Plan The Best Chance of Winning!

You are playing a game of dodge ball with two other people, John and Tom. You're standing in a triangle and you all take turns throwing at one of the others of your choosing until there is only one person remaining. You have a 30 percent chance of hitting someone you aim at, John has a 50 percent chance, and Tom a 100 percent change (he never misses). If you hit somebody they are out and no longer get a turn.

If the order of throwing is you, John, then Tom; what should you do to have the best chance of winning?

This should be you plan to increase chances of winning! 

Planning The Best Chance of Winning!

What was the game?

You should miss the first shot for the purpose.

Remember, about one of your 3 shots (30% accuracy), John's 1 out of 2 shots (50% accuracy) and Tom's every shot is on target.

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CASE 1 : If you target Tom and hit him then John has to hit you. Even if he fails to target you in first attempt he will be successful in his second attempt. 

And since, your first shot was on target your next 2 has to be off the target one of which will give John a second chance.


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CASE 2 : If you target John then Tom will certainly hit you to be winner of the game.

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CASE 3 : Better miss the first shot and then 1 of next 2 shots will be on the target.

Now, John has to target Tom otherwise assuming John as stronger player, Tom will eliminate him immediately. 

    CASE 3.1 : If John hits Tom and eliminates him then it's your turn

                     now and John's next attempt has to be off the target. 

                     So, even if you fail in first try after John's

                      unsuccessful try you can surely hit John in second try.

    CASE 3.2 : And if John misses Tom then Tom will throw John 

                     out of game in his first attempt. Now, it's your turn 

                     and you can target Tom with 50% accuracy.

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This is the best plan to get a chance of winning this game. 

Heavier Vs Lighter Balls

We have two white, two red and two blue balls. For each color, one ball is heavy and the other is light. All heavy balls weigh the same. All light balls weigh the same. How many weighing on a beam balance are necessary to identify the three heavy balls? 

You need only 2 weighings! Click here to know how!

Identifying The Heavier/Lighter Balls

What was the task given?

Actually, we need only 2 weighing. 

Weigh 1 red & 1 white ball against 1 blue & 1 white ball. Weighing result between these balls helps us to deduce the relative weights of other balls.
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Case 1 : 

If this weighing is equal then the weight of red ball and blue ball are different.
That's because the weighing is equal only when there is combination of 
H (Heavy) + L (Light) against L (Light) + H (Heavy). 

Since, one white ball must be heavier than the other white ball placed in other pan, the red & blue balls place along with them must be of 'opposite' weights with respect to the white balls place along with them.

So weigh this red against blue will give us the heavier red (or blue) leaving other red (or blue) as lighter. 

If red (or blue) weighs more in this weighing then the white ball placed with this red (or blue) in first weighing must be lighter than the other white ball placed with blue (or red) ball.  

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Case 2 : Red + White > Blue + White.

The white ball in Red + White must be heavier than the white ball in Blue + White.

The reason is if this white was lighter then even heavier red in the Red + White can't weigh more than Blue + White having heavier white. That is H + L can't beat H + H or obviously would equal with H + L!

So we have got the heavier and lighter white balls for sure.

Now, weigh red from Red + White and blue from Blue + White against other red and blue balls left.

     Case 2.1/2.2 : Red + Blue > or < Red + Blue 

     Obviously, red and blue balls in left pan are heavier (or lighter) than those in other.
    
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     Case 2.3 :  Red + Blue = Red + Blue 

     Obviously, that's because of H + L = L + H.

      But the red is taken from Red + White which was heavier than the Blue + White. 
      Since, L + H (white is heavier as concluded) can't weigh more than H + L
      (other white is lighter as concluded) or L + L, the red in Red + White 
      must be heavier making H + H combination in that pan in first weighing (case 2). 

      So, we got heavier red and lighter blue obviously leaving lighter red 
      and heavier blue in other pan. 

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Case 3 : Red + White < Blue + White.

Just replace Red with blue & vice versa in the deduction made in case 2. 

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Tricky Probability Puzzle of 4 Balls

I place four balls in a hat: a blue one, a white one, and two red ones. Now I draw two balls, look at them, and announce that at least one of them is red. What is the chance that the other is red?

Well, it's not 1/3!

Tricky Probability Puzzle of 4 Balls : Solution

What was the puzzle?

