stigant
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Ok, here's a puzzle that I really liked that appeared in Scientific American.

Three people, A, B, and C are playing a card game. Aces are high, followed by K, Q, J, 10, 9, 8 ... 2. Suits have no bearing on the game. Each player gets a card which they fix to their forheads so that they can't see it, but the other two players can. The person with the highest card wins. In case of a tie, the two (or three) people with the highest card both win.

Now, the trick is, how do they decide who wins? Each player, starting with A (then B, then C, and back to A again and so on until someone is declared the winner), is allowed to make one of the following statements:
"I win" - they know, based on what they can see and what the other players have said, that they have the highest card, and that the other two players have lower cards (no ties).
"I don't lose" - they know that they either win or tie for the win.
"I don't win" - they know that they don't win outright. They may tie for the win, or may lose, but they don't have THE highest card
"I lose" - they don't have the highest card, nor do they tie for the win.
"I don't know" - they can't make any of the above statements with 100% assurity.

Assume all three players are perfect logicians. The following dialogue occurs in one of the games:
A: I don't know.
B: I don't know.
C: I lose.
Who won and what card did they have?

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Hix
Level: Master Delver

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Examine each possible response for players A and B in turn.

Player A has no information about his own card, and thus will not be able to answer "I win" or "I lose" with certainty. He will answer "I don't lose" if he sees a pair of 2's, and will answer "I don't win" if he sees even a single ace. Thus, his answer of "I don't know" tells us that:

(*) B and C do not both have 2's
(*) Neither B nor C has an ace

Player B now has a small bit of information about his own card. Nothing contradicts his having a 3, so he will only be able to answer "I win" if he sees a pair of 2's (then he will know that he does not have a 2 himself). He will only decuce "I don't lose" if C has a 2 (thus B does not have a 2) and A has a 3. Now nothing contradicts his having a K, so B will only be able to decuce "I lose" if he sees an ace, and "I don't win" if he sees a K (both of these because he knows he does not have an ace himself). Thus B's answer of "I don't know" tells us that:

(*) A and C do not both have 2's
(*) If A has a 3, the C does not have a 2
(*) Neither A nor C has an ace
(*) Neither A nor C has a K

The information marked by (*) is the only information Player C has concerning his own card. None of this contradicts his having a Q, so he can only decuce "I lose" if he sees a K or ace. Of course, we already know that no player has an ace, and that neither A nor C has a K. Thus, Player B must have a K, and is the solo winner.

[Edited by Hix at Local Time:02-21-2005 at 09:02 PM]

stigant
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You are correct sir. There were some really devilish problems with the same set up. One has (I believe) 7 or 8 consecutive I don't know's followed by a I Win!


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Hix
Level: Master Delver

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Alice and Zack play the following variation of draw poker, which uses a standard deck of playing cards (4 suits, each with 13 ranked values).

Alice selects any 5 cards from the deck to form her hand, which is placed face-up so both players can see it. Then Zack selects any 5 of the remaining cards to form his hand, which is also placed face-up. Next, Alice discards any cards she wishes from her hand, and selects additional cards from the deck to bring her total number of cards to 5 again. Finally, Zack discards any cards he wishes, and selects additional cards from the deck to bring his total number of cards to 5 again (Zack may not use Alice's discards).

Can either player choose cards in order to always end up with the better poker hand? If so, briefly explain which cards that player must choose to force a win. Otherwise, briefly explain which cards each player must choose to force a draw.

Standard poker hand rankings are used, which can be looked up here.

stigant
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Alice can force a draw since she can draw a royal flush (AKQJ10 of one suit) on the first hand (and discard no cards)

Can she force a win?

She would have to prevent Zach from drawing a royal flush on his first hand. There are a couple of ways to do this.

She could draw 4 Aces (or 4 Ks etc). If she does this, however, Zach can draw 4 of a kind as well, leaving too many straight flushes available for Alice to either upgrade and block him from doing so or maintain her four of a kind and block his straight flush.

She could draw A from one suit, K from another, Q from another, J from another, and a 10. If she does, Zach also draws an A high straight from 4 different suits. Then Alice must draw a 10 high straight flush - no other hand can completely block off straight flushes and be otherwise unbeatable. Then Zack will have a 10 high straight flush available as well.

I don't see any other obvious strategy... drawing a full house (say A's full of K's) allows Zack to draw a four of a kind (in this case, 4 Qs + J), forcing Alice to draw a J-high straight flush which Zack can actually beat (since he can draw a Q high straight flush).

drawing lower hands lends itself to similar counters from Zack.

So I would say Alice can force a draw and no better.

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Maurog
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Actually, I believe the very best move for Alice is four 10's and an A. Now, Bob..er..Zach is in serious trouble - first of all, he has not a single ace-high flush left. Secondly, Alice threatens an ace-high flush the next move, and already has an ace. Clearly Zach has to cut this flush down so he must choose four different suits higher than 10 and another card. As soon as he does that, however, Alice builds a 10-high royal flush and wins. Of course if Zach blocks a low royal flush, by picking four different suits lower than 10 Zach loses because Alice builds an ace-high royal flush and wins. Since Zach can't pick 8 cards, he loses in any case.
Alice is the winner!

