Has anyone tried performing this idea yet? It sounds really improbably and I'm really thinking of setting up a few decks for it.

I've only done it with three decks in identical order and never got around to making up the decks with writing on them. It can't be beaten for the small amount of work involved (just a false shuffle) for such great reactions.

/Tomas

1-(51/52)^52 is the correct answer.

Call the decks A and B. Draw a card from A. The probability that the first card of B does not match the first card drawn from A is 51/52.

Now, drawing the second card from A we want to know the probability that the second card from B does not match. This too must be 51/52 which is the probability that the second card from B resides in one of the OTHER positions - this INCLUDES the previously drawn card from B! There are 51 OTHER positions out of 52 possible locations for the card.

Repeat this process through the two decks and you get (51/52)^52 for the probability that no match occurs from which the correct answer follows.

Generally, using decks with N cards, we will find the probability of a match to be 1-((N-1)/N)^N which tends to 1-1/e as N -> inf.

Tide, I believe the correct answer was given at some point earlier in the thread. The solution you wrote unfortunately isn't correct and you can try your formula with for example 2 cards to see what happens in extreme cases.

/Tomas

Thanks, Tomas! I should have realized that or checked the simpler cases before posting - I did check before I read your post. I was looking for a simple explanation for my pre-precalculus students. Maybe we'll just do the simulation!

Tomas, I've been trying to do this without looking at the faces of the cards. Unfortunately, I accidentally write the correct name on at least one card approximately 63% of the time. Can you tell me why this is?

Jason

Ha!

I tried what Jason did, and I ended up writing the correct name on at least one card 100% of the time.

Turned out I was using one way force decks . . .

Jack Shalom

Very interesting thread! But, I'd like to repeat Mikey-Fly's post of above: Has anybody got any experience performing this in front of real people? How do they feel about all this dealing? How have they reacted? What can we do to maximize the entertainment value, and not get caught up in the mathematical value? Thanks.

Bob, I have only performed the convenient three deck version with no decks signed and only at quiet settings with people I know love analyzing. The results have been incredible, especially the few times they have to re-deal the decks. The first deal-through only seems to be a confirmation of the minimal odds I speak about in the presentation. Not sure what you mean by mathematical value since they never connect this with maths as they make no calculations and the decks are randomized.

I'm not sure I can find this working in any other type of setting. Possible in a mentalism show where things move at that pace.

I'd love to hear what others experience with these tricks. One of these days I'll make one of the marked decks... My guess/hope is that others have tried it by then.

Any feedback from performances is appreciated,

/Tomas

[quote]On 2003-09-25 03:11, TomasB wrote:


Well, you can go from 98 to somewhere near 99.96 by taking into account the "twins" (ie : 3H/3D). There you proceed for another kind of prediction based on these two cards, etc... Or you can use 52 card deck that have two 1H, two 2D, two 3H, etc ! It may go unnoticed if the deal is done real fast

But IMHO it's no use to use the twins or the "double deck" to go from 98 to 99.96 - much trouble for not much result -, but it may be useful to use them to go from 86 to 98, when you choose not to use the "top and bottom not shuffled together" approach, which prevent the spectator to do what he wants with the deck...

I have also two comments about this probability thread :

1) The famous "1-(51/52)^52" mistake. Well,
this is a common mistake because one is used to independent probabilities, while here we're dealing with combination : the fact you did or did not match the first card has some influence on the probability of further matches. Consider 4 cards, with one packet being 1234. If the other packet starts with a 1, there is a 4/6 probability to have a match in the three cards that remain. If the other packet starts with a 2, you only got 3/6 probability (the 1 and the 2 "that'll never be matched again" are an annoyance that doesn't compensate the fact you've got "less room" for 3 and 4...). So the probabilities are not independent.

In fact, "1-(51/52)^52" is the correct solution for the following problem "what is the probability of at least one match after 52 times "let's shuffle and look at the card at the rank the spectactor wants". But it would be incredibly slow to proceed like this ! ;-)

2) About the average number of cards you'll deal before finding the pair. 22 may be indeed the correct answer, but it is a quite meaningless one, alas : this is an "average" with a deviation that is so big that the deviation itself also is meaningless ! It's just like saying that "on the average", the 3H is at the 26th rank of a 52 deck. Actually it can be 1 as well as 52...

Someone found 22 instead of 26 by simulation because of the fact that there is many times several pairs that match, IMO. So choosing the "first match" when you've got a packet with two or three matches makes the average value go a bit down. If, for decks with several matches, you count them all, I'm pretty confident that you'll find 26 : just as for a single card, the place where you start to consider whether a pair of cards match or not as no influence on the result...

I have plans for performing this to friends who want to perform it for specs in a routine. Did that make sense? Soon I will perform this for magicians where it may make the cut for something in a routine (don't worry it's not professional...but I am planning the I.D. kicker with the third deck, although I may use T.W.'s I.U.D. instead. I will keep you informed of reactions.

I think there are many ways you can go with this:

Don't show the backs of the cards and go into a 'marked deck' comedy routine after 'Frequent Miracle'.

