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A magic trick - find out the fifth card if four is given
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Here is a magic trick I saw. My question is how the magician and his partner did it.
Given the simple French deck of cards, with $52$ cards. A person from the audience chooses randomly five cards from the deck and gives it to the partner (the partner works with the magician), without showing it to the magician. Then the partner (who sees the five cards) chooses four cards from the five cards, and gives it to the magician one by one (so the order of the four cards matters). From that the magician knows the fifth card.
The partner and the magician can’t communicate during the trick. How did they do it?
I thought that amoung the five cards there will be two with the same sign (Spades,Hearts,Diamonds or Clubs) and one of these two cards will be the fifth, and the other will be the first card to give to the magician...
combinatorics graph-theory contest-math card-games matching-theory
edited 41 mins ago
greedoid
32.1k114287
asked 2 hours ago
Yeah
382
add a comment |Â
up vote
5
down vote
favorite
Here is a magic trick I saw. My question is how the magician and his partner did it.
Given the simple French deck of cards, with $52$ cards. A person from the audience chooses randomly five cards from the deck and gives it to the partner (the partner works with the magician), without showing it to the magician. Then the partner (who sees the five cards) chooses four cards from the five cards, and gives it to the magician one by one (so the order of the four cards matters). From that the magician knows the fifth card.
The partner and the magician can’t communicate during the trick. How did they do it?
I thought that amoung the five cards there will be two with the same sign (Spades,Hearts,Diamonds or Clubs) and one of these two cards will be the fifth, and the other will be the first card to give to the magician...
combinatorics graph-theory contest-math card-games matching-theory
edited 41 mins ago
greedoid
32.1k114287
asked 2 hours ago
Yeah
382
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Here is a magic trick I saw. My question is how the magician and his partner did it.
Given the simple French deck of cards, with $52$ cards. A person from the audience chooses randomly five cards from the deck and gives it to the partner (the partner works with the magician), without showing it to the magician. Then the partner (who sees the five cards) chooses four cards from the five cards, and gives it to the magician one by one (so the order of the four cards matters). From that the magician knows the fifth card.
The partner and the magician can’t communicate during the trick. How did they do it?
I thought that amoung the five cards there will be two with the same sign (Spades,Hearts,Diamonds or Clubs) and one of these two cards will be the fifth, and the other will be the first card to give to the magician...
combinatorics graph-theory contest-math card-games matching-theory
edited 41 mins ago
greedoid
32.1k114287
asked 2 hours ago
Yeah
382
Here is a magic trick I saw. My question is how the magician and his partner did it.
Given the simple French deck of cards, with $52$ cards. A person from the audience chooses randomly five cards from the deck and gives it to the partner (the partner works with the magician), without showing it to the magician. Then the partner (who sees the five cards) chooses four cards from the five cards, and gives it to the magician one by one (so the order of the four cards matters). From that the magician knows the fifth card.
The partner and the magician can’t communicate during the trick. How did they do it?
I thought that amoung the five cards there will be two with the same sign (Spades,Hearts,Diamonds or Clubs) and one of these two cards will be the fifth, and the other will be the first card to give to the magician...
combinatorics graph-theory contest-math card-games matching-theory
combinatorics graph-theory contest-math card-games matching-theory
edited 41 mins ago
greedoid
32.1k114287
asked 2 hours ago
Yeah
382
edited 41 mins ago
greedoid
32.1k114287
asked 2 hours ago
Yeah
382
edited 41 mins ago
greedoid
32.1k114287
edited 41 mins ago
greedoid
32.1k114287
edited 41 mins ago
greedoid
32.1k114287
32.1k114287
asked 2 hours ago
Yeah
382
asked 2 hours ago
Yeah
382
asked 2 hours ago
Yeah
382
382
add a comment |Â
add a comment |Â
4 Answers
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up vote
2
down vote
Let $S$ be a set of all cards.
So we have a set $A$ of all ordered $4$-couples of the set $S$ and a set $B$ of all $5$ subsets of the set $S$.
Connect 4-couple in $A$ with a subset in $B$ iff all four card from that 4-couple are in that subset. Clearly each 4-couple is connected to $8$ 5-subsets and every 5 subset is connected to $5!= 120$ 4-couples.
This relation give us a bipartite graph $G=(A,B)$. Now this graph satisfies Hall matching condition. Take any subset $X$ in $A$. Then the set of neighbours $N(X)$ satisfies: $$ 120cdot |N(X)| geq 48cdot |X|implies |N(X)|geq |X| $$
So there exist a matching which saturate all vertices in $A$.
answered 45 mins ago
greedoid
32.1k114287
1
This, of course, does not guarantee the existence of a nice strategy for the magic trick; an arbitrary matching might look like a very long list with entries like "If the four cards are $9spadesuit, 3spadesuit, Kdiamondsuit, 9heartsuit$, then the remaining card is $Aclubsuit$." So beyond this, there is some work to be done to describe some matching in a way that a human can reasonably remember.
