Crack And Win $500,000 - Puzzle

You are in a game show with four other contestants. The objective is to crack the combination of the safe using the clues, and the first person to do so will win $500,000.

The safe combination looks like this:

??-??-??-??

A digit can be used more than once in the code, and there are no leading zeroes.

Here are the clues:

1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.

And get moving, I think another contestant has almost figured it out!

Click here for the SOLUTION! 

Crack And Win $500,000 Puzzle - Solution

What was the puzzle?

We know, the safe has a lock having 4 sets of 2 digits as -

?? - ?? - ?? - ??

Since, leading zeros are not allowed any of the set can't be started with 0 like 01, 07, 09 etc. 

Take a look at the clues given - 

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1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.


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STEPS :

1] As per clue (2), digits 2, 3, 4 or 5 aren't allowed in any set. That means only digits 0, 1, 6, 7, 8, 9 are allowed for sure.

2] For (3) to be true, possible combinations of 3rd and 4th sets are - 

10 x 4 = 40
11 x 4 = 44
16 x 4 = 64
17 x 4 = 68
18 x 4 = 72
19 x 4 = 76

The third set can't be 10 for 2 reasons. - 1} As per clue (1), the first set will be 01 and leading 0's are not allowed. 2} The 4th set will have digit 4 which is not allowed.

The third set can't start with 2X, 3X, 4X or 5X as those digits aren't allowed. Moreover, it can't be started with 6X, 7X, 8X as in that case the value of the 4th set will exceed it's maximum possible value of 96 (if digit 2 was allowed) or 76.

Out of all above combinations, only 17 x 4 = 68 and 19 x 4 = 76 are valid combination as rest of combinations have digits that aren't allowed.

So, one thing is sure that the first digit of the third set is 1. And hence the first digit of first set also must be 1 as per clue (1).

3] As per clue (5), the second set can't exceed 20. It can't start with 0. Hence, possible values ranges from 10 to 19. That means, the first digit of the second set is also 1.

4] Now as per clue (4), if you add the first number in the first set with the first number in the second set you will get 8. That means the first number of first set is 8 - 1 = 7. 

As of now, the code looks like : 71 - 1? - 1? - ??

5] If 7 is the first digit of first set then as per clue (1), 7 itself is second digit of the third set.

Now, the code looks as : 71 - 1? - 17 - ??

6] As per clue (3) and rightly deduced as a possible combinations for 3rd & 4th set in STEP 2, the 4th set must be 17 x 4 = 68.

With that, code turns into : 71 - 1? - 17 - 68.

7] Finally, as per clue (6), the second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set. This multiplication i.e. ? x 7 must be equal to 71 + 1 = 72. 
Hence, ? = 9.

The final code looks like : 71 - 19 - 17 - 68.

  

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. 

The Numbered Hats Test!

One teacher decided to test three of his students, Frank, Gary and Henry. The teacher took three hats, wrote on each hat an integer number greater than 0, and put the hats on the heads of the students. Each student could see the numbers written on the hats of the other two students but not the number written on his own hat.

The teacher said that one of the numbers is sum of the other two and started asking the students:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— No, I don’t.

Then the teacher started another round of questioning:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— Yes, it is 144.

What were the numbers which the teacher wrote on the hats?

Here are the other numbers!

Source 

Cracking Down The Numbered Hats Test

What was the test?

Even before the teacher starts asking, the student must have realized 2 facts.

1. In order to identify numbers in this case, the numbers on the hats has to be in proportion i.e. multiples of other(s). Like if one has x then other must have 2x,3x etc.

2. Two hats can't have the same number say x as in that case third student can easily guess the own number as 2x since x-x = 0 is not allowed.

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Now, if numbers on the hats were distributed as x, 2x, 3x then the student wearing hat of number 3x would have quickly responded with correct guess. That's because he can see 2 number as x and 2x on others hats and he can conclude his number as x + 2x = 3x since 
2x - x = x is invalid combination (x, x, 2x) where 2 numbers are equal.

