Can You Solve The 6 Cards Game?
Vložit
- čas přidán 14. 07. 2020
- It takes perfect logical reasoning to solve this puzzle. I thank Joe for the suggestion!
Polish translation by Przemyslav
Znajdź mnie tutaj: forum.cdaction.pl/blogs/blog/...
Reference to TED-Ed talk by Maurice Ashley
• Working backward to so...
Subscribe: czcams.com/users/MindYour...
Send me suggestions by email (address in video). I consider all ideas though can't always reply!
Like many CZcamsrs I use popular software to prepare my videos. You can search for animation software tutorials on CZcams to learn how to make videos. Be prepared--animation is time consuming and software can be expensive!
Why are there comments before the video is published? Get early access and support the channel on Patreon
/ mindyourdecisions
If you buy from the links below I may receive a commission for sales. (As an Amazon Associate I earn from qualifying purchases.) This has no effect on the price for you.
Show your support! Get a mug, a t-shirt, and more at Teespring, the official site for Mind Your Decisions merchandise:
teespring.com/stores/mind-you...
My Books (worldwide links)
mindyourdecisions.com/blog/my...
My Books (US links)
Mind Your Decisions: Five Book Compilation
amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.1/5 stars on 96 reviews
amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 3.7/5 stars on 9 reviews
amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.2/5 stars on 18 reviews
amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.3/5 stars on 19 reviews
amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.6/5 stars on 14 reviews
amzn.to/XRm7M4
Mind Your Puzzles: Collection Of Volumes 1 To 3
amzn.to/2mMdrJr
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 26 reviews
amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.5/5 stars on 6 reviews
amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.1/5 stars on 7 reviews
amzn.to/1NKbGlp
Connect with me
My Blog: mindyourdecisions.com/blog/
Twitter: / preshtalwalkar
Newsletter (sent only for big news, like a new book release): eepurl.com/KvS0r
2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter. - Věda a technologie
Plot twist: After you said yes, his gonna say, “well I don’t.”
TeeHee-ous!
Here's another plot twist. If you say no, he's going to say, "well I do".
*he's.
Just helping you out, nothing else.
This logic seems to fall on the Unexpected Hanging Paradox.
It can't be a 6, so it can't be a 5, so it can't be a 4, etc.
So if there were 1000 cards, I got a 2, and my opponent asked if I wanted to trade cards, should I use this same logic and say no?
I think the answer to the original problem on the video, with 6 cards, is "no", as the video stated. But I don't think the logic used to get to this answer is right.
Play the game with 1,000 cards. I have a 20. Would my opponent ask me to change the cards, if he holds 400? Probably not, because he knew I would only accept with a very small card. But if he held a 30 in a 1,000-Card-game, he could ask too.
I think (!!!) you should accept, if you're holding a card, that belongs to the least 20-25%. But I cannot prove this assumption mathematically.
It's not unexpected hanging.
What he did is completely valid thing to do in game theory.
The problem with unexpected hanging is in the statement of the problem.
I think the point is that 2 is the card just before 3. So if you have 1000 cards the equivalent problem is: do you accept the trade if you got a 499 (1000/2-1).
The aswer is no, because if your opponent propose the trade it means that he have a card under 500 (so less than 499 because you have it), and he know that if you accept the trade that means that you have a card under 500. So at the end if you have the N/2-1 card on a N cards deck, there is no interest to you to accept the trade.
In other hand, if you got the 2 on a 1000 cards game, then accept the trade 😉
The logic holds for 1000 cards, or even a million.....granted, it only "works" if both players are known to be perfect strategists though. A player would NEVER ask to trade his card if it means making his position worse....and that's what happens if they are holding a 2 or greater. The other player will always choose "no" regardless of the question in a perfect strategy.
Using that, we can expand it....let's say you have 100 cards. A player would never ask above 50, as that would be a disadvantage, so eliminate the top 50. The other 50 draw a diagonal down the grid, if your number is greater than theirs, put a "W", otherwise an "L".....you'll notice that one side of the diagonal are all 'W' and the other is all 'L.' Assuming the ONLY time you would ever "want" to ask to change cards is if your "line" of the graph shows more "L" than "W".....for example, holding a "5" shows 4 wins and 45 losses, so if you are thinking that if they say 'yes' then you get to switch all of your "W" for "L" and vice versa in that line. Do that for EVERY line that has more losses than wins (flipping "W" and "L" in the line)...you will find that the first 25 are flipped and the second 25 are not.
.
Now for game theory, we look at the other player's decision on whether they want to accept your offer or not based on this new graph versus the previous position. They look at it from the column perspective though, and they find that choosing "yes" makes their odds worse, not better, for EVERY entry except when holding the "1" card.....so their decision will ALWAYS be a "no" unless they have that card and only that card. So their graph will have only the "1" column flipped and that's it
.
Back to the player who asks the question, now look at the graph, from their perspective, they know they will only say "yes" if they have "1".....this means your odds worsen by 1/(X-1) in asking while holding any other card. So the only card you can even ask for is the "1".....and the other player will say 'no' to everything but the '1'.....and you are holding it, so you lose.
The problem says “The HIGHER card wins the game”, not the card with higher number. So you may have cards with slightly different heights and different numbers, or equal height and different numbers. In the first case, the problem has not got enough constraints, in the second case it’s irrelevant to trade cards or not.
The question is poorly stated:
1) Is it a rule that you can ask to trade the cards or do you have an option to not ask?
2) Can I also ask for a trade if you did ask or if you didn't ask?
3) Is the trading question binding? That is, if I say yes, do you then have to trade your card? If I say no, can you still trade?
4) What happens if I say yes or no?
Mikko Luttinen Was just about to comment this. Poorly stated questions seems to be a running theme in his videos.
Mikko Luttinen Agree the question is poorly stated. But I think it’s like this:
After both players peeked at their own card, player A has the _option_ to offer a trade to player B. It’s not mandatory. So there’s three scenarios: player A doesn’t offer to swap (in this case player B can’t do anything) or he does and then player B can either accept or reject.
The trade question is binding. If A offers to trade and B says yes, then the cards are swapped and the game ends. If B says no, the game also ends.
