KnifeMissile

You are sitting on a mountain of blue and red balls. You have a bag filled with exactly 20 blue and 13 red balls. You decide to play a game and here are the rules.
You will remove two balls from the bag. If they are the same colour, you will add a blue ball into the bag. If they are different colours, you will add a red ball into the bag. Please note that, because you are not placing the two test balls back into the bag, you will always be losing a ball every turn so, eventually, you will be down to a single ball.
Now, what will be the colour of this last ball?

Noob

if you keep taking 2 balls out of the bag and putting 2 balls from the mountain in you will always be sitting up top....

dimbulb

ok, here's some preliminary analysis
Lets break it into 3 events.

Event:
1) Minus 2 Red, Add one blue=> -2R+B
2) Minus 2 Blue, Add one=> -B
3) Minus 1 Red, 1 Blue, Add 1 Red=> -B

When you have 3 balls left, there are 4 possible states
3-1) 3 R
3-2) 2R 1B
3-3) 1R 2B
3-4) 3B

dimbulb

For 2 balls, you have 3 possible states
2-1) 2R
2-2) 1R 1B
2-3) 2B

so using a notation of #Balls-State,

from 3-1, you must pick 2 R and add 1B, so you end up at 2-2.
from 3-2, there are 2 cases: pick 2 red, end up at 2-3
or pick 1R1B and end up at 2-1
from 3-3, you always end up at 2-2
from 3-4, you always end up with 2-3

dimbulb

From 2-1, you end up with 1 blue ball
From 2-2, you end up with 1 red ball
from 2-3, you end up with 1 blue ball...

so it looks like you can end up with both.... hmmm..... i vaguely remember a similar question I did in a Markov Chain course....
unless we can eliminate 2-2 as a possible state....

dimbulb

at any stage, you are deducting either zero or 2 red balls....
and you start off with an odd number of balls.

so it is impossible to end up with an even number of red balls at any stage.

so only 2-2 can exist.
and hence you end up with a red ball.