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Tosca New user Hettimorth Tentenchuihnn, RGA 91 Posts |
R.E. Byrnes - if I understand you correctly, you are describing the following "Any Coin at Any Number" game:
There are two trays on the table, one labelled HEADS and the other labelled TAILS. The spectator is given a coin to flip and as it is flipping, he is asked to say which side of the coin he thinks will land face-up: heads or tails. This is a choice of 1 in 2. As the coin is flipping he is also asked which tray he thinks it will land in: HEADS or TAILS. This is a choice of 1 in 2. There are 4 possible outcomes: heads in HEADS, heads in TAILS, tails in HEADS, tails in TAILS. So he has a 1 in 4 chance of getting an exactly correct answer. This is true. But... the ACAAN scenario is NOT the same as this (even though it is tempting to think that it is). Instead, the ACAAN scenario is equivalent to the spectator first of all DETERMINING precisely which side of the coin will land face up - a choice of heads or tails. Let's say he chooses that the coin will land heads up (in real life he would do this by using a coin with a head on both sides). When he throws the coin up over the trays, he does not know which tray it will land in, but he KNOWS FOR SURE that it will land heads up. So now there are only two possible outomes: heads in HEADS or heads in TAILS. So the probability of his call being correct is 1 in 2. Let's look at this using playing cards. The equivalent of your coin scenario, using cards, is 52 trays, each labelled with the name of a playing card, and a 52-sided die with a playing card printed on each side. The player throws the die up over the trays and sees which side lands face up and which tray the die lands in. As the die is in mid-air, he calls the name of any card and the name of any tray. The chances of him being correct are 1 in 2074. But in ACAAN, it's not a 52-sided die that the player throws up. It's the equivalent of the two-headed coin in the coin game: namely, a precisely-named single playing card. It matters not one tiny bit WHICH playing card that is (in the same way that it didn't matter whether he chose a double-headed coin or a double-tailed coin to determine which side of the coin would land face up, as long as he made sure his call matched that choice). The important point is that it is a SINGLE, PRECISELY-NAMED playing card, and NOT a 52-sided die. And so the player throws up his single, specific, freely-chosen playing card (whatever that may be), and it lands in one of 52 trays. And the chances of it landing in its matching tray is 1 in 52. I hope this helps. |
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ekgdoc Regular user 110 Posts |
For R.E. Byrnes, here is a simple proof that is more mathematical than my previous argument. If a spectator chooses a card and number, then there are 52*52 possible outcomes (the sample space). If a magician pulls out a deck of shuffled cards, then this deck will hold 52 random samples of the (52*52) possible outcomes. Thus, the likelihood that their card is at their number in a shuffled deck is 52/(52*52). QED.
David M. By the way, I have a masters in biostatistics. |
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Tosca New user Hettimorth Tentenchuihnn, RGA 91 Posts |
Also - think of it like this: in the ACAAN scenario, there are 52 trays labelled 1 to 52, and one of them is named by a spectator. The player throws up a 52-sided die (with a playing card on each side) over the 52 trays, and as the die is in mid-air, he calls out the name of any playing card on the die. AT THAT MOMENT, the die instantly transforms from a 52-sided object into the specific, named, single playing card, which flutters down and lands in a tray. The chances of it landing in the named tray are 1 in 52.
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Tosca New user Hettimorth Tentenchuihnn, RGA 91 Posts |
Or - look at it like this: when the magician deals the cards, he is looking to see if the specifically-named card appears at the specifically-named number. This is NOT the equivalent of throwing a 52-sided die over 52 numbered trays, and seeing how it lands. Imagine: a card and tray number are named. The card die is thrown up and lands in a tray. The performer says, "ok - we were hoping to see the die showing the 3 of diamonds in tray 28, but - oh dear - we've actually ended up with the die showing the king of spades in tray 5". That's not the equivalent of ACAAN! In ACAAN, once the 3 of diamonds is named as the target card, and the number 28 is named as the target number, we simply look through the deck to see if the three of diamonds is at position 28 (just 1 out of only 52 possible positions for it!).
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Hideo Kato Inner circle Tokyo 5649 Posts |
Let me return to the initial topic of this thread.
