You wake up and find yourself in a very strange room with a single door. You find a note on the ground surrounded by lots of cards:
There are 66 green and 66 blue door cards on the ground. If you put one blue and one green card into the slots next to the door, the door will open. If you insert two cards of the same color, you will lose a life! I won't tell you how many lives you have, but you will die if you lose all of them!
For most people, this could be completed easily. Unfortunately, you cannot distinguish the color of the cards because you are fully colorblind! Whoever put you there probably knows this. They undoubtedly gave you just enough lives for you to be certain you can get out - if you play their game optimally.
How many tries do you need to guarantee you open the door?
The general strategy I'll be using is just brute-forcing a single card against all others. Pick a card, this'll be the "master" card you'll always use. Then you simply test the master card with another card up to 66 times. If it doesn't work, than you toss out the non-master card, and repeat.
As there are an equal number of green and blue cards (66), we can solve it the same regardless of the first card you chose.
So say your first card is green, then the worst case is that you fail your first 65 tries by getting 65 greens in a row. Then on the 66th try you're guaranteed to have a matching blue card since there are no green cards remaining besides the one in your hand.
The argument applies in exactly the same way if you chose a blue card first, just with blue cards instead.
"How often do you have to break a chocolate apart to end up with 24 pieces?" - The answer is 23 because every act of breaking - no matter how - increases the number of pieces by exactly one
Here in reverse:
Any failed attempt with two cards in effect only tells you that these two cards have the same colour. As they are henceforth equivalent, you can as well dispose one of them - or to stay in the spirit of the chocolate puzzle, glue them together to a larger monochromatic card cluster.
The only move that guarantees that the door opens is when you would upon failure produce a card cluster of 67 or more cards (as a cluster of this size cannot be monochromatic). But producing a cluster of size 67 is easily possible in many different ways with 66 moves, but definitely not with less than 66 moves (cf. chocolate puzzle)
A remark as a thinking outside the box solution:
Most cases of colour blindness are red-green blindness and as far as I know can distinguish blue from green ;)
34. I would try 33 times with pairs of cards that I haven't tried before. Either at least one pair will one green and one blue card, or all the 66 cards that have been tried are of the same color. For the 34th try, I would take one card from the 66 that I used and on from the 66 cards that I haven't used, making sure that at least one of my tries contains two cards of different colors.
the minimum number of tries to open the door is 3. Pick 3 cards. Worst case scenario is: you pick green and blue for the first attempt; then you pick one of the first and test it against the third, which fails. Then, you know that the third card will match the card unpicked on the second attempt and you can open the door.
It is not guaranteed that you can open the door, because there is a chance that you will always pick different colors for the 66 attempts.