Is Persi Diaconis Having Us On?

[Bob Farmer]

In Persi Diaconis's book, Magical Mathematics, written with Ron Graham, he describes an effect at pages 56-57. He outlines the theory but then says, "We will not give further details about the practical performance the interested reader will have to go forward alone."

Well this interested reader tried to and I think Persi may be having us on.

The effect is this: a spectator cuts a deck several times and then deals five cards off the top to five different people.

The magician asks the five participants to group themselves together by suits, but the magician does not know what the suits are. The magician then reveals what each person's card is.

In outlining the method (the deck is stacked), the authors give an example of five spectators, A, B, C, D and E, with cards as follows after the cut and deal:

A 8C B 4D C JD D AH E 10C

The magician tells the spectators to rearrange themselves by suits. In doing so, they end up in this order:

CC DD H

The magician does not know the suits, he just sees the grouping.

The authors claim that with the right stack each set of 5 cards will produce a unique order when they group themselves together and this unique order tells the magician what the cards are (from a cue list).

Let's stop right there. If the magician does not know the suits and gives no instruction as to where the suits are to group together, that CC DD H order could be DD H CC or H CC DD. That is three different possibilities.

So that can't be right.

Then the authors go on to list the stack for 52 cards. For example one grouping might be CDDCC and another CHCHC.

It's obvious that those two groups could both be reordered as CCCDD or DDCCC or CCCHH or HHCCC--and CCCDD looks just like CCCHH, so that can't work--it's not unique.

So, is Persi having us on?

I have worked out a way of doing this that does not use a cue list, but I would be interested in the comments of others who may have been similarly baffled by Persi's claim.

[Joe Lyons]

So, is Persi having us on?

I don't have a copy of that book or the patience to work it out, but does this or this help?

[Bob Farmer]

Joe: Those are great references. Thank you.

Yes, they do help--this is an area I'm very interested in--but Diaconis's claims of a unique grouping is obviously wrong. Yes, you can have unique groupings with Debrujn sequences and related sequences but what he said about having the participants re-sort themselves doesn't work since they can re-sort themselves into more than one pattern with the same cards. Unless the magician is giving instructions ("Clubs to the right".) this doesn't work--and according to Diaconis's conditions--you don't know the suits--so you can't give those instructions since to do so would identify the suits.

He claims to have performed the trick. With the information he has provided that's not possible, so he should prove it, otherwise it's so much mathematical bafflegab. On my analysis so far, I did notice he either missed or suppressed how easy it is to determine the cards if the Joker is one of them (5 cases possible).

[Joe Lyons]

Again, I'm not considering this to the depth that you are - but what about red cards to the left and black to the right? Would that work?

[Joe Lyons]

Let's stop right there. If the magician does not know the suits and gives no instruction as to where the suits are to group together

OK, just read that, nevermind.

[Ian Kendall]

Are the suits not grouped alphabetically? That wouldn't be hard to implement, given that the spectators have to know each others' cards in order to group.

So you would have Clubs, Diamonds, Hearts, Spades.

[Bob Coyne]

Unless I'm misreading his writeup, in decoding the groupings you also know which spectator is in each grouping (since they looked at the cards in a known order).

So for your example, you would get different spectators in each group and use that information to tell which sequence you had.
- CDDCC would be {1,3,4} and {2,3}
- CHCHC would be {1,3,5} and {2,4}

[Brad Henderson]

Without knowing the suits there are only 6 distributive permutations.

(Distributions with duplicate cards)

aa b c d
aa bb c
aa bbb
aaa b c
aaaa b
aaaaa

(aabbb is the same as aaabb)

You’d would need 52 distinct combinations to know what the 5 cards were, I believe. (One for the 5 card sequence starting on card 1, one for the sequence starting on card 2, etc)

[Brad Henderson]

Will look at write up. But if you only had people clump together if the person standing next to them shared the same suit . . .

[Joe Lyons]

Here's a link I found to the problem.
I got a headache verifying B(4) = 15. But doing so did make the trick seem possible.
Good luck Bob.

