Will the Last Ball Always Be the Same Color in This Brainteaser?

[michealsmith]

posting my 1st brainteaser ...hopefully u might enjoy


uve got 20 blue balls and 14 red balls all mixed up ( u also have some spare balls outside the box) in a box u have to take out 2 balls at a time. if they r the same leave them outside of the box and put in a blue spare (the spares r not diffirent in any way to the ones in side already ).if they r diffirent put red back in box leave blue out...question if u keep taking balls out will the last ball always be the same colour ,,if yes wat colour?

a proper detailed answer please

for those who like creative thinking puzzles...
What is " never odd or even " ?

 

[ceptimus]

choice  blue result  red result
b b         -1           0
r r         +1          -2
b r         -1           0
r b         -1           0

Note that after each trial there is one fewer ball in the box than before. So after 33 trials there will be just one ball left in the box.

Reds are only ever removed 2 at a time, and never added, so the number of reds in the box must remain even.

Therefore the last ball in the box after 33 turns will always be blue.

 

[Architeuthis Dux]

I am always intrigued by puzzles and brainteasers. The first puzzle presented involves a box with 20 blue balls and 14 red balls, along with some spare balls outside the box. The goal is to take out two balls at a time and follow a specific rule: if the two balls are the same color, leave them outside and replace them with a blue spare ball. If the two balls are different colors, put the red ball back in the box and leave the blue ball outside. The question is, if you keep taking balls out, will the last ball always be the same color? If so, what color will it be?

To properly answer this question, we need to consider the initial ratio of blue to red balls in the box. Since there are more blue balls (20) than red balls (14), it is likely that the last ball will be blue. This is because every time two balls of the same color are taken out, a blue spare ball is put in, increasing the number of blue balls in the box.

If we assume that we start with two blue balls, the first time we take out two balls, we will have 3 blue balls and 1 red ball in the box. The next time, we will have 4 blue balls and 2 red balls, and so on. This pattern will continue until there are 19 blue balls and 9 red balls in the box. At this point, if we take out two balls, they will both be blue and will be left outside the box, leaving us with 20 blue balls and 8 red balls. The next time we take out two balls, they will both be blue again and will be left outside, leaving us with 21 blue balls and 7 red balls. This pattern will continue until there is only one red ball left in the box. At this point, if we take out two balls, they will both be blue and will be left outside, leaving us with 22 blue balls and 6 red balls. Finally, when we take out the last two balls, they will both be blue and will be left outside, leaving us with 23 blue balls and 5 red balls. Since the last two balls will always be the same color, the last ball will indeed always be blue.

Now, onto the second puzzle - "never odd or even". This is a classic palindrome, which is a word or phrase that reads the same backward