Card Trick Help Please?

Card trick probability help!?

• I had to do a card trick for my statistics class and now we need to explain how our card trick worked through probability... Here is a link to the trick I used: http://www.goodtricks.net/super-card-trick.html I thought it was going to be easy to explain it based on what we've gone over in the text... but it's not as simple as I thought :( The trick is, a spectator picks two cards of his choice (without naming the suit, so it can't be hearts, spades, clubs, or diamonds specifically)... then, he shuffles the cards and gives them back to me. When I spread out the cards, the two cards he had chosen will appear right next to each other. What is the probability that the two particular cards chosen will appear right next to each other? I really thought it was gonna be a simple explanation... but it's turning out to be more complicated than I thought -.- What I've thought is... there's 4 of each value of card; like for a card value of 7, there's diamonds, spades, clubs, and hearts... so if a spectator picks two cards, they're really picking 8 cards total within the 52 cards. But then I get stuck there lol Does this have to do with permutations or combinations or what?? Please help!!!

• Answer:

  here's how to figure out the probability as far as I can figure it: Let's say the spectator picks 5 and 7 Except in cases where the first or last card in the deck is a 5, or where two fives appear next to each other, there are 8 cards in the deck that are next to a 5. So you simply need to figure out the chances that at least one of these will be a seven. I used 11/12 that any one of these cards won't be a seven, then took that to the 8th power to find the probability that none of them would be a seven. When I work this out, it only comes out to be about 50% that it will work. Plus if the first or last card is a 5 or if there are two 5s next to each other, the chances are further reduced. That being said, I tried this trick 5 times and it only worked once. 3 out of the other 4 times there was a card between the two and one out of five they weren't even within 2 cards of each other. Maybe you have better luck. Either way I hope this helps.