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stigant
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Hmmm, nice.

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Hix
Level: Master Delver

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Yes, Maurog, Alice can win by choosing 4 10's on the first move. I didn't do an exhaustive analysis, but Bob/Zack seems to be able to draw if a 10 is available.
Tag.

Maurog
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Hmmm... I'm getting out of puzzles here.
Okay, how about this - which of the following numbers is the odd one out?
Obviously, I expect an elaborate explanation of the reasons. The choices are:

1) 13.
2) 17.
3) 19.

Game on

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stigant
Level: Smitemaster

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13 (although I'm tempted to say they're all odd men out since they're all odd).

Here's my reason:
13^2 = 169
17^2 = 289
19^2 = 361

In the last two cases, the product of 4 and the last digit of the original number gives you the first 2 digits of the square. (7*4 = 28, 4*9 = 36).

Another number with this property is 18 (8*4 = 32, 18^2 = 324)

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Maurog
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Incorrect. For similar reasons I can claim it's 19 because 13^3 = 2197 and 17^3 = 4913 and 21 divides by 3 and 49 divides by 7, but 68 won't divide by 9 (19^3=6859). Finding some obscure correlation is not enough. It has to be pretty solid to be the answer.

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krammer
Level: Smitemaster
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One obvious answer would be that 17 is nine letters, 13 and 19 are only eight letters.

Alternatively, 17 and 13 are both written with four roman numerals (XIII and XVII), 19 is written with three (XIX).

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Hix
Level: Master Delver

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No, No. The obvious answer is 17, as a regular 17-gon can be constructed using only a compass and straightedge; but you can't get a regular 13-gon or 19-gon.

Maurog
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I think we are all aware that the possibilities for correlation between two out of three numbers are infinite. However, my puzzle does have an answer. A solid, good answer with a reasonable explanation that everyone will agree is valid. It's up to you to find it

Good luck.

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Schik
Level: Legendary Smitemaster

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Maurog wrote:
Hmmm... I'm getting out of puzzles here.
Okay, how about this - which of the following numbers is the odd one out?
Obviously, I expect an elaborate explanation of the reasons. The choices are:

1) 13.
2) 17.
3) 19.

I'd guess that 2 is the odd man out, since it's the only even number.

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Hix
Level: Master Delver

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1 is the odd man out, because it is the only perfect square.

Maurog
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Very good, Schik! You were the first to notice the extra three numbers. Sadly, 2 is not the right answer, because "even" is after all nothing more than "divides by 2". If you can claim that 2 is the only one that divides by 2, you can claim 3 is the only one that divides by 3 and so on I'll give you a point anyway though.

Hix, being an "odd man out" means not having an otherwise common feature. If they were all perfect squares except one that would work. C'mon you're very close now.

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Schik
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Okay, 1 is the only one that doesn't have exactly two positive integral divisors.


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Maurog
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Yup, you got it. 1 is the only non-prime number among all listed. Game on.

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Schik
Level: Legendary Smitemaster

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Okay, here's a pretty easy one:

The island of Oyst is inhabited by Smitemasters and Delvers. Smitemasters always tell the truth, and Delvers always lie.

You meet nine inhabitants: Beethro, Bombus, Erik, Mike, Gerry, Neil, Eytan, Clayton and Adam.
Beethro claims that only a Delver would say that Eytan is a Delver.
Bombus claims, "Gerry is a Delver."
Erik says that Beethro and Adam are Delvers.
Mike tells you, "I know that I am a Smitemaster and that Eytan is a Delver."
Gerry says, "Mike and I are both Smitemasters."
Neil tells you that at least one of the following is true: that Eytan is a Smitemaster or that Erik is a Smitemaster.
Eytan says, "It's false that Mike is a Delver."
Clayton says, "It's not the case that Erik is a Delver."
Adam tells you that Mike is a Delver or Neil is a Delver.

Can you determine who is a Smitemaster and who is a Delver?

[Edited by mrimer at Local Time:02-22-2005 at 09:03 PM: just monkeying around]

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Hix
Level: Master Delver

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Maurog wrote:

Hix, being an "odd man out" means not having an otherwise common feature. If they were all perfect squares except one that would work. C'mon you're very close now.

In general, this is a particularly useless distinction to make, unless you are working in some formal system where the common "feature" can only be defined positively. The notions of "nonsquare" or "squarefree" are two very important features that a number may or may not have.

stigant
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why is "being prime" or "not being prime" more interesting than "being divisible by 2" or "insert random interesting numerical fact about a number and its square, or a number and its cube" or even "being one of my 27 favorite numbers"?

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Garlonuss
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Schik wrote:
Okay, here's a pretty easy one:

The island of Oyst is inhabited by Smitemasters and Delvers. Smitemasters always tell the truth, and Delvers always lie.

