Can You Solve The 6 Cards Game?
Polish translation by Przemyslav
Znajdź mnie tutaj:
Reference to TED-Ed talk by Maurice Ashley
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I think this fails in real life situations, like most theories sadly.
What if the opponent is holding a 4? If my opponent is holding a 4 he can strive to get a better number by trying to trade with me, if he assumes I am holding a higher number.
I don't get how it is logical to assume he can only be holding a 1 if he is asking to trade.
The only safe bet you can make is that he will NOT ask for a trade and that's only when he has a 6.
It is reasonable for a chessmaster to ask this question, because he is playing with opponents who are just as capable. But practically you are much safer in considering the emotional barometer of that person if you really want to gamble
I like this video but I do not think you can use IESDS in this case. This is not a game of complete information and therefore you should be using a different solution concept (e.g. https://en.m.wikipedia.org/wiki/Bayesian_game ).
It should be the strategies I have are “accept offer” and “reject offer” and your strategies are “make offer” and “don’t make offer” and each of us can be one of 6 possible types. The way you have it, my strategies are: accept offer and have the number 1, reject offer and have the number 1, accept offer and have the number 2, reject offer and have the number 2, etc. In that case the game would be a 12×12 game and you could use IESDS to solve it. However, I cannot choose to have a number. You and I both have types and we form beliefs about each other’s type (and therefore this is a Bayesian game).
Clearly, everyone watching this video is not acting in a way consistent with the solution you proposed and an alternative solution concept will likely yield a more realistic prediction.
Should have stated that the higher card wins the game before anyone selected. Then the answer would be obvious.
It can't be much of a game if the option to trade was only made one time. Why would anyone play if the option was only given once in a while?
I kinda reasoned it out like this
If you have a 2, then there are 4 numbers that you could lose to
1 is the only number that you could win against
the other player asked to trade cards. The only rational reason this other player asks to switch cards is if his card is a 1, since he knows he will lose
thus, the answer should be no, you should NOT trade cards
This assumes both players use this strategy. What if a player used a simpler strategy – ask to switch if you had 1,2,3 and don't ask to switch if you had 4,5,6?
Sorry but that was not the game you solved for.
The only issue with these kinds of problems is that they seem great and quite intriguing on paper, but when applied in practical life they seem to throw logic out of the town.
no because when we play there are only six card if we pick up any two of them randomly say 5,2 or 3,1 anything there is possibility that my card is below 4,5,6 and other person may also have posibilty that is her card is below 6,5,4,3 so if she ask me to trade there is posiblity that i gamble for that 60% chance of wining
I, should swap them.
Plot Twist:The six was negative.
ok i agree with pretty much everyone arguing that this wouldn't work if there were more cards, and i know exactly nothing about game theory, but my instinct when he asked to trade cards was to think "why would you want to trade unless you knew you were going to lose," and the only way you know you're going to lose is if you have a 1. however, this only works in a perfectly logical system, and not in one where either player bluffs. this sounds like a gambling-type situation rather than a 100% logical one, so i don't think this strategy would work period. but maybe that's just because i don't trust people.
The solution was brilliant
I agree with the strategy.
However, if we look at it logically, the oponent could be holding a 3.
In this case it is logical to ask for a trade, given that there are the number 1,2,4,5 and 6 remaining. Three of those five cards beat the oponents card, so statistically, it makes sense for him to trade. If hes holding a 4,5 or 6 he wouldnt trade.
Therefore, the answer is a 50/50, we either believe he has a 1 and dont trade, or believe he has a 3 and trade.
Ok, but what if you have 1000 cards, you have a 2 and the other asks you to trade cards. Now is a lot more tempting
Wait, if I had 3, I would like to trade because I can recieve 3 that are higher, and 2 that are lower. So the puzzle is incorrect
Logic is incorrect. If other guy took 3, he can think: "Probability that first guy has better card then me is 3/5" so he will want to ask first guy to trade. So cases that other want to trade are: 1 and 3. So chance if 50-50. But if You include a chance of bluffing then it is obvious You should want to trade.
I feel a little bit bad for Presh here. He getting lots of downvotes and comments from people that got it wrong. Your situation is one where your opponent wants to win and his card is so shitty, he's willing to offer you a trade, and the video explains how shitty exactly.
Think Monty Hall but Monty is on a budget and wants the contestant to lose, so he only asks them if they want to switch when they chose the correct door, and otherwise he's like "too bad, you got a goat" (the equivalent of simply flipping over the 6 instead of offering to trade it).
Odds say trade a three. Like your vids, but this one don't fly.
Would you like to trade cards
🅰,🅱,🆎,🅾,🔠&🔤.
There are 6 cards.
And then the judge ordered the execution on a Wednesday, and you were indeed surprised.
