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B. T. Lewis New user London 59 Posts 
I've been reading through this thread and it's fascinating. Tomas, I was wondering if amongst all your calculations you worked out what the likelyhood of getting a match on the value of two cards when dealing alternately through two different suits from a deck? Also, if anyone else know what this probability is, I'd very interested to know. Also, Tomas, eariler on you stated
"Deal cards alternately face up to the table, starting with a card from his deck, counting loud for each card dealt from your _own_ deck." I can see why you need to count for every one of your own cards for the purposes of the third stacked deck/suit, but why does the participant have to deal first? Does he? 

TomasB Inner circle Sweden 1144 Posts 
It was such a long time ago I wrote that, but I can guess it would make it logical the times you get the match when _you_ deal the matching card, since the spectator started dealing.
Since you call out the numbers, it's also logical that the match happens when the spectator deals a card after you called out a number. They might even forget who started dealing due to your counting. If you instead start dealing and start counting it seems like the order of the matches is more specific than if the spectator starts dealing. It'd feel like the match _should_ be when the spectator deals his card. I just wanted to avoid that. As for the question regarding simply hitting a value, I have to simulate that and get back. /Tomas Posted: Nov 15, 2010 3:17am  One million simulations says: If dealing at the same time from two decks, the probability of matching values is 0.984. If dealing alternately the probability of matching values is 0.994. /Tomas 

B. T. Lewis New user London 59 Posts 
Ah, sorry. Guess I wasn't clear. I mean if I have one packet of the 13 hearts, for example, and one of the 13 spades. Assuming both packets are separately shuffled completely fairly, what are the odds of two cards with the same value showing at the same time? (I'm assuming each packet  or at least the participants will be shuffled fairly as one, without shuffling each half of the 13 separately). I really appreciate you working that out for me though, it's just that I'm trying to come up with an impromptu version using 3 suits which I can stack on the fly while I take them out of the deck. The other advantage to this method is that it only takes 1/4 of the time to deal through the cards so if it doesn't work on the first try, it's not such a huge upheaval to go again and use the first time as an example of the unlikelihood of a match of values happening. Thanks anyway though.


TomasB Inner circle Sweden 1144 Posts 
I just wasn't reading correctly.
As mentioned earlier, the series converges quickly so it doesn't change much when adding or removing cards. With 13 cards and simultaneous dealing the probability is still 0.632 and for the alternate dealing the probability is 0.865. One million trials gave those numbers. I'd be very interested in hearing how you envision the trick and if you plan to do anything with the fourth pile or simply put it away. A hit in a greater number of cards seems more improbable, and a prediction matching it also seems more improbable, but you can't beat a onedeck handling for convenience. /Tomas 

B. T. Lewis New user London 59 Posts 
The effect I've got in mind at the moment  though it could easily change in time  is where the participant chooses a suit (say hearts) and I remove it from the deck for them. As they shuffle it, I remove two more suits (thereby leaving the third automatically separated) and they choose which one I should use (lets imagine it's spades). I leave the last 2 suits at the side and after they and I have both shuffled our portions, we deal them alternately, waiting for a match of values.
One arrives, and then the participant counts down to the same position in either of the two remaining suits, and the value is again a match. I ask them to find the matching value in the fourth packet at the same position, but it does not appear. In fact, the card with that value is not even present in the remaining suit. I ask them to reach into my pocket, and of course, there the card is. A bit convoluted in description, perhaps, but I think it could play out quite nicely in performance. I don't know precisely how I would achieve the last part with the vanishing card, but as the wonderful Derren Brown has said, it's best to work from the top down, starting with the effect and working out method afterwards, so that's what I plan to do. Also, I think the effect would be far stronger with three decks as you have previously described, but I like the idea of a one deck version which is potentially impromptu, with a couple of other advantages over the 3 deck version. I also believe that it makes more sense for the performer to start dealing; it just seems more natural to me. If there is a match when the participants card is dealt, then there truly is a match at card number n. If there is a match on your own card, then the match happens at the same time you state "7" or whatever number you're on. Your way (which I'm not trying to be disparaging about, just stating my observations) means that if there is a match on the performers card, you have both these advantages. Which is fine. But half the time, you have neither of these factors; the card is not actually at the same number down, and it does not happen as you announce a number. It just strikes me that if I were a participant in this situation, it would seem more natural to notice that when the performers card was actually dealt and the piles made even that there was no match. Also, my way, if the match happens on a card which is one position out, i.e. on the performers deal, then you can honestly say "the match has occured on my 7th card" with no need to worry about what position their card is at  only that there is a match. By the way, I'm finding it very difficult to articulate just why it feels more natural to me, I hope I'm getting my point across. It probably doesn't really matter  I imagine both ways would work fine in practice. Anyway, there you go. Peace. 