– Misha Lavrov
41 mins ago
Yeah, the question is to find an explicit strategy, it does not ask for existence. Thanks, I'll delete my answer.
– greedoid
38 mins ago
1
No, don't! The proof in your answer is still an important observation :)
– Misha Lavrov
38 mins ago
add a comment |Â
up vote
2
down vote
There is in fact a solution which uses your idea: that some suit will be repeated twice.
Say that among the $13$ ranks within a suit, ordered $A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K$, each rank "beats" the next six ranks, wrapping around when we get to the end. For example, $A$ beats $2,3,4,5,6,7$, and $10$ beats $J, Q, K, A, 2, 3$.
If you have two cards of the same suit, exactly one of them beats the other one. So you should pass, in order:
- A card of the same suit as the missing card, which beats the missing card. This leaves six possibilities for the missing card.
- The remaining three cards, in an order that encodes which of the six possibilities it is.
For the second step, we should order all $52$ cards in the deck somehow; for instance, say that $clubsuit < diamondsuit < heartsuit < spadesuit$ and $A < 2 < 3 < dots < Q < K$. Then the order of the last three cards is one of "Low, Middle, High", "Low, High, Middle", and so on through "High, Middle, Low". Just remember some correspondence between these six possibilities and the values $+1, +2, +3, +4, +5, +6$, and add the value you get to the rank of the first card (wrapping around from $K$ to $A$ of the same suit).
For example, say that the correspondence we chose in the second step is
beginarrayc
textLow & textMiddle & textHigh &+1 \
textLow & textHigh & textMiddle &+2 \
textMiddle & textLow & textHigh &+3 \
textMiddle & textHigh & textLow &+4 \
textHigh & textLow & textMiddle &+5 \
textHigh & textMiddle & textLow &+6
endarray
and you draw the cards $4clubsuit, 5spadesuit, 5diamondsuit, Aclubsuit, Jspadesuit$.
- We have two possibilities for the repeated suit, so let's choose $spadesuit$.
- In the cyclic order in that suit, $5spadesuit$ beats $Jspadesuit$, so the first card we pass is $5spadesuit$.
- We want to encode the offset $+6$, which is the ordering High, Middle, Low.
- So we pass that ordering: after $5spadesuit$ we pass $5diamondsuit, 4clubsuit, Aclubsuit$ in that order, because $5diamondsuit > 4clubsuit > Aclubsuit$.
answered 16 mins ago
Misha Lavrov
40.3k55399
Nice! I was working on the same solution (3 cards give a number between 1 and 6) but couldn't figure a way to convey whether to add or subtract that number. Your "beats" strategy solved that.
– Jens
6 mins ago
add a comment |Â
up vote
1
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It's a cute one. Let's just assume the cards are numbered $1 dots 52$. We draw cards $a_1 < a_2 < a_3 < a_4 < a_5$, and hand out in this order $b_1, b_2, b_3, b_4$. The way we order this set $B=b_1, b_2, b_3, b_4$ already allows to transmit quite some information. Indeed, choosing the orders paramounts to choosing a permutation on the 4 elements of $B$. There are $4! = 4 cdot 3 cdot 2 cdot 1 = 24$ such permutations.
This is not enough still to transmit the $5$th which is in a set of size $52$. Well, we also know that the 5th card is different from those $4$ firsts ones: there is only $52-4 = 48$ possibilities.
I'll leave it to you to complete the strategy. Note that $48/24 = 2$, so we just need to find a trick to only half the search space. Note also that we have still not fixed how we choose the secret card among the set $A$. At last, note at least one of the two inequalities:
- $a_2 leq 25$
- $a_4 geq 58-25$.
answered 56 mins ago
LeoDucas
1846
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0
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For any 4 cards given, we have 24 possible ways of ordering then (4 choices for the first card, 3 choices for the second and so on).
With 4 cards out of the deck, we have 52-4 = 48 possible cards to be the fifth one.
So, it would be impossible to guess the number and the sign of the fifth card only by the order of the 4 cards given.
My guess is: the partner makes some binary non-verbal communication (e.g.: blinking his left or his right eye). Blinking his left eye means the sign is Spades or Hearts and his right eye means Diamonds or Clubs (it isn't 100% correct since we may have up to 26 cards from 2 signs, but it is just an example).
Thus, we are going to have 24 possible ways of ordering 4 cards blinking his left eye plus 24 possible ways of ordering 4 cards bliking his right eye. Summing them, we cover all the 48 possible cards left.