It's not 1/3. It would have been 1/3 if I had taken first ball out, announced it as red and then taken second ball out. But I have taken pair of ball out. So, there are 6 possible combinations.

red 1 - red 2
red 1 - white
red 2 - white
red 1 - blue
red 2 - blue
white - blue 

Out of those 6, last is invalid as I already announced the first ball is red. That leaves only 5 valid combinations.

And out of 5 possible combinations only first has desired outcome i.e. both are red balls.
Hence, there is 1/5 the chance that the other is red 

Whose Number is Bigger?

Ali and Zoe reach into a bag that they know contains nine lottery balls numbered 1-9. They each take one ball out to keep and they look at it secretly. Then, they make the following statements, in order:

Ali: "I don't know whose number is bigger." 

Zoe: "I don't know whose number is bigger either." 

Ali: "I still don't know whose number is bigger." 

Zoe: "Now I know that my number is bigger!"

Assuming Ali and Zoe are perfectly logical, what is Zoe's smallest possible number?

'This' is that smallest possible number!

My Number is the Bigger One!

First you can read what happened?

Recalling what Ali and Zoe said - 

Ali: "I don't know whose number is bigger." 

Zoe: "I don't know whose number is bigger either." 

Ali: "I still don't know whose number is bigger." 

Zoe: "Now I know that my number is bigger!" 

First statement of Ali indicates that she doesn't have either 1 or 9. If she had 1 (or 9) then she would have an idea that Zoe must have bigger (or smaller) number.

Now Zoe is smart enough to know that Ali doesn't have 1 or 9 which is clear from Ali's first statement. Zoe's first statement indicates that she doesn't have 2 or 8 (& neither 1 or 9). If she had 1/2 (or 8/9) then she could have concluded that Ali has bigger (or smaller) number.

Till now Ali has an idea that the Zoe doesn't have 1,2,8,9. So still Ali can't have 2/8 as in that case too she could have made a different statement. Further if Ali had 3 or 7 (and knowing the fact that Zoe doesn't have 1,2,8,9); Ali could have an idea whose number is bigger as 3 is smallest while 7 is biggest among remaining numbers. That means she doesn't have 3 or 7 ( and 1,2,8,9).

From all the statement Zoe can conclude that Ali doesn't have 1,2,3,7,8,9. In short, Ali must have either 4,5,6. 

Now when Zoe says she has bigger number then it must be either 6 or 7 and Ali having 4 or 5. It can't be 5 as in that case Zoe wouldn't be confident as Ali could have 6.

So the Zoe's smallest possible number is 6.

     

Maximize The Chance of White Ball

There are two empty bowls in a room. You have 50 white balls and 50 black balls. After you place the balls in the bowls, a random ball will be picked from a random bowl. Distribute the balls (all of them) into the bowls to maximize the chance of picking a white ball. 

This is the way to maximize the chances!

Way to Maximize White Ball Probability

What was the task given?

Let's distribute 50 black ball in one bowl & other 50 white ball in another bowl.

Then,

Probability (White Ball) = (1/2)(0/50) + (1/2)(50/50) = 0.5.

Now, if 1 white ball is kept in 1 bowl and other 49 white + 50 black = 99 balls in other bowl, then


Probability (White Ball) = (1/2)(1/1) + (1/2)(49/99) = 0.747.

That's nearly equal to 3/4 which is certainly higher than the previous case. And that's the way of maximizing the probability of white ball.

Maximum Runs That Batsman Can Score?

In a one day international cricket match, considering no extras(no wides, no ‘no’ balls, etc.) and no overthrows.

What is the maximum number of runs that a batsman can score in an ideal case ?

Note: Here we assume ideal and little practical scenario. We assume that batsman can not run for more than 3 runs in a ball, as otherwise there is no limit, he can run infinite runs(theoretically) in a ball, as far as opposite team does not catch the ball.”

Could be tricky! Here is correct number! 

Calculation of Maximum Runs by Batsman

What was the question?

It's not as straight forward as it seems at first glance. That is one might think that the maximum score that one can score by hitting 6 on every ball of 50 overs is 50 x 6 x 6 = 1800. 

No doubt, 1800 can be the maximum team score but not the individual score.Since batsman rotates strike every over, here both batsmen share these 1800 runs as 900 to each.

