Your a little ambiguous about what your looking for and what exactly what happens if X triggers. I will go over each and what I think you mean:

Scenario 1: Each Round chance of Event X = P(x). If not Y triggers and thus P(y)=1-P(x). If X triggers it sets P(y) = 0.

For the average number of Ys (which your formula seemed to be going after):

The reasoning for the above is each round there is an 80% chance of the last round not rolling an X, and a 20% chance we do roll an X, thus setting Y to a fixed number. So on round 3 we have 80%(no round 1)*80% chance(no round2)*20%(we got an x this round)*2 (the number of previous Ys so far). With the final one just being the odds of no X ever being rolled*5. This generalizes into:

+

To find out the chance that y would trigger at all, or to put it another way. The chance that Y >=1 is simply 1 - chance that Y never trigers or:

Scenario 2: This is just scenario 1 with

Scenario 3: This is the opposite of Scenario 1. Each round X triggering is P(x) and Y triggering is 1-P(x), but each round is independant of the previous. Thus we need to count all the combinations of ways to get Ys. For example we must count XXYYX as well as YXYXX when counting up y=2. The binomial probability theorem hands this quite well. Here is the link:

Binomial probability - Wikipedia, the free encyclopedia

Assume that binomial probablity theorem is:

Then the average number of Ys is:

and the odds of Getting at least 1 Y are:

Scenario 4: This is just a modified case of Scenario 1 above. We just need to add one more variable, chanceZ. So the new formula for average looks almost the same, except note that P(y) is different:

+

You can determine the average number of Z's by substituting P(z) for P(y) in the above formula. Note for chanceZ =.50 (your case) Average Y == Average Z.

The chance of at y triggering (Y>=1) changes a little as well. Its still 1 - Chance that Y doesn't trigger. But each round Y not triggering is found by:

and thus:

is the chance of at least 1 Y.

Just for clarity your sample case above for Scenario 4 is specifically

and those are the only 2 variable needing to be specified to solve any case

Hope your still reading this thread and it helps.

Tony Chamberlain