Well, This Happened…

Kazzy

One man. Two boys. Twelve kids.

32 Responses

1. Chris says:

Congratulations to you and Zazzy and especially to our future overlord!Report

2. greginak says:

CongratsReport

3. Mike Schilling says:

• Kazzy in reply to Mike Schilling says:

The crazy thing is we planned it this way. Or — perhaps better put — we tried for this one. We just didn’t expect it to actually work as quickly as it did. And, yes, I’ll remind you that I taught 8th grade health this past year.Report

• Chris in reply to Kazzy says:

With all these kids under 5, OT could have a pretty profitable daycare for contributors only.Report

4. James Hanley says:

Stop. Now. Beyond two kids you are outnumbered, and the number of inter-child conflicts increases exponentially.

On the other hand, #3 daughter is delightful and I can’t imagine life without her. So maybe you should keep on doin’ as you’ve been doin’.Report

• Kazzy in reply to James Hanley says:

I’m not-so-secretly REALLY hoping this one is a girl (we plan to find out this time) since I really do want a daughter but think that two might be our limit. If this one is a boy, there will be a strong temptation to try again.Report

• Chris in reply to Kazzy says:

I’m rooting for twins. 😉Report

Congrats Kazzy!

We have some good friends who decided to try for child #2 & got #3 coming as well.

To say that they are a little freaked out is an understatement.Report

• Saul Degraw in reply to James Hanley says:

This feels like a mixed message…Report

• James K in reply to James Hanley says:

@james-hanley

Actually I’m pretty sure the number of sibling conflicts should increase factorially.Report

• Mike Schilling in reply to James K says:

If the conflicts are between two individual siblings, it’s quadratic: n*(n-1)/2.

If your kids form alliances, there are (2**(n-1))-1 possible conflicts that involve two sides with all kids on either one or the other. More if some can abstain, but it’s still exponential.Report

• Alan Scott in reply to James K says:

Assume that each of n children is either on side A or side B or not on a side at all. Each child can have any of these states independently of each other child so the total number of arrangements in 3^n.

However, if no children are on side a, there isn’t a conflict. There are two possible non-side-A states for each child, so that’s 2^n arrangements that aren’t conflicts for that reason. Likewise, there are 2^n arrangements that aren’t conflicts because nobody is on side B. However there’s an overlap there. both of those groups include arrangments where nobody is on side A and also nobody is on Side B. Since there’s only one remaining possibility, that’s 1^n cases–which is simply 1 case, that we must make sure not so subtract twice. So that gives us 3^n -2(2^n) + 1 conflicts.

However, that number treats side A and side B as distinct. If, for example, Sally was on side A and Jimmy was on side B, it would count as a different conflict than if Sally was on side B and Jimmy on side A. We probably don’t want that, so we should divide our number by two to eliminate mirror images.

That gives us a final total of (3^n + 1)/2 – 2^nReport

• James Hanley in reply to James K says:

I’ll let you guys argue out the math, and I’ll just go all fuzzy and say it feels exponential.

More seriously, I think there’s also a factor not caught by the math, which is that any disagreement seems more likely to flare into conflict if there’s a third person around. If 1 has a conflict with 2, and there’s no ally, they might be more likely to back off, sensing less chance for “victory” (whatever they think that might be). But if all 3 kids are present, and 1 senses 3 is on her side on the issue, 1 seems more likely to escalate disagreement into conflict.

Not that my kids are anything other than perfect angels, of course.Report

• Mike Schilling in reply to James Hanley says:

The standard joke is that the third one makes you switch from man-on-man defense to zone.Report

• Chris in reply to Mike Schilling says:

When I was a kid, one of us (not gonna say which, but there are 4 of us) was so bad, my parents usually had to play a box and one.Report

• LeeEsq in reply to James Hanley says:

Just refer the inter-child conflicts to the United Nations. It’ll be as useful as it is with international conflicts but at least your outsourcing the work.Report

• Glyph in reply to James Hanley says:

What the Professor said. Three is WAY harder than two. But #3 is amazing (we just celebrated her first b-day).

Congrats, and prepare to learn why your parents seemed so crazy and cranky all the time!Report

5. Will Truman says:

The Gentlings just keep on coming! Congrats!Report

6. Jaybird says:

It’s a mitzvah!Report

7. James K says:

Congratulations Kazzy.Report

8. Murali says:

Congratulations KazzyReport

9. LeeEsq says:

Ah fertility.Report

10. aaron david says:

Well done Sir!Report

11. Congrats!Report

12. zic says:

Congratulations.

Take time to coddle Zazzy. People seem to thinks second pregnancies are a piece of cake since you’ve already done it before; when the truth is they’re more exhausting because you’ve already done it before. (Mine are exactly two years apart, so I have some similar experience and speak from that experience.)Report

13. Michelle says:

Congratulations!Report

14. Burt Likko says:

Mazel Tov2!Report

15. Kazzy says:

Thank you to everyone for your kind words and support! Most of all thanks to those who did what only we here at OT could do and engage in some complex mathematical debate on exactly how each additional child impacts the work load.

While you’re at it, can you draw up some figures on what managing 9-15 4-year-olds entails? I might be able to use them to pitch my boss on a raise.Report