Overs allowed in the 20?

He levels right back.

AFAIK no, but I can't recall a time they've been requested.

hu is dead everyone is solid btw

Lol what, nobody calls LHE "E".
Agreed. And no qualifier is the devil. Ridiculous game.
This.
Not this.

Stud no qualifier is the best game ever if people don't know how to play it.

fyp

Current problem I'm working on (haven't solved):

Two players are sitting at an empty table with a dealer. But not just any dealer (obv). This dealer, being the creative and generous chap he is, decides to liven up the hearts and the minds of the two young players by offering them a freeroll. So, he pulls out his secret deck of 52 cards, all of which are coincidentally the Ace of Spades, and deals them out on the table, face up, in 4 rows of 13, while explaining, "One of you has to leave the room, and come back in 15 minutes. The other has to stay and watch me flip over cards of my choosing so that some are face-up and some are face-down. But! Don't get too distracted, 'cause despite my seemingly random card-flipping ways, I'm actually focusing on one specific card. And, once I'm done with my flipping, I'll tell you which one it is. Then, and only then, may you touch the cards. But you can only touch one, and you can either choose to flip it over once or not at all. After that, you must leave the room and have no contact with the other player, who's going to come in, sit down, and try to guess, in one try, which card I had in my mind and point to it. If they get it right, you split the contents of my tray; if not, no big deal. Feel free to talk among yourselves first."

can you flip the card over from top to bottom so the card is upside down? I guess that onky works if the card is face down or the back of the deck isn't symmetrical.

Original response: I think this would not work because if we are allowed to "flip" a card in such a way, the dealer would also have that option.

Edit: I wrote a different response and then realized I missed the point that every card was the ace of spades. Now, I think you are on to something, because this is one of the cards in the deck that is not symmetrical.

Wow, now I realize my original response is probably still just as valid as it would be if all the cards were different so I'll type it up again. I shouldn't try to think about these items this early in the day.

This is the best I can come up with: I would say to the other player "When you get back, there will be face up cards and face down cards. Whichever there are fewer of is the minority. The card the dealer was thinking about will be part of the minority." The dealer's best counter-strategy would be to flip over exactly half the cards, in which case I could flip over 1 to give us slightly better than 1/26 odds. If the dealer chose a less optimal counter-strategy, our odds would increase. The worst possible counter-strategy would be to flip over none or all of the cards, in which case we could just flip over the one he was thinking of and have a 100% chance of success.

I KNEW I should've said the 8 of Diamonds.

No, but seriously, I didn't even think of flipping it over that way, though I'm inclined to go with what CL said and assume that the dealer could've dealt them out any which way in that regard. But I suppose if he doesn't think of that as "flipping it over," then you doing that might be considered the act of touching and not flipping over, which would be allowed, even though it wasn't super specified that you couldn't do something like, move the position of a card to a far corner of the table in the act of flipping or not flipping, which is obviously disallowed. So I guess maybe it does "change the position" of the card, although really it might just be changing its orientation - I guess I'm not really sure what those words mean exactly. Either way, if you can figure out a way to win using it, go for it. Just don't change the position of the card, and assume that the dealer didn't necessarily deal them all out with the same orientation.

Nice. 4+% isn't bad. If the dealer arranges a random system... actually I'm not doing that right now. I should probably figure out a way to do something like that fast, though, I guess. Is there some kind of awesome mental shortcut for doing this?

Btw, I still don't have a solution to this.

The best idea I can come up with, so far (it's not exactly the greatest), is to order the possible permutations of boards that can appear so that both players agree on the ordering. Then the first player analyzes the board in front of him and attempts to flip a card such that the permutation number of the new board, when written in base52, has a last digit correlated with the position of the target card.

E.g. In a 4 card game, there are 16 permutations of face-up/face-down cards (I think). If the dealer chose Card#3 (again, agreed upon ordering between players required), player1 would attempt to flip a card to create a new board of either Permutation#3, P#7, P#11, or P#15.

Unfortunately, I can't come up with a method of quantifying how often a randomly arranged board would allow for such a thing given 2^n board permutations, so I'm not sure how often the players would win doing this, but I'm pretty sure I'd be seriously surprised if it were 100%. My brain couldn't handle it apparently, though, so I kinda ended up moving on to a new problem, which actually turned out weirdly similar to this one (hammer/nail, maybe):

10 players vs the dealer, all 10 players being on the same team. The dealer has 10 balls, numbered 1 thru 10, and places them inside 10 opaque boxes, one ball per box. The game consists of 10 rounds, a different player playing each round. The round begins with a player entering the room and pointing at a box, the dealer opening the box and revealing the numbered ball inside, and the dealer then closing the box again, keeping the ball inside. The player will then pick a new box and the same thing will happen all over again. Each player gets 5 picks per round, and then exists the game. They can no longer communicate with players still left to act. The goal of the players is to find the box containing the ball with the number correlated to the current round of the game. I.e. Player1 is trying to find Ball#1, Player2 is trying to find Ball#2, etc. The players win the game if they win 10/10 rounds, and lose if they fail in even one round.

