EDIT: @Suny made me elaborate a lot on the arguments here in my response, and I realized that the post here is quite incomplete. If you’re reading this whole post, I recommend checking out the other one I linked to above for the full picture.
1.0 Introduction
The logic for both choices is clear.
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Everyone knows that if you push red, you are guaranteed to survive. If everyone presses red, everyone survives. All those who press blue willingly choose to risk their own lives.
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Everyone knows that no matter what, some people will press blue. These people will die if more than 50\% press red. Their deaths can be avoided if enough people press blue.
Let’s say you find yourself in the room. No preparation. Just the dilemma, and two buttons. Applying Causal Decision Theory, ask yourself, “what can I cause?” For the first part of this post, I will be assuming causal independence between participants, because given the original scenario, it is implied you are not able to prepare, nor collaborate, with the other participants. The OP says you’re taking a “private vote”, and if cooperation was allowed, that would likely be explicitly declared.
I will call this version of the game the Insulated Game, to separate it from what I call the Preparatory Game, which I will talk about in section 3.0. I also want to mention the Cooperative Game here, where people can interact with each other inside the room with the buttons, and see each other’s decisions. I will not analyze that in this post. The Preparatory Game will be one where people can prepare, but once they enter the rooms, they cannot impact each other at all, upholding causal independence inside the rooms, but not before.
But first, the Insulated Game…
2.0 Is picking red genocide?
A lot of the blue moralists will charge you with the crime of genocide for picking red. They’ll say you’re a part of causing countless people dying. Well, are they right? We need some math to figure that out. Firstly, there’s a mathematical ambiguity in the original problem we must decide on: what happens at 50/50?
Let’s just say that then, a perfectly fair coin is flipped, and it decides whether bluers live or die.
Let’s set some constants and variables/unknowns:
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There’s m participants, which we’ll set to be 8 billion, since that’s roughly the human population of earth at the time of writing.
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Each participant i has some probability p_{r,i} that they will pick red. The average of all those probabilities is P_r.
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Therefore, the average probability of picking blue is P_b = 1 - P_r.
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The final ratio of redders to m after the experiment is R_{\text{final}}
In the next section, we begin by analyzing the central question.
2.1 How likely are the bluers to die?
With m=8 billion, the probability P_r is a really good predictor of what R_{\text{final}} will be, because the outcome variance is extremely low.
Some people may counter by saying that human nature is very unpredictable. I would agree, but this does not help your case.
Let us say that humans are very unpredictable in this respect. That across all i, the average correlation between p_{r,i} and P_r is quite low.
I believe this is true. Humans seem to differ quite a lot on this question. If the distribution of predispositions is even a binomial distribution, it’s probably a messy, multimodal one.
But greater variance in individual probabilities will in fact only reduce outcome variance (Hoeffding 1956). With m = 8 billion, the outcome variance will be extremely low in any case, but the fact that humans are very diverse only decreases the outcome variance.
What does that entail? Well, let’s say P_r = 0.51. If we pick the worst-case scenario for predictability, we make the highly unrealistic assumption that \forall i(p_{r,i} = 0.51), which maximizes the outcome variance. We can then quantify the variance as this:
\begin{align} \frac{P_r(1-P_r)}{8 \ 000 \ 000 \ 000} &= \frac{0.51\cdot 0.49}{8 \ 000 \ 000 \ 000} \\[2ex] &= \frac{0.2499}{8 \ 000 \ 000 \ 000} \\[2ex] &= 0.0000000000312375 \end{align}
In general, we have that R_{\text{final}} \simeq P_r. The probability that all bluers die as a function of P_r could be written as:
P(\text{bluers die}) \approx \Phi \!\left(\sqrt{\tfrac{m}{P_r(1-P_r)}}(P_r - 0.5)\right)
This can be approximated to a far simpler piecewise equation:
P(\text{bluers die}) = \begin{cases} \text{round}(P_r) & P_r \ne 0.5 \\[2ex] 0.5 & P_r = 0.5 \end{cases}
There’s a tiny transition window for P_r-values around 0.5 where P(\text{bluers die}) has some interesting behavior, but this window is so nanoscopically tiny we can just ignore it (unless someone proves that P_r is extremely close to 0.5).
