I would just like to say, he is one of the most dedicated Teaching YT channel i've ever seen. 9 years. wow! this kind of youtube channel should have more than a million and have 100k views per video. I really Admire you!
Spoilers: Let U be the number of face-up red cards on the bottom right. Because the cards are dealt out in pairs, 52 cards is 26 pairs, so 26 face-up cards would be dealt out in total, so excluding the U face-up red cards, that means that (26 - U) face-up black cards were dealt on the bottom left. Let D be the number of face-down red cards in the pile at the top right. There are 26 red cards in total, and the only other place that red cards could be is in the top left pile, so the number of red cards in the top left pile in (26 - U - D). The total number of cards in the top left pile is the same as the bottom left pile (because whenever a black card was dealt face up to the bottom left, a partner card was dealt to the top left), so we already know from before that there are (26 - U) cards in the top left pile. (26 - U) total cards at the top left, (26 - U - D) of them are red, so that leaves (26 - U) - (26 - U - D) black cards at the top left -- which simplifies down to D -- i.e., the same number of red cards in the top right pile.
This is such an awesome math magic trick camouflaged in a very neat way. The 1-5 card swapping doesnt matter. What matters is 26 cards is removed out of play (the opened cards) & The opened cards are counted for red n black, and then closed cards are grouped with those numbers (of red n black opened cards) randomly Then the trick will auotmatically worked itself and it can be easily proven by algebra This is the best way to teach critical thinking. Thank you for what you do, Mr Eddie Woo, you are a truly inspiring teacher
The first thing to notice is that the swapping is a red herring. Since the order of the face down cards is totally random, swapping a few from one pile to another can have no effect on the result.
Right because if you exchange red for black cards then it increases or decreases the amount of each evenly. And if you exchange red for red or black for black, it doesn't change it at all anyway.
1. 26B + 26R = 52 Cards or 26B:26R (B=R) If we take two pairs at a time: Four options BB, RR, BR, RB Notice that whenever we take the pairs: BR, and RB, the B:R ratios do not change. It goes to stand that then when we have a BB pair our ratio looks like this B(-2): R, hence as we know that B=R then for every B(-2): R we must have a B:R(-2). Hence; x[B(-2): R] + x[B:R(-2)] = B:R. Which is to say the number of pairs of BB = RR.
2. From the two upper decks switching any card from one pile to the other can only have four outcomes. A (B B), (RR), (BR) and (RB). Now remember what we wish to observe, the black count above the black deck and a red count above the red deck. Black switching with a black means nothing changes as does red switching with red.[B1:B2:R1:R2] -> [1(B2)1(R2)] = [B1:(B2-1):R1:(R2-1)] Black(above black deck) switching with a Red(above red deck) means the count for both sides is both minus one, similarly switching a Red(above Black deck) and a Black(above Red deck) means the count for both sides is plus one. [B1:B2:R1:R2] -> [1(R2)1(B2)] = [B1:(B2+1):R1:(R2+1)] **Hence no matter how many times cards are swapped B2=R2.
That's a fun little problem! Here's what I think. If you pick two cards of the same color in a row, you'd add one card of that color into the face-down deck. If you pick two cards of different colors in a row, you'd add zero card of the first color into the face-down deck. So we can say: a pair is worth 1 point on either side, a non pair is worth 0 point. It means that if the number of red pairs and black pairs in the scramble is the same, the prediction will be true. Imagine your scramble contains only pairs (13 black pairs, 13 red pairs). Swap two cards of different color, and you'll break one black pair and one red pair. If you swap a card from a pair into a non pair, the number of pairs of each color doesn't change. So any swap you can do is either +1 pair of each color, -1 pair of each color, or no change. Since we know of a case where the number of pairs of each color is equal, and any swap of cards keeps it equal, we know that it has to be equal no matter what the scramble is. Swapping cards at the end is irrelevant. You'll either swap the same color or different colors. Swapping the same color obviously does nothing. Swapping different colors either adds or subtracts 1 point in each deck.
