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.

  

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.