They probably developed some mnemonic to memorize and process it all very fast, but it is mathematically possible.
answered 1 hour ago
Daquisu
464
No eye-bliking/cheating is required ;)
– LeoDucas
55 mins ago
1
You are missing the fact that we also get to pick which card to make the fifth card. Say the order of cards A, B, C, D communicates 24 possibilities, but card E is not any of them. Then maybe instead, we can pass cards A, B, C, E and the 24 possibilities their order encodes include card D.
– Misha Lavrov
38 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $S$ be a set of all cards.
So we have a set $A$ of all ordered $4$-couples of the set $S$ and a set $B$ of all $5$ subsets of the set $S$.
Connect 4-couple in $A$ with a subset in $B$ iff all four card from that 4-couple are in that subset. Clearly each 4-couple is connected to $8$ 5-subsets and every 5 subset is connected to $5!= 120$ 4-couples.
This relation give us a bipartite graph $G=(A,B)$. Now this graph satisfies Hall matching condition. Take any subset $X$ in $A$. Then the set of neighbours $N(X)$ satisfies: $$ 120cdot |N(X)| geq 48cdot |X|implies |N(X)|geq |X| $$
So there exist a matching which saturate all vertices in $A$.
answered 45 mins ago
greedoid
32.1k114287
1
This, of course, does not guarantee the existence of a nice strategy for the magic trick; an arbitrary matching might look like a very long list with entries like "If the four cards are $9spadesuit, 3spadesuit, Kdiamondsuit, 9heartsuit$, then the remaining card is $Aclubsuit$." So beyond this, there is some work to be done to describe some matching in a way that a human can reasonably remember.
– Misha Lavrov
41 mins ago
Yeah, the question is to find an explicit strategy, it does not ask for existence. Thanks, I'll delete my answer.
– greedoid
38 mins ago
1
No, don't! The proof in your answer is still an important observation :)
– Misha Lavrov
38 mins ago
add a comment |Â
up vote
2
down vote
Let $S$ be a set of all cards.
So we have a set $A$ of all ordered $4$-couples of the set $S$ and a set $B$ of all $5$ subsets of the set $S$.
Connect 4-couple in $A$ with a subset in $B$ iff all four card from that 4-couple are in that subset. Clearly each 4-couple is connected to $8$ 5-subsets and every 5 subset is connected to $5!= 120$ 4-couples.
This relation give us a bipartite graph $G=(A,B)$. Now this graph satisfies Hall matching condition. Take any subset $X$ in $A$. Then the set of neighbours $N(X)$ satisfies: $$ 120cdot |N(X)| geq 48cdot |X|implies |N(X)|geq |X| $$
So there exist a matching which saturate all vertices in $A$.
answered 45 mins ago
greedoid
32.1k114287
1
This, of course, does not guarantee the existence of a nice strategy for the magic trick; an arbitrary matching might look like a very long list with entries like "If the four cards are $9spadesuit, 3spadesuit, Kdiamondsuit, 9heartsuit$, then the remaining card is $Aclubsuit$." So beyond this, there is some work to be done to describe some matching in a way that a human can reasonably remember.
– Misha Lavrov
41 mins ago
Yeah, the question is to find an explicit strategy, it does not ask for existence. Thanks, I'll delete my answer.
– greedoid
38 mins ago
1
No, don't! The proof in your answer is still an important observation :)
– Misha Lavrov
38 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $S$ be a set of all cards.
So we have a set $A$ of all ordered $4$-couples of the set $S$ and a set $B$ of all $5$ subsets of the set $S$.
Connect 4-couple in $A$ with a subset in $B$ iff all four card from that 4-couple are in that subset. Clearly each 4-couple is connected to $8$ 5-subsets and every 5 subset is connected to $5!= 120$ 4-couples.
This relation give us a bipartite graph $G=(A,B)$. Now this graph satisfies Hall matching condition. Take any subset $X$ in $A$. Then the set of neighbours $N(X)$ satisfies: $$ 120cdot |N(X)| geq 48cdot |X|implies |N(X)|geq |X| $$
So there exist a matching which saturate all vertices in $A$.
answered 45 mins ago
greedoid
32.1k114287
Let $S$ be a set of all cards.
So we have a set $A$ of all ordered $4$-couples of the set $S$ and a set $B$ of all $5$ subsets of the set $S$.
Connect 4-couple in $A$ with a subset in $B$ iff all four card from that 4-couple are in that subset. Clearly each 4-couple is connected to $8$ 5-subsets and every 5 subset is connected to $5!= 120$ 4-couples.