Now these fine ladies,
They had a plan:
They was out to meet the boys in the band.
They said,
"Come on, dudes! Let's get it on!"
And they proceeded to suck upon our balls WE'RE AN AMERICAN BAND!!!

GoT,

Maybe you can do some kind of checksum algorithm on the rows of cards, because each row is basically a binary number. There are combinations where it's impossible to pick the card, and you can probably only improve the odds of selecting the card. You will probably need sub-algorithms in certain cases where more than one row has a bad checksum, etc.

http://en.wikipedia.org/wiki/Checksum

Hi GoT,

I think I have a solution but it's hard to see how people could execute it in practice !

Suppose we wanted to do this for n cards. There are 2^n states that the cards could be in. Represent this set of states as an n-dimensional hypercube with 2^n vertices. Use a coordinate system to label each vertex so that the vertex at the origin is (0,0,...0) and represents the state where all the cards are flipped up and the vertex at (1,1,1,...1,1) represents the state where all cards are flipped down etc.

Now the problem we have reduces (DUCY?) to finding a collection (S_1, S_2,...S_n) of subsets of vertices in this hypercube such that:

a. S_1,..S_n are mutually disjoint
b. for every vertex v in the hypercube and every S_i, v is either in S_i or at a distance of 1 from some vertex in S_i.

Instead of formally defining how to construct (S_1,..S_n), I'll do it by example and you should be able to generalize. First try it for a 3-dimensional cube. It's worth trying to do it yourself, but you should see that the following collection of vertices suffices: { {001,110}, {100,011}, {010, 101} }

Now extend to 4 dimensions (and higher) as follows: "Draw" a 4-dimensional hypercube in the usual way, as a cube nested inside another cube. Color the outside cube by the above 3-dimensional scheme. Color the inside cube similarly. You're done.

CptR,

Hmm. Cool idea. I've barely just come across checksums, so I don't have any sort of decent grasp on how they work yet, but I should prob take this as a sign that it's time to learn. I've been trying to browse a bunch of info on crypto anyways, and some of it's starting to sink in, so you bringing up an idea like this is kinda making me happy inside.

So yeah. Thanks.

TSB,

Awesome.

Having never worked with hypercubes before (like, at all), there's a pretty good chance I'm misunderstanding a bunch of things (and a guarantee I don't fully grasp the potency of doing this), but I think I managed to draw a picture in my head of what you're saying.

So, you're talking about using this to determine the answer this:

...as opposed to the original problem, correct?

If so, I'm not sure I understand this:

.. especially b.

If more than one vertex is at a distance of 1, wouldn't that give us an unwanted state?

The chances of me completely misunderstanding you are pretty high, since I'd never even thought of shapes like this before, and maybe you don't feel like explaining here or whatever, which is cool. But thanks for thinking/writing this out anyways - I'll be thinking it over for a while and hopefully eventually figure out how it works. Maybe. :P

Okay, I might understand more now...

Or at least misunderstand less.

Basically, while I do see how this would be super useful and is a pretty awesome way to simplify the game, I'm still struggling to see what Player2 is supposed to look at to know what answer to give to the dealer.

Like, say I find an adequate set of subsets. Now we'd have to come up with an agreed upon correlation for each subset within so that each maps to a number 1-52 so that no number is left out?j

But even that wouldn't guarantee that Player1 could alter the board such that the new permutation was contained within our found set AND mapped to the specific card we needed to identify this time, although all boards could be altered to a state within the found set.

Am I completely misunderstanding this?

Oh btw, I get now that you're talking about the original problem. What I'm still unclear on is how the identify or position of the target card gets reliably communicated to Player2. This is a really cool way to map this, though.

Every S_i corresponds to one of the cards. Every dealer state can be mapped to an S_i by flipping one card (distance 1 from a vertex in every S_i) or not flipping any cards (or in S_i).

In the n=3 example given, the first card could map to the 001 set, the second to the 010 set, and the third to the 100 set. So for example, if the dealer sets the 000 state, the player would flip the card of interest. If the dealer would set the 001 state, the player would flip no cards to indicate the first card, the third card (leaving 101) to indicate the second, and the second card (leaving 011) to indicate the third.

Yeah Mntndrew is right. So when player 2 comes in, the cards should be in a state that belong to one of the S_i's and he just needs to figure out which i it is and therefore which card. In the n=3 example, you should verify that whichever state the dealer leaves the cards in, player 1 can get to any of the 3 subsets by flipping at most one card.

I'm not sure this is the intended solution of the puzzle by whoever designed, it seems a little too brute force. But hey, it works !

Maybe if you guys gave an example of a 3-card problem, that would help.

So if the board is 011, which card is it?