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[eHalcyon]

Ah yes, I missed the wrapping around as well, I was going with my start on suits, and following that with OK, what you we do with 4 of a kind, full house, etc.  Then I got down to pairs and realized it was starting to fall apart.

Above, I detail ways to do it with the different suit distributions.  I was only missing a way for 2-2-1 and 2-1-1-1 start.  I've got a way for 2-2-1 though.  You choose between which of the two suit pairs you want to select, and the guesser will know which two suits were available for you to choose from.  With suits having an order, you again can choose to pick the higher or lower suit.

I don't think there's a way to make it work for the 2-1-1-1 start though.

[sudgy]

My favorite card trick:  First, say you want to do a card trick, where someone else picks a card from a deck.  You, somehow (I have a way I always do this), figure out its position in the deck.  You turn cards from the deck, and, when it gets to their card, say, "The next card I turn over is your card."  You flip the card.  You explain how you did this.  Say you will repeat the trick, but shuffle the deck this time.  Take a peak at his card while shuffling.  Turn over cards, and a few cards after his card, say, "The next card I turn over is your card."  Then, turn the card that's already on the table upside down.

That whole first part is just setting it up.

[eHalcyon]

I like Joey's card tricks:

[Tables]

It reminds me of one of my favourite card tricks. Tell someone to pick a card, do something with it so it seems like a genuine trick, then shuffle it in. Tell them you're going to hand them their card. Finish shuffling, then hand them the entire deck of cards.

[Awaclus]

My favorite is the one in which you take a look at the bottom card while shuffling, then shuffle so that the card will end on top, then cut the deck at any point while using your pinkie to keep track of the point, then tell someone to pick any card of their choice and make them choose the card that you have taken a look at. Then, you can tell which card it is even before they hand it back, and if you feel like it, you can pretend to read their mind or something. It doesn't always work, because sometimes the victim will pick the wrong card, and that makes it pretty thrilling, because you'll have to improvise if it goes wrong, and on the other hand, it's also that much more satisfying when it's actually successful.

[Archetype]

Reading descriptions of card tricks makes my head hurt.

[Axxle]

Do it at a bar with random people and just guess a random card, you'll be right 1/52 times and you'll be a mindreader.

[eHalcyon]

Do it at a bar with random people and just guess a random card, you'll be right 1/52 times and you'll be a mindreader.

[Axxle]

Dammit, I bet Simpsons did it too.

[WanderingWinder]

Do it at a bar with random people and just guess a random card, you'll be right 1/52 times and you'll be a mindreader.

You can do a lot better than 1% if you start keeping track of the patterns in what numbers people pick.

[Tables]

On the topic of random stuff, my respect is actually somehow approaching a 1:1 ratio with my post count, which is... kinda surprising for me actually. I mean I get a lot of respect from jokes and stuff but still, it'd be nice if I could actually hit that ratio I suppose.

[qmech]

On the topic of random stuff, my respect is actually somehow approaching a 1:1 ratio with my post count, which is... kinda surprising for me actually. I mean I get a lot of respect from jokes and stuff but still, it'd be nice if I could actually hit that ratio I suppose.

You're not the only one who watches this number.  I can't tell if approaching parity is a sign that post quality is improving, or the inevitable effect of quadratic beating linear growth...

If you take the ratio seriously, memes are the price you pay to allow you to make on topic comments.

[Axxle]

You can do a lot better than 1% if you start keeping track of the patterns in what numbers people pick.

69, dudes!

[Kirian]

You can do a lot better than 1% if you start keeping track of the patterns in what numbers people pick.

69, dudes!

Whoa.

[SirPeebles]

You can do a lot better than 1% if you start keeping track of the patterns in what numbers people pick.

69, dudes!

Whoa.

Same here.  I guess we were 69ing together.

[qmech]

So that's how (electro)magnets work.

[eHalcyon]

I'm incredibly disappointed in the gratuitous use of Democracy to get through Victory Road. I hope they don't stoop to it for the E4.

[blueblimp]

This is a semi-old puzzle, but I recently both remembered it and found out the solution, so I might as well share it.