TomasB Inner circle Sweden 1144 Posts 
I very much like the kicker of the card missing from the last packet. Since magic happens to both last packets it's not much need in having the spectator choose between them. Instead I think this could work as method (if ironed out a bit):
The fourth packet you have in the same order, but with the top card shifted to the face of the packet. This is placed in the card box, just to give it some uniqueness and state that you'll use this last. In the very end of the trick, have the spectator remove the packet from the case and deal cards onto your hand face down, and turn the card at the lucky number face up to display to all. In fact, the last card he dealt in your hand is the matching card so you can palm it as all focus in on him displaying the face of his card. Drop your cards on top of the rest of his cards to have him check all of them. During this you can load the palmed card somewhere, or pretend to take it out of the case or pocket. By "ironed out" I mean that there might be a better way to handle extreme cases. If you secretly pocket the top card of the fourth packet instead of moving it to the face of the deck, the result might be better: If the match is on One, you are already set for the ending. If the match is or 11, 12 or 13 the spectator will note that there are only 12 cards in the packet, which is perfect. When he looks through the cards after you have stolen the correct card, there will only be 11 cards left, of course. But he focuses on finding that specific card and won't note that another value is missing. Add the previous top card as soon as you can in case someone wants to check the packet further. Further ironing out is needed if the number is Two, since it'll be obvious if you palm the only card you hold. I suggest using Lennart Green's Snap Deal to seemingly deal your card to the top of the spectator's packet. Any other ideas along these lines or totally different approaches? /Tomas 

B. T. Lewis New user London 59 Posts 
Hey Tomas, Love the idea of the one card displacement in the cycle  very nice. I can't really add to that too much  you seemed to have covered the possibilities for this particular effect fairly comprehensively.
Posted: Dec 9, 2010 11:51am  Quote:
On 20030911 05:37, TomasB wrote: Tomas, I recently performed this effect for a friend, and since it's essentially your effect I thought I should let you know how it went. Firstly, there was no match after the first deal through the two decks. My friend, however, is a very bad shuffler and this meant that even though there was no match on the first try, there were several runs of 510 cards in a row which all had the same value. This seemed to freak him out anyway, as you might expect. The match came on the second deal at card number 44 (at which point I was starting to worry! Haha). When the third match was revealed he found it extremely powerful and when I asked him how he felt he said something along the lines of "I just don't see how that could ever be possible" which I was happy enough with, so thank you. Some of my beliefs which differ from yours have been reinforced, however. If you deal your card first and say "one", then he deals his card and says "one", you deal the second card and say "two", etc. Then whenever there is a match, the last card dealt will be at the matching position (which was just said out loud by the person who dealt the card) in the third deck. If they deal their card first then this will only happen on half the possible matches. Also, when dealing, my participant started to do it very quickly and this meant it was very difficult to keep track of things. If he is noting the number of the card being dealt at each point I'm sure it would help keep him moving slowly enough to make things easier to handle. Hope you find all of this constructive. Peace, Ben. Posted: Feb 16, 2011 7:22pm  Hey Tomas, got another question, haha. I'm planning on performing this effect but with a completely different methodology (for my own reasons)  What is the likelihood of a match in the first 21 cards? Thanks, Ben. 

TomasB Inner circle Sweden 1144 Posts 
Probability 33.9% to get a match within the first 21 pairs, when dealing simultaneously from two 51card decks.
/Tomas 

B. T. Lewis New user London 59 Posts 
Quote:
On 20030911 05:37, TomasB wrote: I've just a couple of stories for everyone about when I performed this effect for a few people recently. I was back in London over Christmas and decided to show this effect to my girlfriend. We shuffled the cards, and I dealt mine first. She dealt hers, and it matched. She had shuffled the top card exactly back in to position. A one in fifty two chance, so I said "...and if we deal the next card you'll see it might not necessarily be the same"  she dealt the card, and it matched again. This is obviously due to people's tendencies to shuffle cards in blocks. She uncased the third deck and was astounded. Brilliant reaction. I left it there, not wanting to push my luck. Not only that, at a later date I was performing the same piece for my brother, and exactly the same thing happened! The top cards matched again! I thought I might as well try my luck again and we each dealt our next 2 cards  both of which matched! The effect is impossibly strong like this as it seems as though this climax was intended from the start  they shuffle the cards and without you touching them the top 2 or 3 (or more!) cards are an exact match with yours, also including a match with the untouched third deck! Not that anyone can engineer this  I just thought it was a stroke of great fortune that it happened in two consecutive performances. :) 

Dark Knight Loyal user My troops are stationed at 272 Posts 
The odds of showing this trick and having the spectator shuffle the top two cards back into position via careless or block shuffling is....
Three things: 1. In the book, The Magic of Francis Carlyle, there is a description of this twodeck matching card method to identify the random card later to be revealed in an ID or BW. 2. The trick, according to J. Scarne, was a famous "proposition bet" used by the notorious gambler, Titanic Thompson. 3. There is a twodeck coincidence trick in which any card the spectator thinks of in the first deck, an ordinary deck of 52 different cards, is the only card to turn up at the exact same time as its mate in a second deck when both decks are dealt face up by the Cardician and the spectator (or by two spectatorsthe Cardician need not do any dealing). I vaguely think Paul Curry is connected to this trick. I used to do it using a "It Can't Be!" presentation from Barry Richardson's Theater of the Mind. DK 

newguy Elite user 401 Posts 
All of this is most enjoyable, wanted to add this note of historical import.
The staggered dealing idea is great. Unfortunately, it's not new. It was put forth by Bill Elliott, and sent to Howard Lyons for Aziz magazine prior to 1980, under the title "Synchronicity!". It wasn't published until 2002, when the large volume appeared, Ibidem 3, Aziz, & Beyond. 