This relation give us a bipartite graph $G=(A,B)$. Now this graph satisfies Hall matching condition. Take any subset $X$ in $A$. Then the set of neighbours $N(X)$ satisfies: $$ 120cdot |N(X)| geq 48cdot |X|implies |N(X)|geq |X| $$
So there exist a matching which saturate all vertices in $A$.
answered 45 mins ago
greedoid
32.1k114287
answered 45 mins ago
greedoid
32.1k114287
answered 45 mins ago
greedoid
32.1k114287
answered 45 mins ago
greedoid
32.1k114287
32.1k114287
1
This, of course, does not guarantee the existence of a nice strategy for the magic trick; an arbitrary matching might look like a very long list with entries like "If the four cards are $9spadesuit, 3spadesuit, Kdiamondsuit, 9heartsuit$, then the remaining card is $Aclubsuit$." So beyond this, there is some work to be done to describe some matching in a way that a human can reasonably remember.
– Misha Lavrov
41 mins ago
Yeah, the question is to find an explicit strategy, it does not ask for existence. Thanks, I'll delete my answer.
– greedoid
38 mins ago
1
No, don't! The proof in your answer is still an important observation :)
– Misha Lavrov
38 mins ago
add a comment |Â
1
This, of course, does not guarantee the existence of a nice strategy for the magic trick; an arbitrary matching might look like a very long list with entries like "If the four cards are $9spadesuit, 3spadesuit, Kdiamondsuit, 9heartsuit$, then the remaining card is $Aclubsuit$." So beyond this, there is some work to be done to describe some matching in a way that a human can reasonably remember.
– Misha Lavrov
41 mins ago
Yeah, the question is to find an explicit strategy, it does not ask for existence. Thanks, I'll delete my answer.
– greedoid
38 mins ago
1
No, don't! The proof in your answer is still an important observation :)
– Misha Lavrov
38 mins ago
1
1
This, of course, does not guarantee the existence of a nice strategy for the magic trick; an arbitrary matching might look like a very long list with entries like "If the four cards are $9spadesuit, 3spadesuit, Kdiamondsuit, 9heartsuit$, then the remaining card is $Aclubsuit$." So beyond this, there is some work to be done to describe some matching in a way that a human can reasonably remember.
– Misha Lavrov
41 mins ago
This, of course, does not guarantee the existence of a nice strategy for the magic trick; an arbitrary matching might look like a very long list with entries like "If the four cards are $9spadesuit, 3spadesuit, Kdiamondsuit, 9heartsuit$, then the remaining card is $Aclubsuit$." So beyond this, there is some work to be done to describe some matching in a way that a human can reasonably remember.
– Misha Lavrov
41 mins ago
Yeah, the question is to find an explicit strategy, it does not ask for existence. Thanks, I'll delete my answer.
– greedoid
38 mins ago
Yeah, the question is to find an explicit strategy, it does not ask for existence. Thanks, I'll delete my answer.
– greedoid
38 mins ago
1
1
No, don't! The proof in your answer is still an important observation :)
– Misha Lavrov
38 mins ago
No, don't! The proof in your answer is still an important observation :)
– Misha Lavrov
38 mins ago
add a comment |Â
up vote
2
down vote
There is in fact a solution which uses your idea: that some suit will be repeated twice.
Say that among the $13$ ranks within a suit, ordered $A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K$, each rank "beats" the next six ranks, wrapping around when we get to the end. For example, $A$ beats $2,3,4,5,6,7$, and $10$ beats $J, Q, K, A, 2, 3$.
If you have two cards of the same suit, exactly one of them beats the other one. So you should pass, in order:
- A card of the same suit as the missing card, which beats the missing card. This leaves six possibilities for the missing card.
- The remaining three cards, in an order that encodes which of the six possibilities it is.
For the second step, we should order all $52$ cards in the deck somehow; for instance, say that $clubsuit < diamondsuit < heartsuit < spadesuit$ and $A < 2 < 3 < dots < Q < K$. Then the order of the last three cards is one of "Low, Middle, High", "Low, High, Middle", and so on through "High, Middle, Low". Just remember some correspondence between these six possibilities and the values $+1, +2, +3, +4, +5, +6$, and add the value you get to the rank of the first card (wrapping around from $K$ to $A$ of the same suit).
For example, say that the correspondence we chose in the second step is
beginarrayc
textLow & textMiddle & textHigh &+1 \
textLow & textHigh & textMiddle &+2 \
textMiddle & textLow & textHigh &+3 \
textMiddle & textHigh & textLow &+4 \
textHigh & textLow & textMiddle &+5 \
textHigh & textMiddle & textLow &+6
endarray
and you draw the cards $4clubsuit, 5spadesuit, 5diamondsuit, Aclubsuit, Jspadesuit$.
