#411 - Death Match Riddle

In the Wild West, you are challenged into a death match by two bounty hunters nicknamed Golden Revolver (GR) and Killer Boots (KB). You accept the challenge. None of you want to waste any of the bullet and so a certain rules are laid down:
1) All of you will shoot in a given order till the last man standing.
2) Each of you shoots only once upon his turn.
3) If any one of you is injured, the other two will finish him off with iron rod.
4) The worst shooter of all (which is you) shoots first and the best one shoots at the last.

Now, how will you plan things if you know that you hit every third shot of yours, KB hits every second shot and GR hits every shot ?

Death Match Riddle
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The best thing you can do is shooting your first shot in the air.

Suppose if you shoot KB and hit him by luck, you are certainly dead as GR never misses.

If you shoot on GR first and hits him luckily, there is a fifty percent probability that you will die before your next chance.

But if you shoot in the air, KB will definitely shoot on GR as he knows that he is a better shooter. If he misses somehow, then, he is dead and if he hits then, GR is dead.

Now it’s your turn to shoot and you stand a 1/3 chance that you will hit.

This is the best situation you can have. Doing anything else will bring you under worse situation for sure.

#412 - Who Am I Riddle With Clues

Can you tell who am I ?

I am an eight letter word and I am a computer terminology.
The second, third and fourth letters make an animal.
The fourth, fifth, sixth, seventh and eighth letters make a weapon.
The first, second, third and fourth letters can be taken as an outcome of any exam.
The fifth, sixth, seventh and eighth letter combine to form a high end typing software.

Who Am I Riddle With Clues
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Password

#413 - Classical Probability Puzzle

If we roll two dices (six sided normal dice) together.

what is the probability that the first one comes up with a 2 and the second one comes up with a 5?

Classical Probability Puzzle
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The probability will be 1/36.

For the first dice, there can be six possibilities. Similarly, for the second dice as well, there can be six possibilities.
Thus the total possibilities is 6 * 6 = 36.
The outcome we need is that the first comes up with a 2 and the second comes up with a 5. That is possible only in one possibility.
Therefore the required probability is 1/36.

#414 - Fun But Hard Maths Riddle

How high will you have to count before you use the letter A?

Fun But Hard Maths Riddle
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One Thousand

#415 - Mathematical Logical Puzzle

Consider the situation that there is a pond where some flowers have grown up and some bees are hovering over the flowers. Now read the following statements carefully:
1. If each of the bees lands on a flower, then one bee does not get a flower.
2. If two bees share each flower, then there is one flower left.

Now, can you calculate the number of flowers in the pong and the number of bees hovering over them?

Mathematical Logical Puzzle
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There are three flowers in the pond and four bees are hovering over it.

#416 - worlds trickiest riddle

The doctor prescribed you to take one pill from the bottle every half an hour. The bottle is now left with only three pills.

How long do you think you have before you run out of pills?

worlds trickiest riddle
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You have an hour.

The most common answer will be one and a half hour. But remember, when you took the first pill, it was the zero minute.

#417 - Engineer Riddle

Look at the figure below. You have three house and there are three utilities below it: W representing water, G representing gas and E representing Electricity. You have to draw a line that gets each utility into every house without crossing the lines. Can you do it?

Engineer Riddle
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This puzzle is known as the Turan Brickyard Problem and is a classic one. There is no way that they can be connected without crossing the lines in 2 dimension. However, you can do it if you place the scene in a 3d workplace. See to the figure to understand better.

#418 - Hard Conditional Probability Problem

Four friends - Anna, Brian, Christy and Drake are asked to choose any number between 1 and 5.

Can you calculate the probability that any of them chose the same number ?

Hard Conditional Probability Problem
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Let us take this one step at a time.

Let us calculate the probability that Anna and Brian have the same number in their mind.
101/125

Now, there's a 1/5 chance that the numbers will be same and 4/5 chance that the numbers are different.

Let us now include Christy in this data. There can be two cases.
1) Anna and Brian have the same number. In that case, Christy will have only one number to compare.
2) Anna and Brian did not have the same number. In that case, Christy will have two numbers to compare to.

For the first case, the probability will be 5/25. This is if Anna and Brian did have the same numbers.

But if Anna and Brian did not have the same numbers, there is a 2/5 probability that Christy is having the same number (this is because Christy gets to match her number with both Anna and Brian). In that case, we can simply multiply the probabilities.
4/5 * 2/5 = 8/25

Otherwise, if Christy is not having the same number, the probability is 3/5. Now multiplying with the previous chain:
4/5 * 3/5 = 12/25

Now, we can include Drake in our calculations. If we follow the path where Drake's number matches with Anna and Brian, the probability will be 25/125.

Now let us join that with Christy's probability. If Christy's number matches with Anna and Brian and Drakes' number also matches, then the probability will be:
4/5 * 2/5 = 40/125

If Christy's number does not match with Anna and Brian but Drake's number matches with Christy's, the probability will be:
4/5 * 3/5 * 3/5 = 36/125

But if Christy's number does not matches with Anna and Brian and even Drake's does not matches with Christy, then the probability will be:
4/5 * 3/5 * 2/5 = 24/125

Now, we have to tell the probability when all the four friends have same numbers, so we will just add up the probability where all the numbers matches:

25/125 + 40/125 + 36/125 = 101/125.

#419 - What Word Am I Thinking Of Riddle

I am an eleven letter word.
The first, second, third and fourth letter combine to form a bankquot;s name.
The fifth, sixth and seventh letter combine to form a carquot;s name.
The eighth, ninth, tenth and eleventh letter combine to form a mode of transport.

Can you identify what word am I?

What Word Am I Thinking Of Riddle
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CITIZENSHIP

#420 - Cricket Riddle

How many runs at maximum can a batsman score in a normal one day match? Consider the fact that the conditions are ideal and there are no No Balls, no Wide Balls and no Extras in that match.

Cricket Riddle
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If we are considering an ideal case and we have to calculate the maximum score, we will assume that the batsman hits six on every ball. But then, at the end of every over, the strike will change. But we have to make sure, he gets to play maximum number of balls to score maximum runs. Thus, the batsman will hit six in first five balls and will choose to take 3 runs in the last ball so that he retains the strike in the following over.

Total runs per over in that case = 6 * 5 + 3 = 33.

But in the last over (50th over), he don’t have to worry about keeping the strike and thus, he will hit six even on the last ball. Therefore, the maximum runs that a batsman can score:

33 * 49 + 6 * 6 = 1617 + 36 = 1653