The difficulty with talking about people’s strategies for playing this game is that the game is more of a thought experiment rather than a game that actually gets played much. However, as a thought experiment, it has applications. Suppose I call you late in the evening and ask you if you would like a pet boa constrictor. I tell you that they make great pets, and my beloved pet boa constrictor, Squeezy, needs to find a new home NOW. You might have reason to believe that Squeezy is not actually a great pet, so even if you don’t know that Squeezy ate my Great Dane, you would refuse my offer.
Do you want a mathematical answer or a realistic one, because you wouldn't ask that question unless you had a card you WANTED to trade in the first place…
Not 100% sure you'd be holding a 1 necessarily, I could see asking a trade if you were holding a 3 — 50% chance of increasing your card, though I could turn you down.
You FAILED to explain in the statement of the problem that YOU are allowed to REFUSE switching cards. And I saw that ted talk, he didn't create the logic puzzle, he's just obsessed with it, someone else invented it. Also it's way more interesting when it's a lot more than 6. If it's 100 for instance. Should you request a switch if you get 37? If you get 23? 17? 12? It's more interesting with 100 because the game theory answer isn't the real answer. The key is that your opponent doesn't KNOW where your threshold is. Suppose we're playing with 100 cards and you have 3 and little do you know I get 11. Would you not want to ask for a trade because the game theory answer is only trade if you have a 1? You're going to regret not asking for a trade when I win the game 11 to 3. But if you had asked, there's a good chance I would have been happy to trade my 11 away not knowing you actually got something even lower. And if we played an iterated version of this game and we were playing competitions against many other competitors, I may well outperform you by being willing to trade let's say anything 11 or less, though the best threshold is really probably around 6 or 7 I think. It's just like that famous competition Richard Dawkins had with the various algorithms doing the iterated prisoner's dilemma. The winner was a "nice" strategy that never betrayed until the other player did it first, rather than a "mean" strategy that just defected every time.
This is presuming both players use logical thinking and eliminating possibilities. In a simpler and more likely strategy, the player offering the card would only trade if it were more likely to get a higher number. Holding a 4, 5, or 6, that's unlikely, so the player holding a 2 has about a 50/50 chance of improving.
This one somehow blows my mind. The logic makes sense, the maths make sense, yet my brain does not want to fully accept that you should only ask to swap your cards if you have a 1. Especially when the amount of cards gets bigger than what I can comprehend.
By the same logic, not trading a 1 is a strictly dominated strategy, so we eliminate that option, making not trading a 2 a strictly dominated strategy. Therefore, if we have a 2 we should trade.
I'm tempted to disagree with the conclusion / solution, since the other player being a "perfectly rational person" isn't part of the puzzle. In other puzzles, this detail is mentioned (things like "both are mathematicians and employ the best possible strategy", etc) – but that's not part of this puzzle. This means, we do not know if the other person is perfectly rational and have to conclude they likely are not (assuming most people would not be able to solve this puzzle).
I feel it's actually irrational to expect a perfectly rational counter-player, so the answer should still be "yes".
The "yes" answer is easy to prove: play this exact game with any family member or friend, with you asking for the trade. The numbers don't matter, besides 1 or 6 (in that case, play again). Afterwards ask them why they answered "yes" or "no". I doubt any of them would answer with "I applied IESDS".
Game theory, strategy, IESDS… etc., etc. And then there's luck and risk – which doesn't care about any of that.
that's what we call a pro gamer move
A lot of people in the comments are not understanding the problem.
The second player is not gambling, hes trying to win and he uses perfect logical reasoning.
if he got a six he would not ask for a trade.
so if he is asking to trade he doesnt have a six.
If he had a 5 he would assume i dont have a 6 because no one would trade a six.
So the max he can have by asking that question is 4
but if he had 4 he would not ask for a trade because the max he can get is 3, 2 o r 1
if he had a 3 he could go for a 2, 1 or 4 but whoever got the 4 would not accept a trade.
Again by asking for a trade he can only be with a 3 or 1
No one with 4,5 or 6 would accept a trade for a 3 or 1 possibility.
So by asking that he already expect you to have the 2 cuz he got the one.
It doesn't matter if you trade or not, because all the cards are poisoned and I've developed an immunity to the poison. 😁
Unless you are playing against someone who doesn't think. And one more question. If I saw the card as '2' and my opponent didn't say anything, it means 1)he has a higher card or 2)he has a '1' and stay silent to make me think he has a higher card. In such case would I ask to trade cards?
I'm trying to explain this video to my grandfather. He (somewhat) agrees with me that if this was played with 6 cars the optimal move would be not to switch but I can't find the words to explain the logic behind it. I'm also explainign that this logic applies with any number of cards so long you pick a 2. Doesn't believe me.
Well, the opponents don't always use perfect strategy. But sure, not trading is the best option overall.
If opponent has a three he thinks to himself "he wont trade with 4,5,6…" "he will trade with 1,2 and if that happens I lose.