- We have two possibilities for the repeated suit, so let's choose $spadesuit$.
- In the cyclic order in that suit, $5spadesuit$ beats $Jspadesuit$, so the first card we pass is $5spadesuit$.
- We want to encode the offset $+6$, which is the ordering High, Middle, Low.
- So we pass that ordering: after $5spadesuit$ we pass $5diamondsuit, 4clubsuit, Aclubsuit$ in that order, because $5diamondsuit > 4clubsuit > Aclubsuit$.
answered 16 mins ago
Misha Lavrov
40.3k55399
Nice! I was working on the same solution (3 cards give a number between 1 and 6) but couldn't figure a way to convey whether to add or subtract that number. Your "beats" strategy solved that.
– Jens
6 mins ago
add a comment |Â
up vote
2
down vote
There is in fact a solution which uses your idea: that some suit will be repeated twice.
Say that among the $13$ ranks within a suit, ordered $A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K$, each rank "beats" the next six ranks, wrapping around when we get to the end. For example, $A$ beats $2,3,4,5,6,7$, and $10$ beats $J, Q, K, A, 2, 3$.
If you have two cards of the same suit, exactly one of them beats the other one. So you should pass, in order:
- A card of the same suit as the missing card, which beats the missing card. This leaves six possibilities for the missing card.
- The remaining three cards, in an order that encodes which of the six possibilities it is.
For the second step, we should order all $52$ cards in the deck somehow; for instance, say that $clubsuit < diamondsuit < heartsuit < spadesuit$ and $A < 2 < 3 < dots < Q < K$. Then the order of the last three cards is one of "Low, Middle, High", "Low, High, Middle", and so on through "High, Middle, Low". Just remember some correspondence between these six possibilities and the values $+1, +2, +3, +4, +5, +6$, and add the value you get to the rank of the first card (wrapping around from $K$ to $A$ of the same suit).
For example, say that the correspondence we chose in the second step is
beginarrayc
textLow & textMiddle & textHigh &+1 \
textLow & textHigh & textMiddle &+2 \
textMiddle & textLow & textHigh &+3 \
textMiddle & textHigh & textLow &+4 \
textHigh & textLow & textMiddle &+5 \
textHigh & textMiddle & textLow &+6
endarray
and you draw the cards $4clubsuit, 5spadesuit, 5diamondsuit, Aclubsuit, Jspadesuit$.
- We have two possibilities for the repeated suit, so let's choose $spadesuit$.
- In the cyclic order in that suit, $5spadesuit$ beats $Jspadesuit$, so the first card we pass is $5spadesuit$.
- We want to encode the offset $+6$, which is the ordering High, Middle, Low.
- So we pass that ordering: after $5spadesuit$ we pass $5diamondsuit, 4clubsuit, Aclubsuit$ in that order, because $5diamondsuit > 4clubsuit > Aclubsuit$.
answered 16 mins ago
Misha Lavrov
40.3k55399
Nice! I was working on the same solution (3 cards give a number between 1 and 6) but couldn't figure a way to convey whether to add or subtract that number. Your "beats" strategy solved that.
– Jens
6 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There is in fact a solution which uses your idea: that some suit will be repeated twice.
Say that among the $13$ ranks within a suit, ordered $A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K$, each rank "beats" the next six ranks, wrapping around when we get to the end. For example, $A$ beats $2,3,4,5,6,7$, and $10$ beats $J, Q, K, A, 2, 3$.
If you have two cards of the same suit, exactly one of them beats the other one. So you should pass, in order:
- A card of the same suit as the missing card, which beats the missing card. This leaves six possibilities for the missing card.
- The remaining three cards, in an order that encodes which of the six possibilities it is.
For the second step, we should order all $52$ cards in the deck somehow; for instance, say that $clubsuit < diamondsuit < heartsuit < spadesuit$ and $A < 2 < 3 < dots < Q < K$. Then the order of the last three cards is one of "Low, Middle, High", "Low, High, Middle", and so on through "High, Middle, Low". Just remember some correspondence between these six possibilities and the values $+1, +2, +3, +4, +5, +6$, and add the value you get to the rank of the first card (wrapping around from $K$ to $A$ of the same suit).
For example, say that the correspondence we chose in the second step is
beginarrayc
textLow & textMiddle & textHigh &+1 \
textLow & textHigh & textMiddle &+2 \
textMiddle & textLow & textHigh &+3 \
textMiddle & textHigh & textLow &+4 \
textHigh & textLow & textMiddle &+5 \
textHigh & textMiddle & textLow &+6
endarray
and you draw the cards $4clubsuit, 5spadesuit, 5diamondsuit, Aclubsuit, Jspadesuit$.
- We have two possibilities for the repeated suit, so let's choose $spadesuit$.
