@bscutajar someonep93 figured it out... it's like this if p1 (player 1) gets a red card, he guesses red, and let's say p2 gets a black, p1 was wrong so they survive, but now let's say p2 gets a red, p2 needs to simply predict the opposite color of his which is black and since p2 was wrong, they both survive
Clever trick, one player is set to pick the colour of their own card and the other the opposite.
Take, for instance, the player who picks their own colour: if the opposing player has the same colour as them, then the opposing player will have predicted incorrectly, thus surviving. If the opposing player has a different colour then them, they will have guessed incorrectly, thus surviving.
One of the guesses will always be right, but only one.
But the other one needs to know this trick too and you need to decide which one predicts his own color and which one predicts the opposite of the own color, right? so if the other suspect is not your friend you may get killed.but if the other suspect is a total stranger for you when this trick doesn't work, can't you still increase the chance of survival by always choosing the opposite of your own color. cause if you got red the other suspect has 25/52 red and 26/52 black.
An attempt to explain how this works without maths: Considering black cards only, in every case which results in death, the total number of black cards predicted = the total number in hand.
BB = BB
BR = BR
RB = RB
RR = RR
It is thus sufficient to ensure the totals do not match. If one player predicts the opposite of his card, that will create the imbalance. The other player simply needs to preserve the imbalance by predicting according to his own card.
If they told the people on trial everything you told us, the only people that would ever be burned at the stake are those who are not witches. If a witch is defined by the ability to predict the color of the other person's card, then the witch would always choose the opposite color of what they already know to be the other person's card. In this respect, a witch would never be burned. However, any regular person on trial would usually just guess randomly, with a 1/4 chance of getting burned.
I think I've got it. The prediction is based on what colour your previous card was. Let's say I got a red card on the first go, and then i got another red card, I would change my prediction, but if I got a red card on the first go and then I got a black card, I would keep the prediction the same as last. Now to work out why....
on all even number trials singing banana matches the color of his card while the other person picks the opposite of the card they got. on odd number trials they switch roles. that way you will never both guess correctly
I see it but its still weird. In round one, P1 copies, P2 inverts. So if both get R, P2 was wrong. If P1 gets R, and P2 gets B, P1 predicts R and P2 predicts R. P1 was wrong. If P1 gets B and P2 gets R, P1 predicts B and P2 predicts B. P1 was wrong. If both get B, you can see P2 is wrong
So it seems P2 is wrong when they both get the same card colours, and P1 is wrong when the colours differ. This is a bit complicated though, I've explained it in the way that hides the trick.
Not the technique used but this is a possibility...James always picks the same color as the card he picked up & his friend always picks the opposite. That way, if they both have red or both have black, his friend will get the wrong answer. & If James has red and his friend has black (or vice versa) James will pick red (or black if switched) and he will be wrong. These are the only 2 possible solutions as both red/both black can be treated as the same thing, and every time one answer will be wro
@firece174 I think you've got it. They might've mixed up the roles in this particular video to hide the trick.
I had to stop to think about why it works. This might be easier to understand for other people reading this: one person's strategy is that "our cards are the same colour" and the other person's strategy is "our cards are different colours". On average, they will each be right 50% of the time. However, they will never both be correct on the same turn! Their successes will never overlap.
also, the fact that they kept switching roles was only to confuse you and make it harder to guess the trick. real witches would pick a role and stick to it.
i'm not getting 25% for the probability of dying for the case of no previously discussed strategy. if you flip a coin that gives a slight bias to your own card color, as any guesser valuing her life would do, you get a slightly less than 25% probability due to it being slightly more likely that you picked different colors. i'll leave it as an exercise to find the right probability p to assign to "guess same color"
One person predicts the same color as their card, the other predicts the opposite of their card. The player who inverts switches each round. Take this table. Col 1/2 = P1/P2 cards, col 3/4 = P1/P2 prediction. P2 is predicting opposite: (Death occurs if the two inner colors match and two outer colors match) R R | R B = ok R B | R R = ok B R | B B = ok B B | B R = ok Next round, P1 inverts: R R | B R = ok R B | B B = ok B R | R R = ok B B | R B = ok Follow this and nobody will die. :)
k so.... he picks his colour... and you pick your opposite colour... and if he has R and you have R then he picks R and you pick B and you can never have the same :D... i'm happy
On every even trial you have the same color on your prediction as your card , and on every uneven trial you have the opposite color on your prediction as your card. and your friend dose the same but instead he dose opposite of what your doing.
there are four possible outcomes of the cards: BB. BR, RB, RR If you have one person predicting their own card color and the other predicting the opposite, one person will always be wrong!
first turn you make the prediction the opposite of the color of your card and the opposing person makes his prediction the same as the color of his card, and for the next turn they alternate and you predict the same color as your card and he predicts the opposite color of his card
I may be totally wrong, but I think by having a strategic way of switching their guess based on their own card color+ the trial number+the amount of trials left can dramatically increase their odds of surviving.