@@darkhyve9886 Agreed, it has ben so since the viral 6 divided by 2(1+2) period.
Also what is at stake (cost and benefit)? Money cost/reward, bragging right / dishonor, , something else, a combination, or nada? This will depend on whether the 1v1 game is a one-off or part of a session of multiple, which in concert with prior knowledge/ beliefs / suspicions about opponent can affect also optimal strategizing.
@@stromboli183 -- right, but the fact that the trade was _optional_ wasn't explained prior to the invitation to pause the video. The way it was presented, it is perfectly valid to assume that the offer to trade was a _required_ part of the game.
It's akin to giving the same instructions, and then saying "you shouldn't have drawn a card" as a solution, after the instructions seem to at least _imply_ that drawing a card was a mandatory part of the game. If you're explaining game rules to me, I expect to be told when some component of the game is _optional_ and at the players discretion.
This is kind of misleading, you asked if we should trade. I assumed that you always asked that question, I would clarify that you don’t say “would you like to trade cards” all the time.
Right its more about, "If your opponent asks if you want to trade, should you?" and not "Should you trade?". It would be a different problem if a third party always asks if you want to trade, then you really have no clue as to what your opponent has.
I thought the same too
yea i feel this was the problem for me. When he said "I offer to trade with you" I felt it was like a game show host that was just enforcing a rule that happen no matter what. there is probably a way to better communicate the combative nature of the other party
Exactly right. I was expecting the second person to be acting like "the house" in a casino, always asking the question.
What if he did have a 6? Do the two of you just stare at each other without saying anything?
Correct. We weren't given the complete rules of the game to begin with, which is that your opponent can choose to trade or not.
Plot twist: Logic went out the door when it became 2020
Plot twist #2: The cards have random numbers, even negative, not just the integer range 1-6.
@3:19 Assuming he has a five he has no way of knowing if I have the six or not. So it's not a strictly dominated theory. This is the same logical paradox as the unexpected hanging problem. This really comes down to a 50/50. You wouldn't offer to trade on a 4, 5 or 6. Any of those are more likely to win than not. On a 1 or 3, you would offer to trade as it's in your best interests. Since I have the 2 we KNOW you do not. So, you have either a 1 or a 3 and it's a 50/50 shot.
Also, it should be noted that asking "Would you like to trade cards?" is VASTLY different than "I want to trade cards." The first one is a simple question and could be seen as simply fishing for information. As such it really doesn't tell us anything other than you do not have a 6 (as if you didn't it wouldn't matter if I wanted to trade cards or not).
I agree, I thought of the unexpected hanging paradox too.
I think the point was that with the IESDS is that when you remove the strictly dominated strategy, you then look at a "reduced" game/case and then eliminate the strictly dominated strategy from there. in practice, what you are saying sounds logical, however, I assume that the IESDS method is one way of analyzing a situation with "perfect reasoning". Of course, humans are more complex than that, but IESDS, while seemingly non-intuitive, is just a way that humans can create a theoretical, logically sound, best-case strategy for these situations.
I may be wrong, but I assume there are many other ways to logically break down this game theory problem by incorporating additional considerations, which may yield different results/ideal strategies!
Yup. This "reduced game" logic doesnt make any sense to me. There is no reduced game happening
+1 Agree
There idea is that if you have a 5, you know I won't trade a 6, so every result left is worse for you.
I think it should be made more clear here that player 2 *chose* to ask if player 1 wanted to trade cards - it wasn’t just a rule of the game that player 2 *had* to make the offer.
Yes, this is important to clarify. It's a big sticking point in the infamous Monty Hall problem. In the classic Monty Hall problem, the host ALWAYS knows which door has the car, the host ALWAYS reveals a non-car door, the host ALWAYS gives the player the option to switch, and the player knows all of these rules ahead of time. These assumptions are often left unstated, but they should be explicitly stated when presenting the problem. With different assumptions, the problem can have different answers.
Exactly. There is a popular movie where a Professor presents the Monty Hall problem to the class and the protagonist gets the answer correctly, but the Professor presents the problem as though the host doesn't know which door conceals the prize and just happens to open an empty one. Which changes the problem completely and makes it irrelevant whether you switch doors or not.@@matthewmaas9031
Absolutely. I always pause these videos and try to solve the problem for myself, but it's a bit hard when a vital piece of information is left out. And it did need specifying, as there are games where the change option is always offered. (As I notice someone else has pointed out).
Exactly what I came into comments to say. I hated the setup. Luckily this comment was right at the top, and nothing was spoiled for me and I can still play. Thanks.
It's also a hidden rule that the other player is rational. Many, probably most, people would offer a trade on a three or a one. A big problem with economics is that so many models are built assuming people are rational and really, they aren't.
If both sides assume the other has perfect logic, then someone with a 1 would concede defeat rather than offer trade.
Offering to trade is the same as conceding in a rational game. It doesn't matter what you do.
@@super_7710 Not exactly. Offering a trade has a strictly higher win percentage than conceding, so it is the optimal strategy.
I ran a Monte Carlo simulation based on the strategy of only offering if the card is 1-3, and only accepting if the other card is also 1-3, and the result was almost exactly 50/50. If you only offer when your card is 1, then obviously a rational opponent following the same strategy would never accept, so that would also be 50/50. It seems like if both players are following the same strategy, regardless of what that strategy is, there's no way to gain an advantage.
Dumbass me with a card no.5: "you wanna trade?"
some dumass with no.2 trusting answer in the video : "nope"
Nope, i saw a video on mind your decisions
this arguing is kinda ridiculous. imagine this game with 100 cards where you draw, say, a 3. even if the other person has a 2, he would most definitely want to trade it and you would expect him to trade some range of small cards, so your chances are good to get a better card. this IEDS is certainly one way to come up with a rather good strategy which is by construction not dominated, but it also eliminates other strategies that might have a high probability of getting a better card even though it is not guaranteed
the vid is assuming people playing are "perfectly rational", obviously humans wouldnt think through all that, thats how gambling works.