- In the cyclic order in that suit, $5spadesuit$ beats $Jspadesuit$, so the first card we pass is $5spadesuit$.
- We want to encode the offset $+6$, which is the ordering High, Middle, Low.
- So we pass that ordering: after $5spadesuit$ we pass $5diamondsuit, 4clubsuit, Aclubsuit$ in that order, because $5diamondsuit > 4clubsuit > Aclubsuit$.
answered 16 mins ago
Misha Lavrov
40.3k55399
There is in fact a solution which uses your idea: that some suit will be repeated twice.
Say that among the $13$ ranks within a suit, ordered $A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K$, each rank "beats" the next six ranks, wrapping around when we get to the end. For example, $A$ beats $2,3,4,5,6,7$, and $10$ beats $J, Q, K, A, 2, 3$.
If you have two cards of the same suit, exactly one of them beats the other one. So you should pass, in order:
- A card of the same suit as the missing card, which beats the missing card. This leaves six possibilities for the missing card.
- The remaining three cards, in an order that encodes which of the six possibilities it is.
For the second step, we should order all $52$ cards in the deck somehow; for instance, say that $clubsuit < diamondsuit < heartsuit < spadesuit$ and $A < 2 < 3 < dots < Q < K$. Then the order of the last three cards is one of "Low, Middle, High", "Low, High, Middle", and so on through "High, Middle, Low". Just remember some correspondence between these six possibilities and the values $+1, +2, +3, +4, +5, +6$, and add the value you get to the rank of the first card (wrapping around from $K$ to $A$ of the same suit).
For example, say that the correspondence we chose in the second step is
beginarrayc
textLow & textMiddle & textHigh &+1 \
textLow & textHigh & textMiddle &+2 \
textMiddle & textLow & textHigh &+3 \
textMiddle & textHigh & textLow &+4 \
textHigh & textLow & textMiddle &+5 \
textHigh & textMiddle & textLow &+6
endarray
and you draw the cards $4clubsuit, 5spadesuit, 5diamondsuit, Aclubsuit, Jspadesuit$.
- We have two possibilities for the repeated suit, so let's choose $spadesuit$.
- In the cyclic order in that suit, $5spadesuit$ beats $Jspadesuit$, so the first card we pass is $5spadesuit$.
- We want to encode the offset $+6$, which is the ordering High, Middle, Low.
- So we pass that ordering: after $5spadesuit$ we pass $5diamondsuit, 4clubsuit, Aclubsuit$ in that order, because $5diamondsuit > 4clubsuit > Aclubsuit$.
answered 16 mins ago
Misha Lavrov
40.3k55399
edited 7 mins ago
answered 16 mins ago
Misha Lavrov
40.3k55399
answered 16 mins ago
Misha Lavrov
40.3k55399
answered 16 mins ago
Misha Lavrov
40.3k55399
40.3k55399
Nice! I was working on the same solution (3 cards give a number between 1 and 6) but couldn't figure a way to convey whether to add or subtract that number. Your "beats" strategy solved that.
– Jens
6 mins ago
add a comment |Â
Nice! I was working on the same solution (3 cards give a number between 1 and 6) but couldn't figure a way to convey whether to add or subtract that number. Your "beats" strategy solved that.
– Jens
6 mins ago
Nice! I was working on the same solution (3 cards give a number between 1 and 6) but couldn't figure a way to convey whether to add or subtract that number. Your "beats" strategy solved that.
– Jens
6 mins ago
Nice! I was working on the same solution (3 cards give a number between 1 and 6) but couldn't figure a way to convey whether to add or subtract that number. Your "beats" strategy solved that.
– Jens
6 mins ago
add a comment |Â
up vote
1
down vote
It's a cute one. Let's just assume the cards are numbered $1 dots 52$. We draw cards $a_1 < a_2 < a_3 < a_4 < a_5$, and hand out in this order $b_1, b_2, b_3, b_4$. The way we order this set $B=b_1, b_2, b_3, b_4$ already allows to transmit quite some information. Indeed, choosing the orders paramounts to choosing a permutation on the 4 elements of $B$. There are $4! = 4 cdot 3 cdot 2 cdot 1 = 24$ such permutations.
This is not enough still to transmit the $5$th which is in a set of size $52$. Well, we also know that the 5th card is different from those $4$ firsts ones: there is only $52-4 = 48$ possibilities.
