Is Persi Diaconis Having Us On?

[Joe Lyons]

57 is for me. Not 58-59.

Odd, I get this:


If I wanted to perform it I would purchase the book anyway, but I think you're right, Brad - it's best as an app.

[Jonathan Townsend]

Am I correct in concluding that Bob Coyne's idea would require the performer to first memorize the original five people And their order and then compare that to the second grouping? A pen and paper would be required.

And memorize 52 sets and the playing cards in those sets, right?

And then watch out for twins, triplets, and in general people wearing similar clothes among the volunteers

[Bob Farmer]

Assume the five spectators, A, B, C, D and E, start with the card sequence CHCHC (clubs, hearts, clubs, hearts, clubs), respectively, so:

A/Clubs, B/Hearts, C/Clubs, D/Hearts, E/Clubs.

They are told to organize themselves by suit.

[A/Clubs, C/Clubs, E/Clubs] [B/Hearts, D/Hearts]

However, since there are [3X2X1] ways the three clubs people could sort themselves and [2X1] ways the hearts people could sort themselves, that is 12 different ways the group could sort itself. In effect, the resort takes what is unique, a 5-card grouping, and changes it into something that is not unique. And it would seem that a grouping like 3 people here and 2 people there might be duplicated by some other 5-card selection.

So, without more, the conclusion is that without asking some questions, you cannot determine which 5-card group is in play. However, the Diaconis description says no further questions are asked. The only question that is asked--in the form of a statement--is that the "hearts stand together, all the clubs and so on; people with the same suits stand together."

So, I am still baffled. The obvious solution is to ask two questions, one as to color and one as to suit to determine the 5-card group, but Diaconis does not mention this.

[Bob Coyne]

It only works with certain initial setups for the deck. The one given on p57 accomplishes that (as the listing I generated demonstrates). However, if you have different initial setups, then you're right -- there could be multiple solutions. For example, suppose the setup is CHCHCHSHSHSHSH... In that case you might get C{1,3,5) and H{2,4). But you could also get S{1,3,5} and H{2,4}, among others. So there would be no way to uniquely determine the cards from the spectator groupings. In the setup that Diaconis and Graham give, however, the mappings to the cards from the spectator groups are unique.

[Joe Lyons]

And it would seem that a grouping like 3 people here and 2 people there might be duplicated by some other 5-card selection.

Using your example of getting one group with three people and one group with two, you could cut the example stack 9 different ways to come up with that result:

CHHHC (1,5) (2,3,4)
CCDDC (3,4) (1,2,5)
CDDCC (2,3) (1,4,5)
DDCCC (1,2) (3,4,5)
CCHCH (3,5) (1,2,4)
CHCHC (2,4) (1,3,5)
SDSSD (2,5) (1,3,4)
DSSDS (1,4) (2,3,5)
CDCDD (1,3) (2,4,5)

As you can see, there are no duplicates.

I'd really like to see someone do the trick without help from a computer though.

[Brad Henderson]

I think part of the problem is in thinking in terms of suits and not in terms of people.

When you try to fiddle with the problem thinking about suits you get into the non unique sets, but if you approach the problem as a PEOPLE distribution issue, it works.

It isn’t that the diamonds went here and the spades went there (not the least of which reason being you don’t know who has what suits) but where did person 1 and person 3 go and stand, and where did 2 stand. And where is 4 and 5?

There are 52 unique combination of how these people can group themselves.

But you have to account for the individuals In each group, not just the quantity

[Joe Lyons]

Let's assume the math is correct (which it appears to be) and Persi's not having us on.

The real question for me is who (besides Harry Lorayne) would be able to look at the spectators in these groups, remember which order they originally took their cards in and memorize the 52 sets of possibilities that would render the correct 5 card group?

Using the the nine possibilties above and a correctly ordered stack, here are but nine of fifty-two solutions to memorize.

CHHHC (1,5) (2,3,4) = 6c,3h,10h,7h,Ac
CCDDC (3,4) (1,2,5) = Ac,7c,7d,8d,2c
CDDCC (2,3) (1,4,5) = 7c,7d,8d,2c,10c
DDCCC (1,2) (3,4,5) = 7d,8d,2c,10c,5c
CCHCH (3,5) (1,2,4) = 10c,5c,Kh,9c,Qh
CHCHC (2,4) (1,3,5) = 5c,Kh,9c,Qh,4c
SDSSD (2,5) (1,3,4) = Js,Kd,8s,4s,4d
DSSDS (1,4) (2,3,5) = Kd,8s,4s,4d,Ks
CDCDD (1,3) (2,4,5) = 3c,Ad,Kc,2d,6d

[Richard Kaufman]

That was not something Harry could have done.