FrankFindley Special user 763 Posts 
Quote:
On Nov 15, 2010, TomasB wrote: I went ahead and did the exact calculation just for giggles. Tomas wasn't kidding about converging quickly! I think I hit the limit of my software's degree of accuracy. Cards Probability 1 1.00000000000000 2 0.50000000000000 3 0.66666666666667 4 0.62500000000000 5 0.63333333333333 6 0.63194444444444 7 0.63214285714286 8 0.63211805555556 9 0.63212081128748 10 0.63212053571429 11 0.63212056076639 12 0.63212055867872 13 0.63212055883931 14 0.63212055885078 15 0.63212055885002 16 0.63212055885006 17 0.63212055885006 18 0.63212055885006 19 0.63212055885006 20 0.63212055885006 21 0.63212055885006 22 0.63212055885006 23 0.63212055885006 24 0.63212055885006 25 0.63212055885006 26 0.63212055885006 27 0.63212055885006 28 0.63212055885006 29 0.63212055885006 30 0.63212055885006 31 0.63212055885006 32 0.63212055885006 33 0.63212055885006 34 0.63212055885006 35 0.63212055885006 36 0.63212055885006 37 0.63212055885006 38 0.63212055885006 39 0.63212055885006 40 0.63212055885006 41 0.63212055885006 42 0.63212055885006 43 0.63212055885006 44 0.63212055885006 45 0.63212055885006 46 0.63212055885006 47 0.63212055885006 48 0.63212055885006 49 0.63212055885006 50 0.63212055885006 51 0.63212055885006 52 0.63212055885006 

glowball Special user Nashville TN 657 Posts 
I like this thread.
Am thinking about doing the same trick but instead of an exact match, match on the same value and the same suit color (not necessarily the exact suit). My brother David (phd in mathematics) says that would raise the odds of a hit from over 60% to over 80%. He gave me the actual percentage which I forget and the actual formula which I think had something to do with 1 over e for the misses and then subtract from one or something like that. Another option: I don't like carrying multiple decks to do this trick so am proposing to do this trick with one deck and match on value only by cutting the deck in half so that the top half is given to the spectator to shuffle and the bottom half is false shuffled by the magician and there is an extra single prediction paper/card openly set aside face down that has 26 numbered rows that match the magicians bottom half deck. Of course each half of the deck has all 13 values occurring twice. The prediction card could look like this: 1. Black 9 2. Red queen 3. Red 2 4. Black 6 5. Red ace 6. Black 7 7. Black 3 8. Black king 9. Red 10 10. Red 8 11. Black 5 12. Red jack 13. Black 4 14. Black 9 15. Red queen 16. Red 2 17. Black 6 18. Red ace 19. Black 7 20. Black 3 21. Black king 22. Red 10 23. Red 8 24. Black 5 25. Red jack 26. Black 4 To fit all this on one blank card put the first 13 rows in column one and then have a second column that has rows 14 through 26. Or just print this one card off on cardstock via Google documents. Option A: Note that rows 14 through 26 on the prediction card are an exact match to rows 1 through 13 on the prediction card. This means the top half of the deck (the spectator's half) will only have black nines, red Queens, red twos, black sixes etc. Option B: Of course you could construct a deck and prediction card to match the magician's half deck and that way you could have row one say "Black 9" and row 14 say "Red 9" etc giving a better look to the prediction card and better look to each half deck however I think the trick can be presented more impressively if the prediction row matches the spectator's card suit color therefore I believe option A is the better option. Hmmm, well, the prediction card could specify the value and the precise suit of the magician's half when doing option B! Maybe doing option B is better (with the prediction paper/card specifying the exact card value and suit). Note that for both options A and B that rows 14 through 26 on the prediction card can be mixed around versus rows one through 13 (as long as the magician's 26 cards are set in sync with the prediction 26 rows). For "option A" the suit color of the magician's 26 cards will be opposite the prediction color and this is fine). Others have proposed splitting the deck into suits and just matching on value but I think matching on value and suit color is more impressive. Note that the top half of the deck can be premixed or set to have a random look so the deck can be spread face up before giving the top half to the spectator to shuffle. Over 80% of the time you will be successful but if you are not then do as Tomas suggested and do the trick again (the good news is you only have 26 cards to deal through instead of 52). 

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