transaction occurs simultaneously and irreversibly.
- Each party does not trust that the other will honor their side of the transaction.
- Trader A is selling Apples to buy Oranges. Trader B is selling Oranges to buy Apples.
- In practice the traders could be selling/buying any fungible and liquid commodity or currency.
- For simplicity the buy/sell bid ratio of Apples to Oranges is 1:1
- Either mistrusting trader can choose not to trade with the other; but all other market opportunities for trading Apples and Oranges are half as profitable. (i.e. 2 apples for 1 orange & 2 oranges for 1 apple. Perhaps the other traders don't use refrigeration.)
- Trader A has 100 Apples available. Trader B has 100 Oranges available. Neither trader's produce spoils for the duration of this scenario.
So my question is: What class of maths solves this mistrust/opportunity balance?
Would there be a single high-risk high-volume transaction of 100 fruit? Or a gradual step-up of lower to higher volume transactions over time? If there was a fixed-fee cost of each transaction of, say 1 fruit, would that alter the strategy? Was this scenario even adequate enough narrow down the type of maths problem this is?
N.B. I feel that as the factors the traders use regarding risk/reward get more complex, the maths becomes more prominent - hence I didn't put this question in the Philosophy Stack Exchange; but I might be wrong. I also didn't put the question to the Bitcoin Stack Exchange despite being it inspired by recent thoughts on trust relationships in localbitcoins.com trading.
If the way to cheat is "failure to supply", then the exact description of the execution of the transaction is as follows:
1) The traders agree to barter
2) Then they have to "play simultaneously" - i.e. they have to decide to either send the goods or not, without knowing what the other trader will do.
If traders want to trade, it must be the case that each will be somehow better off if he acquires the good of the other while giving up his own good (because say, he prefers to eat oranges that he doesn't have to apples that he does), compared to no trade. But of course, each will be better off if he gets the goods of the other without surrendering his own. So the players must contemplate the "payoffs" (call it $U_A(), U_B()$ they will have in the four possible outcomes of the game. We assume symmetricity of players.
We can schematically represent these payoffs as
A supplies, B supplies : $U_A(SS) = 0.5 \qquad U_B(SS) = 0.5 $
A supplies, B cheats : $U_A(SC) = -1 \qquad U_B(SC) = 1$
A cheats, B supplies : $U_A(CS) = 1 \qquad A_B(CS) = -1$
A cheats, B cheats: $U_A(CC) = 0 \qquad A_B(CC) = 0$
Using the "normal" form of a game, i.e. a $2\times 2$ matrix in our case we have \begin{matrix} & B-supplies & B-cheats \\ A-supplies & 0.5,\,0.5 & -1,\,1 \\ A-cheats & 1,\,-1 & 0,\,0\\ \end{matrix}
The above matrix of payoffs is common knowledge, meaning that its player in isolation looks at this matrix and contemplates what to do.
Note that each player has a pure dominant strategy: to cheat.
If player A cheats, he gets $1$ if player B supplies, which is more than what he would get ($0.5$) if he had supplied and player B had also supplied.
Also if player A cheats and player B cheats player A gets zero, which is better than what he would get ($-1$) if he had supplied and player B had cheat.
So each player will cheat and they will get both zero out of the transaction. But both would have been better if they both had supplied, in which case each would have enjoyed a $0.5$ instead of zero.
This is a variant of the standard Prisoner's Dilemma about which you can find a myriad references easily over the web: because each player cannot trust the other, both end up worse.
Can this result be overcome, and arrive at a "supply, supply" solution? Starting from the wiki article, you will see that if the game can be played many times, then there are conditions under which the "supply-supply" solution can be sustained.