Solution:

The five cards must contain at least one card of the same suit.
Let's assume Rhodes placed one of the two cards of the same suit at the end and the second card was given to Daniel.
In that case, Daniel will know the suit of the card.

Now the real question is how Rhodes made sure that Daniel also knows the number of the card.

Daniel got four cards and one of them will determine the suit, so we are left with 3 cards.
3 cards can be arranged in 3! ways i.e. 6 ways but our card numbers can vary from 1 to 13.
Since we have two cards of the same suit we will make sure that we don't pick the King (13). This leaves us with 12 numbers.

We can also distinguish card from upside down position, therefore, we now have 6 * 2 ways to arrange these card.
"TADA.... We cracked the magic trick".

Let us assume that the smallest card is C1U (upwards position) and C1D (downwards position). Similarly, we have C2U, C2D, C3U and C3D

According to that, the 12 possible arrangements are:
C1U C2U C3U => NUMBER 1
C1U C3U C2U => NUMBER 2
C2U C1U C3U => NUMBER 3
C2U C3U C1U => NUMBER 4
C3U C1U C2U => NUMBER 5
C3U C2U C1U => NUMBER 6
C1D C2D C3D => NDMBER 7
C1D C3D C2D => NDMBER 8
C2D C1D C3D => NDMBER 9
C2D C3D C1D => NDMBER 10
C3D C1D C2D => NDMBER 11
C3D C2D C1D => NDMBER 12

Please note that king (number 13) is not possible as discussed already.

Solution:

I filled a glass full of water and held it over my head for 20 minutes.

Solution:

Just 2

Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

Solution:

It can be solve as shown below.

Solution:

1/3

Explanation
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.

Following are the possible cases for toys distribution:

Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1

In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3

Solution:

zero

I had 3 green socks and 1 Pink sock.

=> chances of pair of green socks = 1/2
chances of pair of mixed socks = 1/2
chances of pair of green socks = NULL

Solution:

Colors:
Y B R G
G R B Y
B Y G R
R G Y B


Signs:
x + - /
- / x +
/ - + x
+ x / -

Solution:

The clerk said that every day the password is CHANGED. Changed is the password.

Solution:

He should shoot at the ground.

If Kangwa shoots the ground, it is Rafael's turn. Rafael would rather shoot at Ferdinand than Kangwa, because he is better.
If Rafael kills Ferdinand, it is just Kangwa and Rafael left, giving Kangwa a fair chance of winning.
If Rafael does not kill Ferdinand, it is Ferdinand's turn. He would rather shoot at Rafael and will definitely kill him. Even though it is now Kangwa against Ferdinand, Kangwa has a better chance of winning than before.

Solution:

2100 miles.

Each vowel equivalent to 500 and each consonant equivalent to -200

Solution:

The driver is actually the wife and mother.