So, in plain English:
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If P_r > 0.5, bluers will almost certainly die.
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If P_r=0.5, it’s undecided.
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If P_r < 0.5, bluers will almost certainly survive.
Given that with no preparation or cooperation, you cannot impact the dispositions of humanity, it seems you, the individual, is powerless to change the outcome…
If P_r > 0.5, it is practically a death sentence to pick blue. And if P_r < 0.5, then the bluers are almost certainly going to live, so it doesn’t matter if you pick red.
So all cases can be described by this little list:
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You save your own life by picking red.
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You do no harm by picking red.
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In an extremely unlikely scenario, you are the deciding vote, and you trigger mass death by picking red.
There’s a subtlety here to consider as well. What if the bluers are already dead by the time you enter the room? Given causal independence, whether the vote is decided before or after you make your decision is irrelevant in a certain sense. But some people care whether their choice comes before or after the fate of the bluers has been sealed.
By that, I mean that, when you enter the room, there’s already been k people who picked red, and k> m/2. If that is true, there is nothing you can do to save the bluers.
In fact, fruitlessly picking blue would only kill one more person…. If the bluers are “already dead” or doomed, when you enter the room, one could argue you actually have a moral duty to pick red.
Because your life is valuable too, and if picking blue would needlessly kill yourself, then picking blue would be bad. But you don’t know what k is… So, we must ask, what is the probability that k > m/2 when you enter the room?
2.2 Are the bluers are already doomed when you enter?
Let’s add to our list in the previous section.
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There have been n participants before you.
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There have been k participants who have already picked red before you.
If k \ge m/2 when you enter the room, then all bluers will die regardless of what you do, their fate is already sealed, and you picking red does not cause anyone to die. You picking blue, however, would cause one more person to die. So, what’s the probability that the inequality above is satisfied?
To simplify the notation, I will say P(k\ge m/2) = P(\text{doom}).
For large enough n-values, we have that k/n \simeq P_r, by the same variance argument as in section 2.1.
But section 2.1 deals with the total number of participants, which is known and is very high. We don’t know what n is.
However, since m = 8 billion, n will tend to be a very large number. There’s a 50\% chance that n \ge 4 billion. There’s a 0.000125\% chance that n \le 10\ 000 participants. If we set n = 10 \ 000, and we assume the worst case to maximize outcome variance, which would be the highly unrealistic scenario that \forall i(p_{r,i} =0.5), the relative standard deviation of the outcome would be sitting at a modest 0.5\%. That means we’d expect a deviation in k of around 50 people. For greater n-values, the relative standard deviation would only get smaller.
So, we can safely assume that n is large enough to make outcome variance negligible, allowing us to start answering the question of what P(\text{doom}) is.
Since we’ve established k/n \simeq P_r, we can apply that to our doom condition k \ge m/2 by dividing both sides with n, getting this:
\frac k n \ge \frac{m}{2n} \implies P_r \ge \frac{m}{2n}
Since n/m = \rho, that is, the ratio of participants that have already decided once you enter the game, we can reformulate the above to a somewhat more intuitive form:
\begin{align} \text{bluers are already doomed} &\iff P_r \ge \frac{1}{2\rho} \\ & \iff \rho \ge \frac{1}{2P_r} \end{align}
Now, the distribution of possible values of \rho is a simple matter given the Principle of Indifference, because that allows us to treat \rho as a random uniform variable in the interval [0,1]. That means we have that P(\rho\ge x) = 1-x. Applying that to our inequality above:
\begin{align} P(\text{doom}) & = P(\rho \ge \frac{1}{2P_r}) \\[2ex] &=\left(1 - \frac{1}{2P_r} \right)^+\end{align}
So, even if P_r = 0.8, the likelihood that bluers are already doomed is just 0.375, or 37.5\%.