If opponent has a one he thinks to himself "he wont trade with 4,5,6 but he might trade 2 or 3, if he doesnt get the catch"
I came to the same conclusion as this puzzle, but I took a different path to get there.
I started by assuming that the opponent chose to trade rather than was compelled to do so. If the opponent had to make the offer, then the problem reduces to pure probabilities, and the problem is easy–too easy for this channel.
So, under what circumstances would the opponent trade? If the opponent has 4, 5, or 6, the move doesn't make logical sense, since it's more likely that the opponent has the superior card. A rational player would not trade away a better chance of victory, therefore a rational opponent would not make such a trade.
What about a 3? It's possible that a naive opponent would make such an offer with a 3, but a more rational opponent would realize that the only way they would benefit is if the player accepted the trade with a 4, 5, or 6, which (for the reasons above) was not likely to happen. While there are a few circumstances where the opponent would offer such a card (such as a double bluff, or mistakenly basing their decision based off of pure probability), it is very unlikely.
An opponent could conceivably offer a 2, but since the player already has the 2, we already know that they don't have this card.
Therefore, the most likely scenario is that the opponent currently holds a 1, and thus it behooves the player to not trade their 2.
It doesn't make any sense
This answer only applies if you play against player who thinks with this algorithm or a computer otherwise when you play against "normal" person you should trade when you have 1 or 2 but if your opponent plays same strategy you never trade unless you have 1 and 2 so in every cases "winner" would be random and even if you trade only 1 sice both cant have 1 this "game" is pointless and trading never happens
I at least would not treat the question "would you like to trade cards?" as a definitive indicator of the other player wanting to trade. They could just want to gather information about what you might have, and if you said you did want to trade cards, they could still refuse. If they said "I want to trade cards, do you?", then I would treat the problem as it was handled here.
Another problem is that the question also presented a new rule to the game, the possibility of trading cards. If I was playing a game like this for the first time, my assumption in this situation would be that the other person asked that to explain the rules on the go, not to actually suggest a card trade. If the rules were familiar to me beforehand, then I would treat the question differently. But it still is not a clear indicator of the other player wanting to trade.
Wrong. You're conflating removing the STRATEGY of trading the 6, with removing the 6 card.
Why wouldn't he possibly want to trade if he has a 3? I think saying yes or no has the same chance of winning.
The logical solution : yes
The real-life solution : no
My thought on this:
First, I eliminated the possibility of player 2 having 4, 5 or 6. This is because there are more cards lower than this number than there are higher, and it wouldn’t make sense to trade them. For a minute I thought that 1 and 3 have equal chances and what P1 chooses doesn’t matter, but then I realized P1 won’t trade 4-6. So only 1 and 2 are possible for what P1 would give in exchange for the 3. So P2 won’t trade the 3. Since P1 already has 2, the only card that P2 can have is 1. 2 > 1. (Or P2 didn’t think that far and had 3 in his hand. You never said that P2 is a perfect logician lol)
Let me correct the question: you see your card is a 2 and the other player looks at his card and quickly asks if you want to trade, which is not part of the rules of the game…. What should you do?
I do not agree. The strategy here is "trading or not." 6 is the given card, not the trading. So "trading 6" cannot be removed as a "dominated strategy."
i feel like the logic on this one doesn't hold up.
Player A has a 2, Player B has an unknown card. Player B offers a trade. We've ruled out the idea that Player B has a 6, because they'd never offer to trade the highest card. So the next number to rule out is 5. But what's stopping Player B from having a 5?
In the video, the question they answered was "If Player A has a 5 and Player B has an unknown card, and Player B offers a trade, should you take it?" And the answer is no, because Player B would never offer to trade a 6, the only card higher than 5. And that makes sense. But the question that needs to be answered is "If Player A has a 2 and Player B has an unknown card, and Player B offers a trade, is it possible that Player B has a 5?" And that question seems to not be answered at all in the video.
What am I missing?
kds
Who says that the other person is totally rational even though I came up with the logic knowing a bit about game theory before encountering this problem.
Fantastic reasoning. But I still find this game a bit ‘meh’. Assume Player A (the one with the 2) and Player B are both perfect logicians. Then simply by asking the question, Player B would communicate to Player A that he is holding a 1, and he therefore knows that Player A would answer ‘No’. Asking the question is the same as saying “I am holding a 1 and I have therefore lost the game”.
So the rational choice for Player B would be NOT to ask the question. If B did not ask the question, A would need to conclude that B either a/ holds a 1 (and does not wish to advertise this fact) or b/ holds a card with a value greater or equal to 3. At this point, the game can only be resolved by showing cards and B would then have lost.
So IF the two players are perfect logicians, the option B has in this game has no value. If he holds a 1, he lost. The fact that he holds a 1 is advertised to the players either by B asking the question to trade or by B not asking the question, leading to a show of hands. B’s option has value only if A is NOT a perfect logician.
I suppose this is the case in real life though so the lessons in game theory that we can take from this are still valid of course.