A while ago I sent a post with a card trick which I explained had a statistical element to it, and asked you to try to work out how it was done. Thanks to those of you who wrote to me with variants on the correct answer.
The rules of the game were that my assistant, Matteo, chose a card at random hidden from me. It happened to be a 5 in the video. I then turned the cards over one at a time and Matteo had to play a counting game. Once he reached the 5th card, he noted its value, which was a 10. So he then counted another 10 cards in the sequence, noted the value of that card, and so on until we ran out of cards. Matteo had to remember the final card in his sequence before the cards ran out, which turned out to be the eight of diamonds. My task as the magician was to predict what Matteo’s final card was, which I did successfully.
Now, there are 2 reasons why this is a statistical card trick.
- It doesn’t always work. It does so with a reasonably high probability, but depending on the configuration of the cards once they are shuffled, won’t always. I’ll be honest: we had to remake the video several times, but that was always due to my incompetence in explaining the trick and not because it ever failed. Still, it won’t always work.
- The second reason it’s a statistical trick is in its execution. The way it works is that I also play the same counting game as Matteo, but starting with the value of the first card I turn over, which happened to be a 10. So, we’re both playing the same counting game but from different starting points. Matteo’s starting point is 5, mine is 10. Although we start from different places, it turns out to be quite likely – though not certain – that the counting sequences we follow will overlap at some point. And once they do overlap, we are then following exactly the same sequence and so will arrive at the same final card.
Technically, the sequences of cards Matteo and I are both following are called Markov chains. These are sequences of random numbers such that in order to understand what the next card might be I only need to know the value of the current card, without knowing the past sequence that took me to the current state. In other words, when Matteo has to start counting 10 cards, it doesn’t matter how he got to that position, just that that’s where he currently is. And I also generate my own Markov chain. With an unlimited number of cards in a pack, the mathematical properties of Markov chains would guarantee that our sequences meet at some point, after which we would be following exactly the same sequence, leading me to have the same final card as Matteo. With just 52 cards in a pack, there’s no guarantee, which is why the trick won’t always work.
The fact that the trick might not work is a little undesirable, but you can increase the chances by counting picture cards as 1 rather than 10. This forces the sequences to change card more often, which increases the chances of our two sequences overlapping.
Markov chains are actually really important building blocks for modelling in many areas of Statistics, which is one reason why I used posted the card trick to the blog. I’ll use a future post to explain this though.
