The New Sudoku Players' Forum

[Cec]

Seems there was interest in the two previous card tricks so here's another one:

As the host of a party, you ask for a volunteer (not an accomplice) to participate in this card trick which requires a telephone book being placed on a table next to a pack of cards laying face down and also a sealed envelope placed on a mantlepiece in the same room.

For this trick, the Jokers plus any two Aces have already been removed from the pack and all the suites in the pack are meaningless. An Ace counts as one, Jack equals eleven, Queen twelve and King thirteen.

The volunteer is asked to cut the pack just once anywhere they wish and, without the host or anyone else seeing the cards, to look at the two cards facing each other and then add the values of these two cards. For example, if the two cards were a Queen (suites are ignored) and an eight then the total value (number) would be twenty.

Just so this is clear, the volunteer picks up any part of the pack say with their right hand which leaves the remainder of the pack face down on the table. The volunteer looks at the value of the bottom card in their right hand, without anybody seeing it, and then looks at the [b]top[/bcard of the remaining pack on the table, again with nobody seeing that card. The volunteer then silently adds the value of these two cards without disclosing this value to anybody and returns the cards held in their right hand to the top of the cards on the table being their original position.

The volunteer is then asked to open the telephone book at the page which corresponds to the sumed value of the two cards. In the above example they would look at page twenty. The volunteer is then asked to closely look at the first surname listed at the beginning of that page and then read out that name to the guests.

You then ask the volunteer to open the sealed envelope previously placed on the mantlepiece which contains only one piece of paper and ask the volunteer to read out the name written on that paper. To everyone's surprise the volunteer confirms both names are the same.

Apart from removal of the joker(s) and any two Aces, the remaining cards in the pack are complete with no markings. However, in fairness to readers to find the solution, the pack of cards was "preset" by the host without the guests or volunteer knowing this but, remember, the pack was randomly cut by the volunteer at the start so, how is the trick performed?. .

Cec

[Cec]

With no responses to date I'll assume readers can't see a solution which was my same conclusion when reading about this trick.

Because the solution requires knowing the sequence of the pre-set cards in the deck then "drip-feeding" hints would still make it difficult for anyone to solve this. Rather than prolong interested people's curiousity I'll reveal solution later today. In the meantime, the following basically explains how the trick is done.

Theoretically, for an "untampered" pack, with the lowest card an Ace (value 1) and the highest card being a King (value 13) then any two cards drawn by cutting the pack could have their sum total ranging from 2 to 26. In reality, it is possible to arrange the reduced pack of fifty cards (Jokers and any two Aces removed as previously mentioned) in such a way that, irrespective of where the pack is cut by the volunteer, there are only two possible answers when adding the values of any two cards facing each other - being either 14 or 15.

Because the host knows numbers 14 or 15 are the only possible answers, he/she could memorize the first listed names at the top of pages 14 and 15 in the phone book. By watching the volunteer to see if they look at the page on the left which would be number (14) or the page on the right (15) the host is then able to "read the volunteer's" mind and disclose the person's name in the phone book.

Alternatively, to avoid embarrassment to the host who could forget the two listed names or their respective order in the phone book, the trick is easier to perform if the two names are written on separate pieces of paper and sealed in separate envelopes which are inconspicuously placed in the room or even hidden in the host's pockets prior to demonstrating the trick. The host will know which sealed envelope to take after discretely observing which of the two pages the volunteer looks at (either 14 or 15) when reading the first name at the top of the page. My apologies for referring to only one envelope being mentioned in my opening post.

The above basically explains how the trick is done other than disclosing the order of pre-setting the cards. I'll allow some more time for others to consider what this sequence could be. Failing an answer I'll reveal card sequence later today.

Cec

[Cec]

Ok.. Sorry for delay.. My phone hasn't stopped ringing asking to release these numbers Excluding any two aces and jokers the deck is preset as follows (suits are irrelevant):

7-8-6-9-5-10-4-J-3-Q-2-K-A-K-2-Q-3-J-4-10-5-9-6-8-7 (Repeat)

The twenty five numbers above are simply repeated in the same order giving the total fifty cards used. It will be seen that no matter which two consecutive cards are chosen - meaning where the deck is cut -the sum of these two cards will be either 14 or 15.

It's 11pm over here and looking forward to watching the Ladies final which is just about to start .

Cec

[Hud]

Cec, I was thinking on the right track but I believe you said in the example that the two cards added up to 20. After that, I gave up. I'll reread the original question to see how it was worded.

OK, it appears you threw out a "red herring" on this. Shame on you for the deception.

[MCC]

Hud, it looks like I'll have to defend Cec here, this is what he posted:

The volunteer is asked to cut the pack just once anywhere they wish and, without the host or anyone else seeing the cards, to look at the two cards facing each other and then add the values of these two cards. For example, if the two cards were a Queen (suites are ignored) and an eight then the total value (number) would be twenty.