Even if P_r = 1, the likelihood that the bluers are already doomed due to the choices of past participants is just 0.5. However, if P_r = 1 (a scenario where no humans have free will and everyone will pick red), then all bluers would be doomed regardless of n, since even if not enough people had already voted red to seal the deal, enough people will. Though, this would be irrelevant, since no bluers would exist, given that P_r = 1. So yeah, that edge case is irrelevant, but it cements the mathematical reality here.
Bluers are probably not already doomed when you enter the room.
If P_r=80 \%, we have that:
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There’s a 37.5\% chance that you’re saving yourself from absolutely certain and pointless death by picking red.
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There’s a 62.5\% you’re not saving yourself from absolutely certain death by picking red, but you are saving yourself from practically certain death.
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There’s a 37.5\% chance that you’re not contributing to the likelihood that bluers die by picking red (since in 37.5\% of cases, they’re already certainly doomed, and you absolutely cannot change it).
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There’s 62.5\% chance that you are, even if microscopically, contributing to the chance that bluers will die by picking red.
I’m not at all proposing that P_r is probably 80%, I’m just showing some elucidating stats here for a high P_r-value.
At least, this is interesting to those who think it matters whether your action comes before the point bluers are certainly doomed, or after.
Personally, I don’t think it matters, because if I pick red before the threshold is met, my “contribution” to the likelihood that bluers die is microscopic. I deal with this in section 2.4.
2.3 Is the game already settled?
Just like we can ask whether the bluers are already doomed, we can ask if they’re already saved. The difference between these scenarios is that in the former, picking blue kills you, but in the latter, it doesn’t matter what you pick.
But figuring out the probability of bluers already being saved is identical to section 2.2, just swapping P_r with P_b. We have that P(\text{salvation}) = \left(1 - \frac{1}{2P_b}\right)^+, and the probability that the game has already been determined once you enter is thus just the probability that either doom or salvation has already happened. That is:
\begin{align} P(\text{determined}) &= \left(1 - \frac{1}{2P_r}\right)^+ + \left(1 - \frac{1}{2P_b}\right)^+ \\[2ex] &= \left(1 - \frac{1}{2P_r}\right)^+ + \left(1 - \frac{1}{2(1-P_r)}\right)^+ \end{align}
Notice that at most one of P_r and 1-P_r can exceed 0.5. So at most one of the two brackets is positive, whereas the other gets clipped to zero by the (\cdot)^+. The formula collapses to:
\begin{align} P(\text{determined}) &= 1 - \frac{1}{2 \cdot \max(P_r,\, 1-P_r)} \end{align}
And \max(P_r, \, 1 - P_r) is really just a measure of how uneven the distribution of redders and non-redders is. The fact that the probability that the game is already “over” when you enter is a function of this uneveness makes sense. But still, taking P_r = 0.8 as an example again, all the same observations from before still hold.
The game is most likely not over once you enter. And yet, we must ask:
How much can you affect it?
2.4 How much do you increase the probability that bluers die?
With the above sections out of the way, we can actually look at how much an individual redder contributes to “the genocide”. If you are standing in that room, there’s a probability P(\text{bluers die}) that all the bluers will die.
By how much does that probability change if you choose to press the red button? That’s a somewhat difficult question, because it depends on whether we’re talking about the epistemic or ontic probability.
The epistemic probability might actually change a lot, because when you first grasp the dilemma, you construct a model of the world in which you might simulate a few people you know, and what their choices are. Sampling across all those simulated people, you see there’s a ratio of redders to the total, R_{\text{simulated}}, and there’s a degree of confidence you have in it. If you pick red, you will change your world model, and thus change the epistemic P(\text{bluers die}), whether that change is more substantive through surprising yourself, or merely an increase in confidence by not surprising yourself.
You might feel tempted to optimize for the lowest epistemic P(\text{bluers die}), but that would be optimizing for good news, which I’ve argued is the same fallacy OBers commit in Newcomb’s Paradox.
