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Future family holiday ideas


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| Forever in progress • Humour • |


Sometimes, when I get to thinking about what an awfully confusing parent I’d be, I invent new and befuddling family holidays to celebrate. The list is below, and they are licensed under a Creative Commons Attribution-NoDerivatives 4.0 International License. Yes, that means you absolutely can use them for profit.

Name: Vanquishing superstition day.
When: Held once a year, on a calendar date that is absolutely not astrologically significant in any way. Except that rule is scratched, because why not, and you can hold it whenever you like. Actually scratch that, you can only hold it on Friday 13th of a Leap Year.
What: It begins with the great ladder walk. 13 ladders are set up in a row, inside. Participants walk under the ladders, holding up opened umbrellas. They must walk through the ladder tunnel any number other than three times. To determine the number, participants roll a dice. If they roll a six—well, that’s beginner’s unluck—and the holiday abruptly ends. Other participants stand outside the ladders, spilling salt on the walkers. Only their open umbrellas protect them. After the ladder walking is concluded, participants must imagine horrible things, say “I hope that doesn’t happen”, and then try very hard not to touch anything wood. The final non-ritual is to smash a mirror with a hammer shaped like a three-leaf clover (made of three upside-down horseshoes). The participant wears a black-cat mask.
Predicted fun level: very high

Name: financially-prudent Santa Claus
When: 24th December
What: Every year, children are allocated a set budget that their parents are willing to spend on Christmas presents, say: $100 per child. Every year, the child must choose between getting Christmas presents now, or having the money invested in an index fund that will mature whenever the child wants, on the condition that it must be spent on a future Christmas present. If they choose the presents, $100 plus the previous years’ interest is spent on the presents. If they choose another deposit, $110 dollars is put into the fund.
Predicted fun level: Scrooge

Name: Sombrero day
When: that’s the point
What: This one is a long game. Every year, on a seemingly random day, the in-the-know members of the family wake up a child with streamers and music, wearing sombreros and looking suitably joyful. The members of the family must act as if they are very disappointed that the child has forgotten the yearly sombrero day. The choice of days must have an underlying patten. Provided that the child works out the pattern and prepares with their own sombrero, they receive the ability to set the next underlying pattern, and the item of head wear. Predicted fun level: uncertain, depends whether kids have very particular sense of humour

Name: Spin-the-wheel day
When: up to you
What: A rotary-style wheel of fortune is created, with, say, 20 possible slots. Children can submit things they would very much like to do, within the bounds of a reasonable budget and time-frame, like: “Our parents do our maths homework for us for a month.” However, they must also submit things they don’t want to do, like cleaning the dishes by hand for a month. For every additional good thing they add, they must add a bad thing. They can also add “neutral” things like: “nothing happens.” It’s up to the kids to decide how much they want to risk to get a return.

Name: Newcomb's horrible holiday
When: never, hopefully
What: There is a reliable predictor (the parent), a team of children, and two boxes designated A and B. The team of children must choose between taking only box B, or taking both boxes A and B. The children know the following:

  • Box A is clear, and always contains a visible single Christmas present.
  • Box B is opaque, and its content has been set by the parent.
    • If the parent thinks the children will take both boxes, then box B contains nothing
    • If the predictor has predicted that the player will take only box B, then the box contains ten Christmas presents

The children don’t know what the parent predicted, or what box B contains. The paradox, for those unfamiliar, is that in taking two boxes, you increase the likelihood (providing the parent is an infallible predctor) that the parent predicted you would take two, in which case you get only one present. If you take one, you increase the chances that the parent predicted you take only one, in which case Box B does contain ten presents, but you won’t get them. The optimum result is to take both boxes, providing the parent predicted you wouldn’t, in which case you get 11 Christmas presents.