I may be totally wrong, but I think by having a strategic way of switching their guess based on their own card color+ the trial number +the amount of trials left can dramatically increase their odds of surviving.
On even numbered trials, you chose a prediction that was the same color as your card, and he chose a prediction that was opposite the color of his card. Then vice versa for odd numbered trials. Is that right?
there is a way of doing if say james changes the board around that could mean james' card was red and so the other player could hear if the board was changed around and adjust his answer accordingly
Your strategy is for someone to always guess the color he has, and the other guesses the opposite color that he has. Say person A always guesses his color and Person B always guesses the opposite. A draws a black and B draws a red. A will say B has black. He is wrong. B will say A has black. He is right, but they are safe. A draws red, B draws red. A says B has red. He is correct. B says A has black. He is wrong. They are safe.
I remember reading something about this sort of logic (using monks with inkdots on their heads), and I think this is somewhat similar. However, I can't think of a way to apply it. (The monks were unable to see their own heads and could not tell each other about the inkdots; they just had to use logic to see if they had an ink dot.)
Who ever starts predicts the opposite color of their card while the other keeps there prediction the same as there colored card, the next time they switch and the other person chooses the opposite while, the original starter chooses the same. I watched the first 5 times and this is how it was done so I thought it was right =P
why would a witch and the otehr witch both want to get the right answer and be burned if there were real witches they wouldt choose the same card thing cad way to find witches
the assumption is that they have to know who's going to say his own color versus who is going to say the opposite. How would the witches know which is which? It still ends up being random doesn't it?
First to draw the card chooses the colour oposite to the one he has,
The next move if you have the same colour as before, you change if you have the same suit
This means if first guy draws two reds, the chances of it being red for the 2nd guy (who draws straight after) are lowered, so he changes sign from black to red
it looks like you altrenate turning your card to the opposite side and to the same side as the color of your card, and when one of you is doing the same color as his card the other is doing the opposite color of his card :) simple
I got it. One predicts the color they have pulled (as if to say both people had the same color) and the other predicts the opposite (as if to say theyre different). That way, if person 1 was right, they both have the same color. person two will be wrong because he predicted the opposite. If person two is right, then the cards are opposite. Person 1 will be wrong because he predicted the two cards will be the same.
@looney1023 Yep. Method is right, but to put the reasoning a bit differently: A predicts they're the same, B predicts they're different. Being mutually exclusive, they can't both be right. Since exactly one will be wrong every time, they live every time.
1. One person chooses the same color as the first card given to him in the initial prediction while the other person chooses the opposite of his own... Hope for the best.
2. Predict same color as the last guess when the next card is a different suit / Flip prediction when next card is the same color
After the first test you can't lose because you can never be in a losing situation- One will always have the same color as his card while the other has the opposite of his own.
very simple trick here, but i see that singinbanana does it a bit more complicated. So this is how it works, both banana and his friend must first agree that one will chose the same color that was given to him, and the other person will choose the opposite of their own you can alternate who picks same and who picks opposite, as long as both are in use, both can agree a pattern to make it more interesting such as; ab.ab.ab or abb.abb.abb abaa.abaa.abaa etc.
Both Banana and his friend switch each time from predicting the same color as that of the card which he drew and the opposite. So if Banana drew a red card on one turn he would predict red. If he then drew another red card on the next turn he would predict black. Also, the friend's sequence was opposite to Banana's. So if Banana was picking same, opposite, same, opposite, the friend would be picking opposite, same, opposite, same. I don't know if the 2nd part was intentional or not.
I think I figured it out. Start out with opposite predictions. If you get the same color card twice in a row, switch your prediction.
Now, this system doesn't work for the first card because it requires at least one card in order to work. So, I assume there is something else to get you past the first round. Such as if you get a black card switch your card. Then you can correct for this switch before continuing on with the previously stated system.