Instead, we must look at how much your choice impacts the actual, ontic probability P(\text{bluers die}). It should be intuitive to anyone who read section 2.1 that the amount you can change the probability is microscopic. My post is already becoming long and all the \LaTeX is lagging the editor, so I will keep it brief. The scenario in which you have the greatest impact on the probability of whether bluers die is exactly the scenario when P_r = 0.5.
Lets say K is the total number of redders, not counting you. If P_r = 0.5, the probability that you impact the outcome is the probability that K = m/2-1 or that K = m/2. In the first case, you turn P(\text{bluers die}) from 0 to 0.5. In the second case, you turn P(\text{bluers die}) from 0.5 to 1. In both cases, your choice changes the probability by a half. In all other possible cases, your choices do not change the probability at all.
So, the probability that your choice affects the outcome is the probability that K is one of those two values, multiplied by a half.
For m = 8 billion and P_r = 0.5, K is approximately normally distributed with mean m/2 and variance m/4. The probability mass at any single value near the mean is approximately:
P(K = m/2) \approx \frac{1}{\sqrt{2\pi \cdot m/4}} = \frac{1}{\sqrt{\pi m / 2}}
Plugging in m = 8 \times 10^9:
\frac{1}{\sqrt{\pi \cdot 4 \times 10^9}} \approx \frac{1}{112{,}000} \approx 9 \times 10^{-6}
The same holds for P(K = m/2 - 1). Putting it together:
\begin{align} \Delta P(\text{bluers die}) &\approx \tfrac{1}{2}\left[P(K = m/2 - 1) + P(K = m/2)\right] \\[2ex] &\approx 9 \times 10^{-6} \\[2ex] &\approx 10^{-5} \end{align}
And remember, this is the absolute best case. It assumes P_r sits exactly on 0.5. If P_r shifts even just to 0.501, then the threshold sits roughly 179 standard deviations away from K’s mean, and the probability of landing on K = m/2 becomes vanishingly small; far, far below 10^{-5}. So in any realistic scenario where P_r isn’t precisely balanced, your influence is even more negligible.
TLDR; your choice does not matter. This is a choice humanity makes, not you.
If we were to look at an expected value here, you could look across all possible values of P_r, see that your decision to blue has a very low chance of saving anyone in all cases, but has a considerable chance of killing yourself in some cases, and you’d conclude pressing red is the highest utility choice… unless you rate making a microscopic contribution to probability that all bluers die a decision of extremely negative expected value, thus tipping the scales.
I want you to consider how many times you contribute microscopically to the countless crises and catastrophes happening across the world, and how you go about your day nonetheless, routines unchanged.
If we build our normative ethics around the microscopic influences of our actions, we’ll have no resources left to handle the macroscopic influences of our actions.
You must choose which hills to die on. You can die on Mount Everest, or you can die on an anthill. Up to you.
Here’s another way to think about it:
If you die because you picked blue, you contributed to the genocide of bluers by having chosen to risk the life of one more person. If you chose red, you probably didn’t contribute at all, in a sense.
3.0 What about the Preparatory Game?
In the sections above, I place you in the room, no preparation or cooperation allowed. But you’re not in that room. You’re here, in real life. If you’ll ever play this red/blue game, and they don’t wipe your memory, you’ll be playing the version where you got to prepare and cooperate with the other participants before the game.
Because, although I can say that you should pick red in the Insulated Game, I might not say that you should promote picking red, because right now, we might be playing the Preparatory Game. Well, of course, it’s highly unlikely that we are, but the thought experiment is valuable nonetheless, due to the translatability of the principles at play here to other, real-life structures.
Firstly, let’s ask Kant. What would his Principle of Universality say? Funnily enough, it doesn’t help. If everyone picks red, that has the same outcome as everyone picking blue, since everyone survives either way. So, given just that, we might as well promote either one!
But, we know that we’ll never be able to promote either one well enough such that everyone picks it. Really, we can take it is a common-sense axiom that no matter what, some people will pick blue.
