A deck of cards contains $26$ black and $13$ red cards, taking out the cards in a particular way, what is the probability the last card left is black

Question

You have a deck of cards containing 26 black and 13 red cards. You pull out 2 cards at the same time and check their color. If both cards are the same color, then a black card is added to the deck. However, if the cards are of different colors, then a red card is used to replace them. Once the cards are taken out of the deck, they are not returned to the deck, and thus the number of cards in the deck keeps reducing. What is the probability the last card left in the deck is black?

my attempt: for both cards are the same colors: (both black)+(both red)

((26/39)(25/38)) + ((13/39)(12/38))=0.5438

the cards are of different colors: (black +red)+(red +black)

((26/39)(13/38))+((13/39)(26/38))=0.4561

please let me know how to continue for the answer for: What is the likelihood the last card left in the deck is black?

Answer 1

Hint: At the end of a turn, the number of red cards is always odd.

Answer 2

The probability of the last card being black is:

0. The last card will always be red.

This is because

The only way to remove any colour is either in pairs or with another colour. If you ignore the mechanics and focus on removing red cards entirely, you can dwindle the red cards down to a single red card by replacing the first 12 with 6 blacks. At this point, no matter what happens, any red that gets drawn will also be replaced with another red. Your final position will therefore always be one red and one black, which will be substituted with a final red.

Alternatively

If you run the black cards down by replacing them in pairs with a single one, you'll end up with one black and the 13 remaining reds. Replacing a pair of each leaves you with 13 reds, which you can then only replace in pairs, finally leaving you with a pile of blacks and a single red. Ultimately this will again reduce to one red and one black, resulting in a final red.

Finally

If you follow the mechanics as stated, no matter how the black cards play out, any reds removed will either be removed in pairs - and replaced with a black - or replaced by another red in a 1:1 ratio. Either way, you'll ultimately end up with one red and one black at the end.