The Prisoners’ Dilemma and Opportunity Cost

Throughout the discussion of the prisoners’ dilemma, I heard people state that they would defect the turn after being defected upon to “make up the points.” This struck me as questionable, as it seems to assume a baseline of five or three for the first turn (the assumption that they “lost” points implies that they started with number of points that they would have had were they not defected upon) but a baseline of zero for the second turn (i.e, that all points could go to replacing losses), which is one of the first things you learn not to do in economics. To show why this method of analysis is problematic, let’s bring the dilemma into bring the values into the real world by making the other party an employer (or series of employers) who, due to greed, malice, or forgetfulness, isn’t guaranteed to pay you every Friday. So, one week he neglects to pay you the agreed-upon market rate of three dollars (you have an inept union). Alternately, you made fifteen dollars and lost three of those dollars somehow (you probably spent it on drugs). Either way, you’re down three bucks. Under the analysis we’re looking at, you’d only need to draw your $3/$15 wage the next week to make that money back, which, as anyone will tell you, is not true, as you’d still be out three dollars. You’d have to make three dollars on top of your normal salary.

At this point, we come to two very important questions: how long would it take to make the money back and how do you measure it? First let us look at how the idea of making money is analyzed in finances and economics:

To figure out whether something is in the red or black is to take the difference between revenue and costs. Of course, there are no actual costs or revenues, just a number, so this is pretty useless even if it could be applied to the question of making money back.

To figure out whether your savings are growing of shrinking, you take the sum of how much you have added to your savings and how the value of those savings has changed over time. In the case of monetary savings, the value change is the davaluation of inflation, so that the net-change analysis is roughly (amount added to account)-(the inflation rate)(the amount it the account) or something. This is generally used in the analysis of interest rates to see if the money in a savings account is actually making you money. As there is no inflation in our game and this analysis is supposed to compare alternatives, this would not be an accurate benchmark.

When assessing stock portfolios, it is standard to compare against “the market,” meaning a market index. The indices are used as a representation of the average of every possible portfolio (read:every possible set of choices) that could have been made. In other words, the “market” is the expected rate of return for any participant. As such, we could measure how many turns are needed to catch up with the standard rate of return, a compelling method given than the game was competative, so that you are assessing your gains against those of everyone else.

Lastly, there is opportunity cost analysis, which measures gains made by a choice as the profit of that choice in excess of the next best choice. This is the standard of most economics analysis, which asks the question of how an action’s results differ from the results that would have come from any other action (including no action). In our case, the analysis would be how many turns it would take for defection to yield enough points that wouldn’t have been obtained anyway to make up for the points lost by being defected on.

For the market rate/baseline analysis, it is first necessary to set the standard earning rate. Because I didn’t write down each person’s score, it is necessary to make several conceptual estimates. Now, the baseline can be set at S=(t+1)(3(x^2)+5x(1-x)+(1-x)^2), where S=average score, x=the probability of a random person cooperating, (1-x)=the probability of a random person defecting, and t=the number of turns you defect to make up points (we’ll be ignoring that until later). This of course assumed that the two actors act independently of one another, a reasonable assumption given that nobody in the class is psychic. We can see that this works by finding the benchmark values. The lowest S, 1, is obtained when x=0, representing when both parties defect and both get 1’s. The highest S, 3, is when x=1, representing when both parties cooperate and get 3’s. When we make the chance of a party defecting (d) and cooperating(c) to be the same, cc, cd, dc, and dd combinations are all equally likely, allowing simple averaging. We already know the values for cc and dd, so let’s average them first. (1+3)/2=2. For both mixed combinations, the result is one participant with five and one with zero, averaging to 2.5. Averaging the two mixed with the two self combos gives us (2+2.5)/2=2.25, which is what the equation gives us for x=.5. First, let’s look at what’s required for you to catch up in one turn of defection. Under ideal conditions, you’ll defect on someone cooperating with you, giving you an overall score of 5 (for now, I’m going to assume you cooperated first turn), which is S(t=1) at x=.634. In other words, you’ll catch up in one turn as long as your counterpart cooperates and the general population cooperates 63% of the time or less. Of course, you and your partner is part of the general population, so let’s make your score Y=(1-x){your first/initial score}+t(5x+1(1-x)){points from defection}. For t=1, S and Y are only the same if x=o, so that you defected and received one point first turn and everyone only receives one point per turn for the whole game. If you defect for two turns, though, you’ll catch up as long as long as x is greater that or equal to 2/3. If I reduced correctly, T’, the number of turns required to catch up with the market equals ((x^2)-4x)/-(x+(x^2)). It seems to work (x=2/3 gives 2, x=0 gives no result) and it shows that, barring x=0, the fastest you can catch up is 1.5 turns and x=1. I’d post an image of the graph, but that would mean making the graph and uploading it instead of simply telling you to get the graphing calculator I know every single one of you owns. Of course, you already know what you got when you defected, so you’d probably want T’=(F-(3x-(x^2)+1))/-(x+(x^2)), where F=the points you received on the turn you’re trying to make up for.

The opportunity analysis is much simpler. How much are you down? If you were defecting when you were defected against, you’re down four (you would have made five if you hadn’t gotten a one). If you were cooperating, you’re down three (from three to zero). If you defect against someone cooperating, you’ll make five, but you would have gotten three anyway, so you’d be up two for that round. If you defect against someone defecting, you’ll make one, which is one more that the nothing you would have gotten anyway. As such, it would take you 1.5 turns of defecting against a cooperator to make up for your cooperation being rebuffed (is this correct usage?) and 2 to make up for co-defection. I’ll let you figure out all the other combinations. Now, there’s no way of knowing what you’ll get, bringing us back to 5x+1(1-x), although we’ll have to change it to 2x+1(1-x). The equation to find how many turns you’d need to make back your loss is T’=D/(1+x), where D=the deficit from the turn you’re trying to make up for (either three or four).

I really should have used this time for studying. No, I won’t show how to figure out how long it’ll take for me to make up for it.


About scalfin

Student blog. Blah, blah, blah.
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