it's not a question of "staring down death", it's a question of understanding the probability.

what is the probability of dying of coronavirus if you're under 50?

well, first you'd have to get it. what is the probability of that? the city that i live in has zero cases. there's some cases in michigan. if we assume there's even 1000 cases in detroit itself, and that it's evenly distributed amongst the population (a generous assumption), then there would be a 1/4200 (0.02%) chance of catching it, for every person you interact with. 

now, let's say you interact closely enough that transmission is possible with 5 people when you go out. then the probably of getting sick is equal to:

P(x1 or x2 or x3 or x4 or x5) = 

P(x1) + p(x2 or x3 or x4 or x5) - p(x1)*p(x2 or x3 or x4 or x5) =
.0002 + 0.9998[p(x2 or x3 or x4 or x5)] =
.0002 + 0.9998[0.0002 + 0.9998(p(x3 + x4 + x5)] =
.0002 + .000199996 + .9998^2[0.0002 + .9998[p(x4 + x5)]] =
.0002 + .000199996 + 0.00019992 + .998^3[0.0002 + .0002 - .0002^2] =
.0002 + .000199996 + 0.00019992 + 0.00039972007 =
0.00099960007 =
0.09996007%

that's not 1%. that's 0.01%. so, that's your chance of getting sick right now, in michigan, in the worst case.

now, if you're under 50, you have about a 0.1% chance of dying given that you have the disease, and you pretty much must already be sick.

so, if A is the probability of getting the disease,
and B is the probability of dying from the disease,
we have

P(B|A) = 0.001, P(A) = 0.00099960007.

as P(B|A) = P(B & A)/P(A),
P(B & A) = P(B|A)P(A) = P(A & B) = P(A|B)*P(B)

so, P(B) = P(B|A)P(A)/P(A|B) 

but, (A|B) = 1.

so, P(B) = P(B|A)*P(A)  = 0.00000099960007 =  0.000099960007%.

those are pretty good odds - if you're young, and the virus is not running rampant. 

what if you were over 70 today, in michigan?

then you're looking at adjusting P(B|A) to around 10%. and, then P(B) = 0.0099960007%. those are still pretty good odds, then, even if you're older.

now, what if 2% of the population were infected, instead of 0.02%?

P(x1 or x2 or x3 or x4 or x5) =
P(x1) + p(x2 or x3 or x4 or x5) - p(x1)*p(x2 or x3 or x4 or x5) =
.02 + 0.98[p(x2 or x3 or x4 or x5)] =
.02 + 0.98[0.02 + 0.98(p(x3 + x4 + x5)] =
.02 + .0196 + .98^2[0.02 + .98[p(x4 + x5)]] =
.02 + .0196 + 0.019208 + .98^3[0.02 + .02 - .02^2] =
.02 + .0196 + 0.019208 + 0.0372712032 =
0.0960792032 =
9.60792032%

now, that's actually 10%. or 1000x the previous amount, roughly.

now, if you're under 50,

P(B|A) = 0.001, P(A) = 0.0960792032
so, P(B) = P(B|A)*P(A) = 0.0000960792 = 0.00960792%

those are still pretty good odds - if you're young.

but, what if you were over 70 today, in michigan?

then you're looking at adjusting P(B|A) to around 10%. and, then P(B) = 0.00960792032 = 0.9%. and, we have our scary stat - that's almost 1%.

now, what about 20% of the people being actively infected, something that's only ever going to happen just right now?

P(x1 or x2 or x3 or x4 or x5) =
P(x1) + p(x2 or x3 or x4 or x5) - p(x1)*p(x2 or x3 or x4 or x5) =
.2 + 0.8[p(x2 or x3 or x4 or x5)] =
.2 + 0.8[0.2 + 0.8(p(x3 + x4 + x5)] =
.2 + .16 + .8^2[0.2 + .8[p(x4 + x5)]] =
.2 + .16 + 0.128 + .8^3[0.2 + .2 - .2^2] =
.2 + .16 + 0.128 +  0.18432=
0.67232

for healthy people, .001*.67232 = .00067232 = .067232%. but, this is an overestimate, because you probably don't have a preexisting condition. that is, you don't really have a .1% chance of dying if you're healthy...

but, the numbers for the elderly are actually pretty realistic.

.67232*.1 = .067232 = 6.7%. and, that's a very scary number.

so, let me tell you - you want the elderly to stay inside, pretty soon. as soon as immunity starts building, their chances of getting sick starts to skyrocket.

so, are you going to tell me i'm a daredevil for taking a chance that low? ok.

but, note that those are better odds than you get from most actual daredevil acts.

so, i don't really think i'm invincible. really. i'm just not as bad at math as the media is

https://www.cnbc.com/2020/03/18/psychological-reason-why-gen-x-is-taking-covid-19-pandemic-seriously.html