The Berglas Effect

[Bill Mullins]

The odds of ACAAN working correctly by random chance are 1 / 52.

Yes, the spec can pick one of 52 cards, and yes, he can pick one of 52 numbers. So the denominator of the fraction is 52 * 52 or 2704. But the numerator is 52 there are 52 different ways it can work out right for the magician. Suppose the deck is in new deck order. If the spec calls out AH and 1, the trick works. If he calls out 2H and 2, the trick works. Etc., etc., until you get to AS and 52. The same logic works for any order of the deck Aronson stack, Berglas stack, Nikola, Si Stebbins, eight kings, random shuffle, alphabetical order, whatever.

There are 2704 possible combinations that the spec can call out. 52 of them "work" in the context of the trick. 52/2704 = 1/52.


But suppose John Bowden predicts in advance, with a sealed envelope, what the results would be if a test was the following:
Person A shuffles a deck into a random order and places on the table. No one touches it.
Person B calls out a card at random say it is the 8C.
Person C calls out a number from 1 to 52 at random say it is 36.
Person D looks at the 36th card in the deck.
All people act independently this is, as much as possible, a fair test.

If John's prediction is "the trick will work", he has a one in 52 chance of being right.
If John predicts 8C and 36, he has only 1/2704 of being right.

The distinction between the two is the reason the odds are confusing. The magician, when performing the trick, is essentially making the first of the two predictions (or, in other terms, he is making the second prediction 52 different ways, and only being tested for one of them).

[El Mystico]

If anyone is still interested: I urge you - get a friend to play 'Suits' with you ('Suits' is a fabulous new game I've just invented). They call a suit, and you, (or a second friend - if you have one) call a number.
You then count down in the deck to see what suit is at that number.
Try this 20 times.
I'll bet you they will be right in about 1 in 4 times.
And not 1/4 * 1/4 = 1/16 times.

(But, if you were to have correctly guessed beforehand what suit they would call, and what suit would be at that number - that would 1/16)

In terms of Joe's example above: whatever John calls, he will be correct 50% of the time.

In the same way, whatever card Richard names, the chances it will be at Matt's position will be 1 in 52.

(Please appreciate thst I make no smutty innuendos about Matt's position)

[El Mystico]

Bill - good argument! Let's see if people understand it!

[Larry Horowitz]

Open letter to Persi Diaconis,

NOW, would be the time to chime in!

[John Bowden]

You know what........I blame the educational system.

Cheers
JB

[El Mystico]

You know what........I blame the educational system.

Cheers
JB

Me too! There are times when i really appreaciate that I got an honours degree in maths and stats!

[Richard Kaufman]

I don't blame the education system: I'm just an idiot at math.

[John Lovick]

For a long time I thought the probability of ACAAN was 1/52*52, but then I had a Ph.D. genius friend of mine explain it to me.

If you were to have a written prediction that said what card the spectator names and the number they call, then the probability is indeed 1/2704, but your prediction is a deck of cards, which covers 52 possibilities, therefore the probability is actually 52/52*52 or 1/52.

(This is why onevery impracticalmethod is to have 52 decks, which would cover all possibilities, and you switch the correct deck in).

Its easier to understand if you imagine only four cards in the deck. There are 16 combinations, but you would only need four decks to cover the possibilities. Heres the illustration. Imagine a deck made of A, 2, 3 and 4. There are 4! (24) deck orders.

A234
A243
A324
A342
A423
A432
2A34
2A43
23A4
234A
24A3
243A
3A24
3A42
32A4
324A
34A2
342A
4A23
4A32
42A3
423A
43A2
432A

The probability that any card is at any position is 6/24 or 1/4, as you can see above (count the number of times A is in position 3, for example).

Any card at any position has 3! possibilities, and the probability is 3!/4!. For N cards it is (N-1)!/N!

With a 52 card deck, that is 51!/52! = 1/52.

[El Mystico]

Thank you, John.

As people become convinced, it will be useful to hear which argument convinced them.

And in the intesret of fairness, for those who aren't convinced by any of this, it will be intereting to hear what aspects of arguments like those presented by John and Bill they don't believe.