Other way, he can think that the student with hat 2x would have guessed own number correctly if I had x on my own hat. Hence, he may conclude that the number on his hat must be 3x.

But in the case, all responded negatively in the first round of questioning. So x, 2x, 3x combination is eliminated after first round.

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That means it could be x, 3x, 4x combination of numbers on the hats.

In second round of questioning, Henry guessed his number correctly.

If he had seen 3x and 4x on other 2 hats then he wouldn't have been sure with his number whether it is x or 7x.

Similarly, he must not have seen x and 4x as in that case as well he couldn't have concluded whether his number is either 5x or 3x.

But when he sees x and 3x on other hats he can tell that his number must be 4x as 2x (x,2x,3x combination) is eliminated in previous round!

So Henry can conclude that his number must be 4x.

Since, he said his number is 144,

4x = 144

x = 36

3x = 108.

Hence, the numbers are 36, 108, 144.

"Get Out of The Hell !"

You’re new to hell, and you’re given a choice: You can go directly to the fourth circle, or you can play simultaneous chess games against Alexander Alekhine and Aron Nimzowitsch. Alekhine always plays black and smokes a pipe of brimstone. Nimzowitsch plays white and wears cuff links made of human teeth. Neither has ever lost.

If you can manage even a draw against either player, you’ll be set free. But if they both beat you, you’ll go to the eighth circle for eternity.

What should you do?

This trick will save your life! 

Freedom From The Hell !

How you are trapped in hell?

Alekhine always plays black and Nimzowitsch plays always white. Obviously, you will be forced to choose white against Alekhine and black against Nimzowitsch. 

Wait for Nimzowitsch's first move and then play the same move on Alekhine’s board. Note how Alekhine responds to move & copy that move on Nimzowitsch's board.

This way, effectively Nimzowitsch is playing against Alekhine and you are just transferring moves between to 2 masters.

Since, they never lost a single chess game, there are high chances that the game between two ends in draw.

Even if Alekhine wins with his black then you are winning against Nimzowitsch as you are copying Alekhine's move against Nimzowitsch's white using your black. Same is case if Nimzowitsch wins.

In short, you will end up with either draw with both or win against at least one (against both is impossible). You will be free in any case.

And in fact, you need not to have knowledge of how to play chess to get freedom from the hell.

Plan an Unbeatable Strategy

Two people play a game of NIM. There are 100 matches on a table, and the players take turns picking 1 to 5 sticks at a time. The person who takes the last stick wins the game. (Both players has to make sure that the winner would be picking only 1 stick at the end) 

Who has a winning strategy?

And what must be winning strategy in the person who takes the last stick looses?

This could be the winning strategy! 

Planned The Unbeatable Strategy!

What is the game?

The first person can plan an unbeatable winning strategy.

CASE 1 : The person picking last stick is winner.

All that he has to do is pick 4 sticks straightaway at the start leaving behind 96 stick. Then, he has to make sure that the count of remaining stick will be always divisible by 6 like 96, 90, 84, 78......6. 

So if the opponent takes away 2 sticks in his first turn, then first person has to take 6 - 2 = 4 sticks leaving behind 90 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.

Now, when there are 6 stick left, even if opponent takes away 5 sticks then 1 stick will be left for the first person.

And even if the opponent picks 4 sticks then first person will take 2 remaining sticks.

CASE 2 : The person picking last stick is looser. 

Now the first person need to take away 3 sticks in first turn leaving behind 97. Next, he has to make sure the count of remaining sticks reduced by 6 after each of his turn. That is, the count should be like 91,85,79,72......7.

So if the opponent takes away 4 sticks in his first turn, then first person has to take 6 - 4 = 2 sticks leaving behind 91 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.

When there are 7 sticks are left then even if the opponent takes away 5 sticks then first person can force him to pick the last stick by picking only 1 stick of remaining 2. 

And if the opponent takes away 4 sticks at this stage, the first person still can force him to pick last stick by picking 2 of remaining 3 sticks.

Conclusion : The first person always has a chance to plan a winning strategy.

The Monty Hall Problem

You’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?”

Is it to your advantage to switch your choice?

