Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

Solution:

1653

Explanation:
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 49th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 49 + 36 = 1653

Solution:

There is a simple logic to solve this question.
The size of the chocolate is 2 x 8. Thus, you need to have 2 x 8 = 16 pieces.
Every time you break the chocolate, you will get one extra piece.
Thus, to get 16 pieces, you must break it (16 - 1) = 15 times.

Solution:

199 - 99

Solution:

The node can be connected as shown.

Solution:

The wise man told them to switch the horses and then race. In this manner, both of them will be willing to race as fast as they can to prove that the other horse (his own horse) is slow.

Solution:

See the colored sticks in the picture. We have moved those sticks in the position given in the second image corresponding with the color.

Solution:

The probability of two numbers being relatively prime is 60%. Thus if you keep playing the bet, you will hypothetically make 20 cents on every play.

Thus, you should accept the bet.

Solution:

Fred Odea won.

Just look at the announcement lines. The first sentence in 'Ok now everybody!' If you take out just the first letter of every word, it will form 'ONE'. In the same manner the second sentence 'The winner of…..' will give form 'TWO' and the third sentence 'The Hardest Riddle Ever Event.' Will form THREE.

Thus the final sentence must form FOUR. There is only one candidate who can suffice with the first two letter and he is Fred Odea. The final sentence of the mastermind must be, 'Fred Odea: Ultimate Riddler!'

Solution:

2, 3, 10 and 10
To maintain the difference of eight and keeping in mind the third and fourth hint, the first and third cards can be 2 and 10 or 10 and 2. But keeping in mind the first hint, we know that the first card is 2 and the third is 10. Similarly, if we see the fifth hint, the difference between the second and fourth card should be 7. Thus we know that the second card is 3 and the fourth card is 10.

Solution:

Four

three placed flat on the table in a triangle(touching each other) and put the fourth one on top of them in the middle.