[El Mystico]

Oh, for what it is worth; here is an argument my daughters understood.
I've got ten cards, numbered 1 - 10.
Olivia picks a card number: from 1 - 10.
Annabel picks a position number: from 1 - 10.

With only ten cards, there is only a 1 - 10 chance of Olivia's card being at Annabel's position.
(Let's be specific: Olivia's card might at at position 1. It might be at position 2. It might be at position 3...need I go on? There are only ten positions it might be at. If anyone thinks there are more, please shout). So the chance it is at the position Annabel is thinking of is 1 in 10.
If anyone thinks it is 1 in 100, please let us know what those other 90 positions are.

But - the chances I predicted that Olivia would choose 6, and that Annabel would choose 8; now, that is 1 in 100.

[erdnasephile]

John:

That is really good--makes sense given the "prediction" is a 52 card deck.

However, I was wondering about this statement: "(This is why onevery impracticalmethod is to have 52 decks, which would cover all possibilities, and you switch the correct deck in)."

Wouldn't you need more decks than 52 to cover all cards in all positions?

[John Lovick]

No. 52 decks would do it, because each deck covers 52 possibilities. Therefore 52 decks would cover all 2704 (52*52) possibilities.

Imagine the first deck is in new deck order. For the second deck move just one card from the bottom to the top, for the third deck move the two cards from the bottom to the top, and so on. Do you see how this rotates each card through each position over the course of the 52 decks?

[erdnasephile]

Aha! Got it--thanks.

The concept of each deck serving as 52 separate predictions is interesting to think about.

[Joe Pecore]

But - the chances I predicted that Olivia would choose 6, and that Annabel would choose 8; now, that is 1 in 100.

But isn't that the differences in ACAAN and CAAN.

CAAN = I'm predicting what number you will choose (1/52).

ACAAN = I'm predicting what number (1/52) and what card you will choose (1/52)?

[Bill Mullins]

But - the chances I predicted that Olivia would choose 6, and that Annabel would choose 8; now, that is 1 in 100.

But isn't that the differences in ACAAN and CAAN.

CAAN = I'm predicting what number you will choose (1/52).

ACAAN = I'm predicting what number (1/52) and what card you will choose (1/52)?

Not quite. If I perform ACAAN and reveal the mystery with a deck of 52 cards, I haven't predicted your number and card. I've made a revelation that confirms your choices. However, I've also got in waiting 51 other revelations, ready to go, which could confirm 51 other pairs of cards/numbers.

[John Bowden]

Joe ............we may be WRONG.

I have my cousin with me who has explained it way better than anyone else here has.

El Mystico's explanation is the one that best made me realize that the mistake I made was in not realizing that there were 52 correct permutations in the 2704 possibilities in any given deck.

With that I think it is now time for my nervous breakdown which will be a consolation to those who tried to explain to me the errors of my mathematical thinking.

Now also let me make one thing clear................JUST because I am wrong now, and I admit I am, it does not mean that I will be wrong in the future. It is a unique event in the history of both magic and mathematics.

Of course there is the possibility that I was just playing around with everyone, now what are the odds of that?

Cheers from the Emerald Isle,
John Bowden.........a very much humbled magician

PS As most of you know from looking at my profile that we don't have street numbers or Postal/zip codes you will realize that numbers are not our strong point.
JB

[Joe Pecore]

Not quite. If I perform ACAAN and reveal the mystery with a deck of 52 cards, I haven't predicted your number and card. I've made a revelation that confirms your choices. However, I've also got in waiting 51 other revelations, ready to go, which could confirm 51 other pairs of cards/numbers.

Ah. That makes sense.

[John Bowden]

Joe what is your excuse.............?

Mine is that I am getting old and this is the first time I ever made a mistake.

Unless of course my memory is going as well and I might be making mistakes every day and I am just forgetting about them.

Cheers all
John Bowden

[Jonathan Townsend]

There was a trick on the market where you seem to have one regular card in a pack of blank both sides - the named card at the named number.

[El Mystico]

A thread with two topics, where all the magicians agree!
I can sleep happily.

[Joe Pecore]

Joe what is your excuse.............?

I think the equation I picked (probability of 2 independent events) did not apply.