[Joe Lyons]

That was not something Harry could have done.

No, I doubt anyone could.

So, is Persi having us on?

Yes, but the math is correct.

[Grant McSorley]

The second set of slides on this page might help: http://www.cardcolm.org/CardColmSlides.html?s=03

Sent from my SM-G960W using Tapatalk

[Bob Farmer]

Hi Grant: Nope of no help, however, it does confirm my theory that knowing the colors tells you the sequence, but that is a standard de Brujn idea. See Leo Boudreau's book, Spirited Pasteboards for the best work in this area.

[Jack Shalom]

Bob, I think there may be a few understandings here. This is not the same as the de Brujn idea.

Persi's 52 card solution is for a very specific kind of stack. There are many ways the stack can be formed but the suits must always look like this:

DDDDDCHHHCCDDCCCHCHCSHHSDSSDSSHSDDCHSSCHSSDHSCHSJCDC

The J stands for Joker

The performer should pick a stack with this structure and memorize that stack (Say, 8D, 2D, 3D, QD, KD, 4C...)

The conditions of the trick are:
1) That five specs have chosen consecutive cards.
2) The performer is aware of who chose the the first card, who chose the second card and so on.
3) The specs group themselves according to suit.
4) The performer is aware of which specs are in which group.

So now for example, let's say the cards chosen are the 5th, 6th, 7th , 8th, and 9th cards in the stack, that is DCHHH.

I'm going to code the selections by suit in four brackets always in CHSD order. So in this case I will code the results as [2] [3,4,5] [] [1]. In other words, spec 2 has a club, specs 3,4,5 have hearts, no one has spades, spec 1 has a diamond.

The good news is that no other set of five consecutive cards has this distribution code

In the case of a Joker being selected, you will tell the spec to make a separate fifth group. You put the code for Joker as the fifth group after CHSD

So for example if the last five cards were chosen, HSJDC, the code would be [5] [1][2][4][3]

So how many five card codes as described above are there? Fortunately, exactly 52.

You can now associate each of the 52 codes with the first card of each five card group. Since you have presumably memorized your stack, you will know what the other 4 cards in the group are as well.

So for example, if the stack begins as given above, 8D, 2D, 3D, QD, KD, 4C, then I will associate the code [] [] [] [1,2,3,4,5] with the 8D. Since I know the stack, by knowing the first card, I know all the following cards.

I hope that's clear. To make this practical, the performer would need to memorize the initial stack; difficult, but not impossible because many here have memorized stacks.

The hard part is that after you've made the list of 52 codes and associated them to the first card of each five card grouping, you've got to find a way to access that. I can't think of a clever mnemonic way to do that, and here I think you're forced to use a crib.

Hope that's helpful in understanding what Diaconis is talking about here.

[Bill Mullins]

To expand on Jack's, a little bit.

You have specs Alice, Bob, Charlie, Donna, Eric.

2) The performer is aware of who chose the the first card, who chose the second card and so on.

That is to say, although the magician doesn't know where in the stack things are happening, he knows that if Alice chose stack # = n, then Bob has n+1, Charlie has n+2, etc.

First, consider occasions in which one of the specs has the Joker (stack # = 49).
If Eric has the Joker, then Alice must have 45, Bob must have 46, Charlie must have 47, etc.

If Alice has Joker, then Bob has 50, Charlie has 51, Donna has 52, and Eric has 1. So the times that someone has the Joker are easy.

But usually that won't be the case. Usually you will have a situation like (for example): Alice + Charlie stand together, Bob + Donna stand together, and Eric stands alone. Let (card that Alice chose = n). Suit of n = unknown1; suit of n + 2 also = unknown1; suit of n + 1 = unknown2; suit of n + 3 also = unknown2. Suit of n + 4 = unknown3.