I'll leave it to you to complete the strategy. Note that $48/24 = 2$, so we just need to find a trick to only half the search space. Note also that we have still not fixed how we choose the secret card among the set $A$. At last, note at least one of the two inequalities:
- $a_2 leq 25$
- $a_4 geq 58-25$.
answered 56 mins ago
LeoDucas
1846
add a comment |Â
up vote
1
down vote
It's a cute one. Let's just assume the cards are numbered $1 dots 52$. We draw cards $a_1 < a_2 < a_3 < a_4 < a_5$, and hand out in this order $b_1, b_2, b_3, b_4$. The way we order this set $B=b_1, b_2, b_3, b_4$ already allows to transmit quite some information. Indeed, choosing the orders paramounts to choosing a permutation on the 4 elements of $B$. There are $4! = 4 cdot 3 cdot 2 cdot 1 = 24$ such permutations.
This is not enough still to transmit the $5$th which is in a set of size $52$. Well, we also know that the 5th card is different from those $4$ firsts ones: there is only $52-4 = 48$ possibilities.
I'll leave it to you to complete the strategy. Note that $48/24 = 2$, so we just need to find a trick to only half the search space. Note also that we have still not fixed how we choose the secret card among the set $A$. At last, note at least one of the two inequalities:
- $a_2 leq 25$
- $a_4 geq 58-25$.
answered 56 mins ago
LeoDucas
1846
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's a cute one. Let's just assume the cards are numbered $1 dots 52$. We draw cards $a_1 < a_2 < a_3 < a_4 < a_5$, and hand out in this order $b_1, b_2, b_3, b_4$. The way we order this set $B=b_1, b_2, b_3, b_4$ already allows to transmit quite some information. Indeed, choosing the orders paramounts to choosing a permutation on the 4 elements of $B$. There are $4! = 4 cdot 3 cdot 2 cdot 1 = 24$ such permutations.
This is not enough still to transmit the $5$th which is in a set of size $52$. Well, we also know that the 5th card is different from those $4$ firsts ones: there is only $52-4 = 48$ possibilities.
I'll leave it to you to complete the strategy. Note that $48/24 = 2$, so we just need to find a trick to only half the search space. Note also that we have still not fixed how we choose the secret card among the set $A$. At last, note at least one of the two inequalities:
- $a_2 leq 25$
- $a_4 geq 58-25$.
answered 56 mins ago
LeoDucas
1846
It's a cute one. Let's just assume the cards are numbered $1 dots 52$. We draw cards $a_1 < a_2 < a_3 < a_4 < a_5$, and hand out in this order $b_1, b_2, b_3, b_4$. The way we order this set $B=b_1, b_2, b_3, b_4$ already allows to transmit quite some information. Indeed, choosing the orders paramounts to choosing a permutation on the 4 elements of $B$. There are $4! = 4 cdot 3 cdot 2 cdot 1 = 24$ such permutations.
This is not enough still to transmit the $5$th which is in a set of size $52$. Well, we also know that the 5th card is different from those $4$ firsts ones: there is only $52-4 = 48$ possibilities.
I'll leave it to you to complete the strategy. Note that $48/24 = 2$, so we just need to find a trick to only half the search space. Note also that we have still not fixed how we choose the secret card among the set $A$. At last, note at least one of the two inequalities:
- $a_2 leq 25$
- $a_4 geq 58-25$.
answered 56 mins ago
LeoDucas
1846
edited 43 mins ago
answered 56 mins ago
LeoDucas
1846
answered 56 mins ago
LeoDucas
1846
answered 56 mins ago
LeoDucas
1846
1846
add a comment |Â
add a comment |Â
up vote
0
down vote
For any 4 cards given, we have 24 possible ways of ordering then (4 choices for the first card, 3 choices for the second and so on).
With 4 cards out of the deck, we have 52-4 = 48 possible cards to be the fifth one.
So, it would be impossible to guess the number and the sign of the fifth card only by the order of the 4 cards given.
My guess is: the partner makes some binary non-verbal communication (e.g.: blinking his left or his right eye). Blinking his left eye means the sign is Spades or Hearts and his right eye means Diamonds or Clubs (it isn't 100% correct since we may have up to 26 cards from 2 signs, but it is just an example).
Thus, we are going to have 24 possible ways of ordering 4 cards blinking his left eye plus 24 possible ways of ordering 4 cards bliking his right eye. Summing them, we cover all the 48 possible cards left.
They probably developed some mnemonic to memorize and process it all very fast, but it is mathematically possible.
answered 1 hour ago
Daquisu
464
No eye-bliking/cheating is required ;)
– LeoDucas
55 mins ago
1
You are missing the fact that we also get to pick which card to make the fifth card. Say the order of cards A, B, C, D communicates 24 possibilities, but card E is not any of them. Then maybe instead, we can pass cards A, B, C, E and the 24 possibilities their order encodes include card D.
– Misha Lavrov
38 mins ago
add a comment |Â
up vote
0
down vote
For any 4 cards given, we have 24 possible ways of ordering then (4 choices for the first card, 3 choices for the second and so on).
