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Zebra,Beaver,Duck,Chipmunk
B,U,7,2
B,U,7,2
B,U,7,2
U
What answers did everyone put?
What answers did everyone put?
Okay so what answer did everyone put to see if we all put the same answer?
Okay so what answer did everyone put to see if we all put the same answer?
I put 7 and U
I put 7 and U
I put U and 2
I put U and 2
I selected U and 2 as they are vowels and even number
I selected U and 2 as they are vowels and even number
U,2
U,2
I think it would be 7 to test if it would be incorrect
I think it would be 7 to test if it would be incorrect
Duck - give your answer
Duck - give your answer
I put U and 2 (forgot to put a number forint so was just deciding which to put)
I put U and 2 (forgot to put a number forint so was just deciding which to put)
So shall we say U and 2??
So shall we say U and 2??
but it says it might be as simple as we think and it seems pretty simple to put U and 2 as that is the vowel and the even number
but it says it might be as simple as we think and it seems pretty simple to put U and 2 as that is the vowel and the even number
*might not
*might not
yes i agreee
yes i agreee
So is it 7 ?
So is it 7 ?
I think if it was U and 2 then you wouldn''t be able to see if the theory was incorrect so i reckon its U and 7
I think if it was U and 2 then you wouldn''t be able to see if the theory was incorrect so i reckon its U and 7
i vote U and 7
i vote U and 7
yeah so you need one of each to see if its right
yeah so you need one of each to see if its right
is it 2 cards or one that should be flipped
is it 2 cards or one that should be flipped
2
2
Ok U & 7
Ok U & 7
so we are agreeing to the fact that we need a letter and a number card?
so we are agreeing to the fact that we need a letter and a number card?
yes
yes
yeah think so bc we need to test the vowel one and the even number one so have too do both to see if its right
yeah think so bc we need to test the vowel one and the even number one so have too do both to see if its right
if we picked both letter cards we could find the rule out though
if we picked both letter cards we could find the rule out though
one vowel, one consonant
one vowel, one consonant
im starting to think you need all the cards to test the rule
im starting to think you need all the cards to test the rule
not just 2
not just 2
I don''t think it matters
I don''t think it matters
I reckon it works with just the U and 7
I reckon it works with just the U and 7
if we turn the U over and there is an even number we test its true
if we turn the U over and there is an even number we test its true
because the logic we''ve applied to selecting 2 of the cards works just as well for any other combination of cards
because the logic we''ve applied to selecting 2 of the cards works just as well for any other combination of cards
there is no where that says we can''t click them all I guess
there is no where that says we can''t click them all I guess
if we turn the 7 over and their isnt a vowel we test its true
if we turn the 7 over and their isnt a vowel we test its true
okay guys vote which one you want to do
okay guys vote which one you want to do
I think 7 and U
I think 7 and U
im gonna vote all
im gonna vote all
All
All
i think U and 7 too
i think U and 7 too
Shall we go for all ?
Shall we go for all ?
no i dont think we need to turn over all the cards
no i dont think we need to turn over all the cards
shall we just put them all bc then that way we are still putting 7 and U but are putting the other ideas in too?
shall we just put them all bc then that way we are still putting 7 and U but are putting the other ideas in too?
i think it works with just turning over two cards
i think it works with just turning over two cards
but which 2 cards
but which 2 cards
7 and U I think
7 and U I think
U & 7
U & 7
coz you can turn over B and 7 and the rule can be proved right or wrong
coz you can turn over B and 7 and the rule can be proved right or wrong
B,U,7,2
Unicorn,Bee,Duck,Chipmunk,Alpaca
3,U,2,V
3,U,2,V
everyone?
there''s one vowel and one odd number, those are the two to flip to confirm
I BELIEVE I HAVE THE CORRECT SOLOUTION.
I BELIEVE I HAVE THE CORRECT SOLOUTION.
I thought it was best to check all the cards since some showed vowels and others consonants and numbers. What did everyone else do?
I thought it was best to check all the cards since some showed vowels and others consonants and numbers. What did everyone else do?
I think it''s U too.
I think it''s U too.
u AND 2 FOLLOW THE RULE
u AND 2 FOLLOW THE RULE
U and 3
U and 3
If one side of a vowel is even, then the other side should be vowel
Yes, it is U and 3
Well U and 2 should follow the rule, but we don''t know unless or until the other side is checked.
Well U and 2 should follow the rule, but we don''t know unless or until the other side is checked.