The logic holds for any number of cards.
if person A draws 2, person B draws 100: A knows B would not ask to trade 100
B draws 99, knows A would not trade 100: A knows B would not ask to trade 99
B draws 98, knows A would not trade 99 knowing B would not ask to trade 100: A knows B would not ask to trade 98
B draws 97, knows A would not trade 98 knowing B would not ask to trade 99: A knows B would not ask to trade 97
B draws 96, knows A would not trade 97 knowing B would not ask to trade 98: A knows B would not ask to trade 96
...and so on all the way down to 3...
A knows B would never ask to trade anything but a 1 so he should not trade.
You have to factor in k level reasoning if you were to try and win the game in real life.
The video assumes both the players are rational and fully aware.
Most people in real life come under k2 or k3 levels which vary the answer and may lead to the correct answer being yes.
It is in perfectly logical game, when both players know that the other is perfectly logical. And then if the other player have anything else but 1 he wouldn't offer trading.
Yeah that's always the catch with these games is that you have to assume both players are logical actors. Thing is that this is rarely the case in real life, so the "perfect" game theory way of playing isn't necessarily the perfect way to play. If you have a 2 in the 100 card game, the odds your opponent has a 1 is only 1%. In a real world scenario, the odds are likely greater he'd make a mistake and trade something like 3-10, which can make that a profitable play.
In my opinion, my opponent probably will have either a 1, either a 3, with equal probablity...
The logic really shows that it is always pointless to ask to trade.
Since we know that the other person did ask, we know the premise of the "perfectly rational" opponent doesn't hold.
Not quite, because an opponent will always ask to trade the 1 even as a perfect strategist on the off chance you aren't.
@@traviscalder If there is any chance that an imperfect strategist is involved, then the logic doesn't hold.
Presh is hiding this fallacy behind a small number of cards.
If there are 1000 cards instead of 6, then even if 90% of the opponents only offer to trade with a 1, the other 10% will give more wins, from accepting a trade with a 2, than losses from the 90%.
The logic only works if there is no "off chance" of someone who isn't a perfect strategist.
If you're rational and the opponent is rational, offering to trade the 1 is the same as not trading at all. Therefore, it doesn't matter what option you choose.
Right. If the opponent knew I was not perfectly rational, he should ask a trade even if he had 2, because I might accept it even if I had 3. This would really work way better with a larger number of cards though, because it's too easy to understand that trading 3 might not be a good idea.
If the opponent knew I was perfectly rational, he would know that I would never accept a trade, and therefore asking for a trade only reveals that they have 1 in advance. It of course depends on the rest of the game whether it's harmful or not. If there was betting involved before revealing the cards, it of course could be harmful.
Well, that only works if you assume both players apply this logic perfectly. 95% of real people (including me tbh) would probably trade any 1-3 because statistically there is a higher chance to get something better.
However, even with that assumption a trade wouldn't be that great in this case, as you might only get 1 or 3, so it's a 50-50 chance.
You failed to explain up front, before the break, that you had the *option* of asking to trade cards or NOT.
Having not explained that, it would be logical to think that asking to trades cards was required - which changes EVERYTHING about this game.
Except that that's just bad game design. Why would you force players to ask for a trade if they don't want to? Since people don't like playing bad games, the only logical conclusion is that offering a trade is optional.
@@realitant but it is not stated to be the case. the fact that this is an assumption that the puzzle-solver has to make is a huge weakness in the puzzle.
@@realitant Imagine it would be just something like this so you ask do you wanna trade and if I say yes then i trade if I say no I don`t next round I have to ask.
Isn't NOT trading a 1 also a perfectly dominated strategy? So if you have a 1, you will definitely ask to trade because not asking will always lose. Now we can eliminate the 1 and look at the reduced game of the cards 2-6. Now not trading the 2 always loses so we can eliminate that option as well... I think you see where I'm aiming at, using the exact same logic as is used in the video you could argue that you should always trade unless you have a 6. So the whole thing doesn't make sense to me. Correct me if I'm wrong.
This sounds like the logic used in the execution, where a prisoner is told he will be executed some day of the week, this week, and he will not know which day until he is led to the gallows in the morning. "Can't be Saturday, because that's the last day if the week, and if I'm alive in morning, I will know this is the day". "Can't be Friday, because it can't be Saturday, so Friday morning I will know", etc. One-by-one the days are eliminated where he can be executed. Conclusion: can't be executed under the conditions stated! Your game is similar. It assumes the opposition is a mathematician, and will make "rational decisions". How do you know that the other player will follow rigorous rules?
Yes, in the Prisoner's Dilemma, the prisoner believes he has talked himself out of the execution and congratulates himself for his clever argument. Then he is utterly shocked when the executioner shows up on Wednesday.
As for the puzzle described in the video, six cards aren't enough to clarify the problem with the logic, so let's say there are twelve. You have the 2 and think you have been very clever to reject the offer to trade, figuring your opponent has the 1. Then your opponent reveals his card, which turns out to be the 3. Did you blunder? I would say so.
@@zanti4132 but it's not the same.
If someone has a 3, it's best for them to not offer a trade in the first place.
@@Somerandomdude-ev2uh If there are 12 cards and I have a 3 I would offer!
@@UnimatrixOne this comes down to how logically your opponent plays, if the opponent does very basic thinking then it is worth trading 3 when it goes upto 12, so fair enough
Exactly!!!
It wasn't clear to me in the question that my opponent had the option of asking whether I want to swap.
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.
I think the puzzle needs to state that the two players are of the same perfect level of rationality (let's call it rlvl) as it is an influential factor in how one player interprets or expects the other player's actions. and THANKS for the video.
This is with the assumption that both the players understand IESDS :D
Presh doesn't, unfortunately, in this case. Trading a 3 isn't strictly dominated, it's actually beneficial over time.
@@tashkiira7838 Imagine you have a 3 and your rational opponent wants to trade his card. Would you trade? I don't think so, because if he wants to trade his card, this means he is NOT 4, 5 or 6.
@@snfnsessizcocugu7884 Therein lies the rub. Presh keeps going on about RATIONAL. Rationality and logic (which is what you guys SHOULD be referring to) are only tangentially related. You might have the excuse of misunderstanding; Presh doesn't as he makes math and logic videos and books for a living. A strictly rational player is going to want to trade a 3, as there are 3 better cards. A logical player might not, due to running through IESDS, and will hold a 3 or 2, when a rational player might attempt a psychological play by offering to trade a 4 if he isn't confident in the card.