Since we’re preparing right now, and basically cooperating, we can ask the question: how can we, humanity, minimize the amount of death in the Preparatory Game? Basically, what should humanity promote right now, as preparation for the event that this happens? However silly it may sound, this is a real ethical question. I see two arguments here:
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We should promote blueing, so that we maximize our chance of saving those who will inevitably pick blue; that is, we maximize our chance of saving everyone!
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We should promote redding, so that we minimize the number of people who die from blueing once humanity probably picks red.
It may seem like the choice between optimistic idealism and pessimistic cynicism / pragmatism. But really, we shouldn’t throw around terms like optimistic and pessimistic before we have some notion of what P_r likely is, and what our ability to change it is.
You see, if P_r = 0.51, it may seem like Option 1 above is viable. But, it is only viable if we actually stand a chance at sufficiently changing P_r. If we’re not able to change the minds of more than 1% of people, then even in the case of P_r = 0.51, we should aggressively promote redding, in order to save as many convertable bluers as possible from needless death.
So, whether we should promote redding or blueing comes down to whether we have the power to move P_r below 0.5. If we don’t, then promoting blueing will likely kill more people, not less.
Let’s denote the probability of our efforts succeeding in causing P_r to go from being greater than 0.5 to being lesser than 0.5, as P(\text{success}). Let’s denote the probability that P_r starts as greater than 0.5 and stays greater than 0.5 as P(\text{failure}).
Let’s say also that the net increase of bluers through those same efforts is B_{net}>0, and let’s denote the number of bluers before any promotion happens as B_{\text{old}}. Let’s also assume that all lives are worth equally much, and we set the value of a life to 1 for simplicity. Then, we have this:
\text{EV}[\text{promote blue}] = P(\text{success})B_{\text{old}} - P(\text{failure})B_{\text{net}}
The expected value is increased by the product of how many bluers existed originally, and how great our chance of success is. But the expected value is decreased by the product of how great the chance of failure is, and how many extra people we thus led into death.
And even if this expected value is positive according to what we believe those numbers are/will be, someone’s expected utility may be negative if their risk aversion is high. In their head, the risk they subject others to by redding (the microscopic probability increase explained in section 2.4) is smaller than the risk they subject B_{\text{net}} people to by promoting blueing.
But, now let’s now look at the expected value of promoting redding. Here, we say that P(\text{oops}) is the probability that the promotion of redding brings P_r from below 0.5 to above 0.5... Yeah, that would be terrible. Also, B_{\text{old}} is still the number of bluers before promotion, and R_{\text{net}}>0 is the net number of new redders.
\text{EV}[\text{promote red}] =P(P_r > 0.5)R_{\text{net}}- P(\text{oops})(B_{\text{old}} - R_{\text{net}})
Yeah, I’d stay away from promoting redding if there’s even the slightest chance that P_r < 0.5, because that could bring the death toll from 0 to not, and that’d be a terrible thing. And yet, in this post, I have promoted redding, by advocating that redding is the right choice in the Insulated Game.
This situation mirrors the Newcomb’s Paradox in many ways (see the link earlier in the post). You see, if I think redding is best in the Insulated Game, it is hard for me to not promote redding in the Preparatory Game.
The reason is, once you entered the room in the Preparatory Game, you can no longer influence other people’s decisions. So, the same probability conclusions from the Insulated Game still apply.
If your promotional efforts during the preparatory stage saved lives, great! But once you’re in the room… there is nothing more that you can do. You cannot save the bluers, but you can guarantee that you keep your own life…
Even in the Preparatory Game, if you agree with redding in the Insulated Game, and you see far enough ahead, then you see there’s no reason to pick blue once you actually reach the room…
Like in the Newcomb Game, you could during preparation sign a contract that imposes a penalty on yourself for redding; a penalty that’s great enough to make you pick blue (and it would need to be very great to overcome your survival instincts). But even so, why would you? In the Newcomb Game, doing so gains you $1M dollars. In this game, doing so gains you only mortal risk.
And yet, if P_r is less than 0.5, I would never promote redding. At least, not beyond something like what I’ve done here, where I’m not really trying to promote redding, but rather to illuminate the ethical and probabilistic structure of the scenario to the tiny percentage of the human population that TPF constitutes, because I think the exercise itself brings a net value.