I think the effect of ACAAN is good if performed almost perfectly like Lu Chen did at EMC. But there seems no method which produces such a perfect effect other than using stooge. Berglas method? His method can produce a near perfect effect sometimes and mediocre effect othertimes. So far I said "perfect method" on premise that magician never touch the deck after the card and number are called. If we don't take that premise, there are several good methods. For an example, if you are mastering Mem work, you can use memorised Instant Deck to situate the called card to the called number simply by Pass. BTW, if we use Inatant Deck, we can produce named card instantly fromm top, from pocket etc. Maybe the OP is suggesting such a direct effect is better than the effect of ACAAN. Hideo Kato |
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velcrowe New user 94 Posts |
Quote:
I could be wrong, but I think this is the source of the miscommunication. You each are assuming a different premise of effect. You are clearly correct that the actual raw probability of ANY card being at ANY position in the deck is 1/52, and that makes sense of the feeling the spectator gets is that the trick is only (or primarily) about the number. If the trick is presented as the performer having previously guessed which card AND which number the spectator would guess, then the probability of that seems to be 1/2704. Magic complicates the way we think about probabilities, as I pointed out in my first post here, because the *REAL* probability is 1/1, seeing as how it's rigged. |
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slyhand Inner circle Good ole Virginia 1908 Posts |
As to whether it is a good trick or not is entirely due to the performance. There are great tricks ruined by crappy performance as well as crappy tricks that are elevated to highly entertaining pieces of magic.
ACAAN can be boring or it could be an exciting trick depending on the skill of the performer. Of course if one hates classical music then no amount of presentation is going to sway that persons taste.
I am getting so tired of slitting the throats of people who say that I am a violent psychopath.
Alec |
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AMcD Inner circle stacking for food! 3078 Posts |
@R.E. Byrnes
Let's illustrate. Say we have a deck made of 4 cards: 1, 2, 3 and 4. How many deck orders we have? Hey it's 4! = 24. Here are the detail: 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 4321 Now, what's the probability say card 1 is at position 2? Just count the cases above, we have 6 cases. Thus 6/24 = 1/4. With formula, when you select one card at one position, you have 3! cases possible, hence 3!/4! for the probability. For N cards, (N-1)!/N!. Now, take 52 cards, you will find what? 51!/52! = 1/52. |
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Herr Brian Tabor Special user Oklahoma City 729 Posts |
Just my two cents.
The problem with Tosca's explanation is this- WE know that, the spectators don't, and I suspect they're not going to go home and work it out. Furthermore, they know we are magicians, and usually will view it as skill, no matter what we'd like to think. I think this is one of those tricks magicians get fascinated with, but no one else. We are intrigued because of it's difficulty, based on our inside knowledge of the trade. Laymen, however, don't have this knowledge, and aren't as impressed with it because they don't have our knowledge. To them, it's just a regular effect, no harder than anything we've done. In fact, the very act of saying it has a 1 in 2074 chance means it can easily be reconstructed later as luck. Let's compare this to something simple like a card change. If you follow quantum mechanics, you know that the probability of a card changing into another card, (and in most cases their card) is such a big number you could wrap it around the earth a hundred times. If you compare the two numbers, 1 in 2057 is relatively tiny. Besides, most laymen have experienced luck in their life, but more than likely, none have experienced an object impossibly changing into another. It's all relative, and the magic is for the spectators, not us. No matter what we know, or like, it's what they like and know that make an effect. It's a good trick, but it's not something they are going to remember the rest of their lives. I've had people years later tell me they remember the sponge balls, or that time I changed their card, etc, but never have I had one say "I remember that time you made my card appear in the deck at a number I named". It's all about perspective. |
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Herr Brian Tabor Special user Oklahoma City 729 Posts |
Just emailed my statistics professor at Marshall University, and the odds are, in fact, only 1 in 52.
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AMcD Inner circle stacking for food! 3078 Posts |
Asking professors or showing off with degrees is useless. You have to explain and convince, which is far better.
Anyway, I did it above . In short, if you ask someone a card and a position you have 2074 possibilities. But, as shown above, there is only 1/52 chance that a card lies at a certain position. They are two different things. What about talking about something else now? |
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Hideo Kato Inner circle Tokyo 5649 Posts |
We don't need a complicated proof of probability as any card at any number possibility is 1/52 without proof.
