Cletus, here are sample

Cletus, I went back and reworked it from scratch.

The formula for probability of being immune is as follows: P(natural) + P(sick&recov) + P(vac) - P(nat + sr) - P(nat + vac) - P(sr + vac) + P(nat + sr + vac).

That formula simplifies because NO ONE can have natural immunity AND get sick and recover. Therefore P(nat + sr) = 0, and P(nat + sr + vac) =0.

So, P(immune) = P(natural) + P(sick&recov) + P(vac) - P(nat + vac) - P(sr + vac).

And since the probability of an individual getting vaccinated is not influenced by natural immunity (how would he know?), then P(nat + vac) = P(nat)*P(vac).

However, many people who got sick know they have natural immunity, and so they decline to get the shot, so P(sr + vac) is not equal to P(nat)*P(vac). Let's agree for the sake of argument that this number is 0.75*P(nat)*P(vac), since many people who we will include in the "sick and recover" category won't know they were even sick, thus.....a stronger factor.

So, with that said, and using your figures, I think:

P(natural) = 0.2, P(s&r) = 100m/320m = 0.34, P(vac) = 0.52*.85%effec = .44, we then get:

P(immune) = 0.20 + 0.34 + 0.44 - (0.20*0.44) - (0.75*0.34*0.44) = 0.78.

That's herd immunity, very little question about it. But I must say, I think two numbers are high.

I don't think 20% of the people have natural immunity (even if they have some immunity where the infection isn't so bad for them, they can still spread it, and thus for herd immunity numbers, they shouldn't count). So, maybe 5-10% have natural immunity is a better number.

Then, 52% is too high also, because even though they get no harm from it, kids still count for spreading the virus. So, I think 52% should be 41% or so.....0.85*0.41 give P(vacc) = 0.350.

If I go with those numbers, we have P(immune) = 0.075 + 0.34 + 0.35 -(0.075*0.35) - (0.75*.075*0.35) = 0.72%, which is still very close to herd immunity.

Either way, one more month to go and it's all over....