suppose that a vaccine is 87.5% effective against a virus with a reproduction rate of pi*, and that everybody is vaccinated. if a promoter wants to put on a concert, how many people can she invite before the probability of transmission is greater than 50%?
you may further assume that the average number of people interacted with by potential concet-goers in the days before coming to the concert is 16/day**, all of whom are of course vaccinated, and that 0.025% of the population is currently infected. however, you may also use different numbers if you'd like, so long as you properly derive them.
show your work.
i added the extra information in to make it intuitive and allow for a simple response that everybody can understand. they're reasonable assumptions. i'm also making assumptions of uniform distribution across the country and whatnot because i wouldn't know where to find robust data, and i think it would just be a waste of time, regardless.
part one: computing the probability of contracting the disease before entry to the concert
so, public health claims that we can pick this up for about 14 days before we notice we have it, and i think that's half overly cautious assumption and half empirical fact, which is a sardonic way to suggest it's probably complete bullshit. nonetheless, that's the accepted "truth", based on what we know about other respiratory viruses in it's class. if we interact with 16 people per day on average then we've interacted with 16*14 = 224 people that may potentially have the virus in the relevant days leading up to the concert. you can argue that's too many in a lockdown, but we're talking about a fully vaccinated population post-lockdown, here.
i'm also assuming that masks are essentially useless and/or not being worn, which is what all of the data pre pandemic suggested. i find it curious, but not very convincing, that a bunch of studies appeared after the pandemic started that completely rewrote the science on the topic, and am at this point writing them off as biased and/or manipulated. i would expect studies redebunking the efficacy of mask wearing to show up any day now. if masks are so effective, why did it take a pandemic to generate data to demonstrate it? why don't surgeons wear little blue masks instead of n95s?
but, it's also just a question of understanding that viruses are very small and "surgical" masks (no surgeon wears those little blue things) are very porous. sometimes you need studies, and sometimes you can just work it out. if a virus can get through the holes in your mask, it's useless - and viruses definitely can get through those little blue masks. those masks are more useful to protect against larger particulate matter, like car exhaust, which is what they're worn for in east asia. it's just not the right tool for the job. i've said before that i'd have no choice but to shut the fuck up if they started handing out n95s, but the science just doesn't uphold the idea that wearing blue surgical masks is going to reduce the spread of a contagious respiratory virus in a meaningful sense. you're looking at a likely 5-10% difference. you can pencil that in, but i'm not going to*.
so, masks are not being presented as a barrier to transmission, because they aren't one, especially not when people are going back to normal activities where they aren't being kept apart by metre sticks or stickers on the ground. and, ideally, we're not wearing them because we don't want to and because we don't have to. i'm trying to model normal social interactions in a fully vaccinated population.
i assumed a 0.025% probability. that number was determined by taking the number of cases reported in ontario on a recent day (where most people are vaccinated, so don't let them tell you transmission is only occurring amongst the unvaccinated, that's nonsense), multiplying it by 10 (a frequently thrown around multiplier) and then scaling it down. if you wanted to do this the way they teach you at school, you'd need a probability distribution function, and we don't have one. if you don't like my probability, insert your own.
but, realize that there's going to be a background infection rate forever, until it gets assimilated into the collection of infections we call "the common cold", which is a bunch of coronaviruses and rhinoviruses, and more or less forgotten about - the virus isn't going to go away. that's what happens in a flu pandemic (which is the functional replacement of one dominant strain of flu with another), and it's what's going to happen here, too.
so, those are my assumptions and where i'm coming from.
so, if we're in contact with 224 people and each has a 0.025% probability of having the disease then what is the probability that we contract the disease before coming to the concert?
rather than bring in a formula that you need a stem degree to understand (with the understanding that i'm dealing with liberal arts majors that have perfectly cromulant degrees but can in fact barely count), i'm going to derive this. but, i'll admit that it's also because i'm a mathematician, i'm not an engineer - i couldn't remember a list of formulas to save my fucking life. i need to derive everything on the fly, and more so than most. i actually used to get shit for it - "just use the fucking formula! stop deriving everything!". but, i can't remember the fucking formula. sorry.
so, you prove something like this using a technique called induction, which is a logical trick based on an infinite chain of logical implications. it's not actually complicated at all, and they do in fact teach it in high school. but, it takes a minute to get your head around it. i should also point out that there are longstanding philosophical critiques of induction as a valid proof method, but i'm actually more of the opinion that your theory of sets would be pretty useless if you couldn't get induction out of it than sympathetic to the idea that induction is in some way controversial. i mean, can you give me a good reason why you can't count forever? i can't come up with one. but, you can look that up if you really want. i'm not going to entertain a critique of induction here, i'm just going to go ahead with it.
