Mathematics desk
< December 24	<< Nov \| December \| Jan >>	December 26 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

December 25[edit]

Can someone please explain this trick? Thanks.[edit]

[1] Imagine Reason (talk) 03:23, 25 December 2010 (UTC)[reply]

Notice how the stack is made from 1:10 to 2:00. The 3 cards always end in the same positions no matter where the piles are cut. This should be easy to see if you ignore the order of operations which is just designed to confuse the viewer. The first card is on top of pile 1. The second card ends on top of pile 2 (which was moved internally by the cut without affecting the second card). The third card ends on top of pile 3 (which was moved internally by the cut without affecting the third card). The remaining is just a rather arbitrary procedure to end up with the 3 fixed positions. PrimeHunter (talk) 04:11, 25 December 2010 (UTC)[reply]

[ec] (For those curious, it is a video of a card trick, 3:23 long.) First consider the second half of the trick, where the cards are dealt alternatingly into two piles -- "up, down, up, down" etc.. Notice that the procedure used is deterministic, and it will always pick out three cards from the deck that started in the 6th, 22nd, and 38th positions from the top. (There's nothing special about those numbers; presumably whoever invented the trick just tried out the procedure with a sorted deck and looked to see which three cards came out at the end.)

The first half of the trick is a convoluted way to put the three chosen cards in the 10th, 26th, and 42nd positions from the top of the deck. Notice in particular that the two cuts of the deck actually did nothing at all. Then by moving the top four cards to the bottom, the deck is set up properly for the second half of the trick. Eric. 82.139.80.114 (talk) 04:21, 25 December 2010 (UTC)[reply]

Mouse in a maze[edit]

Mouse in a maze: Using First Step Analysis, I start in room 1. Rooms are connected by either 2 or 3 doors. Room 11 has 3 doors but one goes to room 12 which has a mouse trap. Room 15 has 2 doors ;one leads to cheese. The game ends if either the room with the trap or the room with the cheese is entered. I am looking for the probability that the mouse gets the cheese before the trap get the mouse.

I have some understandong of the transition matrix. I write the transtion matrix with entries in row i summing to 0 . That is, (for three doors) in room 9 one to 7 one to 8 and one to 10 x9 = -1, x7=1/3,x8=1/3 and x9 = 1/3. I get confused on how to handle the trap room and the cheese room . Also, the single column matrix on the right side has zeros except for one entry. This confuses me. I have software for solving linear systems. Thank you don miller71.20.25.109 (talk) 10:14, 25 December 2010 (UTC)nods[reply]

Once the mouse enters the trap room or the cheese room, it isn't going anywhere else, right? In other words, if the mouse enters the trap room, the probability that the mouse will go to the trap room next is 100%, and the same for the cheese room. Write your transition matrix to reflect that fact. —Bkell (talk) 15:48, 25 December 2010 (UTC)[reply]

Center of simple algebra field[edit]

hi is there the center of any simple algebra field? — Preceding unsigned comment added by Saeedyzdn (talk • contribs) 10:41, 25 December 2010 (UTC)[reply]