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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 25, 2009 09:34 AM |
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Edited by dimis at 09:35, 25 May 2009.
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I can represent the number the guy has in his hand.
It is one of thirteen choices: 2, 3, ..., 9, J, Q, K, A.
So by having the ability to represent 16 choices, I can clearly represent 13 choices.
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The empty set
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 25, 2009 10:06 AM |
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Edited by dimis at 10:18, 25 May 2009.
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Also note by the way that the magician can not perform the trick if he uses more than 1 decks, because the the guy (from the audience) might pick 2 same numbers from the same suit twice (either same or different suit - doesn't matter), and then the magician can only represent 8 choices.
Example with 2 decks: The guy is extremely lucky and picks Ace of hearts, Ace of hearts, 2 of spades, 2 of spades, and something else. Now the assistant can get rid of one of the two duplicates, but not both. 
Edit: Wow! Actually, the above scenario reduces the permutations to half since by swapping the duplicate cards we can not distinguish the difference in the arrangement. Hence we have 12 different arrangements with one duplicate, and as magicians we can not "win" 100% of the time just by one number that needs to be mapped. However, we can guess right with prob. (11/13)*1 + (2/13)*0.5 = 12/13 when such a bad scenario happens (I am not talking here about the expectation before the game starts).
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The empty set
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Ecoris
Promising
Supreme Hero
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posted May 25, 2009 10:27 AM |
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I can't see how you determine the suit.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 25, 2009 10:48 AM |
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I don't. I already said that if I have to determine the suit, I have to rethink.
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The empty set
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Rarensu
Known Hero
Formerly known as RTI
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posted May 25, 2009 12:52 PM |
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The problem isn't 'use 4 cards to determine a 5th random card'. It's 'use 4 cards to determine a 5th card which you had some control over'.
Say the assistant always puts the lowest one in the pocket, and the next lowest card is number 25 or lower on the list of all cards. That means the assistant only has to name a card which is one of 24 cards... which he has enough bits to do! But if the assistant gets 4 cards which are all higher than 25, then he's screwed, and FOG said "Guaranteed".
If you put one in the pocket that is between two others close on each side, and put the lower of those two first, then you get 6 permutations to name a card that is at most one of 12... darn! This probability is a lot better than the one above; you need a very specific layout to foil it. However, not quite probable enough for "guaranteed".
How about using evens and odds? if you have 5-and-0 then give one of 5, if you have 4-and-1 then give one of 4, if you have 3-and-2 then give one of 3... damn... that looks the same as 2-and-3... foiled again. rape that "guaranteed" clause!
Quote:
Write S = 5Q+R, where R is 1,2,3,4 or 5. Put the Rth lowest card of the five cards in the pocket.
the sum of the cards is a multiple of 5 plus 0-4? Then you remove card numbered the remainder (or reverse numbered)? That would mean that there it could be the 0th, a multiple of 5 on the low side, or the 1th, 1 + multiples of 5 in between 0 and 2, etc. That cuts it down to around 15 possibilities. (48/5=10, +5 more possible from the shifting sets squeezing in on each other)
That's very clever! Also I'm mad at you now.
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Sincerely,
A Proponent of Spelling, Grammar, Punctuation, and Courtesy.
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angelito
Honorable
Undefeatable Hero
proud father of a princess
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posted May 25, 2009 01:13 PM |
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You only need 13 choices for the number, and 4 choices for the suit.
So there must be a way to define all possible numbers from 2-13 (2-Ace) with 4 numbers and the suits, and 1 out of 4 suits probably by using the way the 4 cards are arranged on the table.
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Better judged by 12 than carried by 6.
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Ecoris
Promising
Supreme Hero
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posted May 25, 2009 01:25 PM |
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Edited by Ecoris at 13:28, 25 May 2009.
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The procedure I described works.
There are 2,598,960 ways to pick five cards out of 52 and there are 6,497,400 to pick and order four cards out of 52. So it is certainly possible to make one-to-one map from unordered 5-sets to ordered 4-sets which is what we really want to do.
However such a map can't take a given 5-set to any 4-set; the 4-set must be a subset of the 5-set. That's the real problem.
(The existence of such a map is clear due to symmetry, but we need to know one, not just know the existence of one).
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Celfious
Promising
Legendary Hero
From earth
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posted May 25, 2009 02:50 PM |
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HELP please help I cannot sleep until I know!!!
So, I have a time math problem I have
5:12.62
This is 5 minutes, 12 seconds, and 62/100'ths of a second
Whats exact half? Or which . of a second is in the middle?
If you need more clarification please tell me to clarify so I can finally sleep.
