What is the probability that you have more than 3 deaths in an infected family of 6 if each pair of people have a different probability of death?

Question

If say 2 people individually had a .05 chance of survival, 2 individually had a .10 chance of survival, and 2 individually had a .20 chance of survival (pulled these numbers out of a hat)

What is the chance that the family has more than 3 deaths.

The method I have is quite tedious and I was wondering if there was a better way than just going case by case of the combinations and summing the results. I'm not even sure if that gives me the right answer, and I will check later using monte carlo.

Does this have a well known probability distribution? I know some small parts are binomial.

I wanted to know the probability distribution of death in our family, which has people in different age groups, if everyone in the family got covid. Other than by using simulation, I don't know a reasonable way to scale up what I did to include more age groups and more people. So I was curious to know if there is a way to do that, or at least approximate the probability.

Steve · Answer 1 · 2021-06-20T02:44:06.173

Kinda morbid mate. Nevertheless. If the chances of survival are $p_i$ for $i=1,2,\dots,6$, then the chance of

only the $i$-th person surviving is $p_i\prod_{i\neq j}(1-p_j)$
exactly the $i$-th and $j$-th person surviving is $p_ip_j\sum_{\substack{z\neq i\\ z\neq j}}(1-p_z)$
exactly the $i$-th, $j$-th, $k$-th person surviving $p_ip_jp_k\prod_{u\in{1,2,\dots,6}\setminus\left\{i,j,k\right\}}(1-p_u)$

So the probability of at most 3 surivals i.e. at least 3 deaths is the sum of all the above. You can select $i$ in $6$ ways in the first case, $i$ and $j$ in $\binom{6}{2}$ ways the second case, and $i,j,k$ in $\binom{6}{3}$ ways in the third case, then sum them all up. The cases are mutually exclusive as the surivors differ.

Edit: if you fancy Python:

from sympy import *
import itertools  
p_1, p_2, p_3, p_4, p_5, p_6 = symbols('p_1 p_2 p_3 p_4 p_5 p_6')
s = [p_1,p_2,p_3,p_4,p_5,p_6]
pairs = itertools.combinations(s,2)
triples = itertools.combinations(s,3)
prodneg = prod([1-y for y in s])  
def one(x):
    return x/(1-x)*prodneg  
def two(i,j):
    return i*j/(1-i)/(1-j)*prodneg  
def three(i,j,k):
    return i*j*k/(1-i)/(1-j)/(1-k)*prodneg  
allones = sum([one(x) for x in s])
alltwos = sum(two(i[0],i[1]) for i in pairs)
allthrees = sum(three(i[0],i[1],i[2]) for i in triples)  
atleastthreedeaths = allones+alltwos+allthrees

I'm sorry if you are a programmer and I just made your eyes bleed.

Your final formula is the atrocious $$p_{1} p_{2} p_{3} \left(1 - p_{4}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{1} p_{2} p_{4} \left(1 - p_{3}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{1} p_{2} p_{5} \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{6}\right) + p_{1} p_{2} p_{6} \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) + p_{1} p_{2} \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{1} p_{3} p_{4} \left(1 - p_{2}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{1} p_{3} p_{5} \left(1 - p_{2}\right) \left(1 - p_{4}\right) \left(1 - p_{6}\right) + p_{1} p_{3} p_{6} \left(1 - p_{2}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) + p_{1} p_{3} \left(1 - p_{2}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{1} p_{4} p_{5} \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{6}\right) + p_{1} p_{4} p_{6} \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{5}\right) + p_{1} p_{4} \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{1} p_{5} p_{6} \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) + p_{1} p_{5} \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{6}\right) + p_{1} p_{6} \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) + p_{1} \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{2} p_{3} p_{4} \left(1 - p_{1}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{2} p_{3} p_{5} \left(1 - p_{1}\right) \left(1 - p_{4}\right) \left(1 - p_{6}\right) + p_{2} p_{3} p_{6} \left(1 - p_{1}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) + p_{2} p_{3} \left(1 - p_{1}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{2} p_{4} p_{5} \left(1 - p_{1}\right) \left(1 - p_{3}\right) \left(1 - p_{6}\right) + p_{2} p_{4} p_{6} \left(1 - p_{1}\right) \left(1 - p_{3}\right) \left(1 - p_{5}\right) + p_{2} p_{4} \left(1 - p_{1}\right) \left(1 - p_{3}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{2} p_{5} p_{6} \left(1 - p_{1}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) + p_{2} p_{5} \left(1 - p_{1}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{6}\right) + p_{2} p_{6} \left(1 - p_{1}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) + p_{2} \left(1 - p_{1}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{3} p_{4} p_{5} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{6}\right) + p_{3} p_{4} p_{6} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{5}\right) + p_{3} p_{4} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{3} p_{5} p_{6} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{4}\right) + p_{3} p_{5} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{4}\right) \left(1 - p_{6}\right) + p_{3} p_{6} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) + p_{3} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{4} p_{5} p_{6} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) + p_{4} p_{5} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{6}\right) + p_{4} p_{6} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{5}\right) + p_{4} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{5}\right) \left(1 - p_{6}\right) + p_{5} p_{6} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) + p_{5} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{6}\right) + p_{6} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) \left(1 - p_{4}\right) \left(1 - p_{5}\right)$$