After the first draw where one perdicted red and one perdicted black that was the only time where there was a probability of failing. One person knew which color the opposing player would perdict because after the first draw the player would know if the opposing player switched perdictions based on the sound of the card (black side or red side card) shifting. Therefore each player played and due to the condition that they both have to be right to be proclaimed witches each made themselves fail.
This makes it impossible to lose, if one person picks what they have in order to be right the other must have the same color. If this person is always picking the opposite of what he has then it is impossible for him to pick the other person's color if the other person picks thiers.
but it becomes 100% when there are only two cards left. If he counted, lets say that he counted 25 red cards and he pics a black card. then he is 100% certain that the other player has a red card. You can only be 100% certain in the last two cards of the deck.
exactly my point. They did it with 100% accuracy from the beginning, so either they did this video several hundred times while counting cards until they finally got all of them right, or they used another method. Which sounds more likely?
if both witches get the same colour then the "opposite" witch is wrong and if they dont get the same colour the "same" witch is wrong.
of course if they didnt get to talk to each other beforehand then they can still count the number of red and black they have drawn and make predictions accordingly to improve the odds. -drawn tons of black lately? predict black! XD
It makes sense once you think about it in terms of each person's choice.
One person basically says, "we both have the same color card" and another person says, "we have different color cards". Both people can not be correct.
@Error081688 you are correct... if one person has red and the other black then the person with the red would say guess red and the person with black would also guess red making the prediction wrong if they both had the same colour then then one person would be guessing the opposite of there colour making the prediction wrong.
the first one says "predicts" his card, for example BLACK, and if the second one has black, he would predict RED, because he knows that his friends card is BLACK, and if the First one predicts BLACK (for example), and the second one has a red card, he could predict either RED or BLACK, and the will still not lose.
the chance of both guessing right is onone trial only 6,25 %? right?...so like flipping a quarter u look at each time to see the chances..or something lol XD
It's got to be, witch 1 predicts the same colour of her own card and witch 2 the opposite colour. If the colours are the same witch 1 will be right and witch 2 will be wrong and if they're different it'll be the other way around...
This is their strategy but they modified it to make it more confusing. Every trial they alternate who will choose the opposite and who will choose the same.
The first trial singingbanana chooses the opposite color as his own and his friend chooses the same as his own. The second trial they switch and banana chooses the same as his own and his friend the opposite his own. They continue alternating until the end.
I'm guessing they can't see each other's predictions but they can hear the predictor being turned around; then it's just a matter of agreeing on what changing the current prediction means (which is irrelevant, pretty much anything would work to pass the information) or keeping track of each other's predictor's state.
my question is, do YOU have a strategy or could the witches have done the same in totally isolated rooms??
I can tell you our strategy in the video will work in completely isolated rooms. For the sake of the video we only had a thin divider between us, but there was no communication between us.
Both witches have to start with a color offset (red-black or black-red), with this arranged before the performance. Based solely on his/her own previous and current cards, the person makes his/her prediction.
The rule is this: if your current card is the same from your previous card, change the prediction; else, maintain it. In this manner, between the 4 objects (2 cards + 2 predictions), there will always be 3 on the same color and 1 on the different one.
Left player "guesses" same as their own colour in odd rounds. Right player "guesses" same as their own colour in even rounds. Hence in each round one of the players knows what colour the other has plus the other's guess and chooses a guess so that they do not both get it right. However this solution assumes one player hears the other's guess before deciding. You said that in the actual problem the players are in different rooms and decide independently? Not sure i can solve that variation :(
ino ur not gona believe me lol but I got it within a minute :P, so long as one of them puts the same on their board as their card and the other puts the opposite on the board to their card, they can never get both cards matching both boards. They just alternate who uses the same.
Every time, one of them chooses the colour opposite the colour of their card and the other person chooses the colour that is the same as their card.
Then they switch to picking the same colour and different colour so that one person chooses the opposite colour to theirs and the other person chooses the same colour.
I'm well aware of the small likelihood of this occurring. My original statement was tongue in cheek... but I guess the smiley didn't make that obvious enough? :-P
i'm not very good at maths these days as i gave it up for art, but... my guess is that the probability lies within the cards rather than the maths that you blinded us with to begin with oh master of suggestion?!?
solution?
bscutajar 4 weeks ago
@bscutajar someonep93 figured it out... it's like this if p1 (player 1) gets a red card, he guesses red, and let's say p2 gets a black, p1 was wrong so they survive, but now let's say p2 gets a red, p2 needs to simply predict the opposite color of his which is black and since p2 was wrong, they both survive
nothinisworthit 3 weeks ago in playlist More videos from singingbanana
solution? i dont understand any of these comments
bscutajar 4 weeks ago
Clever trick, one player is set to pick the colour of their own card and the other the opposite.