You meet nine inhabitants: Beethro, Bombus, Erik, Mike, Gerry, Neil, Eytan, Clayton and Adam.
Beethro claims that only a Delver would say that Eytan is a Delver.
Bombus claims, "Gerry is a Delver."
Erik says that Beethro and Adam are Delvers.
Mike tells you, "I know that I am a Smitemaster and that Eytan is a Delver."
Gerry says, "Mike and I are both Smitemasters."
Neil tells you that at least one of the following is true: that Eytan is a Smitemaster or that Erik is a Smitemaster.
Eytan says, "It's false that Mike is a Delver."
Clayton says, "It's not the case that Erik is a Delver."
Adam tells you that Mike is a Delver or Neil is a Delver.

Can you determine who is a Smitemaster and who is a Delver?

[Edited by mrimer at Local Time:02-22-2005 at 09:03 PM: just monkeying around]

Beethro = Smitemaster
Bombus = Smitemaster
Eric = Delver
Mike = Delver
Gerry = Delver
Neil = Smitemaster
Eytan = Smitemaster
Clayton = Smitemaster
Adam = Smitemaster


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Hix
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If Eytan tells the truth, then so does Mike; but if Mike is truthful, then Eytan isn't. From this contradiction, we conclude that Eytan lies. Also, Beethro is lying to intimidate anyone who would accuse Eytan of lying. Eytan's lie tells us that Mike lies. So Gerry is lying about Mike, and Bombus is telling the truth about Gerry. Adam tells the truth about Mike, so Erik must be lying about Adam, and Clayton is lying about Erik. Finally, we see that Neil is lying about both Eytan and Erik.

So among our nine inhabitants, we have only 2 Smitemasters (Bombus and Adam), the rest being mere Delvers.

Schik
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Hix has got it. You're up!

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RoboBob3000
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I think Schik was lying about this whole conversation ever taking place.

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Garlonuss
Level: Master Delver

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There is no way that both Eytan and Mike can be the same thing. Mike says "I know that I am a Smitemaster and that Eytan is a Delver." If he tells the truth, then he is a smite master and Eytan is a Delver. If he lies, then he is a delver and Eytan is a smitemaster.

Therefore, if mike is correct and he is a smitemaster and Eytan is a delver, then when Eytan says that mike is a smitemaster (by saying he is not a delver) then that is the crucial contradiction. If mike is lying and he is a delver while eytan is a smitemaster, his statement that Mike is a delver proves he is a smitemaster and that clears up the central conflict of the puzzle.

[Edited by Garlonuss at Local Time:02-22-2005 at 10:00 PM]

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Sometimes they don't make much sense.
Refrigerator.

Maurog
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Positive definitions of the type "These numbers have the feature X" define a group. When you have a group, you can have a number to stand out from the group. Negative definitions of the type "These numbers don't have the feature X" define the complement of a group. A positive definition binds all numbers but one (they all have to belong to the group). A negative definition binds only one number. Therefore, a negative definition is much, much weaker than a positive one. A good example of a negative definition would be "All numbers that are not 8". Yes, 8 is the odd one out in this "group", yet I doubt you will pick 8 as the odd man out given the numbers "2, 4, 8, 16, 17, 32, 64".

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Schik
Level: Legendary Smitemaster

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Garlonuss wrote:
There is no way that both Eytan and Mike can be the same thing. Mike says "I know that I am a Smitemaster and that Eytan is a Delver." If he tells the truth, then he is a smite master and Eytan is a Delver. If he lies, then he is a delver and Eytan is a smitemaster.

The opposite of "A AND B" isn't "NOT A AND NOT B" - it's "NOT A OR NOT B".

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stigant
Level: Smitemaster

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Maurog wrote:
Positive definitions of the type "These numbers have the feature X" define a group. When you have a group, you can have a number to stand out from the group. Negative definitions of the type "These numbers don't have the feature X" define the complement of a group. A positive definition binds all numbers but one (they all have to belong to the group). A negative definition binds only one number. Therefore, a negative definition is much, much weaker than a positive one. A good example of a negative definition would be "All numbers that are not 8". Yes, 8 is the odd one out in this "group", yet I doubt you will pick 8 as the odd man out given the numbers "2, 4, 8, 16, 17, 32, 64".

That's not the point... the condition I gave for 17 and 19 IS a positive definition (ie when you multiply their last digit by 4, you get the first two digits of their square), and 13 fails it. For that matter, being prime could be (and usually IS) stated negatively: Any number that does not have divisors other than itself and 1. Whereas being composite is stated positively: HAS divisors other than itself and 1. My point is that none of these groups (whether stated positively or negatively) is inherently more interesting. Given any group of numbers plus an additional number, I can find a (positive) group definition which all of the original group will satisfy and the addition number will not. This is essentially the basic problem with inductive reasoning. Absent any axiomatic context in which to consider these numbers, nothing interesting can ever be proven about them, only (equally) (un)interesting coincidences can be observed about them.

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