This seems similar to the "probability" theory. I know a similar "trick", it fundamentally works by a very precise manipulation process which at face value seems random without any predictable patterns. However each stage is critical to the previous one, although each step is fundamental to the last which cannot varied as the whole process is relative to the first step to achieve the impossible prediction of knowing what number of black/red cards will be in a randomly shuffled evenly split piles of cards without even looking at them. My trick is take a deck of 52 cards face down and turn 13 up right.now without looking at the cards,split them into 2 piles However u want, but the same amount off upright card's have to identical in each of the 2 piles. Seems impossible right? To start with the the number of upright cards is odd so cannot be evenly split. However it's quite simple...... Take the full deck of 52 cards and simply split them by splitting the top 13 cards 2 give 2 piles now turn the 13 cards u took off the top and flip them upside down now count each pile and you should have the same amount of upright cards in each pile.
I must stay its rather intresting. When I tried it out myself I ended up with each pile having 13 cards each, both on top and bottom. After taking the 2nd step and suffel the 3 cards from to top piles with each other the result was complete symmetry. Not only was the cards in Mr.Woos prediction matching but also the other tow piles with 6 and 6 to 7 and 7 in each pile. Then the odds with ending up with 13 cards in all the piles probably is not very high. Doing the trick once more would most likely give me a different variation.
Interesting point of view is to try manually break this and order the deck is the way it won't work. So first thought could be to create some e.g. black pairs and then put only black&red pairs, but doing this you realize you don't have enough black cards to do it, and amount of red cards which left exactly represent amount of black cards that you used for building first pairs. You cannot spoil it then. I am wondering about this last step to swap random cards and it seems to change nothing, but I am not sure about that, I anylized 1 and 2 cards to shuffle, probably up to 5 it doesn't change anything. Edit: Another way to spoil it, could be to order deck in the way you would only face-up one color, but then you will have 0 b/r cards on boths sides to compare. Edit 2: Swapping cards doesn't do anything, only can increase/decrease amount of counted cards but on both sides at the same time.
There r 26 pairs of cards (left +right) And there r also 26 black and red cards each Let no of upside black cards be m And no of upside red be n Now if their r k pairs on the left then the right would have 26-k pairs We know m+k+26-k-n =26 (that is the no of black cards) => m-n =0 Hence m=n
Let the set of face-up red cards be R with r no. of cards in it. Let the set of face-up black cards be B with b no. of cards in it. Let the set of face-down cards corresponding to R be R1. Let the set of face-down cards corresponding to B be B1. We know, No. of black cards in a deck = No. of red cards in a deck = 26. No. of cards in R1 and B1 together = sum of the remaining red and black cards = (26-r) + (26-b) = 52-(r+b). Since, for each card in R(or B) there is one card in R1(or B1) respectively, the no. of cards in R1=r and in B1=b. So, from the above, r+b = 52-(r+b) which implies that r+b = 26. Now, let the no. of black cards in B1 = b1. Then, the no. of black cards in R1 = total no. of black cards - (sum of black cards in B and B1) = 26-(b+b1) = 26-b-b1. Now, the no. of red cards in R1= total no. of cards in R1- no. of black cards in R1 = r - (26-b-b1). But r = 26-b as r+b = 26. Therefore, substituting the value of r above, we get, no. of red cards in R1= (26-b) - (26-b-b1) = b1 = no. of black cards in B1. (NOTE: This holds true irrespective of the swapping of the cards between R1 and B1.)
I wish Eddie was my math teacher in high school although my teachers did their best to hold our interest hence my love of math and physics!
me too
I would just like to say, he is one of the most dedicated Teaching YT channel i've ever seen. 9 years. wow! this kind of youtube channel should have more than a million and have 100k views per video. I really Admire you!
Spoilers: Let U be the number of face-up red cards on the bottom right. Because the cards are dealt out in pairs, 52 cards is 26 pairs, so 26 face-up cards would be dealt out in total, so excluding the U face-up red cards, that means that (26 - U) face-up black cards were dealt on the bottom left.