[Brad Henderson]

Bottom of page 55 into top of page 56

The order in the groups don’t matter BUT which itemOF the group which stand outside DOES.

So maybe one needs to determine one bit of info through fishing and can then use the list to figure out which set it is???

Looking up more on bell numbers now

[Brad Henderson]

Yes joe. I too am trying to see what the b(4) series looks like. I think it would have been helpful for him to have shown that

[Brad Henderson]

Ok. I found an image on Wikipedia. Don’t know how to post it. It visually shows the 52 possible combinations in a really clear way. And yes, one would need to know which suits were the groups to figure out which sequence it was.

[Brad Henderson]

Look under the section on counting.

https://en.m.wikipedia.org/wiki/Bell_number

[Brad Henderson]

Without knowing the suits there are only 6 distributive permutations.

(Distributions with duplicate cards)

aa b c d
aa bb c
aa bbb
aaa b c
aaaa b
aaaaa

(aabbb is the same as aaabb)

You’d would need 52 distinct combinations to know what the 5 cards were, I believe. (One for the 5 card sequence starting on card 1, one for the sequence starting on card 2, etc)

According to the book you need 5 elements to get to the 52 combo. So while the above is true it’s not correct.

Which makes me wonder - does the SINGLE 5th element somehow add information that contributes to being able to discern sequence without suit knowledge.

By asking if anyone has the joker you can immediate Eliminate 9 cards if they say no OR know it must be a sequence using only those 10 cards.

[Joe Lyons]

A 8C B 4D C JD D AH E 10C

Bob, after reading the effect I don't think this: set is meant as an example.

I think you would have to use a set that fit this: stack.

[Brad Henderson]

Yes. But you run into the same problems

I went through the sequence and you end up with several ‘pods’ that have the same numerical distributions among the sets.

[Bob Farmer]

I have worked out some analysis on this and as soon as I write it up I'll send it around. I took the 52-card stack and listed every single group of 5 cards that occurs. Then I took each grouping and printed each one on a separate blank card. Then I sorted the cards, so for example all the groups that start with two black cards are in one pile, with two red cards in another, etc. But there is no way that resorting each group by bringing suits together works since that produces non-unique groupings. For the trick to work, you have to maintain the order and ask some questions about color and suit. Diaconis says you do not have to. He's having us on.

Sure, the simple solution might be to ask Diaconis, but in the past when I have done this you get no reply. Manners are lacking.

[AJM]

I’m sure Colm Mulcahy (Mathematical Card Magic) or Professor Arthur Benjamin would be able to help if you can track ‘em down.

Andrew

[Brad Henderson]

I have a friend I’m reaching out to. Colm Mulcahy. He may know what’s up.

[Brad Henderson]

Ha. Posts crossed. Colm and I have mutual friend in richard Garriott. Met him at his wedding event. Super nice guy.

[Bob Coyne]

I wrote a simple script to do this. There aren't any duplicates if you run a sliding window over the sequence DDDDDCHHHCCDDCCCHCHCSHHSDSSDSSHSDDCHSSCHSHDHSCHSJCDC

So if you know which spectators are grouped together you then know what sequential set of five cards they looked at.

((1 2 3 4 5))
((1 2 3 4) (5))
((1 2 3) (4) (5))
((1 2) (3) (4 5))
((1) (2) (3 4 5))
((1 5) (2 3 4))
((1 2 3) (4 5))
((1 2) (3 4) (5))
((1) (2 3) (4 5))
((1 2 5) (3 4))
((1 4 5) (2 3))
((1 2) (3 4 5))
((1) (2 3 4) (5))
((1 2 3 5) (4))
((1 2 4) (3 5))
((1 3 5) (2 4))
((1 3) (2 4) (5))
((1 3) (2 5) (4))
((1 4 5) (2) (3))
((1) (2 5) (3 4))
((1 4) (2 3) (5))
((1 2) (3 5) (4))
((1) (2 4 5) (3))
((1 3 4) (2 5))
((1 4) (2 3 5))
((1 2 4 5) (3))
((1 3 4) (2) (5))
((1) (2 3 5) (4))
((1 2 4) (3) (5))
((1 3) (2) (4 5))
((1) (2) (3 4) (5))
((1) (2 3) (4) (5))
((1 2) (3) (4) (5))
((1) (2) (3) (4 5))
((1 5) (2) (3 4))
((1 5) (2 3) (4))
((1 2 5) (3) (4))
((1 4) (2) (3 5))
((1) (2 4) (3) (5))
((1 3 5) (2) (4))
((1 5) (2 4) (3))
((1 3) (2) (4) (5))
((1) (2 5) (3) (4))
((1 4) (2 5) (3))
((1 4) (2) (3) (5))
((1 5) (2) (3) (4))
((1) (2) (3) (4) (5))
((1) (2) (3 5) (4))

