Math behind a card trick

Reply #4

cottoneyedjoe — Tue, 03 Mar 2020 05:48:46 GMT

it's not too much trouble, I like these sorts of probability problems. I did 10 million simulations of a shuffled deck and got a success rate of 74.76% for your trick with wrap around, which makes sense in light of the probabilities of the jack-neighborhood sizes -- a 16-card jack-neighborhood size occurs in only about 31% of shuffles, while the average size is about 13.7 cards. I ran 5 million simulations to approximate the probabilities of getting all the different neighborhood sizes from a mi... [ More ]

Reply #3

ScoobyDue — Mon, 02 Mar 2020 04:26:59 GMT

Wow

That was way better of an explanation than I could have hoped for.

Don't work too hard to calculate all the answers. You've given me all I need to know.

I was always fascinated by the math behind the challenge but could never put a pencil to it.

Reply #2

cottoneyedjoe — Sun, 01 Mar 2020 22:51:07 GMT

Letting the deck cycle through helps. Think of every jack as having a 4-card neighborhood. If the jacks are sufficiently far enough apart so that none of the neighborhoods overlap, then the maximum number of cards that can belong to a jack neighborhood is 16. That means the minimum number of cards that do not belong to a jack neighborhood is 32. Let's call these cards the outskirts.

The probability that none of the 6s are in a jack neighborhood is equal to the probability that all of them ar... [ More ]

Reply #1

ScoobyDue — Sun, 01 Mar 2020 20:04:48 GMT

I guess I should've said it's not infallible but I can't figure out how to edit

Math behind a card trick

ScoobyDue — Sun, 01 Mar 2020 19:02:25 GMT

I saw a card trick once that was pretty cool.

I've since used it hundreds of times to win free drinks at the bar and have often wondered about the mathematics behind the trick.

It's amazingly reliable but not fallible as I have had to buy a few drinks along the way.

The trick goes like this:

Take a shuffled deck and ask the participant to name 2 values of cards (suits do not matter).

Let's say Jack 6.

You then predict that you will find their cards either right next to eac... [ More ]