As you see, twenty was only provided as an example to demonstrate the principle not as the actual number to go by


MCC

[Hud]

MCC, I stand by my claim of foul. The example showed an actual demonstration of the card trick, and 20 was an invalid answer. LOL

[Cec]

MCC, I stand by my claim of foul. The example showed an actual demonstration of the card trick, and 20 was an invalid answer. LOL

Hud, yes.. you're right and I'm pleased you're back on track

"..OK, it appears you threw out a "red herring" on this. Shame on you for the deception."

Hud, I thought I'd share some aspects of this thread which I hope you will also see the "funny" side like me

When posting the trick, rather than be deceiving, I mentioned this ......

"For this trick, the Jokers plus any two Aces have already been removed from the pack" .................."in fairness to readers to find the solution, the pack of cards was "preset" by the host without the guests or volunteer knowing this but, remember, the pack was randomly cut by the volunteer at the start so, how is the trick performed?"

I'm not a professional card player or magician but I enjoy card tricks so tried it out on a couple of friends who were amazed how this trick could be done.. I might add that a "trial run" was suggested to me by a PM to make sure the trick would work without any "hiccups".

When performing the trick I found it hard to remember the two sirnames ( "Agnesi" and "Ahern") printed at the top of each page (14 and 15) of the phone book so I decided it would be easier to write the correct answer (sirname) in a sealed envelope discretely placed on a mantlepiece.

I later realized my mistake using this different tactic because there are two different possibilities for this sirname with the correct name only being determined after the volunteer looks at either of the two pages (14 or 15) of the phone book. Serves me right for overlooking PM's suggested advice to make sure the trick will work exactly as I would perform it.

To overcome the problem of remembering two sirnames I believe the best option is to discretely place two sealed envelopes somewhere in the room or even in each trouser pocket - one envelope has the sirname from page 14 and the other envelope the sirname from page 15. The host of course would only reveal one of these envelopes being the same sirname the volunteer selects from the phone book.

Anyway, with still no responses at that stage and the Ladies tennis final awaiting to cheer me up I tried to keep my sense of humour and maybe got a smile when I posted this :

Ok.. Sorry for delay.. My phone hasn't stopped ringing asking to release these numbers

Cec

[Cec]

Hud, it looks like I'll have to defend Cec here, this is what he posted:

The volunteer is asked to cut the pack just once anywhere they wish and, without the host or anyone else seeing the cards, to look at the two cards facing each other and then add the values of these two cards. For example, if the two cards were a Queen (suites are ignored) and an eight then the total value (number) would be twenty.

As you see, twenty was only provided as an example to demonstrate the principle not as the actual number to go by

Before this thread is finally put to rest, I couldn't forgive myself in failing to acknowledge MCC's defence of my above explanation where I randomly chose two cards (Queen and an eight) as an example of how the value of two cards would be added together. In hindsight and acknowledging Hud's comments I should have chosen two cards as an example having a combined value of either 14 or 15 which of course were the only possible numbers in the "preset deck.

Cec

[udosuk]

Just wanna comment that if you arrange 51 cards in this sequence:

A-K-2-Q-3-J-4-10-5-9-6-8-7-7-8-6-9-5-10-4-J-3-Q-2-K-A-K-2-Q-3-J-4-10-5-9-6-8-7-7-8-6-9-5-10-4-J-3-Q-2-K-A

Then only 1 ace is needed to be discarded. The beauty of the original 50-card sequence is that you can cut it (wrap around) an infinite number of times and the property still remains...

Off to see Federer/Nadal now...

[MCC]

Just wanna comment that if you arrange 51 cards in this sequence:

A-K-2-Q-3-J-4-10-5-9-6-8-7-7-8-6-9-5-10-4-J-3-Q-2-K-A-K-2-Q-3-J-4-10-5-9-6-8-7-7-8-6-9-5-10-4-J-3-Q-2-K-A

Then only 1 ace is needed to be discarded. The beauty of the original 50-card sequence is that you can cut it (wrap around) an infinite number of times and the property still remains...

If you use this arrangement, then after the first cut (placing the bottom half of the deck on top) you'll end up with two Aces together. So you need to lose another Ace, i.e., two Aces for the trick to work.


MCC

[Cec]

"... The beauty of the original 50-card sequence is that you can cut it (wrap around) an infinite number of times and the property still remains..."

That's an interesting observation udosuk and one which, as I now see, would permit an improved presentation of the trick as the guests could observe the host shuffle the cards in this manner (wrap around shuffle) prior to asking the volunteer to randomly cut the pack. This aspect wasn't mentioned in the book where I read about the trick and, in my opinion, participants viewing a card trick that isn't shuffled at the start of the trick are more likely to be sceptical. Nice observation and appreciated.

Cec