Take, for instance, the player who picks their own colour: if the opposing player has the same colour as them, then the opposing player will have predicted incorrectly, thus surviving. If the opposing player has a different colour then them, they will have guessed incorrectly, thus surviving.
One of the guesses will always be right, but only one.
Probably already posted but yah;
SlothScripts 1 month ago
This has been flagged as spam show
What someonep93 said, and in this video they are alternating roles.
zantiertasa 2 months ago
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zantiertasa 2 months ago
sorry...I forgot you need guess wrong to survive so predict always your own color
JamiJami91 2 months ago
But the other one needs to know this trick too and you need to decide which one predicts his own color and which one predicts the opposite of the own color, right? so if the other suspect is not your friend you may get killed.but if the other suspect is a total stranger for you when this trick doesn't work, can't you still increase the chance of survival by always choosing the opposite of your own color. cause if you got red the other suspect has 25/52 red and 26/52 black.
JamiJami91 2 months ago
An attempt to explain how this works without maths: Considering black cards only, in every case which results in death, the total number of black cards predicted = the total number in hand.
BB = BB
BR = BR
RB = RB
RR = RR
It is thus sufficient to ensure the totals do not match. If one player predicts the opposite of his card, that will create the imbalance. The other player simply needs to preserve the imbalance by predicting according to his own card.
I hope that wasn't even more confusing...
roddy0 3 months ago
Solution?
AnthonyWilder 4 months ago
It's confusing but I agree with vampiracy
CTFxCMarleyZoe 5 months ago
If they told the people on trial everything you told us, the only people that would ever be burned at the stake are those who are not witches. If a witch is defined by the ability to predict the color of the other person's card, then the witch would always choose the opposite color of what they already know to be the other person's card. In this respect, a witch would never be burned. However, any regular person on trial would usually just guess randomly, with a 1/4 chance of getting burned.
vampiracy 5 months ago
I think I've got it. The prediction is based on what colour your previous card was. Let's say I got a red card on the first go, and then i got another red card, I would change my prediction, but if I got a red card on the first go and then I got a black card, I would keep the prediction the same as last. Now to work out why....
schoolmornings 5 months ago
on all even number trials singing banana matches the color of his card while the other person picks the opposite of the card they got. on odd number trials they switch roles. that way you will never both guess correctly
PouringRain21 7 months ago
Comment removed
GuyMcpersonLol 7 months ago
I see it but its still weird. In round one, P1 copies, P2 inverts. So if both get R, P2 was wrong. If P1 gets R, and P2 gets B, P1 predicts R and P2 predicts R. P1 was wrong. If P1 gets B and P2 gets R, P1 predicts B and P2 predicts B. P1 was wrong. If both get B, you can see P2 is wrong
So it seems P2 is wrong when they both get the same card colours, and P1 is wrong when the colours differ. This is a bit complicated though, I've explained it in the way that hides the trick.
guitardudeguy00 8 months ago
Well, if p1 guesses the same color as p1's card and p2 guesses the color not on p2's card they will always survive.
All three distinct cases using this tactic:
RR gives guess BR - they survive
BB gives guess BR - they survive
BR gives guess BB or RR - they survive
someonep93 9 months ago 11
Not the technique used but this is a possibility...James always picks the same color as the card he picked up & his friend always picks the opposite. That way, if they both have red or both have black, his friend will get the wrong answer. & If James has red and his friend has black (or vice versa) James will pick red (or black if switched) and he will be wrong. These are the only 2 possible solutions as both red/both black can be treated as the same thing, and every time one answer will be wro
firece174 10 months ago
@firece174 I think you've got it. They might've mixed up the roles in this particular video to hide the trick.
