|
|
B. T. Lewis New user London 59 Posts |
Hi.
Just wondering what the odds are of getting the different possible values when performing the Kruscall count and adding up the values of the 'key cards' afterwards. For example, you have performed the basic effect and the cards are in identical order face up on the table. You ask them their thought of number and then redo the count (and are able to do it much quicker this time since they don't need to count in their heads) but separate the key cards as you go. After this you total the sum of their values. What possible results are there for this total and what are the odds of each? I have no idea how to calculate this and would therefore be very grateful to anyone who helps me, with the added bonus that you can use these results for your own purposes (if you so wish!) Cheers, Ben. |
Scott Cram Inner circle 2678 Posts |
The actual math isn't easy, but you can get a good approximation of the probability of success with this formula:
1-((((x^2) - 1)/(x^2))^n) (x = average value of the cards, n = total number of cards) It helps if you always start on the first card. Of course, you should also make sure that you have a way to handle failure. BTW, with a full randomly shuffled deck, there's about a 58% chance that every one of the first 10 cards will all lead to the same ending card. |
S2000magician Inner circle Yorba Linda, CA 3465 Posts |
I think you may be complicating things a bit here. If they thought of, say, 7, and ended on, say, card 48, the sum of the intermediate cards will be 41 (= 48 - 7), and the sum of all of the cards will be 48 + x, where x is the value of card 48.
I'm out of town till December 18, and I don't have the Monte Carlo simulation I programmed for the other question you asked; it's at home. When I get back I'll happily modify it and compute some probabilities for you. |
TomasB Inner circle Sweden 1144 Posts |
I think this sounds like a finished trick that has a slightly higher probability of succeeding than the Kruskal Count itself, which is about 84% according to http://www.voofie.com/content/48/using-p......ability/
So if I understand you correctly: 1. Just make a number up and do the Kruskal Count secretly along with the spectator. 2. Add 52 to the value of the card you end on. 3. Count backwards from this number until you reach the last card of the deck. 4. As soon as the spectator names his number, pretend to scan the spread inhumanly quickly and announce the number as the sum of all the cards he landed on. No need to add the number he named since you claim to sum up all the cards he landed on. 5. Or simply check out http://www.themagiccafe.com/forums/viewt......&forum=2 and you can predict such a sum even before the spectator starts the count. /Tomas |
TomasB Inner circle Sweden 1144 Posts |
Ooops, I think I wrote step 4 wrongly. You should claim that you will add all those cards along with the number he named. That way you won't have to do any further manipulation of the number.
Hope I got that correct. /Tomas |
The Magic Cafe Forum Index » » Magical equations » » Another 'What are the odds' thread (0 Likes) |
[ Top of Page ] |
All content & postings Copyright © 2001-2024 Steve Brooks. All Rights Reserved. This page was created in 0.02 seconds requiring 5 database queries. |
The views and comments expressed on The Magic Café are not necessarily those of The Magic Café, Steve Brooks, or Steve Brooks Magic. > Privacy Statement < |