You can't use IESDS to eliminate the headgames. Headgames will win battles and wars when nothing else will, and game theory is taught to officer candidates specifically so that they know the effect of headgames. Cao Cao held a city with a single lute player against an army based on headgames when the opposing general used IESDS--correctly!--to determine there was a trap and refused to take the city. (Cao Cao's army was off attacking the same enemy elsewhere. He ordered the gates opened and the garrison to join his army in order to preserve the men involved, and was actually expecting to surrender. Cao Cao's reputation as a tricky general was so great at the time that any rational man in charge of an army would have assumed that Cao Cao playing a lute on top of the city wall with the gates open would have to be a trap..)
@@tashkiira7838 Well explained, but I didn't really get how the psychological games apply to this card game. could you explain why a logical guy would offer to trade his 4?
Just remember it could also be a bluff. There are also other factors such as rules of the game, intimidation, psych outs, a persons tendencies
As someone correctly pointed out, you wouldn't say no to a trade if you had a '2' in a '1000' card game. Pretty intuitive, because the correct way (I believe) to solve this problem is by using probability.
In a 6 card game, I have a 2. If he has a 6(or 5 or 4), for him, the probability that I have a higher card is 0( or 1/5 or 2/5). So trading is not profitable for him (as all chances are less than 50%), and hence he wouldn't ask if I wanted to trade.
If he had a 3( or 1) the probability that I have a higher card is 3/5 (or 1), and so he would ask the question.
As we know all of this, and we know that he asked, we can deduce that he either has a 3 or a 1, which creates a 50-50 chance of us winning/losing. So there's no definitive answer.
Note that this will only happen when you have the card one below the halfway card.
In all other cases, following the same logic, you have to calculate probability of profit from a trade. (If you're below 25%, do the trade, if you're above 25%, don't, given that he asks for a trade)(you can derive this through the same line of reasoning)
Anyways, if asking for a trade is mandatory no matter what card one has, then there's no strategy, except do the trade if you have a card below the halfway mark.
I came to the same conclusions that you did. This looks like a 50/50 to me.
But Even if the person had card 3 he could have offered a swap because there are 2 cards less than 3 but 3 cards more than 3.
Why we didn't consider this?
If you had a 4 5 or 6 would you accept your opponent's offer to trade? Probably not, right? Because he wouldn't be trading away a better card than you have. So if you have a 3, and you offer a trade, that same logic will apply to your opponent. He'll trade you a 1, but certainly not a 4 5 or 6.
Oppenents also have brain ^^
I thought that too and still cannot understand the resolution
@@NathyIsabella an opponent asks himself the same question. 6,5,4 wouldnt want to trade. 3 he wouldnt want to trade because of 6,5,4 wouldnt want to trade. 2 would want to trade but only 1 would accept it.
@@NathyIsabella imagine you had 3, knowing that noone who has a 4or5or6 wants to trade, would you want to trade?
Bro someone is gonna make an anime and this is gonna be apart of a fight where that one dude is trying to out smart the other I can just see it now
You better watch Kakegurui then
Rather like the Coyote in the Road Runner cartoons czcams.com/video/PkBu-mnXOAk/video.html
@@fatehboufertala730 pleb tier.
Akagi and kaiji are the real shiet this isn't too far from the restricted janken with 4 of each.
He could just as well had a 3 and judged that chance would give him a higher card. You theory collapses like a house of cards!
The real question is "how likely is my opponent to bluff?" If the opponent wants you to take their 1s, they need to occasionally offer 2-5.
I think the algorithm is poorly explained. I thought the right player always had to ask about trading the cards. I didn't understand it was his choice to ask or not.
You cannot use IESDS in this case. The only strictly dominated strategy is if a 6 is being held. As others have mentioned, if the card is a 5 or lower, it is NOT strictly dominated.
If the card is not a 6, you will have to rely on probabilities. For 4 and 5, there is a better than .5 probability that it is a winning card. The only cases in which a trade would be offered is if a 1 or 3 is held. Of those cases one (3) would be a win to trade, one (1) would be a win NOT to trade. So there is no definitive winning play.
There’s no reason to ask to trade on a 5 either. If your opponent’s card is higher than yours (where a trade would be good), they have a 6 and would never trade. If it’s lower, they might trade. So it only loses.
Similarly there's no reason to ask to trade on a 4. If your opponent has 6 they won't trade, and if they have 5 they'll know you don't have 6 because you wouldn't ask if you did, so they still know you have less than them.
No use of Gougu Theorem... 😂😂
I solved the riddle, but you should mention that the other guy plays perfectly.
I got curious and simulated this in python. My game dealt both players a card between 1 and 100 (because I wanted more scope than just 6 cards) players independently choose to swap or keep their card based solely off of it's number, if both choose to swap then they do so. I set up a competition with 10 players:
Player 1 - would swap any card below 90
Player 2 - would swap any card below 80
(continue in the same way...)
Player 9 - would swap any card below 10
Player 10 only swaps on a 1 - as suggested by game theory.
After 1,000,000 random games against each other, the player with the highest total win percentage was Player 7 (swaps any card below 30), with 55.8%.
However, the only player with a positive win % (albeit very small) against all other players was Player 10. So this is the strategy that will beat any other strategy on average.
I found it quite an interesting exercise - it helped me make better sense of the video, two perfectly logical players should only ever swap a 1 (which means they never swap) but it goes to show that the 'best' strategy depends on the other players. For example, I re-ran the test with only the players who would swap below 50 and Player 9 - who would swap 10 or below came out best.
This phenomena happens only once in a blue moon...
What?! I am not talking about the Game or the Game theory. I am talking about Presh's video being illogical.
Before watching the video: Trade. If you get a 1, 2 or 3 you should trade (because the average value of the cards that aren't yours is higher than the value of your card), and vice versa.
After watching the video: The entire problem is posed incorrectly. I'd say about 50% of the videos on this channel suffer from the same thing: extremely sloppy wording or outright missing information that ruins the puzzle. Be better.
What if the opponent has a 3 and hopes that I might be holding a card greater than 3?