you'll probably remember this when i start doing it. i'm not going to get formal with it, either, i'm just going to scrawl it out.
but, the idea is this:
if you can show that something is true for n=k (usually k=0 or k=1) and then also show that if it's true for n=k then it follows that it's true for n=k+1 then you've shown that it's true for every m>k and less than infinity (that is for every number you can mash into a keyboard, like 65479365473021654321).
i know that the symbols scare people, but it's not that bad. when you're in high school they make you do this in a series of steps, but it obscures the logic of it in favour of rote formalism, and really isn't the kind of thing you do in real math. we can figure this out constructively fairly easily.
if you've shown something is true for n=0, and you can show that whenever it is true for n=k it is also true for n=k+1, then you get an infinite train of logic. that the statement is true for n=0 means it's true for n=1 (because 1=0+1), which means it's true for n=2 (2=1+1), which means it's true for n=3 (3=2+1), and you just keep going until you get to n=643547302564734570643756347564289320658932 and on and on and on and on and on and on. you can then use the formula to plug whatever n you want in, and it will always be true.
where the philosophical qualms come in is the question of the infinite chain, but we're not even dealing with limits, here - we have a finite n, 224. so, we could constructively demonstrate that the claim is true for all n<=224, if we really wanted to. nonetheless, let's get to it.
i can't go through the axioms of probability theory, here, but you can look it up yourself, if you'd like. i just googled this blindly, and am posting it without analyzing it:
so, suppose you have interacted with just one person in the two weeks before going to the concert. if we do away with all the questions of transmission variables, assume the disease is so contagious that you more or less catch it on contact (it is airborne) and then just focus on the issue of vaccine efficacy (87.5%) then the probability of contracting the disease is the the probability of coming into contact with the disease, which we've pegged at 0.025%, which is 0.00025 (because you divide it by 100 to get from a percentage to a probability), multiplied by the probability that the vaccine didn't work, which is 100%-87.5%, which is 0.125, when divided by 100 to get from a percentage to a probability:
.125*0.00025 = 0.00003125 = 0.003125%.
so, your vaccines are going to reduce the transmission rate significantly, so long as you stay inside by yourself with the windows closed. but the point is that we want to go out and play.
now, suppose you've interacted with two people in the two weeks leading up to the concert. then, you might have caught the disease from the first person, or caught the disease from the second person or...caught the disease from both? hrmmn. you can't really catch the disease from both, can you? and, it's certainly attractive to cut out the intersection in the sum. see, this is another reason why working things out is useful - the intersection doesn't actually make any sense in context, does it? i guess maybe you could catch two strains, but we're just concerned about the binary question of transmission, here.
so, if you've interacted with two people you could catch it from the first person (p=0.00003125) or you could catch it from the second person (p=0.00003125) and the idea that you could catch it from the first person and the second person (p=0.00003125^2) seems silly.
0.00003125 + 0.00003125 = 2*0.00003125 = 0.0000625.
now, suppose you've interacted with three people in the two weeks leading up to the event. then, you could catch it from any of the three people, and we're doing away with the intersection. it should be obvious that the answer is 3*0.00003125 = 0.00009375.
if we were doing this via induction (which is mostly useful to deal with the intersection term we've discarded), we'd want to go through a process of showing that if it's true for n=k (k*p) then it's also true for n=k+1 ((k+1)*p), but the elimination of the intersection term makes it trivial.
(edit: i'm going to provide an alternate derivation that does what i was intending to do, and it will quickly be apparent that the difference is trivial.
let's re-examine the interaction with two people question. if you've interacted with two people, you could catch it from the first person (p=0.00003125) or you could catch it from the second person, if you didn't catch it from the first person (p=(1-0.00003125)*.00003125). this is a scaling down of the second probability, by introducing the idea that you could only catch it from the second if you didn't catch it from the first. but, the fact that the probability of catching it is so low means the probability of not catching it is very high and that factor will tend to 1 very fast. the idea that you could catch it from the first person and the second person (p=0.00003125^2) is still silly. then,
0.00003125 + (1-0.00003125)*.00003125 = 0.00006249902.
