I don't think that's completly correct. But let me ask for clarification; Are you wanting to calculate the odd of getting AT LEAST ONE leggo, given the probability of getting a leggo at each independent event? If so that is a simple calculation.
If you open 12 shards with a 12% chance of obtaing a legendary, the odd of at least one legendary are;
Odds of pulling a leggo from any single shard = 0.12
Odds of NOT pulling a leggo from any single shard = 1 - 0.12 = 0.88
Odds of not pulling a leggo from at least one of x (12) number of shards = 0.88^x = 0.88^12 = 0.216.
Odds of pulling at least one leggo from 12 shards with each independent event having a 0.12 odd of pulling a leggo = 1 - odds of no leggo = 1 - 0.216 = 0.784 or 78% chance of at least one leggo from 12 shards during a 2x event. Pretty good odds. If you do get a leggo it could be on the first, the twelth, or any other shard. But 1 in 4 on average will not get a leggo, about 1 in 4 may get 2 or more leggos.
That part is pretty straight forward. But there are a few complicatons. First, are all 12 shards pulled during a 2x? If so this should hold. But if you pull 3 during no even, 2 during a 2x, 2 during a 10x, and the rest during a 2x, then the shards no longer have the same probability so the formula will not hold.
The bigger complicaction is the mercy system, once you begin adding 2% to the odds with each succesive event, the events are no longer independed. You have the problem above compounded by lack of independence. The formulas here are not straight forward and would require more time than I'm willing to put into it.