The opponent would know you would only accept the trade if you were holding a card less than 3, and therefore would not offer it.
I beg to differ on the logic of this one. Suppost the amount of cards is huge, saying a million. It makes no sense.
For people confused why Presh's logic holds:
- Let's represent OUR game states by a 6x6 grid where the columns are labeled with your card and the rows have the opponent's. The diagonal doesn't exist because you can't have the same card. Each point in the grid has W or L depending if your column beats the opponent's row. The Ws and Ls would be flipped if it was THEIR game states.
* 1 2 3 4 5 6
1 X W W W W W
2 L X W W W W
3 L L X W W W
4 L L L X W W
5 L L L L X W
6 L L L L L X
- Asking to switch cards when you have card "i" is the same as picking column "i" and flipping the Ws to Ls and Ls to Ws. Clearly we want to end up with more Ws than Ls after switching so we can eliminate columns 4, 5, and 6 as decisions we'd ever take. Technically both players are allowed to ask to switch and both players are aware who asked.
* 1 2 3
1 X W W
2 L X W
3 L L X
4 L L L
5 L L L
6 L L L
- Here's where I think Presh didn't explain that well. Assuming your opponent is also equally rational, we know they'd also eliminate the 4, 5, and 6 decisions in THEIR game states grid, which corresponds to rows 4, 5, and 6 in OUR game states grid. Since they'd never play those decisions, we can eliminate it from our grid as well.
* 1 2 3
1 X W W
2 L X W
3 L L X
- Now we can apply the same reasoning given our previous beliefs and eliminate column 3.
* 1 2
1 X W
2 L X
3 L L
- But wait we assume the opponent is as rational as us, so we can eliminate them ever playing row 3.
* 1 2
1 X W
2 L X
- Ok now importantly still believing everything we've believed up to this point, I'd obviously never play column 2.
* 1
1 X
2 L
- I could eliminate row 2 as well, but I can already notice here, I only gain by switching so I'd definitely ask to switch. Thus I'd only switch if I have 1. Symmetrically, the opponent would only ask to switch if they had 1. In the video, since we're both operating under this rationale, the opponent knows we don't have 1 since we didn't ask to switch regardless of what card they have, and we know the opponent has 1 since they asked to switch regardless of what card we have.
Someone else can work out what happens if you take the more intuitive approach of asking to switch if you're below 4 and the opponent either does the same or plays the strategy from the video. Please correct me if I made a mistake, otherwise I hope it helped!
Very good!
Wow solved a very similar question not too long ago. Great brain teaser haha! Keep up the good work!
Can you share the brain teaser which you solved
Kartik Bagoli it was a card trading brain teaser I found from a collection~
My first thought was, "Anybody who's seen their card and wants to swap with me must have a pretty lousy card!" I would never offer to swap a 4, 5 or 6, and I already have the 2, so I suspect my opponent must be holding either a 1 or a 3. In which case, switching cards has at best only an even chance in my favour and probably less, since there is no way I would _not_ offer to swap a 1, but I might hang on to a 3. So there is _at worst_ an even chance of winning if I refuse to switch.
I just woke up, saw this video and now feel heavy again.
Great solution, but that is under the assumption that our competitor also knows this theory or is intelligent enough.
So, incase any of you guys are playing this game, please do consider the mentality of your competitor.
Good point, although he did technically say we are playing against him
yes
The solution fails because of the algorithm. It is recursive, and no exit condition was given.
The algorithm as it was diagrammed would find the 1 as a dominate strategy as well and kick it out, then there would probably a stack overflow since there would be nothing left for it to compare and loop infinitely.
I guess
There's similar one prisoners hanging riddle
And it has similar flawed logic thinking.
@@puliverius lol correct
There is some discrepancy on the rules. You can also interpret the game like this. Both players get one card and they each peek at their own card. An imaginary dealer asks if they want to swap. *Player A must answer first.* If he says no, the game ends. If he says yes, then player B must answer. If he says no, the game ends. If he says yes, they swap their cards and the game ends.
After the game ends, they both reveal their card and the higher card wins.
Now suppose you play this game with a deck of 1000 cards instead of 6.
• If you are player A, in which case would you say Yes?
• If you are player B, and player A said Yes, in which case would you say Yes?
This hurts my brain a bit.
If opponent has 6, he doesn't offer to trade because he already wins.
If opponent has 5, he doesn't offer to trade because the only card he wants, the 6, will not be traded.
If opponent has 4, he is hoping for 5 or 6 to come his way. The 6 won't because nobody gives that up. I will only trade my 5 if I think my opponent has the 6. But he has offered to trade, so he can't have the 6. He holds the 4, hoping I have the 5 and hoping that I'm hoping he has the 6. He wants me to think he has the 6, but he has offered to trade, which he never would do with that card. He's depending on me hoping that he has the 6, but he has ruled out that possibility by offering to trade. I won't give up my 5 for his 6 because he can't have the 6.
If opponent has 3, my head explodes. He is hoping I have 4, 5 or 6. If I have 6, there is no point in him asking to trade as he knows I will refuse. He is hoping I have 4 or 5 and hoping I'm hoping he has something higher. He doesn't offer with the 6 so I can't be holding 5, hoping he has 6 and then receiving an offer to trade. But do I have 4? I can't hope he has 6, because he wouldn't offer to trade. He has 3, and he hopes I have 4, and he hopes that I hope he has 5. If he has the 5, would he offer to trade? He wouldn't waste his time because the only card he wants, I would not accept to trade. So he can't have the 5 if he is offering to trade. He knows that I know he can't have the 5.
I need to lie down now.
Simpler reasoning: opponent would never offer 4-6 as it can only hurt them (with your reasoning), I would never take an offer holding 4-6 for the same reason. Opponent wouldn't offer 3 either because he knows I would only take the offer with 1-2. Since I had 2, they must have 1!
The crazy thing is that this logic seems hold even for 1000 cards...
As with most other game theory problems, this depends on the players playing optimally, which doesn't usually happen irl (if anyone's confused abt this)
I think I know what they're trying to say:
If an opponent dares to offer a trade, then they MUST have a lower-valued card; thus trading is a bad idea ONLY if they offer.