now, what about n=3? you could catch it from the first person (p=0.00003125) or catch it from the second, if you didn't catch it from the first (p = (1-0.00003125)*.00003125) or catch it from the third if you didn't catch it from the first two (p = ((1-0.00003125)^2)*.00003125)). so, the sum is:
0.00003125 + (1-0.00003125)*.00003125 + ((1-0.00003125)^2)*.00003125 = 0.00009374707
this is a sum in the following form: p*(1-p)^n, which is a geometric series.
when n=3, we can note that,
1) s = p + (1-p)*p + (1-p)^2*p
2) s*(1-p) = p*(1-p) * (1-p)^2*p + (1-p)^3*p (multiplying both sides by 1-p, generally shortened as 'r')
then,
1 - 2 can be written as
s - s*(1-p) = p + (1-p)*p + (1-p)^2*p - p*(1-p) - (1-p)^2*p - (1-p)^3*p <--->
s[1 - (1-p)] = p + [(1-p)*p - (1-p)*p] + [(1-p)^2*p - (1-p)^2*p] - (1-p)^3*p
you'll notice that the middle terms cancel, leaving:
s*p = p - (1-p)^3*p
dividing both sides by p,
s = 1 - (1-p)^3
this is where the induction step i'm going to skip comes in, but it's easy enough to show that, for n=k,
s= 1 - (1-p)^k
then, for n=224,
s = 1 - (1-p)^224 = 1 - (1-0.00003125)^224 = 0.00697566568
the number i previously derived was .007.
if there was a way to make sense of removing the intersection, it would likewise impact the sum in a very minimal sense as even p^2 is negligible. but, the better way to look at it is really that the intersection is 0 because the probability of catching it from the second given that you've already caught it from the first is 0. that is, p(b|a) = 0 ----> p (a ∩ b) = 0. but, we'll skip conditional probabilities - which are still, nonetheless, high school math.)
so, i don't think constructing a summation formula is necessary here after all - if we have a probability of p of catching the virus from each of the 224 people we come into contact with, then we have a probability of 224*p of catching the virus, altogether, which follows directly from the third axiom of probability theory.
16*14*.125*0.00025 = .007 = .7%
so, there's a close to 1% chance that you're going to bring the virus to the concert, after being vaccinated, without even knowing it.
now, let's stop for a moment to consider that 87.5% efficacy is actually pretty optimistic, right now. studies are coming out suggesting that the vaccines may only be 50% effective against the new strains. one study suggested 39% efficacy.
what if we plug in 50% efficacy instead?
16*14*.5*0.00025 = 0.028 = 2.8%.
a 3% chance of unknowingly bringing the virus to the concert, after being vaccinated, is not what you're being told about the usefulness of vaccines, is it?
16*14*(1-.39)*0.00025 = 3.416%, and those numbers will increase to a maximum of:
16*14*(1-0)*0.00025 = 5.6%.
but, realize that the background rate should be presented as a function of the vaccine efficacy, and will increase (exponentially) as the vaccine efficacy decreases, so this model is a facile underestimation - when vaccine efficacy falls that far (as it probably already has), the background rate will shoot up very quickly (which is no doubt what's currently happening).
a real world number right now is probably something more like this:
16*14*.5*0.001 = 11.2%
you obviously want rapid-testing then and not vaccine passports, right?
the point is we're going in the wrong direction with this, and increasing vaccination rates will not help until we get updated vaccines that are distributed widely enough to prevent further mutation and/or enough old people die that we're not worried about it anymore (which is how viruses operate, historically).
but, i'm presenting best-case optimistic scenarios here to make a point, so let's stick with the smaller number of 0.7%. it's more than large enough to make the point.
part two: computing how many people can be allowed at the concert to reduce the probability of the virus entering the space to be less than 50%
a coin toss is of course a much higher level of danger than public health wants right now, but we'll look at that as i work it out
you could be naive about it and just say that if everybody has a .7% chance of bringing the virus then we need to add .7 to itself until we get to 50 to get the answer. then,
x*.7 = 50 <----> x = 50/.7 = 71.42857.
so, you're looking at about 70 people allowed at the concert before transmission is more likely than not.
(edit: as i did before, we could look at a geometric series.
so, the probability of the first person bringing transmission is .007, the probability of the second person bringing transmission (and the first person not bringing transmission) .007*(1-.007), the third term is .007*(.993^2) and so on. then, the sum is still a geometric series and the kth term is 1 - (1-p)^k, with p = .007. so,
1 - (.993)^k > .5 <--->
.5 > .993^k <---->
kln(.993) < ln(.5) <---> [where ln is the logarithm of the constant e, named after euler, 2.71828....)
k < ln(.5)/(ln.993) <--->
k< 98.6740464467
it's a little bigger, but the point remains the same)
and, if you want to reduce the number to 10%, which is still more than public health would want, you're looking at:
x*.7 = 10 <----> x = 10/.7 = 14.2857.
