B _ T H
W _ L L
H _ V _
T H _
S _ M _
_ M _ _ N T .
L _ T ' S
S _ Y
T H _ R _ ' S
1 0 0 M L
_ F
T _ _
_ N D
1 0 0 M L
_ F
C _ F F _ _
_ N _ T _ _ L L Y
_ N D
T H _
T _ _ S P _ _ N
H _ L D S
1 0 M L
_ F
L _ Q _ _ D .
A F T _ R
T R _ N S F _ R R _ N G
1 0 M L
_ F
L _ Q _ _ D
F R _ M
T H _
T _ _
T _
T H _
C _ F F _ _
C _ P
_ N D
_ N _ T H _ R
1 0 M L
F R _ M
T H _
C _ F F _ _
T _
T H _
T _ _
C _ P ,
T H _
Q _ _ N T _ T Y
W _ L L
R _ M _ _ N
T H _
S _ M _
-
1 0 0 M L
_ F
C _ F F _ _
_ N D
T _ _
M _ X
_ N
T H _
T _ _
C _ P
_ N D
1 0 0 M L
_ F
C _ F F _ _
_ N D
T _ _
M _ X
_ N
T H _
C _ F F _ _
C _ P .
S _
L _ T ' S
S _ Y
_ F T _ R
T H _
M _ X _ N G
_ S
D _ N _
T H _ R _
_ S
9 9 M L
_ F
T _ _
_ N D
1 M L
_ F
C _ F F _ _
_ N
T H _
T _ _
C _ P .
T H _
1 M L
_ F
T _ _
H _ S
T _
B _
_ N
T H _
C _ F F _ _
C _ P .
T H _ T
M _ _ N S
T H _ T
T H _
C _ F F _ _
C _ P
H _ S
9 9 M L
_ F
C _ F F _ _
_ N D
1 M L
_ F
T _ _ .
I N
_ T H _ R
W _ R D S ,
_ F
T H _ R _ ' S
( X ) M L
_ F
T _ _
M _ S S _ N G
F R _ M
T H _
T _ _
C _ P ,
T H _ N
T H _ R _
H _ S
T _
B _
_ N
_ Q _ _ L
_ M _ _ N T
_ F
_ F
C _ F F _ _
M _ S S _ N G
F R _ M
T H _
C _ F F _ _
C _ P Clue
BOTH WILL HAVE THE SAME AMOUNT. LET'S SAY THERE'S 100ML OF TEA AND 100ML OF COFFEE INITIALLY AND THE TEASPOON HOLDS 10ML OF LIQUID. AFTER TRANSFERRING 10ML OF LIQUID FROM THE TEA TO THE COFFEE CUP AND ANOTHER 10ML FROM THE COFFEE TO THE TEA CUP, THE QUANTITY WILL REMAIN THE SAME - 100ML OF COFFEE AND TEA MIX IN THE TEA CUP AND 100ML OF COFFEE AND TEA MIX IN THE COFFEE CUP. SO LET'S SAY AFTER THE MIXING IS DONE THERE IS 99ML OF TEA AND 1ML OF COFFEE IN THE TEA CUP. THE 1ML OF TEA HAS TO BE IN THE COFFEE CUP. THAT MEANS THAT THE COFFEE CUP HAS 99ML OF COFFEE AND 1ML OF TEA. IN OTHER WORDS, IF THERE'S (X)ML OF TEA MISSING FROM THE TEA CUP, THEN THERE HAS TO BE AN EQUAL AMOUNT OF OF COFFEE MISSING FROM THE COFFEE CUP FROM (1) AND (2), THERE ARE 6 POSSIBILITIES. POSSIBILITYABIGAILBRIDGETCLAUDIA P1COFFEECOFFEECOFFEE P2COFFEETEATEA P3TEACOFFEECOFFEE P4TEATEATEA P5TEACOFFEETEA P6TEATEACOFFEE THEN, FROM (3), P1 AND P5 ARE ELIMINATED. AND FROM (4), P2 AND P5 ARE ELIMINATED. SO BASED ON THE REMAINING POSSIBILITIES P3, P4 AND P6, WE KNOW THAT ABIGAIL ALWAYS ORDERS THE SAME DRINK (TEA) AFTER DINNER THE PROBLEM IS THAT THE QUESTION IS CLEVERLY PHRASED TO CONCEAL WHAT IS REALLY GOING ON. LET'S LOCATE ALL THAT MONEY. THERE ARE TWO WAYS TO THINK ABOUT HOW MUCH MONEY IS OUT THERE TO BE FOUND.THE WAY THAT THIS QUESTION IS TRICKY IS THAT IT COMBINES THAT TWO WAYS. THE FIRST WAY IS THIS. HOW MUCH MONEY DID THE THREE MEN PAY ORIGINALLY? THE SECOND WAY IS THIS. HOW MUCH MONEY DID THEY END UP PAYING? SO IF IT IS THE FIRST WAY, THEN CLEARLY THE TOTAL WE NEED TO ACCOUNT FOR IS THE $15. SO LET'S SEE WHAT HAPPENS TO THAT $15. THE CHEF GETS $10, THE WAITER GETS $2 AND THE GUYS GET $3 BACK. THAT ADDS UP FINE. NOW LET'S LOOK AT THE SECOND WAY. HOW MUCH MONEY DID THEY END UP PAYING? WELL $12, AND $10 OF IT WENT TO THE CHEF AND $2 TO THE WAITER. NOW, THAT ADDS UP TOO. THE PROBLEM WITH THE QUESTION IS THAT THE THE $2 THAT THE WAITER TOOK IS CONTAINED IN THE $12 THAT THEY END UP PAYING SO WE SHOULDN'T EXPECT THEM TO ADD TO ANYTHING MEANINGFUL THE SHEEP WILL SURVIVE. IF THERE WERE 1 LION AND 1 SHEEP, THEN THE LION WOULD SIMPLY EAT THE SHEEP. THE SHEEP WILL NOT SURVIVE. IF THERE WERE 2 LIONS AND 1 SHEEP, THEN NO LION WOULD EAT THE SHEEP, BECAUSE IF ONE OF THEM WOULD, IT WOULD SURELY BE EATEN BY THE OTHER LION AFTERWARDS. THE SHEEP WILL SURVIVE. IF THERE WERE 3 LIONS AND 1 SHEEP, THEN ONE OF THE LIONS COULD SAFELY EAT THE SHEEP, BECAUSE IT WOULD TURN INTO THE SCENARIO WITH 2 LIONS, WHERE NO ONE CAN EAT THE SHEEP. THE SHEEP WILL NOT SURVIVE. IF THERE WERE 4 LIONS AND 1 SHEEP, THEN NO LION WOULD EAT THE SHEEP, BECAUSE IT WOULD TURN INTO THE SCENARIO WITH 3 LIONS. THE SHEEP WILL SURVIVE. CONTINUING THIS ARGUMENT, THE CONCLUSION IS AS FOLLOWS: IF THERE IS AN EVEN NUMBER OF LIONS, THEN NOTHING HAPPENS AND THE SHEEP SURVIVES. IF THERE IS AN ODD NUMBER OF LIONS, THEN ANY LION COULD SAFELY EAT THE SHEEP AND THE SHEEP WILL NOT SURVIVE. THIS IS SIMILAR TO THE UNEXPECTED HANGING PARADOX