My first is in alphabet, my third is in Guinness, my sixth is in donut and my eighth is in elephant. My fifth doesn’t appear in walrus, my second isn’t in stealth, but my seventh is in tremendous and my fourth is in horse. I can create peace.
I am pronounced as one letter but written as three, There are two of me, I am single, I am double, I can be blue, brown, black or green. I can be read from right to left or left to right and am still the same.
Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have a younger child as well,” she replied.
What is the probability that both of her children are girls?
1/2 probability. This has been know to cause raging debates and is known as one of the variations of the Boy or Girl paradox. This variation is more straightforward because knowing the position of the child leaves only two possibilities – the other child is a boy or a girl, each of which have a 1/2 probability.
A princess is as old as the prince will be when the princess is twice the age that the prince was when the princess’s age was half the sum of their present ages.
This one took a while to figure out and there are numerous valid ways of finding the answer.
Here is the solution I came up with
I created the following table from the riddle:
Current
Future
Past
Princess
x
2z
(x+y)/2
Prince
y
x
z
I then created three equations, since the difference in their age will always be the same. d = the difference in ages x – y = d 2z – x = d x/2 + y/2 – z = d
I then created a matrix and solved it using row reduction.
x
y
z
1
-1
0
d
-1
0
2
d
.5
.5
-1
d
It reduced to:
x
y
z
1
0
0
4d
0
1
0
3d
0
0
1
5d/2
This means that you can pick any difference you want (an even one presumably because you want integer ages). Princess age: 4d Prince age: 3d
Ages that work
Princess
Prince
4
3
8
6
16
12
24
18
32
24
40
30
48
36
56
42
64
48
72
54
80
60
To see other solutions check out the comments from when I posted this on my blog.