I had to stop to think about why it works. This might be easier to understand for other people reading this: one person's strategy is that "our cards are the same colour" and the other person's strategy is "our cards are different colours". On average, they will each be right 50% of the time. However, they will never both be correct on the same turn! Their successes will never overlap.
allenu 6 months ago
also, the fact that they kept switching roles was only to confuse you and make it harder to guess the trick. real witches would pick a role and stick to it.
quintopia 10 months ago
i'm not getting 25% for the probability of dying for the case of no previously discussed strategy. if you flip a coin that gives a slight bias to your own card color, as any guesser valuing her life would do, you get a slightly less than 25% probability due to it being slightly more likely that you picked different colors. i'll leave it as an exercise to find the right probability p to assign to "guess same color"
quintopia 10 months ago
MadManMarkAu 11 months ago
@MadManMarkAu Bleh, YouTUBE decided to mess up my formatting, I'm sorry. Nothing I can do about that.
MadManMarkAu 11 months ago
lool you can see his card every time?
MaidMujan 1 year ago
k so.... he picks his colour... and you pick your opposite colour... and if he has R and you have R then he picks R and you pick B and you can never have the same :D... i'm happy
fatescure 1 year ago
i'm going to figure this out i swear it
fatescure 1 year ago
Just say purple.
mtflyer05 1 year ago
sorry i fogot he said they half to choose just the color
thepimp537 1 year ago
you chose your own card cause they dont have it
thepimp537 1 year ago
On every even trial you have the same color on your prediction as your card , and on every uneven trial you have the opposite color on your prediction as your card. and your friend dose the same but instead he dose opposite of what your doing.
there are four possible outcomes of the cards: BB. BR, RB, RR If you have one person predicting their own card color and the other predicting the opposite, one person will always be wrong!
Mrdavidhansson 1 year ago
kcaj777 got it first methinks.
ebutemetube 1 year ago
It looks like they are alternating switching.
first turn you make the prediction the opposite of the color of your card and the opposing person makes his prediction the same as the color of his card, and for the next turn they alternate and you predict the same color as your card and he predicts the opposite color of his card
EojRepus 1 year ago
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I may be totally wrong, but I think by having a strategic way of switching their guess based on their own card color+ the trial number+the amount of trials left can dramatically increase their odds of surviving.
aznboiij0sh 1 year ago
I may be totally wrong, but I think by having a strategic way of switching their guess based on their own card color+ the trial number +the amount of trials left can dramatically increase their odds of surviving.
aznboiij0sh 1 year ago
You kinda cheated by using the same deck over and over.Try your strategy with 20 re-shuffled decks :)
MeetYourMeaker 1 year ago
@MeetYourMeaker no, not a cheat... it doesn't matter how many decks you use... this will always work.
gtq838 1 year ago
Left guy: odd numbered turns predict the same as your own card, even numbered turns predict the opposite of your own card
Right guy: odd numbered turns predict the opposite as your own card, even numbered turns predict the same as your own card
On each go there are four possible outcomes of the cards: BB. BR, RB, RR
If you have one person predicting their own card colour and the other predicting the opposite, one person will always be wrong!
You just alternate to cover it up!
eskimokid 1 year ago
On even numbered trials, you chose a prediction that was the same color as your card, and he chose a prediction that was opposite the color of his card. Then vice versa for odd numbered trials. Is that right?
Upfallv2 1 year ago
i think it has something to do with the cycle of your predictions because it went like
both the same
1 switched
1 switched
1 switched
both switched
and it went like that for a while so thats how i think you did it.
dakingofblingg 1 year ago
there is a way of doing if say james changes the board around that could mean james' card was red and so the other player could hear if the board was changed around and adjust his answer accordingly
Everyvideo123 1 year ago
Your strategy is for someone to always guess the color he has, and the other guesses the opposite color that he has. Say person A always guesses his color and Person B always guesses the opposite. A draws a black and B draws a red. A will say B has black. He is wrong. B will say A has black. He is right, but they are safe. A draws red, B draws red. A says B has red. He is correct. B says A has black. He is wrong. They are safe.
mikechis101 1 year ago
u saw each others cards first then u made the call
messitup 1 year ago
I remember reading something about this sort of logic (using monks with inkdots on their heads), and I think this is somewhat similar. However, I can't think of a way to apply it. (The monks were unable to see their own heads and could not tell each other about the inkdots; they just had to use logic to see if they had an ink dot.)
fireedud 1 year ago
Who ever starts predicts the opposite color of their card while the other keeps there prediction the same as there colored card, the next time they switch and the other person chooses the opposite while, the original starter chooses the same. I watched the first 5 times and this is how it was done so I thought it was right =P
XiTerrorXi 1 year ago
why would a witch and the otehr witch both want to get the right answer and be burned if there were real witches they wouldt choose the same card thing cad way to find witches
hawkfake 1 year ago
the assumption is that they have to know who's going to say his own color versus who is going to say the opposite. How would the witches know which is which? It still ends up being random doesn't it?