Picking a number and picking a card are two events which does apply and is 1/52 * 1/52.

Picking any card at any number is just one event (it does not matter if I pick the card and you pick the number or if you pick both the number and card).

- Joe

[Jonathan Townsend]

IMHO it's the suggestion, perhaps implication, that you must have known both items is what drives the larger denominator in this discussion.

[erdnasephile]

...If I perform ACAAN and reveal the mystery with a deck of 52 cards, I haven't predicted your number and card. I've made a revelation that confirms your choices. However, I've also got in waiting 51 other revelations, ready to go, which could confirm 51 other pairs of cards/numbers.

Yes, that's a good way to think of it.

To me, the easiest way to understand the probabilities is that if I am making a single guess at the card/position, the odds are 1/2704; however, the deck of 52 cards represents 52 separate guesses made all at the same time, so the odds drop to 1/52.

[erdnasephile]

The other thing I think is interesting is I suspect that ACAAN would have been perceived as less likely to happen by chance than Scotty York's Quintuplicate Coincidence by most audience members.

However, the odds (as calculated by Chip Denman) for the outcome of the latter effect are 712,757 to one.

[Pete McCabe]

The odds of Gemini Mates are 2,450 to 1.

[erdnasephile]

Hi, Pete:

I was quoting the odds listed in the York instruction manual for this effect. After reading your post, I calculated it out myself and concur with you.

The text says that Chip Denman is the manager of the Statistics Laboratory at the University of Maryland so I'm assuming he knows what he is doing here.

I suspect that he's taken some license and multiplied out the odds of multiple individual events that are actually fulfilled all at the same time.

Even given that, I think my earilier point holds: Gemini Twins is statistically less likely that ACAAN, although to me at least, I would have thought otherwise.

[Dave V]

And you believed him? It's called "patter." At the time he published my handling, he was working in a bar. If he did his job right, the patrons weren't exactly thinking clearly enough to challenge him over the math.

[erdnasephile]

Point well taken-- the non-patter claim in the instructions that: "And by the way, the probabilities are quite accurate--if they seem counterintuitive, well, that's probability for you!" threw me. :grin:

(Do you use any of those odds in your patter, Dave?)

[Richard Kaufman]

John of the Emerald Isle: you can always do what I do. If the question has anything to do with math, I immediately defer to ANYONE else in the vicinity.

[Pete McCabe]

The odds of the traditional Out of this Worldwith aces used as indicatorsis one in 2^48 or 281,474,976,710,655 to 1.

[Jonathan Townsend]

The odds of the traditional Out of this Worldwith aces used as indicatorsis one in 2^48 or 281,474,976,710,655 to 1.

How did you get that number?

If there are fifty two cards, of which twenty six are red and twenty six are black - ... and you deal them down into two piles of twenty four cards - then one could as well argue that there are not forty eight choices to make but only twenty four. The odds for twenty four choices are 2**24 (or 16,777,216) to 1 against.

[Smurf]

"...Now on to Smurf you are correct for CAAN but for ACAAN there are two unknowns. One unknown is which card will be chosen and the other which number will be chosen. Two equal probabilities.
The fact that the card is already in the deck does not diminish in any way the number of cards in the deck any more that knowing that it is in one of fifty two positions diminish the number of possible positions. It is still 52*52 and that is still 2704 and I am nearly blue in the face telling you that. I'm turning into a smurf myself."

John,

I'm glad you have seen the light so you can catch your breath and I can retain my status as the only Smurf on the Genii forum.

[Joe Pecore]

Would this be a factual statement for a magician to say at the beginning of an ACAAN effect?:

Deal down four cards and turn it over to reveal the card at that location which happens to be the Queen of Clubs and say "The odds of someone choosing both the Queen of Clubs (1/52) AND the number 4 (1/52) is 1 in 2,704."

[Bob Coyne]

The odds of the traditional Out of this Worldwith aces used as indicatorsis one in 2^48 or 281,474,976,710,655 to 1.

How did you get that number?

If there are fifty two cards, of which twenty six are red and twenty six are black - ... and you deal them down into two piles of twenty four cards - then one could as well argue that there are not forty eight choices to make but only twenty four. The odds for twenty four choices are 2**24 (or 16,777,216) to 1 against.