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What are you up to
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angelito
Honorable
Undefeatable Hero
proud father of a princess
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posted May 25, 2009 03:17 PM |
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2.36.31
Western European Summertime 
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Better judged by 12 than carried by 6.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 25, 2009 03:32 PM |
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@Rarensu: Every number is a multiple of 5 plus 0-4. I don't see how this information alone cuts down your possibilities.
@Ecoris: If you have an upper bound of 16, then use the four cards as bits and define the mapping.
So, if you can determine the suit, I have a new problem to find a simple way for this mapping.
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The empty set
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Celfious
Promising
Legendary Hero
From earth
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posted May 25, 2009 03:49 PM |
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Thanks Blizzard and Angelito.. Not only did you tell me the answer but you helped me figure out how easy it is to do for the future.
... I hope .. lol
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What are you up to
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Ecoris
Promising
Supreme Hero
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posted May 25, 2009 05:06 PM |
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Quote:
@Ecoris: If you have an upper bound of 16, then use the four cards as bits and define the mapping.
Converting to bits seems unnecessary. There are 24 ways to order the four cards and there is an obvious way to order these ordered 4-sets. E.g.
(51,40,11,3) > (51,40,3,11) > (51,11,40,3) > (51,11,3,40) > (51,3,40,11) > (51,3,11,40) > (40,51,11,3) > etc. > (3,11,40,51).
Lets continue the example and say the magician was presented with the ordered 4-set (51,11,40,3). The sum is 105. The possible values of the last card are therefore: 1, 7, 13, 18, 23, 28, 33, 38, 44, 49. If we let the greatest element of the ordering above correspond to the lowest possible value and so on, the result is 13.
The algorithm is not that simple, but with some training you could probably become quite fast.
Of course you also need to be able to convert usual card values into integers.
Of course there is room for some optimization, but the above is already good.
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted June 01, 2009 02:15 AM |
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Reading through the posts it was difficult to see if anybody got the complete answer. (btw the assistant did look at the cards before picking them) Anyway, here goes:
Basically you assign every card a number, from 1 to 52. It doesn't really matter how. There's 4!=24 ways to order any four cards in the deck, so the assistant can pinpoint 24 different cards just by ordering the 4 cards in a certain way. That still leaves 24 options though (52 minus 4 minus 24). So the magician simply adds the magic number to the lowest card the assistant sends back and he can pinpoint the hidden card.
Remember the assistant can pick any card he wants, all he needs to do is pick a card that lies within 24 of the lowest card amongst the 5. If there's a situation like this: 2, 35, 37, 38, 39, then the assistant will send back the 2 knowing the magician will "roll over" the magic number.
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Binabik
Responsible
Legendary Hero
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posted June 01, 2009 02:48 AM |
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I just saw this.
"bring the 4 cards back to the magician, place them face down on the table"
If face down vs face up isn't determined until the assistant sends the cards back, then it doubles the 24 to 48 which will cover all possibilities.
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted July 09, 2009 08:24 AM |
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awrightey, ready for some mental calisthenics?
An orc slavemaster boards a rowing ship with 8 long chains in his hand. On the ship there's 8 gnomish rowers divided into 4 port (left) rowers and 4 starboard (right)rowers. The slavemaster holds out his hand and instructs each rower to take one end of a chain in each hand. The rowers are allowed to grab their ends from their side only, they aren't allowed to reach over to the other side and grab an end. The slavemaster announces that if they form one and only one circle, he will let them all go. Of course he is playing with them, he has no intention of letting anybody go. Still though, the gnomes get excited because they think their chances at freedom are actually quite good.
So the question is: What are the chances that the gnomes will form one circle?
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Binabik
Responsible
Legendary Hero
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posted July 09, 2009 09:41 AM |
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Edited by Binabik at 10:09, 09 Jul 2009.
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Off the top of my head I'd say 1:8, but that's probably wrong. There's a question of combinations vs permutations which I'm too lazy to think through.
I'll explain my crude logic after others have a chance.
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Warmonger
Promising
Legendary Hero
fallen artist
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posted July 09, 2009 12:58 PM |
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1/4. There are 24^2 different permutations of chains forming a circle and four times more permutations of gnomes grabbing chains in random order.
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted July 09, 2009 01:27 PM |
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My swift guess:
COS = 7!/8^8 = 0.03 %
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What will happen now?
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted July 10, 2009 08:26 AM |
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no correct solutions so far. I'll give you a hint, it's larger than you expected....
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted July 10, 2009 08:29 AM |
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Ok let me get the details right - he will let them each pick an end in each hand, but he won't let go of any of the chains before all goblins have picked their two ends - right on the highlighted part?
And would like to change my guess from above to:
COS = 60 * 7!/8^8 = 1,8 %
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What will happen now?
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