A group of 10 logicians have been challenged by a puzzlemaster to play a game.
"I have here two sets of 10 cards, each with cards numbered 1 through 10. As you can see, one set is colored blue and the other is colored red. I will deal a blue card and a red card to each one of you. The red card can be made available for everyone to see, but the blue card should be kept private to you. Make sure you do not communicate with your associates regarding the value of your blue card."
One they complied, he continued:
"Your goal is to guess the value of the blue card possessed by the person who has a red card with value identical to that of your own blue card. Your guess is made by raising your hand and saying a single number out loud when I call on you. Be warned! You may only speak if you are making a guess. Everyone is disqualified if anyone attempts to communicate in a subversive way. I know who has what card, so I can check the validity of your guesses. I will award 5 points for every correct guess and subtract 1 point for every incorrect guess. It goes without saying that someone who has already guessed correctly can no longer make guesses. You will see your points change in real time on the scoreboard. Good luck!"
What is the minimum possible final score if every logician plays optimally to maximize the final score?
Is it possible to generalize to N people? (Correct guess would be worth N/2 points)
Hidden Cards
Moderators: jestingrabbit, Moderators General, Prelates
Hidden Cards
Last edited by Cradarc on Mon Nov 21, 2016 12:40 am UTC, edited 1 time in total.
This is a block of text that can be added to posts you make. There is a 300 character limit.
Re: Hidden Cards
Cradarc wrote:Your goal is to guess the value of the blue card possessed by the person who has a red card with value identical to that of your own blue card.
Just running thought the numbers 1 to 10 will reach this goal. The logicians must have some other goal as well. I suspect they should be trying to maximise the score (by making as few incorrect guesses as possible), but you don't actually state this.
If I were playing, I would start by making a "guess" of my own blue number. I would then stay silent until I heard someone else say my red number, and if that turned out to be an incorrect guess, I would then "guess" their red number. If everyone else did the same, then the minimum score would be 40 or thereabouts (one wrong guess and one correct guess for everyone).
Cradarc also wrote:Make sure you do not communicate with your associates regarding the value of your blue card. [...] Everyone is disqualified if anyone attempts to communicate in a subversive way.
Does my strategy violate these rules? I'm communicating the value of my blue card, but it's not by a backchannel, it's blatant and in the clear.
Re: Hidden Cards
Sandor wrote:I suspect they should be trying to maximise the score (by making as few incorrect guesses as possible), but you don't actually state this.
Oops, yeah. It was implied, but I should have stated that.
Sandor wrote:If I were playing, I would start by making a "guess" of my own blue number. I would then stay silent until I heard someone else say my red number, and if that turned out to be an incorrect guess, I would then "guess" their red number. If everyone else did the same, then the minimum score would be 40 or thereabouts (one wrong guess and one correct guess for everyone).
1. They didn't have time to converse after hearing the rules, but I guess you can argue they can each individually deduce the optimal strategy. This is a loophole to the game though (see point 3).
2. Your method doesn't guarantee 40 points
Let P(N) denote the person with red card = N. and B(N) denote the blue card value of P(N).
Consider the scenario where:
B(2j) = B(2j+1)
B(2j+1) = B(2j)
With your strategy, a person guesses correctly every 2 turns. But what happens after 5 turns?
Everyone who has yet to guess correctly needs to guess the blue card of one of the people who has already guessed correctly (and can no longer speak).
At this point, the first guess has to be a random choice from the 5. The next person guess a random choice from 4, etc.
3. A better strategy (if we're assuming they are allowed to make fake guesses to communicate and all choose this strategy) is to simply wait until all but one person has said their blue number, then win back all the points. That would actually get you 41.
This is a block of text that can be added to posts you make. There is a 300 character limit.
Re: Hidden Cards
It makes sense for the logicians to at least follow this rule: if you know that you can guess correctly, then raise your hand immediately. Otherwise, wait a little to give those who can guess correctly a chance, and then raise your hand a little later if nobody else does (the time you wait should be slightly randomised, so that the uncertain logicians don't all guess at once). I hope that this doesn't count as sneaky communication: for me, sneaky communication would be more like waiting a number of seconds corresponding to the number on your blue card before raising your hand.
