You can calculate Pi by dropping toothpicks on a wood floor

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This is a neat application of a math puzzle that a teacher showed us in a college math class. The puzzle was to work out the probability of that toothpick intersecting a line when dropped. I was thrilled when I figured it out years later.


This is not a way of calculating pi. It is a way of estimating pi using a Monte Carlo-type method.
And yes, I am a grumpy pedant.


When Daddy steps on a toothpick and jams it up under his big toenail, you can learn new vocabulary too!


Everyone knows pi is 22/7. Don’t need no toothpicks.

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A much better pi is 355 / 113. Easy to remember: the inverse of 113/355. Good to more digits than you’d expect.


Everyone knows pi is 245850922/78256779. Just as easy to remember as 3.14159265358979323846…


I usually remember it as

3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117068 over 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.

Much easier to remember when the denominator is a clean power ten.


Probably would be a lot easier to simulate on a computer then. Of course, most of the library functions that would be called in such a simulation rely on M_PI in some way.

Came to this thread hoping to find someone reminding us this was a probabilistic approximation.

Was not disappointed :smiley:



Usually one of the first things one does in a monte carlo class – approximate pi.

Edit: first page of my monte carlo book…

Hmm. I guess you don’t really need more than sqrt(x).

Not if you use an infinite number of toothpicks, Mr Smarty.


3+ 1/(7 + 1/(15 +1/(1 +1/(292 +1/(1 +1/(1 +1/(1 +1/(2 +1/(1 +1/(3 +1/(1 +1/(14 +1/(2 +1/(1 +1/(1 +1/(2 +1/(2 +1/(2 +1/(2)))))))))))))) or thereabouts…

I like this problem as an excellent example of a math problem that becomes easier when you generalize it. The more general problem is: if you have a curved piece of wire of length L, and you toss it on this floor (where the lines have spacing 1), what’s the expected number e(L) of lines crossed? First you argue that e(A+B)=e(A)+e(B), and from that and continuity, that e(A) = cA for some c. Then let L be a circle of diameter 1 to figure out that e(pi) = 2 (it crosses one line, twice). Hence e(1) = 2/pi, which was the original question.

Isn’t a Monte Carlo a type of sandwich?

(Monte Cristo, japhroaig!! Cristo!!)


That’s not a sandwich - that’s a heart-attack.


Don’t we all agree here that you should be able to choose a death that is suitable for you?


A weapon of mass destruction that could be deployed in less than 45 minutes - let Tony Blair know.

A very good idea. I have decided that the most suitable death for me would be to die of a heart attack due to the sexual attentions of one of my many mistresses, while travelling on one of my sailing ships from one of the ports of my 2000-year-old empire to one of my holiday palaces, in or about the year 4016.
Perhaps I should start a kickstarter. Sandwich - lacks ambition.


and perfectly straight toothpicks and perfectly paralell lines. Possibly perfectly zero thickness too.

Oh, you can forget about pseudorandom generators…