# An Australian Dining Phenomenon

In an early draft of my forthcoming book, *The Maths Behind*, which will be available wherever good books are sold from September, I believe, I took the following unprovoked dig at Australia:

“… it crashed into the ocean about 1,600 miles to the west of Perth, Australia.

There’s nothing there.

There’s nothing much in the ocean about 1,600 miles to the west, either.

I felt a bit bad about it - apart from cricket in the 1990s and the music of Jason Donovan, Australia has never done anything untoward to me - and it didn’t make the final edit. However, now I learn about the horrors of Alternate Drop Service, I feel like writing to my editors and demanding they reinstate the jibe.

In the civilised world, when you go out for a meal in a sophisticated venue such as a burger van or an aeroplane, you are offered a choice of meals. That’s a vital part of the dining experience.

In posh Australian venues, which I’m assured exist, that’s not always how it’s done. Food may be served alternately: for example, the first person to be served might be given chicken, the second pasta, the third chicken again, and so on.

Read that again, if you can bear to.

Not only do they do this, *some people defend the practice*, the flaming drongos. “Oh, but you can swap with someone else!”

### Right. Let’s analyse this

Suppose - generously ((This is the best-case scenario for alternate drop service.)) - that a random diner is equally likely to prefer chicken or pasta.

**The probability of everyone getting what they want without any swaps** is fairly clearly $2^{-10}=\frac{1}{1024}$. But that’s ok, we can always swap… Assuming there are five chickenites and five pastafarians at the table.

**The probability of there being five chickenites and five pastafarians at the table** is $\frac{\nCr{10}{5}}{2^{10}} = \frac{63}{256} \approx 24.6\%$. More than three quarters of the time, someone is forced to suffer through the wrong meal.

On average, **the expected number of people getting the wrong meal** is around an eighth of the diners. That’s worked out as $\frac {5 \times \nCr{10}{0} + 4 \times \nCr{10}{1} + … + \left| 5-k \right| \nCr{10}{k} + … + 5\times \nCr{10}{10} }{2^{-10}} = \frac{1260}{1024} \approx 1.23$ for a table of 10.

Even if the table is correctly split, **the expected number of swaps required** is 2.5 - meaning half of the people at the table will need to change their meal.

In short: I don’t give a Castlemaine XXXX if it keeps the cost of the meal down. I think it belongs in the ocean about 1,600 miles to the west of Perth.