(edit: 1 - .993^k > .1 <-----> k < ln(.9)/(ln(.993)) <----> k < 14.9987603026)
now, what if we use the biggest numbers instead of the smallest ones?
x*11.2 = 50 <-----> x = 50/11.2 = 4.464 (edit: ln(.5)/ln(1-.112) = 5.8353809287)
x*11.2 = 10 <----> x = 10/11.2 = 0.89 < 1 (edit: ln(.9)/(ln(1 - .112)) = 0.88699595259)
i guess you'd have to book a three-piece?
however you want to interpret this, one thing is very clear - a large gathering, like a rock concert or a sports event, which is where these passports are being most aggressively promoted, is almost certainly going to feature substantive transmission, even in a fully vaccinated audience.
if k = 10000,
1 - (1-.007)^10000 = 1 - epsilon ~ 1.
the subtracted term is too small for a computer to calculate without floating point error - the likelihood of transmission is certain.
and, i guess that the reproduction number, in context, isn't that important. i mean, we can use it to try to calculate how much transmission might occur at an event, but it's not that relevant in calculating the likelihood of transmission occurring in the first place. so, we can skip the mini lesson on basic differential equations, then.
so, are the passports justified by science? is that a proportional, reasonable reaction? the answer is "no", quite clearly.
and, this is grade 11 - grade 12, tops - mathematics. it's not esoteric, it's not outside of the realm of what most people should be able to calculate. it should be easy for most people to understand.
if you are an adult, and you like to present yourself as educated, and you cannot follow this, i strongly advise you enroll yourself in a numerical literacy course. i'm not trying to make fun of you, but the situation demonstrates the point: we need numerical literacy for democracy to be functional.
it's the same basic problem on the fake left we've seen for decades.
you suck at math.
part three: computing the probability of an individual catching the virus at a concert with k people
i should clarify that the previous calculation calculated the likelihood that somebody would bring the virus to a concert in a fully vaccinated population, not the likelihood that transmission would occur in any specific person at a concert in which the virus is brought to. the number calculated is the number of people required to bring the virus into the concert with a specific level of certainty.
so, there is an underlying assumption that transmission will occur in a large event so long as the virus makes it's way into it. and, the numbers i'm throwing around - from 70 to 10000 - are enough that the question is pretty much a triviality. i don't think i need a mathematical argument to suggest that if you bring a virus into an event with as little as a handful of people then it's going to spread.
if you want to calculate the likelihood of transmission then occurring at the concert, you'd have to calculate the individual probabilities of catching the virus at the concert and add them up in the same way, which i've avoided doing because it largely shifts the nature of the question - i'm not trying to provide the probability that you will catch or circulate the virus, i'm trying to provide the probability that transmission will occur at the event. the latter question is of concern to public health, whereas the former mostly is not.
nonetheless, if you have a 12.5% percent chance of catching the virus when exposed to it (after vaccination) and there's a 0.7% chance you'll catch it from any specific person at the concert, what's the chance that you'll catch it from any specific person, when fully vaccinated?
the answer is .007*.125 = 0.000875.
so, that's your new p.
now, how many people are you going to be in contact with at that concert?
you're probably going to interact with roughly five people in front of you and roughly five people behind you in line to get into the concert. there's going to be 2-3 people at the door. there's 2-3 people at the bar or concession stand. if you're lined up in rows at the concert, you could have ten people in front of you, ten people behind you and five people to either side of you that are close enough to breath or expel air on you. and, let's say you walk by 100 people in total as you're moving around.
then, that's:
(5 + 5) + 2.5*2 + 10*3 + 100 = 145. about 150.
in a larger concert, you'll be around more people.
so, using the geometric sum previously derived,
1 - (1-p)^k = 1 - (1-0.000875)^150 = 0.12305188948
so, even fully vaccinated, you're looking at about a 12% chance - as an individual - of catching the virus at a concert in which you interact with at least 150 people, if you ignore the effects of transmission at the concert. a better model would increase p with k.