I think this is a problem that we could get insight into with some simulation. Something like a program that would generate agents that make or accept trades on various numbers, pair them up randomly with each other, and see who beats what.
And I think there’s a simpler way to get to the solution here: The first strategy most people would come up with is to make or accept trades if your card is in the lower half of the numbers (1-3), since the opponent’s card is more likely to be better. You would initially accept if you had a 2, but look at what your opponent is thinking. If they offered, they must have either a 1 or 3. However, if they had a 3, they’d know that you either had 4-6 (where you’d refuse a trade) or 1-2 (where a trade would be bad), so they would have no reason to offer a trade. Thus, they must have a 1 to offer it, and you should not accept.
Wouldn’t the expected probability of how much the opponent knows about game theory and how likely they would use a similar logical strategy be an influence in wether you would be willing to trade the card? I could be up against an opponent with a 3 or 4 willing to bet that I’d have a 4 (assuming he had a 3), 5 or a 6, right?
One of the worse videos I’ve seen. I’d elaborate but there’s enough here to make my case.
Wouldn’t you be able to apply the three doors game trick to this problem, thus making it advantageous to trade?
Tried. Opposing player had a 3 and was trading up because he expected me to have a higher card.
*Similarly:* each of a group of people writes down a secret number between 1 and 100 on a piece of paper; the numbers are then revealed. To win the game one needs to have written down the number which is closest to *half the average* (arithmetic mean) of all the numbers written. Using strict logic, every person partaking would have written the number 1 - but in a real life situation it is doubtful whether that would happen, or even whether writing 1 would win the game.
Was not expecting tô actually knowing how to do the challenge, feeling a bit smarter now :3
There are two things going on here, one relating to whether B asks the question in the first place and one whether A says yes
A holds the 2. B asks “do you want to swap”
1. If B had the 4,5 or 6 she would not have asked to swap in the first place as there’s a greater than 50% chance that she would get a lower card
2. If B had the 3: She thinks that A would say no if he had the 4,5 or 6 for the same reason, so would only say yes if he had a 1,2 meaning she would lose out, so does not ask the question.
3. That means that B only asks the question if she has a 1 so A must say no thanks.
I think this works uniquely for 6 cards because 3 are above 50% (4,5,6) and the other two possibilities are either side of the two.
I agree
@@toonhkuitjes6382 Plenty are arguing it still applies with a million cards - I'm trying to reason differently but it's not the straightforward argument you would think t should be.
Another one of these clever puzzles ^^
You already know that your opponent would not possibly switch their 4,5, or 6 (obviously) and so if you have a 3, making the offer to switch (and them accepting), will inevitably leave you with a worse hand (1 or 2).
So, when your opponent is making you this offer and you hold 2, they can't possibly have a 3 (for the reason above), leaving them with only one option: 1.
The explanation in this video could have been better framed without relying on the IESDS...
You blew my mind with that.
The problem that I have with this puzzle is that it is not stated that the trade offer is optional. If the game requires that the offer be made, it offers no insight into the other player's decisions.
*Corollary* :
_If your own solution to a MindYourDecisions puzzle_
_sounds too good to be true, it probably is!_
I think this is an ill defined problem. If the swapping of cards was part of the game then this would not be good logic as they might have to consider this question.
If this is not part of the game, then the reasoning in this video is good since we’re assuming both of them are logical. However, I still wouldn’t go as far as saying that only the “1 card” would be a reasonable choice to ask for a switch. If the person had cards 1 or 2 then that might be the only times it would be reasonable to ask for a change since there’s a 50% chance of getting a 4, 5, or 6. So even if you had a three, there’s only 50% chance that the other person could’ve gotten a higher number than you. Then again, due to this seemingly paradoxical nature, it could be argued that if you had a 3 then that would also be a time that is reasonable to ask for a change since there’d be a 3/5 probability that the card in other persons hand would be greater than their own.
Regardless, it depends on what card you have. If, in the case of this not being part of the game, it is 2, then your chances are arguably 50/50 so don’t take it. If you have a three, definitely don’t take it, and if it’s a 1 take it. But anything greater than a three would be a no since there’s no reason to change cards as the odds are stacked against you at that point.
So I guess, if it’s not part of the game, or part of the game but just isn’t a necessary question, then it’s always a no unless you have the “1 card”. The reason for it also being the case if it’s not necessary is because the only reason to ever ask that question is to maximize your chances of winning.
From the logic of 'Expectation' it would be 19/5 which is > 3 anyway. So you must answer 'yes'.
In chess, ALWAYS consider the other players motives and moves first. After you have covered those possibilities, to the best of your abilty, then you might consider, trading, attacking the centre, taking space, etc. It is never a good idea to say, "Oh well; let's see" ... that's a game loser!
This was a cool vid! Thanks, Presh!
Analyses the sweat rate and heartbeat of opponent.
Proceeds to blunder Queen.
@@romilgoel4191 🤣 💪😎👍
Really similar to the problem of support and resistance levels in technical analysis of stocks as in IESDS poses some sort of a syllogism problem. There's an unethical strategy I can think of out of this. The opponent asks to trade cards, while having a card with less than 50% winning chance (1 or 3). Suppose the opponent has 3, If you answer yes, that would reveal the information that you would have 1 or 2 because if you have higher cards it wouldn't statistically make sense to do it, and just then call off the trade and open the cards directly and win the game.
At first even I thought 1,2,3 should ask to trade and 4,5,6 shouldn't, but now that I've thought over it, it all makes sense and it's just too cool!
The main thing to keep in mind is both the players are perfect at strategy which is why this works
Why would a player not want to trade a 3? They would have a 60% chance of trading to a better card.
@@clayton97330 If you have a 3 and the _other_ guy initiates the trade, he can't have a 6, since he wouldn't trade that, and he couldn't trade a 5, since the only better option than that would be a 6, which _you_ wouldn't give away. If he had a 4, the only _other_ higher card (aside from 6) that *you* might have would be a 5, but if you had a 5 and the other guy offers you a trade, clearly he can't have a 6, so you wouldn't trade a 5 either, making it pointless for him to offer his 4, since he can only receive a lower card.