The only place in the whole stack where this can happen is Alice has 17 (Hearts), Bob has 18 (Clubs), Charlie has 19 (Hearts), Donna has 20 (Clubs), and Eric has 21 (Spades). If Alice had any other stack number, then Alice, Bob, Charlie, Donna and Eric would be grouped differently. There is a unique grouping of the 5 specs for any place in the stack that Alice could land.

[AJM]

This is going to be next to impossible to achieve seamlessly with either brute force memory or mnemonics (unless you are someone like Dominic O'Brien)

Using a crib sheet, a 'Miracle Poker Machine' type calculator or an app that does some sort of VLOOKUP on an indexed list is only going to detract from the effect.

Andrew

[Jack Shalom]

One more very important point that I should clarify. While the specs group themselves according to suit, amazingly, the performer does *not* need to know which group belongs to which suit. That is, in Bill's example above of cards 17 through 21, the code if we arranged it by CHSD suits would be [BD] [A,C] [E] []. But we don't have to worry about confusing it with a re-arrangement among the suits.

In other words, no where else in the stack will you find an arrangement with the same four groupings of people, regardless of suit.

So suppose, for example, you see all 5 people grouped together. The only cards that are keyed to one group of [A,B,C,D, E] is the first five diamonds.

In a different stack, yes, they could be all hearts or spades and so on, but in this stack there are only five diamonds together. The same is true for all other combinations of five in this stack; they are unique.

This is really important to understand because it means a much more powerful effect--you ask the specs of each suit to stand together, but you do *not* ask them their suits. This, to me, is the feature that makes the whole trick worthwhile in the first place.

[AJM]

This is going to be next to impossible to achieve seamlessly with either brute force memory or mnemonics (unless you are someone like Dominic O'Brien)

Using a crib sheet, a 'Miracle Poker Machine' type calculator or an app that does some sort of VLOOKUP on an indexed list is only going to detract from the effect.

Andrew

Seems a great effect in theory but perhaps not in practice.

The fact that the spectators are asked to select consecutive cards must immediately arouse suspicion...

[Joe Lyons]

Conversely, you could stack the deck, let the spectators cut to their heart's content, not memorize their order, not ask them to group together, glimpse the bottom card on the deck and perform your miracle...

[Brad Henderson]

Would look exactly the same to the audience.

Peek a card/peek a cue sheet.

I think we are more enamored with the method than we are concerned with maximizing the effect.

[AJM]

Peek a card in a nano-second.

Pore over a crib sheet with coded entries for twenty minutes or so (at least if my eyesight is anything to go by) while asking the spectators to chat amongst themselves.

[Bob Farmer]

Alternate methods for doing this not using the Diaconis stack do not advance the solution, unfortunately.

Jack taking your example: [BD] [A,C] [E]. That could also be arranged as [E] [A,C] [BD], or [A,C] [E] [BD].

That is three different arrangements for the same cards. The spectators could decide to arrange themselves in a different order, though still sorted as requested. Let's call the order 2, 2, 1, for two spectators, two spectators and 1 spectator.

If 2, 2, 1--no matter what order it is in--is unique then we have a solution. If we see any 2,2,1 order we know the cards.

I haven't checked to see if this is the case.

[Joe Lyons]

If 2, 2, 1--no matter what order it is in--is unique then we have a solution. If we see any 2,2,1 order we know the cards.

I haven't checked to see if this is the case.

Don't know if you're saying the same thing or not Bob but 2,2,1 isnt unique.

2,2,1 with bd together, ac together and e by itself is unique.

[Bob Farmer]

Joe: I was going for the grouping not the individual components of each part. If that works, then you don't need to know the individual parts.

[MagicbyAlfred]

I'm going to speak to Persi about all of this when he comes over for his Math lesson this week...

[Joe Lyons]

Joe: I was going for the grouping not the individual components of each part. If that works, then you don't need to know the individual parts.

I understand, Bob, I just don't think it will work. Their are too many ways to cut that stack that will produce 221. (Or 212 or 122 or whatever other way they decide to stand). I don't think it works without knowing the parts (not the suits).

[Edward Pungot]

Alternate methods for doing this not using the Diaconis stack do not advance the solution, unfortunately.

But it does help take the blinders off the workhorse.

Ed Marlo's "A Miracle With Cards"
(Ibidem No.8. p.147).

[Jack Shalom]

Alternate methods for doing this not using the Diaconis stack do not advance the solution, unfortunately.