You and a partner have a deck of cards, and want to show off a neat trick. First, a third person chooses 5 cards from the deck of 52. You look at those cards, remove 1 of them, and place the other 4 on the table in any order. (You must place all 4 cards, I believe the puzzle is much easier if you don't have to.) Your partner must then say what card you removed from the 5.

Come up with a scheme that makes this trick possible.

Wrote up a solution to this before reading Tables's solution. It's pretty much the same idea, but doesn't use suits.

There is an easy and hard part to this.

The easy part first. Agree ahead of time on a correspondence between cards and integers 0,1,...51. Then the basic ordering is to lay out the cards in numeric order, and there are 4!=24 ways to reorder that. So that can communicate 24 different messages to your partner.

The problem is that there are 48 cards remaining in the deck apart from the ones you placed. So we still need to signal which half of the remaining deck the omitted card was in. And we have to do it without changing the order that we play the 4 cards, since we already used that information.

Here's the best idea I can come up with to communicate which half of the deck to look in, though it's a little clunky. Your partner will follow the rule of picking the maximum of the 4 cards you play (according to the agreed correspondence with integers), then counting upwards, skipping over cards in the played 4, wrapping around if exceeding 51.

Given the 5 cards you start with, here is how you communicate the appropriate signal by removal. Let A<B<C<D<E be the initial 5 cards. We will only choose to remove D or E, so we can ignore A,B,C and pretend D and E are 2 cards in a 49 card deck. If there were a >=24 card gap from D up to E _and_ a >=24 card gap from E up to D, there would be >=50 cards in the remaining deck, more than 49. So either counting up from D will reach E within 24 counts or counting up from E will reach D within 24 counts.

If counting up from D works, delete E. (Your partner will start from D, by the rule.) If counting up from E works, delete D. (Your partner will start from E, by the rule.)

[blueblimp]

Sorry to double post, but one interesting thing about the puzzle is that a solution still wastes a lot of possible outputs.

Inputs are 5 cards chosen from 52. That's (52 choose 5).
Outputs are 4 cards chosen from 52, ordered. That's (52 choose 4) * 4!.

# outputs / # inputs = (52 choose 4) * 4! / (52 choose 5) = 5! / 48 = 5 / 2.

Of course, having enough possible outputs doesn't immediately guarantee a solution is possible, since certain sets of inputs might not have enough outputs between them. (This comes down to bipartite matching theory.) If you go just by counts though, it suggests it might still be possible with a 124 card deck.

[WalrusMcFishSr]

Is it possible to do with fewer cards? If not, what if you're allowed to not play all of them, for extra information?

[blueblimp]

Is it possible to do with fewer cards? If not, what if you're allowed to not play all of them, for extra information?

Using the original rules, it isn't possible even you're given 3 cards and need to remove one. In that case,

# outputs / # inputs = (52 choose 3) * 3! / (52 choose 4) = 4! / 49 = 24 / 49,

so there simply aren't enough possible outputs to account for every input.

If you're allowed not to play all of them, the analysis technique I'm using doesn't apply anymore, because even if you can figure out what the removed card was, you might not know the entire 5 card input set, meaning the function from inputs to outputs no longer needs to be injective (i.e. no two inputs mapping to the same output).

[SirPeebles]

Such a strange mixture of emotions.  There is a math problem that I've been working on for years.  I think I may have now solved it, which is exciting, but there is still that worry that I have a gap, like every other time I've thought I solved it.

[Kirian]

Such a strange mixture of emotions.  There is a math problem that I've been working on for years.  I think I may have now solved it, which is exciting, but there is still that worry that I have a gap, like every other time I've thought I solved it.

42.  The answer is 42, and always was 42.

[qmech]

Such a strange mixture of emotions.  There is a math problem that I've been working on for years.  I think I may have now solved it, which is exciting, but there is still that worry that I have a gap, like every other time I've thought I solved it.

At that point I tend to go home for the day, or do something else.  There's always the chance that tomorrow you find a mistake, but for one night you feel like a champion.

Equally exciting is going to bed thinking something's broken and waking up with a solution.  Or at least entertaining nonsense.

[SirPeebles]

Wow.  I just finished that lecture series on string theory, and the last lecture really blew my mind.  Physics is so much more beautiful when you see the mathematics behind it.