If you have a 3 and _you_ initiate the trade, if the guy you want to trade with had a 6, he wouldn't trade obviously. If he had a 5, he'd know _you_ don't have a 6, otherwise you wouldn't ask for a trade, so he'd never trade his 5. If he had a 4, the only higher cards would be 5 and 6. He knows you wouldn't trade a 6 if you had one, he also knows you can't have a 5, since only the "winning card" 6 would be higher, so offering a 5 for trade wouldn't make any sense. Meaning if he had a 6, 5 or 4, he'd never trade those. So if he agreed to trade your 3, he'd have either a 1 or a 2.
I'm quite tired though, so correct me if I made any mistakes ^^
As many comments have pointed out in some way or the other, the premise that the game with the reduced set of cards is independent of the overall game is just not well justified.
Nice one!
A common flaw in most mathemathicians probems is that you forget to state the rules of the game (is asking for a trade optional or obligatory) and forget to describe your opponent (perfect mathematical reasoning or any common player). This puzzle hinges on trading being optional and your opponent being mathemathically reasonable, but is presented as if asking for a trade is part of the rules and your opponent is just anyone.
In addition: if your card is a 1, you could also conclude that only a 1 would ask for a trade, and therefor would never get a yes to that question. The game has a rule that allows trade, but will never be used, because mathematically nobody would want to appear weak. It's a better game if the players HAVE to offer the trade, which introduced a risk and an element of bluffing to the game. it's more mathematically logical for the game to have enforced trading offers, so as you described the puzzle it is in fact more logical for the player with a 2 to accept the trade.
Not everyone plays so strictly. The gambling element is what makes games fun. However, this is a great video, Presh. Game theory is an interesting topic!
Then there's that guy who got 6 and asks to trade anyway to screw with you
Nice! We actually get a puzzle this time and not another exercise from a geometry textbook!
What if you trick the other person playing by pretending you have a one?
It would also help to know the rules and intentions of the game. I thought there was a mandatory trade phase.
The logic in the video is correct, but, unlike most of Presh's videos, I don't think it's explained particularly well.
Here's a stab at another explanation:
Assuming:
i. both players are perfect logicians
ii. either player may offer a trade, but neither is obliged to
iii. once a trade has been offered, the other player is free to choose to accept or reject the trade
iv. a trade only happens if both players agree to it, that is, a player offers a trade and the other accepts
Both players realise that they should always trade away a 1, since 1 is the lowest card, so trading it would result in a guaranteed win every time. Therefore, both players will always offer to trade away a 1, or accept a trade when offered if they are in possession of the 1.
Conversely, they both realise that trading away a 6 is a terrible strategy, since 6 is the highest card, so trading it would result in a guaranteed loss every time. Therefore, neither player is ever going to offer to trade away a 6, or accept a trade when offered if they are in possession of the 6.
Knowing that their opponent would never trade away a 6, it doesn't make sense to ever offer to trade, or accept a trade, if they are holding the 5.
Knowing that their opponent would never trade away a 5, it doesn't make sense to ever offer to trade, or accept a trade, if they are holding the 4.
Knowing that their opponent would never trade away a 4, it doesn't make sense to ever offer to trade, or accept a trade, if they are holding the 3.
Knowing that their opponent would never trade away a 3, it doesn't make sense to ever offer to trade, or accept a trade, if they are holding the 2.
We are left with the optimal strategy for both players being to only ever offer or accept a trade if they are holding the 1. But because their opponent would never offer or accept a trade in this scenario, no trades ever take place, all games are resolved with the players comparing their originally selected cards, and in the long run both players win approximately 50% of the time. If either player deviates from this strategy it leads to them winning less than 50% of time on average.
What if Presh were not in fact a perfect logician and were to deviate from the optimal strategy described above? Let's say he decides to offer a trade if he's holding a 2 or lower (as he correctly surmises that if he's holding the 2, the odds are that your card is higher than a 2). In this case, it's still not in your interest to deviate from the optimal strategy of only ever accepting trades on 1.
If he offers to trade when he's holding a 1 or a 2, your best strategy remains only ever accepting the trade if you are holding the 1. Because if you're holding the 2 and you accept the trade, you'd be trading away a 2 for a 1. In this case, trades will only ever happen when Presh is holding the 2 and you are holding the 1, which has odds of 1/30, and you'll win approximately 53% of the time.
Let's assume that Presh deviates even further from the optimal strategy and decides to offer a trade if he's holding a 3 or lower. Even in this case, it's still not in your interest to deviate from the optimal strategy of only ever accepting trades on 1. In this case, trades will happen when Presh is holding either the 2 or the 3 and you are holding the 1, which has odds of 2/30. Sticking to this strategy, you'd win on average almost 57% of the time. These odds don't change if you change your strategy to accept trades when you're holding the 2; in this case, trades would also happen when Presh is holding either the 1 or the 3 and you are holding the 2 - but only one of these trades is favourable to you and either is equally likely. And the odds drop back to 50:50 if you decide to accept trades when you're holding the 3, as you are then both following the same strategy again.
So there's never any advantage to you in ever deviating from the original strategy of only trading on 1.
What if I had a 4? does my opponent still have a 1?
wouldn't it make sense to ask for a switch if you have any number lower than 4? (regardless of my number)
Then you both flip your cards open, and the other guy has a 3.
Going with GTO works in theory, but not in practice when people don't act optimally and sometimes just consciously make bad moves for fun, or because they think they are good moves.
With 6 cards it's not as obvious, but if you were to play a million cards (1 to 1,000,000) and drew a 2 and your opponent offered to trade, would you do it? I would do it in a heartbeat.
Pretty sure the vast majority of the population would trade cards in the hundreds, or even in the thousands. The least 'thinking' people would probably trade cards like 100k or 200k, just thinking "it's one of the weak cards so the odds are with me".
By trading a 2, your odds of winning against an average person would probably be >99.9%.
agree
Presh what kind of programme do you use to animate your solutions
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.
I loved this puzzle but It is not clear why laws of probability do not applay if the player sees the card she picks. (That’s what it is said in the video). Of course the possible outcomes will change but it is clear that probability can always be used to model the problem. Also the theory of dominated/dominating strategies apply to iterated games. This is important. In the video it is never said that the game is iterated. If it is, it should be said if each player will have the chance to ask the other one for trading cards.
what if my opponent has a 3 and poor reasoning skills?
