[bodyboarder]

Hi... I had to do a probability card trick for my statistics class and now I have to explain it :(

The trick is, I ask you to pick 2 cards, without naming the suit (hearts, spades, diamonds or clubs). That narrows down the different cards to 13 different cards (2 through 10, and King, Queen, and Jack). After you pick 2 cards (like 5 and 7, or Queen and 2), I spread out all 52 cards, and according to the website I got the trick from, there's about a 9 out of 10 chance that the two cards you chose will appear right next to each other somewhere in the deck of 52 cards.

Here is a link to the trick: http://www.goodtricks.net/super-card-trick.html

How do I explain the probability that the 2 out of 13 possible cards you chose, will appear right next to each other in a deck of 52 cards? I can't simply say what the website says, that 9 out of 10 times they will appear next to each other. I need to explain why it happens about 9 out of 10 times...

Someone please help because this is a nightmare and I need someone to help me end it :(

This may be a totally illogical answer, but this is what I came up with after hours and hours of stressful thinking (cuz I'm that dumb lol)... there are a total of 78 combinations of 2 taken out of 13 different cards. So, a spectator has 78 different ways of choosing 2 cards out of 13 different cards.

Since there are 4 groups of 13 in the deck of 52 cards, I multiplied 78 by 4 and got 312 combinations for the entire deck.

Then, there are 1,326 different combinations of 2 out of 52 different cards... divided by 4 is 332 per group of 13... I don't know, someone please help lol

[ebaines]

Have you actually tried this trick to see if it works 90% of the time? I don't think the web site is correct - I believe it overstates the probability that the deck will contain at least one occasion of the two named card values being next to each other. By my reckoning this will work only about 53% of the time, not 90%. Here's why:

Suppose the named cards are Ace and King. And let's assume that the A's are spread through the deck (no two A's are next to each other, and neither the first nor last card is an A). If you look through the deck for the A's each one has two card that are next to it. So there are a total of 8 cards that are next to all the A's, and if any one of those are a K you win. The probability of any of those 8 cards being a K is equal to one minus the probability that none of them are a King. The probability that none of the 8 cards is a King is:

44/48 x 43/47 x 42/46 x 41/45 x 40/44 x 39/43 x 38/42 x 37/41 = 0.47.

So the chance that at least one of the cards is a King is 1 - 0.47 = 0.53. This is no where near 90%.

Note that this calculation ignores the occasion where the A is either the first or last card in the deck - if you take that possibility into account the odds go down, as there are now only 7 (or perhaps only 6) cards that lie adjacent to the Aces. It also ignores the possibility that the deck may have two or more aces in a row - which again lowers the chance of a King being adjacent - or the possibility that the Ace's have only one card between them. So I conclude that the probability of two named cards being next to each other is actually a bit less than 53%.

[jcaron2]

I did a really quick simulation of this "trick" on a computer, where it's simple to do a large number of runs. After a trial of 1 million runs, the two cards ended up next to each other only 48.7% of the time, right in line with what ebaines said.

No wonder you had a difficult time figuring out the math to reach an answer of 90%!