Jumping in Puddles

My Weekly Top 5 Artists

Tue, 27 Dec 2011 08:29:27 +0000

My Weekly Top 5 Artists

Sun, 25 Dec 2011 01:31:27 +0000

Albums of the Year 2011

Fri, 23 Dec 2011 18:31:00 +0000

This year has been a truly excellent year for music, with loads of great records, brilliant singles and some fantastic gigs. Here are my top ten albums of 2011; each one is a gem.

My Album of the Year

Bon Iver Bon Iver

The Rest of the Top Ten

(in no particular order)

Burst Apart The Antlers
The Rip Tide Beirut
Hurry Up, We’re Dreaming M83
Tamer Animals Other Lives
Zonoscope Cut Copy
Slave Ambient The War on Drugs
I Am Very Far Okkervil River
Barton Hollow The Civil Wars
Into The Murky Water The Leisure Society

Why not sample them all through a nice, easy Spotify playlist?

My Weekly Top 5 Artists

Tue, 20 Sep 2011 16:15:47 +0100

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Tue, 13 Sep 2011 10:22:28 +0100

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Thu, 18 Aug 2011 19:01:16 +0100

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Wed, 10 Aug 2011 08:18:19 +0100

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Tuesday Teaser #36

Tue, 09 Aug 2011 22:19:01 +0100

Look at the diagram below:

The problem is this: can you draw a continuous curve, without taking your pen off the paper, that crosses each line once and only once? So, for example, one attempt might look like this:

Unfortunately, this attempt misses one line, which I’ve highlighted above.

Hint: If you need a hint, notice the similarities with the very famous Seven Bridges of Königsberg problem.

Answer: Tuesday Teaser #35

Tue, 09 Aug 2011 22:17:33 +0100

In order to answer this teaser, we need to calculate the probability of the three man jury coming to the correct verdict.

In order to come to a correct verdict, two of the three must independently reach the correct verdict. This happens when either:

The two “serious” jurors come to the correct verdict; or
The two “serious” jurors disagree, but the “flippant” juror comes to the correct verdict.

The first case has probability p*p = p^2. The second has probability 1/2 * [p(1-p) + (1-p)p].

Summing the totals, the total probability becomes p, exactly the same as that of the one man jury. This seems like a curious result given that the one man jury intuitively feels as though it should be more serious (and hence reliable) than the three man jury.

My Weekly Top 5 Artists

Wed, 03 Aug 2011 02:45:51 +0100

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Tuesday Teaser #35

Tue, 02 Aug 2011 23:47:31 +0100

Suppose there is a three man jury which has two members who each independently have probability p of making the correct decision, and a third member who flips a coin for each decision. In this three man jury, the majority rules.

Also, suppose there is a one man jury with a probability p of making a correct decision.

Which of these juries offers the best probability at arriving at the correct decision?

Answer: Tuesday Teaser #34

Tue, 02 Aug 2011 23:38:00 +0100

This puzzle is pretty tough: the best way to think about it is to generalize what happens if a company hires one more person than it already has.

Let’s let D(n) be the expected number of working days for a company with n employees. If the company has 1 employee, it is clear that there is always 1 unique birthday. So D(1) = 364.

Now, inductively, adding an extra employee will have a D(n)/365 chance of reducing the day count by 1, and a [365-D(n)] chance of staying the same.

So: D(n+1) = D(n)*[365-D(n)]/365 + [D(n)+1]*D(n)/365 = D(n) - D(n)/365

Using our initial starting point of D(1) = 364, we can derive that D(n) = 365 * (364/365)^n

So the total expected number of man days worked by n employees in a year is then: E(n) = n * 365 * (364/365)^n

And the maximum E(n) is given when the derivative is equal to zero. Here, that occurs when n = 365. It’s a nice little coincidence that the number of days in the year coincides with the maximal number of employees.

For a little bit more detail around this solution, there is a comprehensive spreadsheet available on the web that you can have a play around with.

My Weekly Top 5 Artists

Sat, 23 Jul 2011 11:09:59 +0100

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Tue, 05 Jul 2011 16:24:45 +0100

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Tue, 28 Jun 2011 00:06:28 +0100

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Fri, 24 Jun 2011 05:51:29 +0100

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Tuesday Teaser #34

Tue, 21 Jun 2011 21:22:43 +0100

A very tough, but extremely interesting, probability problem for my first teaser in a while.

Let’s forget the standard working week and assume that companies decide when people have to work using a completely different method. Let’s suppose that they have to give a holiday to every single employee at the company whenever any one employee has a birthday. Other than those holidays, employees work every day. Forget leap years; let’s assume that there are 365 per year.

If a company wants to maximize the expected total number of man-hours worked per year, how many employees should it hire? (To ensure randomness, assume companies cannot discriminate in hiring based on someone’s birthday.)

Answer: Tuesday Teaser #33

Tue, 21 Jun 2011 20:40:40 +0100

This can be solved using the classic trial and error method, but we can also do it mathematically by using the information to set up a couple of simultaneous equations.

Let’s call the number of dogs x, the number of cats y, and the number of mice z.

The first thing we’re told is that there must be 100 animals, so x + y + z = 100.

The second thing is that the 100 animals must cost $100 in total. So 15x + y + 0.25z = 100.

The mathematically-minded will have noticed that this is only two equations for three unknowns. Fortunately, however, there is one other piece of information that we can use - namely that the three unknowns, x, y and z must be whole numbers.

Subtracting the two equations gives: 14x = 0.75z, which simplifies to 56x = 3z.

This is crucial, as we can now deduce that x must be divisible by 3. Given that all dogs cost $15 and the total cost must be $100, this means that there are only two options for x, the number of dogs: 3 or 6. If we were to buy 6 dogs, that would leave only $10 remaining to purchase 94 animals, so we can conclude that x=3.

This immediately allows us to conclude that z=56, from which it easily follows that y=41.

So we must buy 3 dogs, 41 cats and 56 mice.

My Weekly Top 5 Artists

Fri, 20 May 2011 10:05:22 +0100

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Wed, 27 Apr 2011 18:28:03 +0100

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