However, if the batsman on strike runs 3 runs on the last ball of the over then he can hit 5 more sixes in next over as strike is rotated back to him in next over. He can continue in this way till 49th over. And in 50th over he can hit 6 sixes on 6 balls.

Maximum Individual Score = 49 x [(5x6)+3]  + 36 = 1617 + 36 = 1653.

In this case, the batman at the non-striker end scores 0 runs as he doesn't get strike on a single ball.

Who Will Be Not Out?

It is a strange cricket match in which batsman is getting bowled in the very first ball he faced. That means on ten consecutive balls ten players get out.

Assuming no extras in the match, which batsman will be not out at the end of the innings?  

Know that lucky player!

Source 

"He Will Be Not Out!"

What happened in the match?

First let's number all the players from 1 to 11 as Batsman 1, Batsman 2, Batsman 3 & so on with last player as Batsman 11. 

Now let's take a look at what must have happened during 1st over.

1st Ball : Batsman-1 got out
2nd Ball : Batsman-3 got out
3rd Ball : Batsman-4 got out
4th Ball : Batsman-5 got out
5th Ball : Batsman-6 got out
6th Ball : Batsman-7 got out

Batsman 8 comes in 

Batsman 2 is still standing at non-striker end watching fall of wickets. Remember in the match all batsman are bowled out so no change in strike because of run out or before catch is taken.

At the end of first over, the strike is rotated and Batsman 2 comes on strike while Batsman 8 at the non striker end.

Now here is what happens in second over.

1st Ball : Batsman-2 got out
2nd Ball : Batsman-9 got out
3rd Ball : Batsman-10 got out
4th Ball : Batsman-11 got out

So the only batsman left NOT OUT is Batsman 8 standing at the non-striker end.
 

The CryptArithmetic Problem

Can you solve the below alphametic riddle by replacing letters of words by a numbers so that the below equation holds true?

BASE +
BALL
---------
GAMES
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Find numbers replaced letters here! 

Source 

The CryptArithmetic Problem's Solution

What was the problem?

Let's first recall the given equation.

  BASE +
  BALL
---------
GAMES
----------

We are assuming repeating the numbers are not allowed. 

Let's first take last 2 digits operation into consideration i.e. SE + LL = ES or 1ES (carry in 2 digit operation can't exceed 1). For a moment, let's assume no carry generated.

10S + E + 10L + L = 10E + S .....(1)

9 (E - S) = 11L

To satisfy this equation L must be 9 and (E - S) must be equal to 11. But difference between 2 digits can't exceed 9. Hence, SE + LL must have generated carry.So rewriting (1),

10S + E + 10L + L = 100 + 10E + S

9 (E - S) + 100 = 11L

Now if [9 (E - S)] exceeds 99 then L must be greater than 9. But L must be digit from 0 to 9. Hence, [9 (E - S)] must be negative bringing down LHS below 100. Only value of E - S to satisfy the given condition is -5 with L = 5. Or we can say, S - E = 5.

Now, possible pairs for SE are (9,4), (8,3), (7,2), (6,1), (5,0). Out of these only (8,3) is pair that makes equation SE + LL = SE + 55 = 1ES i.e. 83 + 55 = 138. Hence, S = 8 and E = 3.

Replacing letters with numbers that we have got so far.

    1---------
  BA83 +
  BA55
---------
GAM38
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Now, M = 2A + 1. Hence, M must be odd number that could be any one among 1,7,9 (since 3 and 5 already used for E and L respectively).

If M = 1, then A = 0 and B must be 5. But L = 5 hence M can't be 1

If M = 7, then A = 3 or A = 8. If A = 3 then B = 1.5 and that's not valid digit. And if A = 8 then it generates carry 1 and 2B + 1 = 8 again leaves B = 3.5 - not a perfect digit.

If M = 9, then A = 4 (A = 9 not possible as M = 9) and B must be 7 with carry G = 1.Hence for first 2 digits we have 74 + 74 + 1 = 149.

Finally, rewriting the entire equation with numbers replacing digits as -

    1
---------
  7483 +
  7455
---------
14938
----------

So numbers for letters are S = 8, E = 3, L = 5, A = 4, B = 7, M = 9 and G = 1.