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The Magic Cafe Forum Index » » Magical equations » » A new probabilistic magic question. (0 Likes) Printer Friendly Version

owen.daniel
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It seems as though there has been quite a lot of interest in probabilistic questions about cards, and how they can be applied to magic tricks. So now I thought I'd contribute my own addition to the topic, and see what people can come up with from the magical point of view. In this case the mathematics is already done for you (no need for simulations etc, the formulae are known), and in addition you'll all be familiar with the problem its based on, but the question is can we come up with magical uses for the mathematics!

Mathematically minded magicians will perhaps remark during a trick that the probability of two shuffled decks being in exactly the same order is 52! ~ 8e-67, a remarkably small number. Now one can ask the question how many decks would we have to have in a row before there's a good chance that two of them are in exactly the same order. This question is the same as the well known Birthday problem: how many people have to be in a room so that there is a good chance that two of them share a birthday. The problem is normally stated so that the probability of this happening is a half, and the correct answer is 23.

As many of you will know, the formula for the probability that n people all have distinct birthdays is

365! / (365 - n)! x 365^n.

We can use this formula for our pack of cards question, except now we have to replace 365 with 8e67. But now the formula is somewhat unusable, why? because we have a number of the form ( 52! )!, and that number really is big. This is a silly way to approach the problem however, and fortunately general bounds are known for the problem when 365 is replaced by arbitrary N, (see the wikipedia page about the birthday problem). One accurate approximation is

sqrt( 2 N log(2) ),

In the case N = 52! this is still a big number: we'd have to have 10^34 packs of cards before there was more than 50% chance of having two decks in exactly the same order (where they have been uniformly shuffled, and not taken fresh out of an unopened box). This lead me to think about how small the deck has to be before we are likely to start getting matches.

Instead of asking for a match of the whole pack, what if we asked for the order of one of the suits to match the order of another suit? i.e. we run through the deck separating the cards into suits, and then look to see whether any of the 4 suits are in the same order. Unfortunately the approximation above says you would need over 92000 suits for two of them to match with a probability of a half.

What if we went a bit smaller. If we take out the A, 2, 3, 4 of each suit, then we would only need 5.7 suits for there to be a 50% chance of a match. And if we wanted only the A, 2, 3 then in a normal pack of cards there will be a match with a probability of 72%.

Ok, so what can we do with these properties. Well matching the order of the A, 2, 3 of two of the suits isn't really all that impressive, so lets move back to the previous case of 4 cards. Now we note the following: split the cards into three groups 2, 5, 8, J is one group, 3, 6, 9, Q is a second, and 4, 7, 10, K is a third (so the aces have been left out). Then there are now 12 sets of these groups in the pack (one of each group in each suit). So now when we divide the deck into these groups the probability that the order of at least two groups is the same is 0.965, where I mean the same in the sense that the 8, J, 5, 2 of clubs may have come out in that order, and the 10, K, 7, 4 of hearts came out in that order.

One less convoluted form of this would be with ESP symbols. If we create a deck of 55 ESP cards: 11 different back designs, and for each back design a set of the 5 different symbols. Now if the cards are shuffled and then separated into their colours, the probability that two of the 11 packets match is just under a half, 0.47ish.

Now the question is, can any of this be of any use? The obvious approach to using this would be to show some sort of coincidence effect... look you shuffled the cards a lot, and still these sets match... But the probabilities really don't work in our favour: when there are sufficiently many sets to make a match likely, the fact that two sets match doesn't look all that impressive (this is actually rather disappointing, given that the original birthday problem is so unexpected).

So is there a way to use this principle covertly, not drawing attention to the probabilities of the matches?

I'm curious to see what people can come up with.

Owen
TomasB
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Since the number of ways to order things is so untuitively large, you have it working the opposite direction of the unintuitively low number of samples needed for the Birthday Problem. At best you can probably create a trick that makes it seem like a reasonable thing is happening.

I really like your ideas here though and your insight that not all deck orders have to be exhausted before two shuffled decks have been found to be in the same order. Most people use 52! as the number of decks needed, which you have shown is not true for a match to be probable.

I will keep thinking about your proposed problem here, but so far I can only see the two properties (number of ways to order and the Birthday Problem) working against eachother.

Great to see you on these parts of the forum, by the way.

/Tomas
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