With 4 cards out of the deck, we have 52-4 = 48 possible cards to be the fifth one.
So, it would be impossible to guess the number and the sign of the fifth card only by the order of the 4 cards given.
My guess is: the partner makes some binary non-verbal communication (e.g.: blinking his left or his right eye). Blinking his left eye means the sign is Spades or Hearts and his right eye means Diamonds or Clubs (it isn't 100% correct since we may have up to 26 cards from 2 signs, but it is just an example).
Thus, we are going to have 24 possible ways of ordering 4 cards blinking his left eye plus 24 possible ways of ordering 4 cards bliking his right eye. Summing them, we cover all the 48 possible cards left.
They probably developed some mnemonic to memorize and process it all very fast, but it is mathematically possible.
answered 1 hour ago
Daquisu
464
No eye-bliking/cheating is required ;)
– LeoDucas
55 mins ago
1
You are missing the fact that we also get to pick which card to make the fifth card. Say the order of cards A, B, C, D communicates 24 possibilities, but card E is not any of them. Then maybe instead, we can pass cards A, B, C, E and the 24 possibilities their order encodes include card D.
– Misha Lavrov
38 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For any 4 cards given, we have 24 possible ways of ordering then (4 choices for the first card, 3 choices for the second and so on).
With 4 cards out of the deck, we have 52-4 = 48 possible cards to be the fifth one.
So, it would be impossible to guess the number and the sign of the fifth card only by the order of the 4 cards given.
My guess is: the partner makes some binary non-verbal communication (e.g.: blinking his left or his right eye). Blinking his left eye means the sign is Spades or Hearts and his right eye means Diamonds or Clubs (it isn't 100% correct since we may have up to 26 cards from 2 signs, but it is just an example).
Thus, we are going to have 24 possible ways of ordering 4 cards blinking his left eye plus 24 possible ways of ordering 4 cards bliking his right eye. Summing them, we cover all the 48 possible cards left.
They probably developed some mnemonic to memorize and process it all very fast, but it is mathematically possible.
answered 1 hour ago
Daquisu
464
For any 4 cards given, we have 24 possible ways of ordering then (4 choices for the first card, 3 choices for the second and so on).
With 4 cards out of the deck, we have 52-4 = 48 possible cards to be the fifth one.
So, it would be impossible to guess the number and the sign of the fifth card only by the order of the 4 cards given.
My guess is: the partner makes some binary non-verbal communication (e.g.: blinking his left or his right eye). Blinking his left eye means the sign is Spades or Hearts and his right eye means Diamonds or Clubs (it isn't 100% correct since we may have up to 26 cards from 2 signs, but it is just an example).
Thus, we are going to have 24 possible ways of ordering 4 cards blinking his left eye plus 24 possible ways of ordering 4 cards bliking his right eye. Summing them, we cover all the 48 possible cards left.
They probably developed some mnemonic to memorize and process it all very fast, but it is mathematically possible.
answered 1 hour ago
Daquisu
464
answered 1 hour ago
Daquisu
464
answered 1 hour ago
Daquisu
464
answered 1 hour ago
Daquisu
464
464
No eye-bliking/cheating is required ;)
– LeoDucas
55 mins ago
1
You are missing the fact that we also get to pick which card to make the fifth card. Say the order of cards A, B, C, D communicates 24 possibilities, but card E is not any of them. Then maybe instead, we can pass cards A, B, C, E and the 24 possibilities their order encodes include card D.
– Misha Lavrov
38 mins ago
add a comment |Â
No eye-bliking/cheating is required ;)
– LeoDucas
55 mins ago
1
You are missing the fact that we also get to pick which card to make the fifth card. Say the order of cards A, B, C, D communicates 24 possibilities, but card E is not any of them. Then maybe instead, we can pass cards A, B, C, E and the 24 possibilities their order encodes include card D.
– Misha Lavrov
38 mins ago
No eye-bliking/cheating is required ;)
– LeoDucas
55 mins ago
No eye-bliking/cheating is required ;)
– LeoDucas
55 mins ago
1
1
You are missing the fact that we also get to pick which card to make the fifth card. Say the order of cards A, B, C, D communicates 24 possibilities, but card E is not any of them. Then maybe instead, we can pass cards A, B, C, E and the 24 possibilities their order encodes include card D.
– Misha Lavrov
38 mins ago
You are missing the fact that we also get to pick which card to make the fifth card. Say the order of cards A, B, C, D communicates 24 possibilities, but card E is not any of them. Then maybe instead, we can pass cards A, B, C, E and the 24 possibilities their order encodes include card D.
– Misha Lavrov
38 mins ago
add a comment |Â
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