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Combinatorics. Ithaca Teaching Journal, Day 4

God Bless YouI often use this example, last night given as homework: if you have a bottle of bourbon, one of scotch, and another of beer, how many ways can you pour a tasting? The order in which you drink the spirits and the fruit of the barely will affect the taste of each. Start with the scotch and the beer will taste differently than if you had started with the beer. And so on.

There are three choices for the first drink. Once that is chosen, there are two choices for the second. And once that is imbibed, there is only one choice left. That makes 3 x 2 x 1 = 6 arrangements, which has the special mathematical name 3!, read “3 factorial.”

Now add in the mix Count Balboa (or similarly named), a Spanish rioja. There are now four choices for the first drink; after that three choices for the second, and so on. Answer: 4! = 4 x 3 x 2 x 1 = 24 arrangements.

Since by now we’re getting good at this, enter into the lists a syrupy Chardonnay, imported from some vague Eastern European country. Then, for good measure, include a by-now rapidly warming rosé produced at a Finger Lakes winery.

And since we’re on a roll, we might as well include a red wine with an unreadable label given as a gift for client services, and then another bottle, which by this time is the most delicious red you have ever tasted in your life. Of course, we mustn’t forget—we must never forget—Jameson’s, a splendid Irish whiskey.

How many choices now, huh? Huh?

I don’t care. See you tomorrow.

10 thoughts on “Combinatorics. Ithaca Teaching Journal, Day 4 Leave a comment

  1. We now have one choice, the Jameson’s because of prior conditioning not captured by the maths.

  2. Back in the day, we had switched from an analog equipment monitor to a digital one. We were getting about one false alarm per day and it was shutting the system down.

    The problem boiled down to: given 100 trials, what is the chance of having fewer than 70 successes. Realizing that there was no way my VIC-20 would handle 100!, I wrote a clever little program in Basic, proved that we would get about one error per day just based on random chance and shipped the result off to head office. The head office guys shipped the work to the equipment maker and their guys, not paying attention to what I had done, programmed the equations into their mainframe. It crashed. It turns out that the mainframe couldn’t handle 100! either.

    If you pay close attention to what you are doing, you don’t need much computing power. If you don’t pay close attention, or if you are clueless, all the computing power in the world won’t help you. Hmm, makes me think of climate models.

  3. I choose to solve this through massive amounts of experimentation. And to continue until I am certain that the solution is verified as repeatable.

  4. Bob,
    Had the same problem calculating binomial coefficients on a computer. You have to be sneaky because the numbers get large really fast if you attempt to do the calculation directly.

    Anyway, I’ll have what the Pope is drinking. If it’s good enough for him, it’s good enough for me.

  5. ish good lesson (*hic*), Prof. B. you f’got s’really N! + 1 com — comb — you know what I mean. Drink ’em all at once on first tasting. Best if done as a shooter like when drinking low grade tequila.

  6. dearieme, that is Joseph (then) Cardinal Ratzinger now better known as Pope Benedict XVI (he has also since ditched the red and black getup for all white). He is known to have a penchant for fine German wheat beers.

  7. In the spirit of carefully designed experiments with meticulously assembled evidence regarding Irish Whiskey, one will inevitably run into a distinction between the following:

    1. Catholic Irish Whiskey
    2. Protestant Irish Whiskey

    Shedding the biases and estimating roughly the prior probabilities about the superiority of any of them will allow to establish the initial necessary number of repetitions. It will undoubtedly increase. Each iteration will allow to fine tune or update the prior. Jameson! Bushmills! Jameson ! Bushmills ! Tullamore! Powers ! Bushmills ! Kilbeggan ! In mid-stream we decided to ignore the initial question as absurdly irrelevant and embark on more definitive task. After a while we will be able to suspect that the preferences may form a particular distribution. So how many observations will we need to estimate that distribution? Wait ! Doesn’t it depend on the type of the distribution. Somewhere here one may suspect that we are getting ourselves into a circular argument between the suspected distribution type and the number of necessary observations to prove it. Should we alter the hypothesis regarding the distribution? Pareto-Levy ? Oh, that would be too good.

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