Don't name both decks the same way, and use I.D. to find three cards...the match, the name on card one, and the name on card two (there are some problems here but that's what makes it fun right?).

Or, not showing the back again find both the card in ID, and two cards representing it's position (face cards are zero, for a two digit number (or one if it is single digit). Alternatively a creative cut can place the card you find, at the position of the match (although this means you have to count the cards in an odd manner) in the original deck. In other words, the third card is found in ID but at the same position as the match was made in the original decks.

We will be experimenting this weekend.

M

Posted: Aug 22, 2004 11:20am
Ok, it was experimented with and made the cut. There is something we are quite happy with now...the stacks allow a single riffle a la gilbreath. Thus they can shuffle theirs as many times as you like, and I can shuffle mine once (maybe with some false as well), then we use the ID kicker, and move into a royal flush deal routine. This is shaping up nicely. The magicians I am working with are very happy that this maintains a stack and is a good opener to some gambling work. As an opener, it appears to show that probabilty works in mysterious ways for the rest of the evening, thus preparing our audience for the vastly improbably gambling routines they will see for the rest of the evening.

I will write more when we sit down to do a 'dress rehearsal'. So far so good...nice work on probabilities and ideas friends. I am very happy with this thread and all the cooperation involved.

M

Here is my approach:


Setup:
I take away all the card of one seed (e. g. Clubs) from the deck and spread them face up on the table, maintaining the random order in which they occur; then, I take the cards of the other seeds and set them in the same order of the Clubs ones, arranging them in packet. I cut the four packets so that the face card would be different for each seed and I re-assemble the deck.

I show th deck to the audience saying. "Here is a well mixed deck... er, in fact it is not so well mixed: the cards are separated in seeds". I ribbon spread face up and separate the four packets. The fact that they are cutted differently, and they are four, keeps enough undetectable the fact that the relative order of the cards is the same.

"we will play a game of coincidence in four. You, sir, will keep the "previsional packet". You can choose which one is it! (let's say A choose Clubs. I have to notice which card is in the face of the chosen packet).

"Well: now you and you (addressing to B and C) can take the packet you prefer; I'll take the remaining". Each of us now will shuffle his cards. I false shuffle, and then cut my packet at the same bottom card of A, that results in an identical packet.

"the probability that in a well mixed packet occur a coincidence dealing together the cards are really low. Let us try with you (addressing to A"

Now I deal together with A, counting aloud. Let us now imagine that a coincidence occurs on the third card (e. g. an Ace)

"Well: in spite of the odds, a little coincidence happened". Both of us have got an Ace, on third position". I take the spectator' ace and put in on the table.

I do the same with B; let us imagine a coincidens on the seventh card, a Queen).

"And here is another little coincidence: after an Ace in third position, now a Queen in seventh". I take spectator's queen and put it on the table.

Now I ask A to place his third card face down on the Ace and the seventh on the Queen, and I turn the two cards face up: two perfect matches.

A quick question on odds and probability, a subject that gives me anxiety attacks. If two people have 3 objects, (subject #1s objects are identical to subject #2s objects) what are the odds that both subjects would arrange their objects in the same order? How does on figure that out?

The probability that they will arrange their 3 objects in the same order is 1/6

One way is to simply list all the possible orders of 3 objects:

ABC
ACB
BAC
BCA
CAB
CBA

Since there are six possible arrangements, the probability of matching an arrangement is 1 out of 6.

Jack Shalom

Of course, if they arrange the objects in a circle, there are only two possible arrangements.

0pus

Thanks. I didn't think the probability was the same as the permutation, but I guess I was wrong.

While I am here...anyone mind checking a probability of OOTW? I have a debate going on another board, so another pair of eyes, checking the work for errors, always helps.

To clarify...this OOTW results in complete packets of the same colour. All red on one side, all black on the other. Here's the paper:

http://ohmslaw.org.uk/cards.pdf

So before anyone posts a link to another paper with a correct answer...please make sure it is the probability of the same OOTW. I am more concerned that this paper is correct BTW (or why it isn't correct), than I am that you have another proof.

Looks to me that you are very correct in your reasoning. Just to check it I made this thought experiment: A deck of cards can be arranged in 52! ways. The number of ways to get all reds or all blacks on top is 2[red or blacks]*26![number of ways to arrange the black portion]*26![number of ways to arrange the red portion] so the result is 2(26!)^2/52!. This should be multiplied by the probability that the spectator actually will deal 26 cards in each pile to get the case OOTW simulates.

One major flaw I find in OOTW is very clear now. The spectator can only separate the cards in colors _if_ he happens to accidentally deal exactly 26 cards in each pile. Having two piles next to eachother it's clear when he hasn't got the same number of cards in each pile and in those cases the result is illogical. I'd favour not using the complete deck so they are not sure of the number of reds and blacks among the cards being dealt.

/Tomas

Spot on Tomas. I have always noticed that spectators never keep track of the number of cards. The piles are always uneven, which means the probability is 0.
I have always thought this very funny, and comment on it when I perform this trick. I think it makes it even more magical

Thanks for the feedback. One day I will find that perfect spectator who only deals 26 into each pile.

M