Will you switch or stay with your door?

Note : Monty Hall was the host of game show called 'Let's Make a Deal' on which above puzzle is based. (Source-Wikipedia)



This is what you should do! 

Winning The Monty Hall Game Show

What was the game show?

Suppose you always choose DOOR 1. Then host will open DOOR 2 or DOOR 3 behind which car is not there.

If the car is behind the DOOR 1, then host will open the DOOR 2 or DOOR 3. And if you switch to remaining DOOR 3 or DOOR 2, you will find goat behind it & you will loose.

And if the car is behind the DOOR 2, then host will be forced to open DOOR 3. Now, if you switch your choice to DOOR 2 then you will win the car behind that door.

Again, if the car is behind the DOOR 3, then host has to open the DOOR 2 behind which goat is there. And now if you switch your selection from DOOR 1 to DOOR 3, then you will be winning the car.

So out of 3 possibilities, in 2 you will be winning this game show if you switch your choice. The probability of winning the game show is 2/3.

And if you stay with your first choice, then probability of having car behind selected door is 1/3.

To conclude, it's better to switch the choice as it increases the probability of winning the game show from 1/3 to 2/3. 
 

The Number Game!

Let’s play a game. You name an integer from 1 to 10. Then we’ll take turns adding an integer from 1 to 10 to the number our opponent has just named, giving the resulting sum as our answer. Whoever reaches 100 first is the winner.

You go first. What number should you choose?

This is how you can be winner!

Winning The Number Game!

What was the game?

Here the player whose number 'forces' sum to fall in range of 90-99 will be ending on losing side. 

That means, somehow if you 'force' the total at some point to 89 then opponent has to fall in the range of 90-99 with his number. 

To get 'door' to total 89 you have to force the previous sum to 78 so that opponent is forced to open a 'door' for total 89 for you.

And so on backward you have to make stops at 67, 56, 45, 34, 23, 12, 1.

So you have to start with 1 & achieve all above milestones.

Let's verify our conclusion. Suppose you started with 1.

You       Sum      Opponent    Sum

1            -             8              9
3           12            5              17
6           23            7              30
4           34            9              43
2           45            4              49
7           56            10            66
1           67            3              70
8           78            2              80
9           89            4              93
7           100           

YOU WIN!

Unfair Game of Strange dice

Katherine and Zyan are playing a game using strange dice. Each die is a cube with six sides. Katherine's die has sides numbered 3, 3, 3, 3, 3, and 6. Zyan's die has sides numbered 2, 2, 2, 5, 5, and 5.

To play the game, Katherine and Zyan roll their dice at the same time and whoever rolls the higher value wins. If they play many times, who will win more frequently, Katherine or Zyan?

This person will be winning more! 
 

Advantage in Unfair Game of Strange Dice

What was the game?

Shortest Way : 

Katherine's die has sides numbered 3, 3, 3, 3, 3, and 6. And Zyan's die has sides numbered 2, 2, 2, 5, 5, and 5. That means whenever Zyan rolls a 2 then Katherine will always win. The probability that Zyan rolls 2 is 3/6 = 1/2. So in half of the cases, the Katherine will be winner of the game. Moreover, whenever, Zyan rolls a 5, Katherine can win if she rolls a 6. In short, in more than half cases, Katherine will win hence she has more advantage in this game.


Precise Way : 

Katherine will win if

1. Zyan rolls a 2 with probability 3/6 = 1/2

2. Zyan rolls a 5 (probability 3/6 = 1/2) and Katherine rolls a 6 (probability 1/6). So the probability for this win = 1/12.

Hence, Katherine has got 1/2 + 1/12 = 7/12 chances of winning.

Zyan will win if he rolls a 5 (probability 1/2) and Katherine rolls a 3 (probability 5/6) with probability = 1/2 x 5/6 = 5/12.  

That proves, Katherine has more chances (7/12 vs 5/12) of winning this game. 

Another Way : 

There can be 36 possible cases of numbers of cubes out of which in 15 cases Zyan seems to be winning and in 21 cases, Katherine is winning. Again, Katherine having more chances of winning (21/36 = 7/12) than Zyan (15/36 = 5/12). 

