A modestly fun probability question from AskMetaFilter: If you shuffle a standard deck of 52 cards, and draw them 2 at a time until the deck is exhausted, what is the expected number of pairs (two cards of the same value) that you will draw?

I haven't looked at the answers on AskMe; I long ago learned not to get drawn into any discussions of probability questions on the internet, ever.

https://ask.metafilter.com/357711/How-many-pairs-in-a-deck-drawn-two-cards-at-a-time

https://aditya-r-m.github.io/twisty-polyhedra/

Pretty for sure, but it'll take a while to get a feel for how comfortable the controls are.

Probably Solution

I haven't seen this before in general, and in particular, I haven't seen it in The Times.

I introduced \(k=\left(5/3\right)^x\) and deduced \(k^2-k-1=0\) leading to \(k=\left(1+\sqrt{5}\right)/2\) and so \(x=\log_{5/3}\left(\left(1+\sqrt{5}\right)/2\right)\).

My calculator says that's about 0.94 which agrees with my a priori prediction (intermediate value theorem) that there should be a single real root between 0 and 1, closer to 1.

http://homepages.gac.edu/~jsiehler/games/pentomino-set.html

"In this post, I am going to show you how to write Fizzbuzz in the programming language Fractran."

https://www.youtube.com/watch?v=O4UpNSlzKAM

- Academic Homepage
- http://homepages.gac.edu/~jsiehler/

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Joined Jan 2019