3 IS NOT AN EVEN NUMBER, IT IS ODD
3 IS NOT AN EVEN NUMBER, IT IS ODD
2 doesn''t follow the rule, because the rule DOES NOT state that a constant can not be an even number
2 doesn''t follow the rule, because the rule DOES NOT state that a constant can not be an even number
I''m aware 3 is odd, but if vowels have even numbers on the other side, you''d have to check 2 too.
I''m aware 3 is odd, but if vowels have even numbers on the other side, you''d have to check 2 too.
in theory there could be a consenent behind 2, but to check the statement, 3 needs to flip to confirm
in theory there could be a consenent behind 2, but to check the statement, 3 needs to flip to confirm
@ Unicorn, EXACTLY. Which is why if you turn over the 3 card and you get a vowel, it violates the rule, which is why you have to turn it over
@ Unicorn, EXACTLY. Which is why if you turn over the 3 card and you get a vowel, it violates the rule, which is why you have to turn it over
it''s definitely U and 3
it''s definitely U and 3
everyone agree?
everyone agree?
2 would be included too, since the rule is all cards with vowels on one side have an even number on the other, so 2 is even, so it should have a vowel on it too
2 would be included too, since the rule is all cards with vowels on one side have an even number on the other, so 2 is even, so it should have a vowel on it too
so maybe U 2 and 3?
so maybe U 2 and 3?
I still think it''s U
I still think it''s U
All cards with vowels on one side have an even number on the other.
All cards with vowels on one side have an even number on the other.
I agree with chipmunk U and 3
I agree with chipmunk U and 3
I still say 2 too. in addition to u and 3.
I still say 2 too. in addition to u and 3.
2 WOULD HAVE TO HAVE A VOWEL IT IS AN EVEN NUMBER
2 WOULD HAVE TO HAVE A VOWEL IT IS AN EVEN NUMBER
you don''t need 2 because it only mentions vowels, it makes hard rule against consenents
you don''t need 2 because it only mentions vowels, it makes hard rule against consenents
@ Unicorn and Bee
@ Unicorn and Bee
Maybe its U and 2 then
Maybe its U and 2 then
it should have a vowel, but we don''t know unless we check
it should have a vowel, but we don''t know unless we check
Read the rules, constants can be an even card
Read the rules, constants can be an even card
Exactly, Chipmunk gets it
Exactly, Chipmunk gets it
but since 2 is even, it should have a vowel on the other side
but since 2 is even, it should have a vowel on the other side
so we should check it
so we should check it
@ Bee, NO
@ Bee, NO
AGREE BEE
AGREE BEE
we doen''t need to know what 2 has
we doen''t need to know what 2 has
This is the only rule: All cards with vowels on one side have an even number on the other.
This is the only rule: All cards with vowels on one side have an even number on the other.
2 could be either a vowel or consenant
2 could be either a vowel or consenant
U
U
the rule is even number on one side, vowel on the other. so to check the rule, check the 2
the rule is even number on one side, vowel on the other. so to check the rule, check the 2
No, a constant can have an even number too!
No, a constant can have an even number too!
you''re wrong if you check the 2
you''re wrong if you check the 2
it''s only U and 3
it''s only U and 3
they prove the rule out of the 4 cards shown
they prove the rule out of the 4 cards shown
so would 2
so would 2
since it is an even number
since it is an even number
The rule DOES NOT say that a constant has to be even or odd!
The rule DOES NOT say that a constant has to be even or odd!
BEE, consenants can have eve or odd numbers,
BEE, consenants can have eve or odd numbers,
if there are two sides to a card, and if one side has a vowel, there''s an even number on the side, then 2 should be checked. it''s an even number.
if there are two sides to a card, and if one side has a vowel, there''s an even number on the side, then 2 should be checked. it''s an even number.
WE ARE MERELY CHECKING THE RULE OF WHAT IS GIVEN PROVES THE RULE. 3 WILL NOT PROVE THE RULE BECAUSE IT IS NOT EVEN..
WE ARE MERELY CHECKING THE RULE OF WHAT IS GIVEN PROVES THE RULE. 3 WILL NOT PROVE THE RULE BECAUSE IT IS NOT EVEN..
It would make a stronger test to prove the rule to include the 3 (u, 2,3) than just u and 3
It would make a stronger test to prove the rule to include the 3 (u, 2,3) than just u and 3
bee, it''s a trick, you only go by the rule as stated, think critically
bee, it''s a trick, you only go by the rule as stated, think critically
So what do you say to check Unicorn?