Three Hat Colors Puzzle

A team of three people decide on a strategy for playing the following game.  

Each player walks into a room.  On the way in, a fair coin is tossed for each player, deciding that player’s hat color, either red or blue.  Each player can see the hat colors of the other two players, but cannot see her own hat color.  

After inspecting each other’s hat colors, each player decides on a response, one of: “I have a red hat”, “I had a blue hat”, or “I pass”.  The responses are recorded, but the responses are not shared until every player has recorded her response.  

The team wins if at least one player responds with a color and every color response correctly describes the hat color of the player making the response.  In other words, the team loses if either everyone responds with “I pass” or someone responds with a color that is different from her hat color.

What strategy should one use to maximize the team’s expected chance of winning?

These could be the strategies to maximize the chances of winning!

Three Colors Hats Puzzle - Solution

What's the puzzle? 

There can be two strategies to maximize the chances of winning in the game.

STRATEGY - 1 :   

There are 8 different possible combinations of three color hats on the heads of 3 people. If we assume red is represented by 0 & blue by 1 then those 8 combinations are - 

Here only 2 combinations are there where all are wearing either red or blue hats. That is 2/8 = 25% combinations where all are wearing hat of same color and 6/8 = 75% combinations where either 1 is wearing the different colored hat than the other 2.  In short, at least 2 will be wearing either red or blue in 75% of combinations.

Now for any possible combination, there will be 2 hats of the same color (either blue or red). The one who sees the same color of hats on heads of other two should tell the opposite color as there are 75% such combinations. That will certainly increase the chances of winning to 75%.

STRATEGY 2 :    

Interestingly, here 3 responses from each member of team are possible viz RED (R), BLUE (B) and PASS (P). And every member can see 3 possible combinations of hats on the heads of other 2 which are as 2 RED (2R),  2 BLUE (2B) and 1RED:1BLUE (RB). See below.

Let's think as instructor of this team. We need to cover up all the possible 8 combinations in form of responses in the above table. 

For every possible combination, at least 1 response need to be correct to ensure win. But out of 9 above, 3 responses of 'PASS' are eliminated as they won't be counted as correct responses. So we are left with only 6. Let's see how we can do it.

First of let's take case of 2R. There are 2 responses where A sees 2 RED hats (000,100). We can't make sure A's response correct in both cases. So let A's response for this case be R. So whenever this 000 combination will appear A's response will secure win.

After covering up 000, let's cover up 001. For that, C's response should be B whenever she sees 2 red hats on other 2. And only left response P would be assigned to B.

So far,we have covered up these 2 combination via above responses.  

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Now, let's take a case of 2B. A can see 2B hats whenever there 011 or 111 appears. Since A's R response is already used previously, let B be her response in the case. So the combination 111 will be covered up with A's response.  

B can see 2B hats in case of 101 or 111. Since 111 is already covered above, to cover up 101, B should say R whenever she sees 2 BLUE hats on the heads of other 2. With this only response left for C in case of 2B is P.

 
With these responses, we have covers of so far,

 
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After filling remaining 1 possible response in response table for every team member in case of 1 RED and 1 BLUE hat, 

 B's response as BLUE in this case will ensure win whenever 011 or 110 combination will appear. Similarly, C's response as a RED will secure win whenever 010 or 100 appears as a combination.

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In this way, there will be at least 1 response correct for every possible 8 combinations. This strategy will give us 100% chances of winning this game!

The above table shows who is going to respond correctly for the given combination ( the block of combination & correct response are painted with the same background color).
 
SIMPLE LOGIC : 

The same strategy can be summarized with very simple logic. 

There must be someone to say RED whenever she sees 2 RED hats; someone should say BLUE and remaining one should say PASS. Similarly, one has to say BLUE; other should say RED & third one should say PASS whenever 2 BLUE hats are seen. Same logic to be followed in case of 1 RED and  1 BLUE hats seen. But while doing this, we need to make sure responses are well distributed & not repeated by single member of team (See table below).