the actual positions of the cards are 100% chosen by the casino/gameshow... so there isnt random weighting involved- each could be as equally likely as the next

I think we are agreeing loudly on this point. The positions of the colors are random [unknown to us] and the combinations of so many reds and so many greens --

it doesn't matter that we used different numbers -- permit an enumeration of all possible,

equally likely distributions.What's not 50-50 is the chance that any one particular cup is

red or

green.

Count the number of times in my 26 equally likely color distributions that

the first cup [for example] is

red. It's 15. Not 13. --

or is green. It's 11. Not 13.

Next, you see that in

7 of the 15 red-removed cases any 2 of the remaining 4 cups have the same color.

In

5 of the 11 green-removed cases any 2 of the remaining 4 cups have the same color.

If all you are saying is that

7/15 is different from

5/11, then the discussion is over and we can get some sleep.

[Perhaps erroneously] I took your Case 2 question to mean in

**all **cases when a cup is picked and removed from the group

[not a

**particular **case, like a red cup is picked first] does your chance of picking two matching colors

differ from your chance of picking two matching colors when all 5 cups are there.

My answer to that question is that the chances do not change.

This follows from the observation you can make by inspecting the 26 equally likely distributions of 2 colors among 5 cups.

Namely, that

whatever two cups you inspect, 12/26 of the time they will be the same color.

That statement is true whether

any or all of the

other three cups were removed initially or not.

That is, there are no circumstances that force you to pick 2 particular cups that do

**not** have a 12/26 chance of having the same color,

simply because

**every** pair of cups in the 26 equally likely cases has a 12/26 chance of a color match.

Finally, it appears that if you eliminate the 5-red case [requiring at least 1 green]

you will decrease the total distributions from 26 to 25,

and the colors-will-match distributions from 12 to 11.

That would change the answer for both Case 1 and Case 2 to 11/25 = 0.44.