This means that anyone whose two cards match will guess right at the start, since they know that they can guess correctly just by naming the number on both their cards. To explain my strategy from this point on, I need a little notation. First, note that B is a permutation of the numbers from 1 to 10. Let's call the inverse permutation C. Let's say some uncertain logician P(i_{0}) is called on to make a guess. Then she should guess the number on her blue card. This guess will be wrong, of course, but now the other logicians know the value of B(i_{0}). Let i_{1} = C(i_{0}). Then P(i_{1}) can make a correct guess, namely B(i_{0}). This is all as in the previously described strategies. But we can go further. Now it is public knowledge that B(i_{1}) = i_{0}, and so, letting i_{2} be C(i_{1}), the logician P(i_2) can now make the correct guess i_1. This makes it public knowledge that B(i_2) = i_1, and allows P(i_3) to make a correct guess, where i_3 = C(i_2). And so on. In this way, the logicians corresponding to a whole cycle of the permutation B will all guess correctly (in reverse order around the cycle). Eventually the whole cycle will have been eliminated, and only uncertain logicians will remain.
We can add an extra trick to slightly improve this: if the logicians reach a point where some cycles have been eliminated and at most 3 logicians remain, then they can all guess correctly. First, they know that there are no 1cycles remaining (since logicians in 1cycles guess correctly right at the start). So they either form a 2cycle, if there are 2 of them, or a 3cycle, if there are 3. They know who comes after them in the cycle, and so they can work out who has which blue card and all guess correctly.
If the logicians follow this strategy, then there can only be at most 4 incorrect guesses. So in the worst case, the logicians get 46 points with this strategy.
This means that anyone whose two cards match will guess right at the start, since they know that they can guess correctly just by naming the number on both their cards. To explain my strategy from this point on, I need a little notation. First, note that B is a permutation of the numbers from 1 to 10. Let's call the inverse permutation C. Let's say some uncertain logician P(i_{0}) is called on to make a guess. Then she should guess the number on her blue card. This guess will be wrong, of course, but now the other logicians know the value of B(i_{0}). Let i_{1} = C(i_{0}). Then P(i_{1}) can make a correct guess, namely B(i_{0}). This is all as in the previously described strategies. But we can go further. Now it is public knowledge that B(i_{1}) = i_{0}, and so, letting i_{2} be C(i_{1}), the logician P(i_2) can now make the correct guess i_1. This makes it public knowledge that B(i_2) = i_1, and allows P(i_3) to make a correct guess, where i_3 = C(i_2). And so on. In this way, the logicians corresponding to a whole cycle of the permutation B will all guess correctly (in reverse order around the cycle). Eventually the whole cycle will have been eliminated, and only uncertain logicians will remain.
We can add an extra trick to slightly improve this: if the logicians reach a point where some cycles have been eliminated and at most 3 logicians remain, then they can all guess correctly. First, they know that there are no 1cycles remaining (since logicians in 1cycles guess correctly right at the start). So they either form a 2cycle, if there are 2 of them, or a 3cycle, if there are 3. They know who comes after them in the cycle, and so they can work out who has which blue card and all guess correctly.
If the logicians follow this strategy, then there can only be at most 4 incorrect guesses. So in the worst case, the logicians get 46 points with this strategy.

 Posts: 60
 Joined: Fri Jun 15, 2007 2:33 am UTC
 Location: Multiple Places (only one now that you read this...)
Re: Hidden Cards
I see your 46 and raise you 47 with the following strategy:
Of course, this strategy only works if all the logicians come up with it together. Some sort of superrationality is needed for this to be useful, I think.
Spoiler:
Of course, this strategy only works if all the logicians come up with it together. Some sort of superrationality is needed for this to be useful, I think.
Re: Hidden Cards
Spoiler:
Who is online
Users browsing this forum: Google Feedfetcher and 10 guests