Oh god, I'm sorry I made you type all that out. Thank you so much. If I had the reputation right now, I would give you an upvote. — Nicholas Dow, Jun 20 '21 at 02:46
I didn't type any of that, that's what the Python code is for :P — Steve, Jun 20 '21 at 02:47

score 1 · Answer 2 · answered Jun 20 '21 at 14:43

Consider the polynomial $$f(x) = (.05 + .95 x)^2 (.10 + .90 x)^2 (.20 + .80 x)^2$$ When expanded, we have $$f(x) = 0.000001 +0.000064 x+0.00159 x^2+0.01948 x^3+0.123865 x^4+0.387144 x^5+0.467856 x^6$$ The relation of this expanded polynomial to the original problem is that the coefficient of $x^n$ is the probability that there will be exactly $n$ deaths, so the probability that there will be zero deaths is $0.000001$, the probability of exactly one death is $0.000064$, the probability of exactly two deaths is $0.00159$, etc.

$f(x)$ is an example of a generating function. If you find this approach interesting and would like to learn more about generating functions, there are many resources in the answer to this question: How can I learn about generating functions?

Oh, I have never heard of generating functions. Thank you so much. — Nicholas Dow, Jun 21 '21 at 05:19

score 0 · Answer 3 · answered Jun 20 '21 at 07:21

I slightly modified Steve's answer to specifically fit what I was asking for

by changing this line

p_1, p_2, p_3, p_4, p_5, p_6 = symbols('p_1 p_2 p_3 p_4 p_5 p_6')

to this

p_1, p_2, p_3, p_4, p_5, p_6 = symbols('p_1 p_1 p_2 p_2 p_3 p_3')

you will get:

$$2 p_{1}^{2} p_{2} \left(1 - p_{2}\right) \left(1 - p_{3}\right)^{2} + 2 p_{1}^{2} p_{3} \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right) + p_{1}^{2} \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right)^{2} + 2 p_{1} p_{2}^{2} \left(1 - p_{1}\right) \left(1 - p_{3}\right)^{2} + 8 p_{1} p_{2} p_{3} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right) + 4 p_{1} p_{2} \left(1 - p_{1}\right) \left(1 - p_{2}\right) \left(1 - p_{3}\right)^{2} + 2 p_{1} p_{3}^{2} \left(1 - p_{1}\right) \left(1 - p_{2}\right)^{2} + 4 p_{1} p_{3} \left(1 - p_{1}\right) \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right) + 2 p_{1} \left(1 - p_{1}\right) \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right)^{2} + 2 p_{2}^{2} p_{3} \left(1 - p_{1}\right)^{2} \left(1 - p_{3}\right) + p_{2}^{2} \left(1 - p_{1}\right)^{2} \left(1 - p_{3}\right)^{2} + 2 p_{2} p_{3}^{2} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right) + 4 p_{2} p_{3} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right) \left(1 - p_{3}\right) + 2 p_{2} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right) \left(1 - p_{3}\right)^{2} + p_{3}^{2} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right)^{2} + 2 p_{3} \left(1 - p_{1}\right)^{2} \left(1 - p_{2}\right)^{2} \left(1 - p_{3}\right)$$

which suits the three groups of two in the question.

What is the probability that you have more than 3 deaths in an infected family of 6 if each pair of people have a different probability of death?

3 Answers3