wormed7600 1 year ago
First to draw the card chooses the colour oposite to the one he has,
The next move if you have the same colour as before, you change if you have the same suit
This means if first guy draws two reds, the chances of it being red for the 2nd guy (who draws straight after) are lowered, so he changes sign from black to red
SousukePanic 1 year ago
it looks like you altrenate turning your card to the opposite side and to the same side as the color of your card, and when one of you is doing the same color as his card the other is doing the opposite color of his card :) simple
natethegreatBlitzMax 1 year ago
I got it. One predicts the color they have pulled (as if to say both people had the same color) and the other predicts the opposite (as if to say theyre different). That way, if person 1 was right, they both have the same color. person two will be wrong because he predicted the opposite. If person two is right, then the cards are opposite. Person 1 will be wrong because he predicted the two cards will be the same.
looney1023 1 year ago
@looney1023 Yep. Method is right, but to put the reasoning a bit differently: A predicts they're the same, B predicts they're different. Being mutually exclusive, they can't both be right. Since exactly one will be wrong every time, they live every time.
vpaulsmithjr 1 year ago
1. One person chooses the same color as the first card given to him in the initial prediction while the other person chooses the opposite of his own... Hope for the best.
2. Predict same color as the last guess when the next card is a different suit / Flip prediction when next card is the same color
After the first test you can't lose because you can never be in a losing situation- One will always have the same color as his card while the other has the opposite of his own.
thehunto 1 year ago
you always guessed the opposite of the color you pulled?
TheSlightOfHand 1 year ago
One person would pick the opposite of the card they had and the other would pick the same. This is alternated every round of cards.
ziggybomb200 1 year ago
very simple trick here, but i see that singinbanana does it a bit more complicated. So this is how it works, both banana and his friend must first agree that one will chose the same color that was given to him, and the other person will choose the opposite of their own you can alternate who picks same and who picks opposite, as long as both are in use, both can agree a pattern to make it more interesting such as; ab.ab.ab or abb.abb.abb abaa.abaa.abaa etc.
metalchango 2 years ago
Both Banana and his friend switch each time from predicting the same color as that of the card which he drew and the opposite. So if Banana drew a red card on one turn he would predict red. If he then drew another red card on the next turn he would predict black. Also, the friend's sequence was opposite to Banana's. So if Banana was picking same, opposite, same, opposite, the friend would be picking opposite, same, opposite, same. I don't know if the 2nd part was intentional or not.
sleepybrownbear 2 years ago
they take turn one person chooses the their prediction the same colour as their cards and the other chooses the opposite.
lukeyryr 2 years ago
I think I figured it out. Start out with opposite predictions. If you get the same color card twice in a row, switch your prediction.
Now, this system doesn't work for the first card because it requires at least one card in order to work. So, I assume there is something else to get you past the first round. Such as if you get a black card switch your card. Then you can correct for this switch before continuing on with the previously stated system.
amcnea 2 years ago
Comment removed
amcnea 2 years ago
After the first draw where one perdicted red and one perdicted black that was the only time where there was a probability of failing. One person knew which color the opposing player would perdict because after the first draw the player would know if the opposing player switched perdictions based on the sound of the card (black side or red side card) shifting. Therefore each player played and due to the condition that they both have to be right to be proclaimed witches each made themselves fail.
EXEveryNote 2 years ago
He did the opposite of what he had.
His friend used the same as what he had.
This makes it impossible to lose, if one person picks what they have in order to be right the other must have the same color. If this person is always picking the opposite of what he has then it is impossible for him to pick the other person's color if the other person picks thiers.
packered 2 years ago
they picked the opposite color of the color card they had ;D
izn3rdyhomeboy 2 years ago
count the numbers of red and black.
if the deck is bran new and you get a black card the probability is more that the other person has a red card
herman9737 2 years ago
the probability is a little bit weighted, but not much. Certainly not 100%, right?
Error081688 2 years ago
no.
but it becomes 100% when there are only two cards left. If he counted, lets say that he counted 25 red cards and he pics a black card. then he is 100% certain that the other player has a red card. You can only be 100% certain in the last two cards of the deck.
herman9737 2 years ago
exactly my point. They did it with 100% accuracy from the beginning, so either they did this video several hundred times while counting cards until they finally got all of them right, or they used another method. Which sounds more likely?