But let me be honest. Even if we showed that P_r is less than 0.5, and I partook in promoting blueing, I would still pick red if I ever found myself in that room. You can say that’s hypocritical or selfish, but I’d say that I did all I could by promoting blueing given the P_r < 0.5 reality, and then once I entered the room, I maximized my expected value by eliminating the risk to myself as opposed to eliminating some microscopic increase in the risk to bluers my choice might cause.
This seemingly paradoxical stance of mine reminds me of a quote:
“There is no God, but don’t tell that to my servant, lest he murder me at night.”
It’s misattributed to Voltaire, but that’s irrelevant to my point. The idea is that some falsities are helpful when believed by the masses. I personally believe that it is a falsity that one should blue inside that room, and yet if P_r < 0.5, it is a falsity I hope most people believe.
However, if P_r > 0.5, then whether I’d like people to believe in this falsity depends on whether I believe the promoters of blueing stand a chance at converting enough redders to blue.
But, I am skeptical of that. We redders are some cold bastards. So, if P_r > 0.5 and we don’t stand a chance at changing that, then the falsity that you should pick blue inside that room is a highly dangerous one.
What people ought to believe here depends on circumstances. What is actually true… does not.
4.0 Conclusion
The above is far from comprehensive. My post lagged so much I had to move it to a separate document in order to continue writing. That was my cue to leave it at its current length. If I had more time and energy, I’d go a lot deeper into the mathematics and ethics of this whole thing. I actually think this red-vs-blue debate is trickier than the Newcomb Game, because there’s an ethical weight to this dilemma.
I think Jamal raised a very important point near the start of the discussion. When T_Clark asked what the benefit of picking blue was, Jamal replied with this:
I disagree with the framing that you “save” anyone, given my probabilistic CDT understanding of the situation. But far more important than the question of “saving”, is the question of empathy. Jamal mentions children and the mentally impaired, which is a vital reminder. Some redders make an inhuman allusion to some idea of bluers almost bringing it on to themselves, almost “deserving” it. Now, granted, there’s probably some pseudo-intellectual, hyper-judgemental, asshole bluers out there that are, to some degree, bringing it on to themselves…
But the vast majority of bluers, I believe, are a mix of the most selfless among us, and the most vulnerable among us (the children and mentally impaired people). No respectable redding position would ever imply that the we shouldn’t even think about preventing bluers from dying, because “they’re bringing it on themselves”. That would be comparable to saying that you shouldn’t save a child from being run over because they’re playing football close to a dangerous street.
As such, I don’t think it affects the calculus in the slightest the fact all risk to bluer life is self-imposed by the bluers themselves. A lot of the risks we face in life are self-imposed, and in this scenario, it’s not like the bluers chose to set up the entire event either. They just chose to do what they thought best in a very difficult situation. So, I don’t think there’s any ethical grounds to blame bluers for the deaths of bluers. Nor do I think there’s any ethical grounds to blame the redders for the deaths of bluers either. The blame lies on the hypothetical event organizers, and on them alone.
Now, at no point during this post did I offer an estimate on what I think P_r is. The primary reason why I didn’t is because I think getting any kind of good estimate on that would take a lot of research, and presenting that research here would be out of scope for my post (though maybe I can make a different post).
The secondary reason is because I doubt whether there’s any data out there that could give us any good estimate at all.
It should go without saying that an experiment where you pose this dilemma as a hypothetical to a bunch of people is lacking the most important aspect of the real deal: the fear for one’s own life, and the fear for other people’s lives. You cannot accurately simulate that. A lot of self-professed bluers would be redders in the real situation, and some self-professed redders might be bluers in the real situation, perhaps finding themselves overcome with the fear for the lives of their loved ones that they feel they might be endangering by redding.
So yeah, I highly doubt there’s any good data on this, nor do I think we could ever ethically gather good data on it.
Anyways, that’s just my two cents.