Let D be the number of face-down red cards in the pile at the top right. There are 26 red cards in total, and the only other place that red cards could be is in the top left pile, so the number of red cards in the top left pile in (26 - U - D). The total number of cards in the top left pile is the same as the bottom left pile (because whenever a black card was dealt face up to the bottom left, a partner card was dealt to the top left), so we already know from before that there are (26 - U) cards in the top left pile. (26 - U) total cards at the top left, (26 - U - D) of them are red, so that leaves (26 - U) - (26 - U - D) black cards at the top left -- which simplifies down to D -- i.e., the same number of red cards in the top right pile.
well explained
This is such an awesome math magic trick camouflaged in a very neat way.
The 1-5 card swapping doesnt matter.
What matters is 26 cards is removed out of play (the opened cards) &
The opened cards are counted for red n black, and then closed cards are grouped with those numbers (of red n black opened cards) randomly
Then the trick will auotmatically worked itself and it can be easily proven by algebra
This is the best way to teach critical thinking.
Thank you for what you do, Mr Eddie Woo, you are a truly inspiring teacher
Now I know every time I shuffle cards I am making history
You teach how my 8th grade math teacher taught and he was the best teacher I ever had in my entire school life. You just gained a new subscriber :D
Magnífico! Thanks for sharing it.
If Mr Woo was my teacher, my life would be so different... These students are so lucky.
I loved It.
The first thing to notice is that the swapping is a red herring. Since the order of the face down cards is totally random, swapping a few from one pile to another can have no effect on the result.
Right because if you exchange red for black cards then it increases or decreases the amount of each evenly. And if you exchange red for red or black for black, it doesn't change it at all anyway.
That's not what Ms. Direction told me
1. 26B + 26R = 52 Cards or 26B:26R (B=R)
If we take two pairs at a time:
Four options BB, RR, BR, RB
Notice that whenever we take the pairs: BR, and RB,
the B:R ratios do not change.
It goes to stand that then when we have a BB pair our ratio looks like this B(-2): R, hence as we know that B=R then for every B(-2): R we must have a B:R(-2). Hence; x[B(-2): R] + x[B:R(-2)] = B:R. Which is to say the number of pairs of BB = RR.
2. From the two upper decks switching any card from one pile to the other can only have four outcomes. A (B B), (RR), (BR) and (RB). Now remember what we wish to observe, the black count above the black deck and a red count above the red deck. Black switching with a black means nothing changes as does red switching with red.[B1:B2:R1:R2] -> [1(B2)1(R2)] = [B1:(B2-1):R1:(R2-1)]
Black(above black deck) switching with a Red(above red deck) means the count for both sides is both minus one, similarly switching a Red(above Black deck) and a Black(above Red deck) means the count for both sides is plus one.
[B1:B2:R1:R2] -> [1(R2)1(B2)] = [B1:(B2+1):R1:(R2+1)]
**Hence no matter how many times cards are swapped B2=R2.
Thank you.
That's a fun little problem! Here's what I think. If you pick two cards of the same color in a row, you'd add one card of that color into the face-down deck. If you pick two cards of different colors in a row, you'd add zero card of the first color into the face-down deck. So we can say: a pair is worth 1 point on either side, a non pair is worth 0 point.
It means that if the number of red pairs and black pairs in the scramble is the same, the prediction will be true.
Imagine your scramble contains only pairs (13 black pairs, 13 red pairs). Swap two cards of different color, and you'll break one black pair and one red pair. If you swap a card from a pair into a non pair, the number of pairs of each color doesn't change. So any swap you can do is either +1 pair of each color, -1 pair of each color, or no change. Since we know of a case where the number of pairs of each color is equal, and any swap of cards keeps it equal, we know that it has to be equal no matter what the scramble is.
Swapping cards at the end is irrelevant. You'll either swap the same color or different colors. Swapping the same color obviously does nothing. Swapping different colors either adds or subtracts 1 point in each deck.
🙏
Very nice proof 🙂
I am sure this will be very awosome .
This seems similar to the "probability" theory. I know a similar "trick", it fundamentally works by a very precise manipulation process which at face value seems random without any predictable patterns. However each stage is critical to the previous one, although each step is fundamental to the last which cannot varied as the whole process is relative to the first step to achieve the impossible prediction of knowing what number of black/red cards will be in a randomly shuffled evenly split piles of cards without even looking at them. My trick is take a deck of 52 cards face down and turn 13 up right.now without looking at the cards,split them into 2 piles However u want, but the same amount off upright card's have to identical in each of the 2 piles. Seems impossible right? To start with the the number of upright cards is odd so cannot be evenly split. However it's quite simple......