Actually neither of these is correct since the probability of picking successive cards of the same color decreases as correct choices are made and cards are removed from the pool. It would only be 1/(2^24) if the 24 choices were made from an infinite deck.

Given a fixed deck size, the first choice would be 24/48, the second would be 23/47, the third 22/46,..., the last choice would be 1/25. So the probability of selecting 24 of the required color would be 24*23*22*...*1/(48*47*46*...*25) = 24!*24!/48!

Another way to look at it is that the formula for the number of combinations of n items chosen k at a time is n!/((n-k)! k!), so the probability of getting any particular combination (eg all reds or blacks) would be the inverse of that.

[Jonathan Townsend]

I'm not convinced the selection (n choose m) model is applicable in this case. The cards are dealt into one of two piles. Each choice red/black, right/wrong seems more of 2**n item.

Or perhaps letting them move cards between the piles after dealing as they please to make them into 24/24 is what makes it a selection (therefore n choose m) situation?

Help?

[Bob Coyne]

Jonathan, just consider a very minimal case where you have two red cards (r1, r2) and two black cards (b1, b2).

There are twelve possible, equally likely outcomes: (b1,b2) (b1, r1) (b1, r2) (b2, b1) (b2, r1) (b2, r2) (r1, b1) (r1, b2) (r1, r2) (r2, b1) (r2, b2) (r2, r1).

Two of those pairs hit the jackpot -- all reds or all blacks, depending on what you're looking for. So the probability is 1/6 (as per the n choose k formula), not 1/4 (which you would get if the choices were independent (1/2^n) and didn't influence each other...as with an infinite deck).

[Richard Kaufman]

The actual author of this mathematical conundrum is Wilber Sanders.

[Dave V]

Point well taken-- the non-patter claim in the instructions that: "And by the way, the probabilities are quite accurate--if they seem counterintuitive, well, that's probability for you!" threw me. :grin:

(Do you use any of those odds in your patter, Dave?)

No, not really. It's more of an approximation. First phase: [color:#CC33CC]"There are about ten different deck patterns making it around 10:1..."[/color] Same line for the second card. [color:#CC33CC]"Chances of both of you choosing a matching back[/color] [color:#FF0000]at the same time[/color] [color:#CC33CC]and the odds start to multiply".[/color]

Next phase: [color:#CC33CC]"minus the two choices, the odds are 50:1 for each of you that you chose the mate to your card. Those odds double if you both choose your matching cards[/color] [color:#FF0000]at the same time"[/color]

[color:#CC33CC]"But wait... there's more! The odds that each of you chose matching backs, and matching faces, and that you both chose exactly those cards[/color] [color:#FF0000]at the same time[/color] (see a pattern yet? The excitement in your voice goes up with each explanation) [color:#CC33CC]"is... well, trust me it's big"[/color]

Then you go into the reveal [color:#CC33CC]"But here's the kicker. After all that[/color] (restate the odds if you think necessary) [color:#CC33CC]what are the odds that you chose the only black cards"[/color] (ribbon spread the cards across the table in your best grand fashion)[color:#FF0000]"in an all red deck!"[/color]

[John Lovick]

Given a fixed deck size, the first choice would be 24/48, the second would be 23/47, the third 22/46,..., the last choice would be 1/25.

The last card (like every one before) is a fifty-fifty shot. How can you say the last choice would be 1/25?

[Jonathan Townsend]

Given a fixed deck size, the first choice would be 24/48, the second would be 23/47, the third 22/46,..., the last choice would be 1/25.

The last card (like every one before) is a fifty-fifty shot. How can you say the last choice would be 1/25?

John, the last card in that picking process would be the only red card among the twenty four black cards.

If this were a straight forward deal down and then reveal process I'd be with you about the red/black fifty/fifty analysis. What happens is that after they deal cards as they please you get them to make packets of 12 (times 2, so 24) by moving them around. Once you have them moving cards rather than dealing it's a "choose" situation - 48 cards choose the 24 red ones. The rest are by default the 24 black cards. so 48 choose 24 it is, IMHO.

:)