(the naive way would be 0.000875*150 ~13%)
i would not advise trying to calculate k based on a preferred s, because the specific risks to an individual actually wouldn't increase beyond the number of people they could feasibly interact with. how many people can somebody interact with at a concert? 200? 300? then, it doesn't matter if there are 10,000 people at the concert - what matters is how many people you share air with, which is much smaller.
if you suppose that the maximum number of people you could interact with is k=300, s ends up as:
1 - (1-0.000875)^300 = 0.23096201146.
a 25% chance isn't what you signed up for, right?
part four: computing the maximum number of people that can attend a concert such that the probability of transmission occurring is less than 50%
now, if you want to add this up, and you don't want to fix k, you'll need to take a summation of a sequence of identical functions of k. those are big words to some, but it's just taking the existing idea to the next level of abstraction - you're just taking another sum, but the components of the sum are themselves sums. that's done all the time in math, it's not crazy at all.
in fact, it's common even to take three sums. now, we're living dangerously, right? get me some more coffee. three sums? pftt. she's making this up.
in the end, you still want the formula of 1 - (1-p)^k, but you want to plug in what the probability for p is, based on k. so, you need to replace p with a different set of terms that includes k. i'm going to lose people here and that's fine - this is extra credit shit, now.
we're maybe doing second year math, here. i don't recall seeing these ideas explicitly until first year algebra, and also first year calculus. that said, the concepts are still high school level, it's just built up a little in a way that you wouldn't typically see in high school. there's maybe some advanced counting (combinatorics) ideas that i'm glossing over that could be taught in third year, but you wouldn't need 2nd year pre-reqs for them.
so, let's start this off by constructing a sequence of sis in place of the previous s:
q = si = 1 - (1-p)^k <----the probability that the ith individual will catch the virus in a crowd with k people, if each has a probability of p of spreading it
now, how do we add these sis up? they should all be the same, right? i needs to the index variable, but in the end we're adding up k individuals. so, both sums need to be over k. but, this is no longer disjoint, as the conditional probabilities are no longer zero - we have to keep track of the intersections. so, we need to build a different type of series. let's go back to the initial induction step:
if k=1, there's 2^1 possibilities - it's just one individual, so they catch it or don't and it's just q
if k=2, there's 2^2 possibilities, and
- you could have neither catch it
- you could have only the first one catch it (q) (not dependent on the other)
- you could have only the second one catch it (q) (not dependent on the other)
- you could have them both catch it (q^2) <---independence
so, p(s1 + s2) = p(s1) + p(s2) - p(s1 ∩ s2) = q + q + - q^2 = 2q - q^2 = kq - q^k
if k=3, there's 8 possibilities:
- nobody catches it (000) - 0
- 3 single catches (100, 010, 001) - 3q
- 3 iterations of 2 catches, one non-catch (110, 011, 101) 3q^2
- all three catch it: q^3
so, the sum is 3q - 3q^2 + q^3.
yeah, see - you're going to laugh at me but i totally forgot that this was the binomal theorem until i wrote a few lines down.
k=1: (-1)^(1-1)*(kc1)*q^k
k=2: (-1)^(1-1)*(kc1)*q^1 + (-1)^(k-1)*(kck)q^k
k=3: (-1)^(1-1)*(kc1)q^1 + (-1)^(2-1)*(kc2)q^2 + (-1)^(k-1)*(kck)q^k
then k=4 has sixteen possibilities,
- 0000
- 1000, 0100, 0010, 001 - 4q - (4c1)q
- 1100, 1010, 1001, 0110, 0101, 011 - 6q^2 (4c2)q^2
- 1110, 1101, 1011, 0111 - 4q^3 (4c3)q^3
- 1111 - q^4 = (4c4)q^4
so, k=4: (-1)^(1-1)*(kc1)q^1 + (-1)^(2-1)*(kc2)q^2 + (-1)^(3-1)*(kc3)q^3 + (-1)^(k-1)*(kck)q^k
so, it's clear what's going on here, and the summand is, over j=1..k,
(-1)^(j-1)*(kcj)*q^j
that -1 is annoying, and i need to get it in the form of k-j to use the binomial theorem. so, i can multiply by (-1)^2+k-k and take the 2-k out of the sum, leaving k + j -1. then, i can add j-j to that, resulting in k + j -1 + j -j. rearranging,
k + j -1 + j - j =
k- j + j + j -1 =
k -j + 2j - 1
i can then take (-1)^2j = 1 out (-1 squared is always one, no matter how many additional times you raise it), leaving
k - j - 1
then, i can take the remaining -1 out and leave it with the 2-k, leaving 1-k.