Jack taking your example: [BD] [A,C] [E]. That could also be arranged as [E] [A,C] [BD], or [A,C] [E] [BD].

That is three different arrangements for the same cards. The spectators could decide to arrange themselves in a different order, though still sorted as requested. Let's call the order 2, 2, 1, for two spectators, two spectators and 1 spectator.

If 2, 2, 1--no matter what order it is in--is unique then we have a solution. If we see any 2,2,1 order we know the cards.

I haven't checked to see if this is the case.

Bob, I addressed this already in my last post. In the given stack carefully worked out by Persi there is only ONE set that corresponds to [E] [A,C] [BD] or any of its arrangements. Anytime you see person A and person C together, and person B and person D together, and person E separately, then it MUST be the 5 card group starting with card 17. That's the beauty of the stack.

When I get some time I will list all 52 partitions so you'll see how it works out.

It' s not that we know from just the 2, 2, 1 distribution, but which people are in those groups. We don't care though, how they stand within those groups.

[Jack Shalom]

Maybe it will help if we do an internet version.

Someone here please choose from the stack above five consecutive cards from consecutive specs who we'll label A, B, C, D, E.

Now suppose the specs stand together in suit order.

Just tell me who is standing with whom in any order you like--not the suits!--and I'll tell you which cards were picked.

[Bill Mullins]

I think we are more enamored with the method than we are concerned with maximizing the effect.

Surely you jest. That is not how magicians operate . . .

[Joe Lyons]

When I get some time I will list all 52 partitions so you'll see how it works out.

Please do, I'd like to see them, it would help with this exercise. Working out the 15 partitions for B(4) taxed me.

It' s not that we know from just the 2, 2, 1 distribution, but which people are in those groups. We don't care though, how they stand within those groups.

I get it. I know the math works and the fifty-two partitions are unique. I know you have to know which spectators are in which group and why you do.
Thanks to Bob Coyne's list I am even getting an inkling as to how Diaconis came up with the stack, and I'm not a math guy.

Maybe it will help if we do an internet version.

I have no doubt whatsoever that you will be successful.

My only question: does this have a real-world performance application and if so how? I am curious, not argumentative.
Thanks,
Joe

[Jack Shalom]

Let's not kid ourselves: the Diaconis book is a book meant to introduce some interesting math concepts. It happens to use some magic tricks as illustrative examples.

Clearly this effect uses a cannon to swat a gnat.

But there is much to be learned from following through an example like this. Understanding the method fully could lead to other more plausible uses.

In any magic effect there are weak points that must be dealt with. If you use a stack and then utilize a peek to determine the cards taken, clearly the peek is a weak moment. It can no longer be hands off. It can only appear to be so. And the taking of consecutive cards must be accounted for as well.

In Persi's trick, there are several weak points. The biggest one is associating the 52 Bell partitions with each of the cards. Otherwise, the problems are the same as above. IF we can overcome that condition, then I think we have a better trick. There's no peek to catch so you set up the trick so that it *emphasizes* that no peek can occur. I agree that a machine, computer or crib makes the trick way too suspicious. No, the only answer here is to find a reasonable mnemonic system that associates each of the partitions with each of the cards. I have no idea if there is such a thing. But if there is, then it becomes a pretty good method.

[Jack Shalom]

FYI:

STACK:
DDDDD CHHHC CDDCC CHCHC SHHSD SSDSS HSDDC HSSCH SHDHS CHSJC DC (Use this stack. I think I had a missing card above!)

PARTITION GROUPINGS. (each partition number corresponds to the first of five consecutive cards. For example partition #6, [A,E] [B,C,D] tells you that if the specs group themselves that way, then the cards chosen must be the five cards beginning with card #6, i.e. CHHHC
1. [A,B,C,D,E]
2. [A,B,C,D] [E]
3. [A,B,C] [D] [E]
4. [A,B] [C] [D,E]
5. [A] [B][C,D,E]
6. [A,E][B,C,D]
7. [A, B, C] [D,E]
8. [A,B] [C, D] [E]
9. [A][B,C][D,E]
10. [A,B,E] [C,D]
11. [A] [B,C] [D,E]
12. [A, B] [C, D,E]
13. [A] [B,C,D] [E]

