Last night was the last of the eight nights of Hanukkah. There is all sorts of fun math in Hanukkah, such as figuring out how many candles you need in a box to cover all eight nights of Hanukkah, which will be one less than a triangular number. There are many fun combinatorics problems to construct and explore as well. How many different color combinations can be constructed if you have n different colors of candles available?

Today's Albany

*Times Union*had a great mathematics of Hanukkah story, A one-in-trillions dreidel game, about a remarkable string of luck by a first-time dreidel player, who managed to spin the four-sided top 68 consecutive times without ever seeing the losing side come up and winning the entire pot 56 of those spins. This inspired his great-nephew, a Princeton sophomore studying operations research and financial engineering, to pull out his calculator to compute the astronomically large odds against such a long streak of good luck.

There are several ways to model this probability calculation and it would be interesting to discuss the pros and cons of each approach. One simple possibility is to compute the likelihood of never losing in 68 spins, which would be 1 - (3/4)

^{68}, which works out to 1 in 22.5 trillion. You can make the odds even more astronomical if you also consider the likelihood that he actually wins the pot on 56 of those 68 non-losing spins, because the remaining three sides are equally likely to come up, and only one of those spins yields the entire pot to the spinner.

Of course, this kind of analysis gives rise to interesting philosophical discussions of the sort that physicist Richard Feynman raised in his book,

*The Meaning of It All: Reflections of a Citizen Scientist,*when he talked about the probability of seeing a particular license plate in a parking lot as well as the work of Stanford mathematician Persi Diaconis on the mathematics of coincidences.

Moving onto Christmas, we are currently in the midst of the fabled "Twelve Days of Christmas," as memorialized in the song that starts with one gift given on the first day (December 25, "a partridge in a pear tree"), three gifts given on the second day (December 26, "two turtle doves" plus another "partridge in a pear tree"), and so on through the twelfth day (January 6, which happens to be our next math circle meeting date!)

John Cook at the Endeavor blog has a great post on the mathematics of the Twelve Days of Christmas. He observes that the total number of gifts received each day is a triangular number and also notes that the cumulative number of gifts received through the end of each day is a tetrahedral number. He has written up several nice proofs demonstrating that this is true in general, and he also includes a wonderful link and illustration from the Math is Fun blog.

The Math is Fun blog's illustration at right is a great way to illustrate the triangular/tetrahedral nature of the 12 Days of Christmas song for your younger friends and relatives. The image shows the situation for the first five days of Christmas, with the top layer representing the number of presents given on the first day (1), the next layer representing the number of presents given on the second day (3), the next layer the presents given on the third day (6), the next layer the presents given on the fourth day (10), the next layer the presents given on the fifth day (15). The cumulative number of presents given on the first through fifth day is the number in the entire five layer tetrahedron, or 1+3+6+10+15 = 35. You can extend this indefinitely, of course. If you stop after 12 days, you will have a 12-layer tetrahedron with a cumulative total of 1+3+6+10+15+21+28+36+45+55+66+78 presents, which turns out to be 364, a very nice number, since it is one less than the number of days in a typical year!

Another fun fact to share with your younger friends and relatives: take the number of cumulative gifts received during the 12 days and multiply it by the number of days in 2012 and you get a nice opportunity to discuss factoring differences of squares since 365

^{2}- 1

^{2}= (365-1)(365+1).

There are wonderful mathematical possibilities to explore in every religion and culture--including the modular arithmetic of the various calendar systems generally designed to reconcile discrepancies between lunar and solar numbering systems, which move many holiday observations around relative to one another. It's also interesting to note the frequency with which calendars in so many widely divergent calendars use a 7-day week. This is mathematically very convenient, since the number nearest to 365.24 with a conveniently large number of factors is 364, which is divisible by 7.

The Hebrew calendar has a 19-year cycle with a leap month in seven of those years. The Islamic calendar has a 30-year cycle with a leap day added to the final month in 11 out of those 30 years. The Gregorian calendar commonly used in the west appears to have a four year cycle with a leap day every four years, but it's actually more complicated than that.

New years festivals are observed at many different times in different calendars, starting with the Jewish new year, Rosh Hashanah, which arrives in early fall. The Hindu new year celebration happens on their festival of lights, Divali, in mid-fall. The Chinese New Year generally comes later in winter than the Gregorian new year on January 1. The early Roman calendar (before Caesar came along and reformed it) began its new year in March, had only ten lunar months, and then had a mysterious unlabeled winter period of 61 day that were apparently not considered to belong to any month. Caesar changed the New Year to January 1, but later Christian rulers moved it to March, before Gregory moved it back to January.

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