If you’re planning a Christmas party and you’re the sort of miserly Scrooge type that doesn’t want to waste any money on excess Christmas crackers, science has got you covered.
A trio of those party-loving statisticians has worked out the formula for the optimum number of crackers for a party to ensure that everyone wins one prize, while nobody gets a second bite at the cherry. And they only needed to simulate 10,000 parties, ranging in size from two to 50 guests, to do it.
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According to their calculations, for ten party guests the host will need to stock up with 15 Christmas crackers and then militantly ensure that the pulling of crackers takes place according to strict rules.
“Only five guests are required before there is already a greater than 90 per cent chance that someone will go home without a win. Indeed, this probability increases rapidly with the radius of your circle of friends,” the boffins pointed out in their article.
“The obvious solution to this unfortunate (but not uncommon) event is for the game to continue between individuals who have not yet won. We call this the “Santa” strategy, as it ensures that prizes are generously distributed to everyone.
“Two questions then arise. Firstly, how long should the host allow for this entertainment to ensure that the goose is not (over)cooked? And secondly, how many spare crackers should be on hand to ensure the Christmas spirit prevails throughout the evening?”
Of course, you could just let your guests carry on pulling crackers until all were satisfied, as in the “Santa strategy”, but for efficiency, cost-saving and ensuring the dinner doesn’t get carbonised, the Scrooge-led approach is best.
Each guest gets paired off with another to pull one cracker and then the loser of that sharp tug gets to pair up with another loser, until just one person is left - who pulls a lonely cracker with themselves to get the last prize.
“We can easily see why a Scrooge-like character would love this set-up. There is little left to chance as both the number of crackers required (N) and the number of rounds (⌈log2(N)⌉+1) are fixed quantities,” the statisticians say.
“Compared with the Santa approach, Scrooge requires the maximum number of rounds, as each round contains the maximum number of losers. However, Santa requires twice the total outlay of crackers over those rounds.”
Merry Christmas Reg readers! ®