I’ll share some thoughts on the problem. I, for one, like the problem as it pertains to how framing can influence how we view presumably equivalent situations.
I am not sure I find the highly contextualized scenarios convincing. There is the blender thing to favor red but we can find one that favors blue. One goes like this:
“It’s an election between the fascist party and the B party. If the fascists win, they’ll kill the non-fascists and if the B party wins, nobody gets killed. Non-fascists are anyone who voted for the B party. Who should you vote for?”
And here, red would be voting for the fascists and “just stay out of the blender” becomes “just vote for the fascists”. I don’t really want to discuss if this particular version represents the problem “correctly”. That’s not the point; my point is that in both cases, the scenarios add non-trivial context to push you towards a certain outcome.
I’ll say that for people who boast about red being the obvious rational choice, they do very little analysis. Let’s try to do just that.
I’ll map it like a prisoner’s dilemma. You don’t know what the others will decide but you can’t influence that so you can already lay out three possibilities and the outcomes:
- Blue will lose even if you vote blue
- Blue wins iff you vote blue
- Blue wins regardless of your vote
Let’s say there are 2X+1 participants (an odd number to simplify things). We get this table:
|
Less than X voted blue |
exactly X voted blue |
More than X voted blue |
| Blue |
-1 |
+ X |
0 |
| Red |
+1 |
- X |
0 |
Explanations
One good thing to note is that your choice is causally independent from the cases outlined.
If not enough people voted blue without you, it’s obvious that if you vote blue, you are not saving anyone and in fact only dooming yourself compared to if you had voted red. If it is already the case that more people voted blue then your vote doesn’t matter.
If you are the deciding vote, then the first thing to realize is that you survive whatever you vote for. The only difference is that voting blue saves the other participants too.
It’s important to realize we are not saying anything about you causing the death of the others or being responsible for it, it’s just the difference in what happens in each scenario.
This is already taking some assumptions. For example, I am assuming that you value your life the same way you value the life of any individual. But to amend this, you would just add a coefficient c in the first column (Everyone else counts for 1 and you count for c, presumably c > 1).
The first thing to realize is that unlike some problems where it can be said that the rational solution is obvious, here no option dominates the other. This means that it is not true that whatever case you are in, one solution is always better than the other. So, what’s left to do is to assign probability to the cases and estimate the expected value of each choice.
These probabilities depend on how you believe people will behave. One simple but, in my opinion, flawed way to do things, is to take Michael’s way of doing:
I don’t think it’s true that one ought to consider it a coin toss but we can see what happens if we consider it a coin toss.
So, assuming everyone has 50% chance of voting blue, what is the probability of less than X participants voting blue? Well the simplest is to first compute the probability that we get exactly X participants voting blue, it is:
P_m = \frac{\binom{2X}{X}}{2^{2X}}
Explanations
Basically, we have 2X coin tosses. The total number of outcomes is 2^{2X}. Each coin can either give blue or red so 2 possibilities and you have 2X coins.
Now we want the outcomes with exactly X blue. This means we must choose X positions out of the 2X tosses to be heads (the remaining X will automatically be tails). The number of ways to do this is given by the binomial coefficient: \binom{2X}{X}
The interesting thing about this formula is that it decreases when X increases. So, we have that and the probability of having less than X blue votes is the same as the probability of having more than X blue votes so it’s:
P_d = \frac{1 - P_m}{2}
So, the expected values of voting blue and red (assuming others vote randomly) are:
E_b = -P_d + X P_m = \frac{P_m - 1 + 2XP_m}{2} = \frac{(2X+1)P_m - 1}{2}\\
E_r = P_d - X P_m = \frac{1 - P_m - 2XP_m}{2} = \frac{1 - (2X+1)P_m}{2}
Great, what do we do now? We try to find for which values of X is E_b greater than E_r. First thing we can see is that if (2X+1)P_m is greater than 1 then E_b will be positive and E_r negative so E_b will be greater than E_r. And I won’t go into details, but it is always true that (2X+1)P_m \geq 1.
We can check for some values.
If X = 1 (so 3 participants), P_m = 0.5 so 3 \times 0.5 = 1.5 > 1.
If X = 2 (so 5 participants), P_m = 0.375 so 5 \times 0.375 = 1.875 > 1.
So, it is always the case that the expected value of voting blue is greater than that of voting red if you assume other participants have 50% chance of picking blue. So blue is the rational choice. The reason is that there is already a non-negligible chance, your vote doesn’t matter (more blue) and when it does matter, the amount of people you could save if you are the deciding vote weights far more, despite the fact that you are probably not the deciding vote.
Now what if we don’t believe people have a 50% chance of picking blue, how does that change the analysis? The math gets a bit more complicated and I’ll perhaps touch on that in a following post but this is enough for now.