[Brad Henderson]

Ah. So you not only look at the groupings but also which which person(s) Are in each grouping. So if I see the person who selected the first card standing alone with two groups of two that’s different from seeing the person who selected third card from standing with two groups of two. And tnose are different depending on which volunteers are in each of those groups of two.

An arrangement of three and two isn’t just a an arrangement of three and two. It’s WHICH spectators are in the group of three and the group of two.

That’s the key, correct?

[Joe Lyons]

That’s the key, correct?

Ah, a nice, easy, no sleight effect.



(No offense, Andrew )

[Bob Coyne]

Right. Diaconis describes the procedure as "five people take consecutive cards" after the deck has been randomly cut. So the performer knows the order they looked at the cards. As a result, the resulting groups are not defined solely by their sizes but by WHICH spectators are in each group. With the appropriate initial ordering of the cards, that's enough to determine the cards they each looked at.

Note: the listing I made above doesn't include the wrap-around case, which you can get depending where the deck is cut. But the same principle applies, and the same setup works.

[Max Maven]

Look under the section on counting.

https://en.m.wikipedia.org/wiki/Bell_number

First published use of Bell numbers in magic: "Belling the Catch" in Ibidem 3, Aziz & Beyond, 2003.

[Brad Henderson]

Now THIS is The kind of thing for which an app would be ideal

Enter the groupings/distribution of the specs and it lets you know the 5 cards.

[Jonathan Townsend]

Agreed Brad. A smartphone app to slide around four of five markers into rows and then let you swipe into a screen with the cards... wishful thinking.

[Brad Henderson]

Ha. Posts crossed. Colm and I have mutual friend in richard Garriott. Met him at his wedding event. Super nice guy.

Actually it was a different math magic guy I knew through richard. He died. So he won’t be offering any assistance.

[Brad Henderson]

Agreed Brad. A smartphone app to slide around four of five markers into rows and then let you swipe into a screen with the cards... wishful thinking.

One marker in each corner of screen and one in middle. Feels easy to group with few and direct/clear strokes.

[Brad Henderson]

My original thought was it’s a secret app/cue sheet but this could work:

You do the sliding under pretense of opening the mind reading app.

They take the phone and press the mind reading button showing on the screen and the phone reveals card.

They just have to present themselves to the app in order

[Bill Mullins]

Brad -- you'd know Colm from G4G, I'd think.

[Brad Henderson]

He and I have just exchanged a couple of emails. I don’t recall meeting him at the one I attended but I could be mistaken.

[Jack Shalom]

Real Men don't need cribs.

[Bob Farmer]

Bob Coyne's idea seems right. I'll work on it. You don't need a cue sheet or a smartphone. Use a second deck of cards and one each card you have the order and the cards.

[Joe Lyons]

Note: the listing I made above doesn't include the wrap-around case, which you can get depending where the deck is cut. But the same principle applies, and the same setup works.

But shouldn't the listing include 52 sets because b(5)=52?

[Bob Farmer]

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.

[Joe Lyons]

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?

[Joe Lyons]

Here's a link I found to the problem.

Strange, pages 57-58 aren't available anymore...

[Brad Henderson]

57 is for me. Not 58-59.