So if you'd expand this game to 100 cards, would you still not trade the 2?
That's what I was wondering as well
I am fairly certain that no. I can see that with 6 cards, an opponent that's intelligent enough wouldn't ask, and the way the video presents it implies that it works for any number of cards.
With 100 cards, the cutoff point would have to be a fair bit lower than 50 because the top 2 cards at least can be eliminated for obvious reasons. Since no one in about the top half would trade, the remaining game would have to consider the lower half, but cannot be reduced by the same logic because that would be forgetting that it is just looking at the lower half of a game.
For that reason, and as another commenter has already said, I would expect the cutoff to be lower than at 25% of the highest card, but would have to think about it more to find out if that way of thinking is forgetting something and where exactly the point at which you would want to start trading is, or how many cards you need to answer the question with yes if you have a 2.
it's all very well and good, but you're adding rules to the game as you reveal the solution!!
If the other card was a three, a trade would appear to have a 60% chance of getting a higher card, so it would make sense to ask for a trade.
Yeah but the opponent would only accept the trade if he has a 3 or lower, so a 1 or 2, so asking for a trade with a 3 is never a good deal.
I think thats what Presh failed to explain in this video
Meanwhile, the other guy flips over his card and it was a 3, so you lose. He didn't watch the video beforehand, and his reasoning was that there were 3 cards better than his and only two worse. He felt his odds were better by trading.
I used to play a card game called, Tripoli.
In this game, the dealer gets TWO HANDS. If they like the hand they're dealt, they can sell the "unused" hand (unseen). If they dislike their hand, they can take the unseen hand, and sell the "USED" hand (seen).
Of course, there are many more cards in each hand, but, I doubt even Mr. Talwalkar would care to tackle such a divergent strategy situation.
Buying a "used" hand also means you have at least one opponent who knows your hand.
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?
Reminded me of 2 things. Rolling for Chance in Yahtzee. First roll, keep 5's and 6's second roll, keep 4,5 and 6's. Also an over in cricket where you can bowl ONE Bouncer and it is a good idea to. Will be obvious if you leave it till the sixth bowl, so then it will be obvious you're not bowling it on the 5th bowl coz then the batsman would know you're bowling it on the 6th and then the 4th... 1st. So you never end up bowling one! In fact, after seeing the whole of this video, it is exactly the same as the Cricket analogy.
for those confused, lets work through if the guy whos asking had a 4. Assume "we" are the one whos asking. If we had a 4 we simply wouldn't ask the question. This is because the only reason we would ask the question is if the other had a 5 or 6. the other guy clearly would decline if he had a 6. If he had a 5, he would want a 6 in return, which clearly we dont have, since we asked the question(if we had a 6 we wouldnt ask). So therefore if he had a 5 he wouldnt trade either. so theres no point in asking if we had a 4. Use the same logic on if we had a 5 or 3 and thats why we would only ask to trade if we had a 1.
Obviously, we wouldnt do this irl, this is just assuming perfectly rational people.
Good question!
thanks for the video,
but when you asked for the solution, you didnt specify that the 2. player can choose to ask for trading cards and not have to ask.
with the assumption of a choice by player 2 the game developes into the game, wich solution you privide.
without the assumption the general case, you described as false would be the correct one, I think.
Please add Vietnamese subtitles, I'm from Vietnam and I love your channel.❤️
He'll do that, as soon as he learns Vietnamese. It might take a while; a long while…
@@user-vn7ce5ig1z Actually I just need him to translate the subtitles with google translate and then copy. Although not very accurate but it would be better not to have. Learning Vietnamese is very difficult, so i think he no need to learn Vietnamese. 😂
And then the judge ordered the execution on a Wednesday, and you were indeed surprised.
Only people with a 1 would want to trade, yet at the beginning we had a 2 and wanted to trade
Yeah, it's really a flawed assumption to assume your opponent would only offer to trade with a 1.
This is an incorrect application of IESDS.
Anyone doubting this solution can simulate the outcome, if Player A trades at only 1 and Player B also at 2. Player A will have a win rate of >50%. Also if you repeat this for 1000 cards and A trades only at 1 and B at anything to say 30, then again Player A will have a win rate > 50%.
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.
I can't understand, why my should I eliminate 4.Can you explain me once?
Suppose there were 10 cards, and he has a 3 and I have a 2. I know for sure he doesn't have a 2 and he knows for sure I don't have a 3. If I assume he acts strictly based on the odds, I know that he would be willing to trade only if he had a low card. So can I not assume that he has a 1, 3, 4, or 5 for sure? So then do I not have a 3/4 likelihood of winning that trade?
In his scenario, if we assume he trades only if the odds are in his favor, and he is offering to trade, then we can assume he has a 1 or 3. If he has a 1, he would surely want to trade. If he has a 3 though, would he want to trade? I think so. He would have a three out of five chance of receiving a better card would he not? So then if that is the case I know he has either a 1 or a 3, right? Is this not correct?
Watch the Maurice Ashley TED-Ed talk for a 30-second explanation (starting at 4;38 in that video) that took Presh over 3 minutes to explain. And he never showed off with words like "strictly dominated strategy."
No, if I had a 3, I would also ask if my opponent wants to trade. So I have a 2, he has a 1 or 3, the chances are 50/50.
Nope. You're actually lowering your chances to win if you offer to trade a 3. Without such offer your chances to win are 2/5, but if you offer, the opponent will likely accept only if he has 1 or 2, so your chances of winning are pretty much down to zero.
But if you had a 4 or higher, you'd keep that 4 or higher, right? So don't you think your opponent would also keep a 4 or higher? So can't you rule out them having a 4 or higher by their willingness to trade?
And there's me now thinking for the DOUBLE PSYCHOLOGY😂
This is a nice example. Here is a related one. Suppose you are selling your card to the highest bidder. You draw the 2. You will receive higher bids the higher your card. You can show your card to potential bidders or keep it hidden. Should you show it? (What if instead of 6 cards there were 100?)