so, the following is now out of the summand: (-1)^1-k and the following is now in the summand:
(kcj)*(-1)^(k-j)*q^j
by the binomial theorem, that leaves us with:
(-1)^(1-k) * (-1 + q)^k (1 > q, as q is a probability)
let's clean that up:
(-1)^1*(-1+q/(-1))^k = -1*(q-1)^k
(those pesky -1s)
now, q-1 < 0 (because q is a probability), so all these nice absolute values get messy in the inequalities. but we want:
-1*(q-1)*k > 0.5
let's plug q in and see if it helps:
-1*([1 - (1-p)^k] -1)^k =
-1*[-(1-p)^k]^k
-1*[(-1)^k]*[(1-p)^(k^2)]
(-1)^k+1 * (1-p)^k^2
ok, so this is only going to be valid for every second k, then. we can just pull that out and set k+1 = 2w. to be clear, it will be a less precise answer, but only by a triviality. we're only interested in setting a loose bound, and we don't want to deal with the even ks. we can then check by substitution to make sure we didn't overshoot.
then, k = 2w-1 and the answer becomes
(-1)^2w * (1-p)^[(2w-1)^2]
the first part disappears, leaving
(1-p)^ (2w-1)^2 > 0.5 <----->
(2w-1)^2*ln(1-p) > ln(0.5) <--->
(2w-1)^2 > ln(0.5)/ln(1-.000875) <--->
2w-1 > abs val [ln(0.5)/ln(1-.000875)]^.5
then we can plug k back in to get
k > 28.1393244802
as k has to be odd, i need k=29.
why is that number so much smaller? because the individual probability of catching the virus given that we know it's already there is a lot higher than the probability that it might enter the area in the first place.
this is a more challenging calculation and i'm only 75% sure i did it right, myself. don't feel bad if you didn't follow it. the math isn't hard, but it's a bit conceptually abstract.
so, the conclusion is that you can invite up to 70 people before there's a 50% chance that the virus gets there, but you can't even get to 30 people before there's a 50% chance of actual transmission.
conclusions
i'm going to respond to a potential critique - why did i calculate the likelihood of catching the virus over two weeks? why didn't i just use the background rate?
because i'm trying to get the point across that the virus is circulating at a high rate.
but, if we take the formulas i derived, what happens if you just plug the estimated background rate of 0.025% in?
- ln(.1)/(ln(1-0.00025)) = 9209.18903145 [bound for 90% chance of transmission]
- ln(.5)/(ln(1-0.00025)) = 2772.24213421 [bound for 50% chance of transmission]
- ln(.9)/(ln(1-0.00025)) = 421.389380178 [bound for 10% chance of transmission]
these are higher numbers, but they don't undermine the argument - it would still suggest that any kind of substantive event is going to feature transmission, even if people are fully vaccinated.
what if the background rate ticks up to .1%, if it's not already there?
- ln(.1)/(ln(1-0.001)) = 2301.43360847 [bound for 90% chance of transmission]
- ln(.5)/(ln(1-0.001)) = 692.800549179 [bound for 50% chance of transmission]
- ln(.9)/(ln(1-0.001)) = 105.307826616 [bound for 10% chance of transmission]
remember: this is the size of a fully vaccinated crowd that will lead to the subsequent likelihood of transmission occurring within it, based on a small background rate, and with no attempt to calculate the likelihood of patrons picking it up before the show. i think it's an inferior calculation, but it gets the point across - this is much smaller than a stadium show, and more in the realm of the kind of small bar shows i'd normally go to.
the vaccine passport is pointless.
i'm not trying to argue that you shouldn't get vaccinated. i don't think it's an appropriate step for myself at my age, as i think i'm better off fighting the virus in the wild, but i'm not against other people vaccinating themselves, if they choose to - and i've never argued otherwise.
the point is that vaccine passports are stupid, and if you insist on being anal about it then you want to do rapid testing, instead.
and, please take the proper precautions, if you need to - don't put blind faith in a technology that the science underlying it doesn't justify.
so, what are the conclusions?
- at risk people need to accept that there is a new normal in place where they can't live the way they used to anymore.
- rapid testing makes more sense than vaccine passports. i mean, if you're serious about public health - if it's not just a pr gloss to increase consumer confidence.
- there is no utopian outcome. the society needs to move on.
additions:
*: on the morning of nov 7, i came across the following article that upholds my estimates on the efficacy of mask use, although it frames it in the opposite manner: https://www.nature.com/articles/d41586-021-02457-y