Poison Tea Riddle
Once upon a time, in a dark, stormy night, there were a young merchant and his wife traveling into the mysterious woods by their carriage. Seemingly lost, they sought out a shelter and finally found a small shack nearby. After the soaking lovers knocked on the front door, an old lady politely invited them in and let them warm themselves before the fireplace. Then she served 2 cups of tea and handed 2 pots of sugar cubes to her guests, the husband and wife thanked for her hospitality before both adding 1 sugar cube into the tea independently and drinking it.
Old lady: Now there's 43% chance that one of you shall die and one shall live.
Soon afterwards, the wife dropped to her knees, gasping for air, while the bewildered husband looked back at the old woman, who now revealed herself to be an evil witch and told him that she had dipped some sugar cubes with deadly poison. The poor man begged for his loved one's life, so the witch challenged him.
Witch: I will give you an antidote if and only if you can tell me the number of poison sugar cubes before you had your tea.
The merchant counted all sugar cubes and saw that there remained 29 cubes in his pot and 19 in another, but all those looked identical and so impossible to tell which were poisoned. Anxiously, he asked for more clue, and the witch slyly agreed to give it if he would give her something she would need in the future. In desperation, he made a promise.
Witch: It would be more likely for both of you to die than both of you to live when you drank that tea
Originally, how many sugar cubes were poisoned in total?
The answer is 35.
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1 solution
Since the pots ended up with 1 9 and 2 9 cubes in them, they must have contained 2 0 and 3 0 cubes to begin with.
Suppose there were originally m and n poisoned cubes in the pots with 2 0 and 3 0 cubes, respectively.
By the witch's first statement, exactly one of the two people took a poisoned sugar cube.
− P ( poisoned from 2 0 -pot, non-poisoned from 3 0 -pot ) : − P ( non-poisoned from 2 0 -pot, poisoned from 3 0 -pot ) : 2 0 m ⋅ 3 0 3 0 − n = 6 0 0 3 0 m − m n 2 0 2 0 − m ⋅ 3 0 n = 6 0 0 2 0 n − m n
Then applying the probability the witch mentions:
P ( exactly one person took a poisoned sugar cube ) 6 0 0 3 0 m + 2 0 n − 2 m n 1 5 m + 1 0 n − m n m n − 1 5 m − 1 0 n m n − 1 5 m − 1 0 n + 1 5 0 ( m − 1 0 ) ( n − 1 5 ) = 1 0 0 4 3 = 1 0 0 4 3 = 1 2 9 = - 1 2 9 [ Now we notice the left side is similar to the expansion of ( m − 1 0 ) ( n − 1 5 ) ] = 2 1 = 2 1
Thus ( m − 1 0 ) and ( n − 1 5 ) are (not necessarily positive) factors of 2 1 .
The factor pairs of 2 1 are ( - 2 1 , - 1 ) , ( - 7 , - 3 ) , ( - 3 , - 7 ) , ( - 1 , - 2 1 ) , ( 1 , 2 1 ) , ( 3 , 7 ) , ( 7 , 3 ) and ( 2 1 , 1 ) .
The respective solutions for ( m , n ) are ( - 1 1 , 1 4 ) , ( 3 , 1 2 ) , ( 7 , 8 ) , ( 9 , - 6 ) , ( 1 1 , 3 6 ) , ( 1 3 , 2 2 ) , ( 1 7 , 1 8 ) and ( 3 1 , 1 6 ) .
We can reject any solutions with negative m or n , as well as any with m > 2 0 or n > 3 0 .
This leaves the possibilities ( 3 , 1 2 ) , ( 7 , 8 ) , ( 1 3 , 2 2 ) and ( 1 7 , 1 8 ) .
The first two would give a higher probability of both people living than both people dying (see note below), so due to the witch's final statement, they are also rejected.
In either of the remaining two cases, we get that the total number of poisoned sugar cubes is m + n = 3 5
Note: It's not difficult to individually calculate this probability for the last two rejected possibilities, or we could use the following algebra:
P ( both people die ) 2 0 m ⋅ 3 0 n m n 3 m + 2 n > P ( both people die ) > 2 0 2 0 − m ⋅ 3 0 3 0 − n > 6 0 0 − 3 0 m − 2 0 n + m n > 6 0
which disqualifies those two possibilities.