Hideo Kato |
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MeetMagicMike Inner circle Gainesville Fl 3501 Posts |
I actually love this kind of discussion. It's not easy to change some ones mind once they dig in.
So here is yet another attempt to explain it: In ACAAN* there are 52 cards all of which are capable of being winners. Someone names a card and then tries to guess it's position. No matter which card they pick they can win if they guess it's location. Interestingly, after the effect has concluded you would be correct in saying "What you have just witnessed will only happen 1 out of 2074 times. However these are not the chances of success in ACAAN because there are 52 winning scenarios even though you only witnessed one. The chances of that particular outcome are 1 in 2074 but the chances of A successful outcome are 1 in 52. *The Effect of ACAAN is simply this: A deck is placed in view. A spectator names any card. Another spectator names any number from 1 to 52. The cards are dealt (and counted) face up onto the table, and the named card is found at the named number (position) in the deck. - (Thanks Tosca) |
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AMcD Inner circle stacking for food! 3078 Posts |
Without proof? Well, why some people don't get it then ?
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R.E. Byrnes Inner circle 1206 Posts |
One last try. The spectator isn't choosing from among only the 52 pairings in in the deck. Rather, he's choosing from all 2704 possible card/number pairings. A-1,A-2,A-3 ....K50, K-51, K-52 Thus, from the outset, there are 2704 different thing the spectator can say.OnSet 52 of those total 2704 possible pairings are in the deck, though. Accordingly, the great majority of the possible pairings that fouls be named, aren't even in the deck that's being used. Much as there are four ways a consecutive coin flip can turn out (not two, not three), there are 2704 possible card-number pairings a spectator can offer, as he is not constrained to pick from among only the 52 in a particular deck.
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Herr Brian Tabor Special user Oklahoma City 729 Posts |
I wasn't "showing off" with degrees, I don't have a degree. I just asked someone who does, so at most I was showing off my ability to use e-mail.
As far as "proof" - Once position of the card is determined, there is a 1 in 52 chance that the number will be correct, and vise versa. In other words, if you pick a number first, that number becomes a constant, and the card identity is a variable, and vise versa. You don't multiply a variable with a constant to get a statistic, only two variables. One constant is represented as 1. The variable in this case is 52, giving us 1 in 52. The only way mathematically for the probability of the ACAAN trick to be 1 in 2074 is if you count the probability of it happening twice in a row! If math isn't proof I don't know what is lol. Besides, my original point is that laymen aren't going to think about this effect like this anyway, and to them it can be dismissed as luck. |
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Herr Brian Tabor Special user Oklahoma City 729 Posts |
Another way to get the odds of 1 in 2074 is to have predicted the number they would name as well, making both variables.
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R.E. Byrnes Inner circle 1206 Posts |
My error. I assumed a specific outcome, and thus talked everyone in circles. Sincere apologies all around for getting heated and blundering ther very premise of the trick.
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MeetMagicMike Inner circle Gainesville Fl 3501 Posts |
R.E. Byrnes Wrote:
Quote:
Rather, he's choosing from all 2704 possible card/number pairings. A-1,A-2,A-3 ....K50, K-51, K-52 Thus, from the outset, there are 2704 different thing the spectator can say. Perfectly said Quote:
52 of those total 2704 possible pairings are in the deck, though. Perfectly said So, 2704 possbile pairings. 52 of them "win". Therefore: Possibility of winning = 2704/52=52 (This is ACAAN) Possibility of winning with a particular pairing = 2704 (This is ACAAN where the card has a red back) So we agree, R.E Byrnes? (If not which part do you not agree with?) |
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R.E. Byrnes Inner circle 1206 Posts |
To be clear, I thought about this the wrong way, for all the ways noted, and then some. I was flat out wrong. I appreciate that being tolerated so long. Now I need to go find of if it really is true that Nobel Prize winners were similarly arrogant and insistent about the wrong solution to the Monty Hall Problem. even if they only won in literature or peace, it would be some consolation. Again, apologies to all those to whom I was dismissive. For what it's worth, you planted the seed that something was off, enough so that I was going to figure out one way or the other, with the objective of owning up to it if it turned out I was wrong. Unequivocally wrong.
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