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NB: This is based on a puzzle over at BrainBashers.com.

The setup

You and a friend play a card game called Green Jack. The deck consists of 16 cards, divided into 4 cards in 4 colors (standard playing card abbreviations here, A = Ace, K = King, Q = Queen, J = Jack):

  • $\color{red}{\rm A,\,K,\,Q,\,J}$
  • $\color{blue}{\rm A,\,K,\,Q,\,J}$
  • $\color{orange}{\rm A,\,K,\,Q,\,J}$
  • $\color{green}{\rm A,\,K,\,Q,\,J}$

Cards are ranked as listed, A > K, K > Q, Q > J except the $\color{green}{J}$ which beats everything. If two cards have the same face value, then the color determines the win:

  • $\color{red}{A} > \color{blue}{A} > \color{orange}{A} > \color{green}{A}$

except, again, the Green jack, which beats everything.

The game is played as follows: you are dealt one card face up, and your friend is dealt one card face down. Your friend then makes three true statements, and you have to work out who has the higher card, you or your friend.

The hand

You are dealt the blue king ($\color{blue}{K}$). Your friend makes the following three true statements:

  1. My card can beat an orange queen ($\color{orange}{Q}$).
  2. Knowing this, if my card is more likely to be an ace (A), then it is a queen (Q); otherwise it is a king (K).
  3. Knowing the above, if my card is more likely to be a queen, then it's $\color{red}{red}$; otherwise it is $\color{blue}{blue}$.

Who wins the game?

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  • 1
    $\begingroup$ What do you mean by 'more likely'? Given what constraints? $\endgroup$ – Rand al'Thor Dec 26 '14 at 15:26
  • $\begingroup$ Answering your query requires me telling you how to solve it which would be bad, I would think. $\endgroup$ – Kyle Kanos Dec 26 '14 at 15:28
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    $\begingroup$ @adamsmith: I suppose if you truly wanted to be pedantic, that would suffice. I suppose most people are rational and would play the game fairly, speaking of the cards instead of whatever they wanted. $\endgroup$ – Kyle Kanos Dec 27 '14 at 0:35
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    $\begingroup$ @KyleKanos I think the bigger puzzle would be finding three questions that would accurately and logically define which player would win! :) $\endgroup$ – Adam Smith Dec 27 '14 at 0:40
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    $\begingroup$ I'd upvote your question, but "more likely" is vague enough that I wasn't able to solve the puzzle myself because I thought you meant something else. In particular, I thought statement 2 meant that if all the cards that player 2 could have, over half of them were aces. This interpretation lead to me concluding the other player also had a blue king. If you clarify that "more likely" means "more aces than any other individual face value, but not necessarily a majority," I'll upvote. $\endgroup$ – Kevin Dec 30 '14 at 7:11
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Constraint 1 leaves 10 possibilities: red A, red K, red Q, blue A, blue Q, orange A, orange K, green A, green K, green J.

Among these 10, 4 are aces, 3 are kings, 2 are queens, and 1 is a jack. If I'm interpreting the meaning of 'more likely' correctly, this means the card is 'more likely' to be an ace (than anything else), so constraint 2 means it's a queen. Now constraint 3 tells us it's red.

So your friend has the red queen, and you win.

I'm probably not interpreting this right though!

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  • $\begingroup$ That is the correct interpretation of "more likely" here. Have a pretty green check mark. $\endgroup$ – Kyle Kanos Dec 26 '14 at 15:36
  • $\begingroup$ @KyleKanos Thanks! :-) But having now looked at your link to Brainbashers, I think their phrasing of the 'more likely' thing was a lot clearer. $\endgroup$ – Rand al'Thor Dec 26 '14 at 15:39
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    $\begingroup$ Going back, I think it is more clear with the "knowing this" aspect. I've updated my question to include that phrase. $\endgroup$ – Kyle Kanos Dec 26 '14 at 15:57
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An alternate interpretation of "most likely" (before the "knowing this" edit):

Statement 1 is as rand al'thor stated.

Since statement 2 and statement 3 are true, then the card must be either a Queen or King, and Red or Blue, regardless of the interpretation of "most likely". Since the Blue King is in play, the 3 possibilities are the Red Queen, the Blue Queen and the Red King.

From this, the card is unlikely to be an Ace (0% probability) and likely to be a Queen (66% probability), so the mystery card is the Red King and you lose.

Of course, the "knowing this" edit throws off this interpretation, but now the intended answer is trivial.

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    $\begingroup$ ... and hence this interpretation must be wrong. $\endgroup$ – Rand al'Thor Dec 26 '14 at 15:57
  • $\begingroup$ I edited to create a non-contradictory answer. $\endgroup$ – McMagister Dec 26 '14 at 16:33
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I'm thinking the other guy got a Red Queen, meaning you win.

Alright, I'll elaborate. The first truth eliminates some possibilities. It leaves you with everything that CAN beat an orange Queen. These are Red Ace, King, and Queen, Blue Ace, King, and Queen, Orange Ace and King, Green Ace, King, and Jack.

You already know that the other guy can't have a Blue King, as that's the card you drew. So you're left with (and I'll just abbreviate) RA, RK, RQ, BA, BQ, OA, OK, GA, GK, and GJ.

The second truth, since the card IS more likely an Ace, tells that the card is a Queen. This leaves us with only two choices: RQ and BQ. [(It's more likely an Ace because there is a larger pool of Aces being the possible card) 4:3:2:1 Ace:King:Queen:Jack]

The third truth simply finishes. Since the card IS more likely a King, making it a Queen, then the card is Red.

Red Queen.

I didn't know we needed explanations and such with these. Hopefully this one'll suffice.

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  • $\begingroup$ This is what I put in my answer 9 hours ago, only without explanation. $\endgroup$ – Rand al'Thor Dec 27 '14 at 1:20
  • $\begingroup$ I suppose you got this account for the kofia hat? $\endgroup$ – Kyle Kanos Dec 27 '14 at 2:41
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    $\begingroup$ Would it be possible for you to edit in a more detailed explanation as to why you believe this answer is correct? By site policy, answers need to contain detailed explanations, not just be guesses, otherwise they may be deleted. Thank you! $\endgroup$ – Doorknob Dec 27 '14 at 4:21
  • $\begingroup$ After reading al'Thor's answer, I think we have the same explanation. Is there another way to solve the problem? $\endgroup$ – Charles Dec 27 '14 at 22:38
  • $\begingroup$ @Charles No worries! Sometimes it happens. You're free to delete your answer (there's a delete button underneath this), though it's fully optional. If you feel you can edit to explain it better than the existing answers, or if you feel that it already does, then you're free to leave it! $\endgroup$ – Aza Dec 28 '14 at 0:09

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