Take the full deck of 52 cards and simply split them by splitting the top 13 cards 2 give 2 piles now turn the 13 cards u took off the top and flip them upside down now count each pile and you should have the same amount of upright cards in each pile.
I must stay its rather intresting. When I tried it out myself I ended up with each pile having 13 cards each, both on top and bottom. After taking the 2nd step and suffel the 3 cards from to top piles with each other the result was complete symmetry. Not only was the cards in Mr.Woos prediction matching but also the other tow piles with 6 and 6 to 7 and 7 in each pile. Then the odds with ending up with 13 cards in all the piles probably is not very high. Doing the trick once more would most likely give me a different variation.
We are making History doing Eddie Woo maths work/Homework.
EVEN WORDS IS SO HARD TO REACH IN RIGHT PLACE FOR ME. EVEN THOUGH THERE IS PHONE.IF U HELP POOR PPL MY PBLMS WILL GET LOW A BIT.
Is there a part 2?
Interesting point of view is to try manually break this and order the deck is the way it won't work. So first thought could be to create some e.g. black pairs and then put only black&red pairs, but doing this you realize you don't have enough black cards to do it, and amount of red cards which left exactly represent amount of black cards that you used for building first pairs. You cannot spoil it then. I am wondering about this last step to swap random cards and it seems to change nothing, but I am not sure about that, I anylized 1 and 2 cards to shuffle, probably up to 5 it doesn't change anything.
Edit: Another way to spoil it, could be to order deck in the way you would only face-up one color, but then you will have 0 b/r cards on boths sides to compare.
Edit 2: Swapping cards doesn't do anything, only can increase/decrease amount of counted cards but on both sides at the same time.
I like your approach of trying to manually break the result -- leads really elegantly to a proof when you realize it can't be done
Hello 😀👍🙏
Part 2??
That number is called "Unvigintillion"
Gosh i love maths now
Maths is magic
I am dying to study from you in-person 😭😭
same
22
the point at @6:42 is probably false due to birthday paradox right?
YOU ARE SO FUCKING SMART
There r 26 pairs of cards (left +right)
And there r also 26 black and red cards each
Let no of upside black cards be m
And no of upside red be n
Now if their r k pairs on the left then the right would have 26-k pairs
We know m+k+26-k-n =26 (that is the no of black cards)
=> m-n =0
Hence m=n
Let the set of face-up red cards be R with r no. of cards in it.
Let the set of face-up black cards be B with b no. of cards in it.
Let the set of face-down cards corresponding to R be R1.
Let the set of face-down cards corresponding to B be B1.
We know, No. of black cards in a deck = No. of red cards in a deck = 26.
No. of cards in R1 and B1 together = sum of the remaining red and black cards = (26-r) + (26-b) = 52-(r+b).
Since, for each card in R(or B) there is one card in R1(or B1) respectively, the no. of cards in R1=r and in B1=b.
So, from the above, r+b = 52-(r+b) which implies that r+b = 26.
Now, let the no. of black cards in B1 = b1.
Then, the no. of black cards in R1 = total no. of black cards - (sum of black cards in B and B1) = 26-(b+b1) = 26-b-b1.
Now, the no. of red cards in R1= total no. of cards in R1- no. of black cards in R1 = r - (26-b-b1).
But r = 26-b as r+b = 26. Therefore, substituting the value of r above, we get,
no. of red cards in R1= (26-b) - (26-b-b1) = b1 = no. of black cards in B1.
(NOTE: This holds true irrespective of the swapping of the cards between R1 and B1.)
Brunettes are great but there's something about redheads. Well maybe that's just me.
Wait... this video ends on a cliff hanger?!
The title says "1 of 2" for a reason...
@@TheHuesSciTech Yes, I should have taken that as a hint. But honestly - who expects a cliff hanger in math?
@@Schmidtelpunkt Anyone calculating the breaking point of steel cables :)
Comment below to be included maybe
if you subscribe i will report you
wilo
will e mean whllo