14. [A,B,C,E] [D]
15. [A,B,D] [C,E]
16. [A,C,E] [B,D]
17. [A,C] [B, D] [E]
18. [A,C] [B,E] [D]
19. [A,D,E] [B] [C]
20. [A] [B,E] [C,D]
21. [A,D] [B,C] [E]
22. [A,B] [C,E] [D]
23. [A] [B,D,E] [C]
24. [A,C,D] [B,E]
25. [A,D] [B, C, E]
26. [A,B,D,E] [C]


27. [A,C,D] [B] [E]
28. [B,C,E] [A] [D]
29. [A,B,D] [C] [E]
30. [A,C] [D,E] [B]
31. [A] [B] [C,D] [E]
32. [A] [B,C] [D] [E]
33. [A,B] [C] [D] [E]
34. [A][B][C][D,E]
35. [A,E] [C,D] [E]
36. [A,E] [B,C] [D]
37. [A,B,E] [C] [D]
38. [A, D] [B,E] [C]
39. [B,D] [A] [C] [E]
40. [A, C,E] [B] [D]
41. [A,E] [B,D] [C]
42. [A, C] [B] [D] [E]
43. [B, E] [A] [C] [E]
44. [A,D] [B,E] [C]
45. [A,D] [B] [C] [E]
46. [A,E] [B] [C] [D]
47. [A][B][C][D][E]
48. [C,E] [A] [B] [D]
49. [B,D] [C,E] [A]
50. [A,C] [B,D,E]
51. [A,C,D,E] [B]
52. [B,C,D,E] [A]

[Jack Shalom]

*11. should be [A, D,E] [B,C]

[Bob Farmer]

Jack, I think you've nailed it. Roadwork required. Now here is a simple way to hide the cue list you need. Use a blue-backed deck and a smartphone. Claim that the deck, since it's blue, is Bluetooth enabled and can read minds. After they have resorted themselves, bring out the smartphone and ask if anyone is a Libra. You then supposedly enter this every important information into the app. Then ask whether anyone was born outside the U.S. and enter that information.

With two "entries" and two lists you should be able to nail the cards.

What is needed is some sort of list app where you enter the order and it automatically brings up the cards.

[AJM]

I've been having a look for 'list' type apps that have the functionality required however I've drawn a blank so far.

May be better with a database app where an indexed table (or tables) could be created and a query written to extract the required record when the inputs have been entered.

Andrew

[Edward Pungot]

Tape the crib within the pages of Magic for Dummies for easy reference.

Then toss them a rubber banded one-way deck.

[Bob Farmer]

Edward those are excellent ideas. I'm sure there is thread where they might be useful.

Andrew: yes, a database app might work.

[Jack Shalom]

For those of you keeping score, I offer my very very humble beginning for a mnemonic set.

[A,B,C,D,E]--> you see ONE big group so it must be associated with the number 1
[A][B][C][D][E]--> the individual line up reminds me of a RocK slide, so 47 (mnemonic alphabet)
[B,C,D,E] [A]--> Only A is left out, but the first shall be last, so 52
[A,C,D,E] [B]--> Only B is left out, but the second shall be next to last, so 51
[A,B, D, E] [C]--> Only C is left out, the middle guy, so the middle number is 26
[A,B,C,E] [D]--> Only D is left out, the DooR slammed on him, so 14
[A,B,C,D] [E]--> The other letters of the alphabet are all ahead of E, so he's in SECOND place, 2.

The rest, as they say, we leave as an exercise for the reader. (I always wanted to say that).

[Bob Farmer]

Jack: If using my smartphone idea (above) to store the cue list, you can show the spectators what the phone is showing you. Since you claim it's a mind-reading app, after you've got the cards, switch to a picture of a brain, hopefully one where there are flashes of light here and there. Claim this is a combined image of the spectator's brain and the flashes are the "clues" to the cards.

[Bob Farmer]

For the iphone app, has anyone used this:

https://apps.apple.com/us/app/airtable/id914172636

[Jack Shalom]

I'm not much of a phone app guy, but it seems clear to me that this can be done with two clicks:

1. Choose from a list of the seven "styles' of partition (e.g 4-1, 3-1-1, 2-2-1, etc.)

2. Once that is narrowed down, it shows a list of the lettered possibilities within that "style." The largest list will be 15 for 2-2-1, which is still manageable I think.

That would be the basic structure, but maybe someone can come up with something more efficient. And then it can be dressed up in many ways as Bob suggests.