Error081688 2 years ago
They are obviously both witches and have found a way to beat the test
CogitoErgoCogitoSum 2 years ago 72
haha that's what I said in my first comment
Error081688 2 years ago
it was under my other profile though, Tom
Error081688 2 years ago
Without alternating whom guesses hes own color and who guesses oposite, or one constantly guesses wrong and one right you cant guarantee the outcome.
The effect of the game can clearly be shown by randomly picking cards from the deck, it works.
Would have been nice to see a concept where one didnt alternate. Its not impossible, its just takes longer to achieve :-)
grodahlen 2 years ago
if both witches get the same colour then the "opposite" witch is wrong and if they dont get the same colour the "same" witch is wrong.
of course if they didnt get to talk to each other beforehand then they can still count the number of red and black they have drawn and make predictions accordingly to improve the odds. -drawn tons of black lately? predict black! XD
Alfaomegabravo 2 years ago
i think you could have set up the deck somehow...
TheAceOfSpades100 2 years ago
well the solution is there, but unless you have time to talk it out with the other person, you won't necessarily survive every time
sillybilly452 2 years ago
I got it. There is 2 conditions to your strategy.
1. On the first draw, one person predict his own color, the other person predicts his opposite color. (This is determined before the trial.)
2. When you draw a "repeat" color, switch your prediction. (If you drew a red card last turn, then you switch your prediction.)
I solved it like this: Draw Pred. Draw Pred.
RB RR or BR BB is the first draw according to condition 1.
continued.....
froboykujo69 2 years ago 3
You can only lose with RR RR, BB BB, RB BR, BR RB.
However, according to condition 2 only ONE letter from each group MUST change every turn.
Using Condition 1 as the first prediction, there is no way by changing it to a losing combination even if you keep running the trial.
Also, Condition 1 prevents a loss on the first turn. (channel fireball.... nvm)
I hope that is sufficent enough to explain the strategy.
froboykujo69 2 years ago 2
(edit)
there are two more starting draws RB BB and BR RR, but it still works.
froboykujo69 2 years ago
why do they pick the color AFTER they see their cards?? i dont get it.. shouldn't they predict the card colour before they see the cards?
localbroadcast 2 years ago
It makes sense once you think about it in terms of each person's choice.
One person basically says, "we both have the same color card" and another person says, "we have different color cards". Both people can not be correct.
Error081688 2 years ago 3
@Error081688 you are correct... if one person has red and the other black then the person with the red would say guess red and the person with black would also guess red making the prediction wrong if they both had the same colour then then one person would be guessing the opposite of there colour making the prediction wrong.
gtq838 1 year ago 2
It's quite simple.
The guy who created the trial is an idiot; it should have been the other way around. You are both witches and this proves it.
:)
TomStockman3 2 years ago 3
One player predicts same colour, one player predicts opposite colour to own card. I created a set:
boolies: {true, false}
Then wrote a program to show all four possible outcomes:
for b1 in boolies do for b2 in boolies do print("c1", b1, "p1", b1, "c2", b2, "p2", not b2)$
Here's the output:
c1 false p1 false c2 false p2 true
c1 false p1 false c2 true p2 false
c1 true p1 true c2 false p2 true
c1 true p1 true c2 true p2 false
So at least one of the witches always wrong.
ThBearsFriend 2 years ago 2
example:
the first one says "predicts" his card, for example BLACK, and if the second one has black, he would predict RED, because he knows that his friends card is BLACK, and if the First one predicts BLACK (for example), and the second one has a red card, he could predict either RED or BLACK, and the will still not lose.
easy and simple, and yet again interesting :)
1MAKEDONEC1 2 years ago
the chance of both guessing right is onone trial only 6,25 %? right?...so like flipping a quarter u look at each time to see the chances..or something lol XD
CheatProjectDotCom 2 years ago
omfg i have no idea lol
FingerboardFingers13 2 years ago
It's got to be, witch 1 predicts the same colour of her own card and witch 2 the opposite colour. If the colours are the same witch 1 will be right and witch 2 will be wrong and if they're different it'll be the other way around...
a simple but mystifying puzzle :)
scotland7yard 2 years ago
i agree
cowan913 2 years ago
yea seems like thats what theyre doing
jduran400 2 years ago
This is their strategy but they modified it to make it more confusing. Every trial they alternate who will choose the opposite and who will choose the same.
