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If you can partition everyone into independent groups, a good first approximation (with lots of such groups) would be then to add the means and variances across independent groups and then treat the sum as normal (perhaps with continuity correction). A more accurate approach would be to simulate the process, but this is an unnecessary level of precision, since there's so many layers of approximation already - it's like being told the dimensions of a room to the nearest foot and then calculating how much paint you'll need to the nearest milliliter - the additional precision is pointless.

If you can partition everyone into independent groups, a good first approximation (with lots of such groups) would be then to add the means and variances across independent groups and then treat the sum as normal (perhaps with continuity correction). More accurate approach would be to simulate the process or to compute the distribution exactly via numerical convolution; while both approaches are straightforward, this is an unnecessary level of precision for this particular application, since there's so many layers of approximation already - it's like being told the dimensions of a room to the nearest foot and then calculating how much paint you'll need to the nearest milliliter - the additional precision is pointless.

The first is amenable to calculation, either by normal approximation or via simulation. The second might be simulated under various assumptions, either specific to the people, or by considering some distribution of dependences. (The third item is more difficult.)

If you can partition everyone into independent groups, a good first approximation (with lots of such groups) would be then to add the means and variances across independent groups and then treat the sum as normal (perhaps with continuity correction). A more accurate approach would be to simulate the process, but this is an unnecessary level of precision, since there's so many layers of approximation already - it's like being told the dimensions of a room to the nearest foot and then calculating how much paint you'll need to the nearest milliliter - the additional precision is pointless.

[Once you have your combined distribution that incorporates such group-dependencies, you might then wish to apply any sources of overall joint dependency (such as severe weather) -- or you may wish to simply insure against or even ignore such eventualities, depending on circumstances.

Edited to address followup questions in comments:

If I understand your phrasing correctly, for the family of 4, you have a 50% chance each of either 4 people or none coming. That's an expected number of 2, certainly, but you'd want to have some idea of the variability around the expectation as well, in which case you probably want to keep the actual situation of 50% of 0/50% of 4.

If you can partition everyone into independent groups, a good first approximation (with lots of such groups) would be then to add the means and variances across independent groups and then treat the sum as normal (perhaps with continuity correction).

So imagine (for simplicity) we had four groups:

[A more accurate approach would be to simulate the process, but on the full problem rather than the cut down example this is probably unnecessary since there's so many layers of approximation already.]

The first is amenable to calculation, either by normal approximation or via simulation. The second might be simulated under various assumptions, either specific to the people, or by considering some distribution of dependencies. (The third item is more difficult.)

If you can partition everyone into independent groups, a good first approximation (with lots of such groups) would be then to add the means and variances across independent groups and then treat the sum as normal (perhaps with continuity correction). A more accurate approach would be to simulate the process, but this is an unnecessary level of precision, since there's so many layers of approximation already - it's like being told the dimensions of a room to the nearest foot and then calculating how much paint you'll need to the nearest milliliter - the additional precision is pointless.

So imagine (for simplicity) we had four groups:

[A more accurate approach would be to simulate the process, but on the full problem rather than the cut down example this is probably unnecessary since there's so many layers of approximation already.]

Once you have your combined distribution that incorporates such group-dependencies, you might then wish to apply any sources of overall joint dependency (such as severe weather) -- or you may wish to simply insure against or even ignore such eventualities, depending on circumstances.

Edited to address followup questions in comments:

If I understand your phrasing correctly, for the family of 4, you have a 50% chance each of either 4 people or none coming. That's an expected number of 2, certainly, but you'd want to have some idea of the variability around the expectation as well, in which case you probably want to keep the actual situation of 50% of 0/50% of 4.

If you can partition everyone into independent groups, a good first approximation (with lots of such groups) would be then to add the means and variances across independent groups and then treat the sum as normal (perhaps with continuity correction).

So imagine (for simplicity) we had four groups:

1. group A (1 individual) - 70% chance of attendance

2. group B (1 individual) - 60% chance of attendance

3. group C (family of 4) - 0: 0.5 4: 0.5 (if anyone stays home, none will come)

4. group D (couple of 2) - 0: 0.4 1: 0.1 2: 0.5 (i.e. 50% chance of both, plus 10% chance exactly one will come, e.g. if the other has work commitments or is ill)

Then we get the following means and variances:

      mean   variance
  A    0.7     0.21
  B    0.6     0.24
  C    2.0     4.0
  D    1.1     0.89

 Tot   4.4     5.34

So a normal approximation will be pretty rough in this case, but would suggest that more than 7 people would be pretty unlikely (on the order of 5%), and 6 or less would occur roughly 75-80% of the time.

[A more accurate approach would be to simulate the process, but on the full problem rather than the cut down example this is probably unnecessary since there's so many layers of approximation already.]