The first trial singingbanana chooses the opposite color as his own and his friend chooses the same as his own. The second trial they switch and banana chooses the same as his own and his friend the opposite his own. They continue alternating until the end.
jesselee34 2 years ago
I'm guessing they can't see each other's predictions but they can hear the predictor being turned around; then it's just a matter of agreeing on what changing the current prediction means (which is irrelevant, pretty much anything would work to pass the information) or keeping track of each other's predictor's state.
my question is, do YOU have a strategy or could the witches have done the same in totally isolated rooms??
scotland7yard 2 years ago
I can tell you our strategy in the video will work in completely isolated rooms. For the sake of the video we only had a thin divider between us, but there was no communication between us.
singingbanana 2 years ago 3
@singingbanana
So.... are you not going to explain it? Or do I have to be intelligent? :)
GenVK 10 months ago
This has been flagged as spam show
** SPOILER **
Both witches have to start with a color offset (red-black or black-red), with this arranged before the performance. Based solely on his/her own previous and current cards, the person makes his/her prediction.
The rule is this: if your current card is the same from your previous card, change the prediction; else, maintain it. In this manner, between the 4 objects (2 cards + 2 predictions), there will always be 3 on the same color and 1 on the different one.
Survival % = 50
283cirht 2 years ago
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283cirht 2 years ago
Comment removed
283cirht 2 years ago
Great puzzle!
wafflethug 3 years ago
I think this has made me think more than any other thing I've ever seen on the Internet, and I still didn't figure it out. It shouldn't be allowed.
pubwebmaster 3 years ago
Left player "guesses" same as their own colour in odd rounds. Right player "guesses" same as their own colour in even rounds. Hence in each round one of the players knows what colour the other has plus the other's guess and chooses a guess so that they do not both get it right. However this solution assumes one player hears the other's guess before deciding. You said that in the actual problem the players are in different rooms and decide independently? Not sure i can solve that variation :(
mikehoops 3 years ago
ino ur not gona believe me lol but I got it within a minute :P, so long as one of them puts the same on their board as their card and the other puts the opposite on the board to their card, they can never get both cards matching both boards. They just alternate who uses the same.
Phil05R 3 years ago 2
I think that one works! Singingbanana, can we have a filmed proof, please?
pubwebmaster 3 years ago
perfect mate
Jammylegend1 3 years ago
I think i have the solution!
Every time, one of them chooses the colour opposite the colour of their card and the other person chooses the colour that is the same as their card.
Then they switch to picking the same colour and different colour so that one person chooses the opposite colour to theirs and the other person chooses the same colour.
kcaj777 3 years ago 20
@kcaj777 but do u know y that works?
namasr 5 months ago
Comment removed
namasr 5 months ago
@kcaj777 wut? :P
DijakJumper 4 months ago
I've worked it out - brilliant!
Fantastic variation on an already
good trick. Elementary maths,
no tricks, outstanding!
riderofgiraffes 4 years ago
The boy is a genius. He gets it off his dad
burnel55 4 years ago
I reckon you just re-filmed it until you got a run of right answers :-)
tarmin 4 years ago
Keep guessing :)
singingbanana 4 years ago
I was astonished to observe German soldiers walking about within rifle range behind their own
line. Our men appeared to take no notice. I privately made up my mind to do away with that sort
of thing ; such things should not be allowed. These people evidently did not know there was a 10
war on. Both sides apparently believed in the policy of live and let live.
Supermanisafaggit 1 year ago
Do the math, as the Americans would say.
The number of re-films required is enormous.
There's no chance that can be made to work,
so there must be something else at work.
riderofgiraffes 4 years ago
I'm well aware of the small likelihood of this occurring. My original statement was tongue in cheek... but I guess the smiley didn't make that obvious enough? :-P
tarmin 3 years ago
My conclusion? Eminent mathemeticians are
terrible 'snap' players!
Uncle Len.
louielw6 4 years ago
oh my god you even stumped my dad!!
lemb1983 4 years ago
are they ALL 9 spades ?
robgrime1 4 years ago
See you Wednesday -- bring the Lottery numbers with you
Mum & Dad
robgrime1 4 years ago
i'm not very good at maths these days as i gave it up for art, but... my guess is that the probability lies within the cards rather than the maths that you blinded us with to begin with oh master of suggestion?!?
i'm gonna ask my dad... he'll know
lemb1983 4 years ago
i actually love you james!
lemb1983 4 years ago