You could pull 3 Ancients and get 3 Lyssandras....but you're far more likely to pull thousands and not get one. It'a all RNG except guaranteed leggo events, login rewards, and token trader, I think. 

I've pulled three Minayas over 22 months in the game, just from sheer blind luck (or the opposite of it :D)

This is a math problem but in this instance given Lys has bever been a guaranteed champ, etc. so the player you encountered was some combination of a spender and hit an unlikely outcome to some degree in pulling three.

trust me, you're barely scratching the surface. i've come across someone who had pulled no fewer than 11 kymars.

beating them was rather satisfying.

The answer I guess is 3 shards, in theory

😆

had a good chuckle at this, thank you 

I have 3 Noble, after almost 4 years, if that helps.

I am waiting for @kramaswamy.kr to post the probabilities.  

With RNG it could be 3 shards or thousands without ever getting a Lego.  

i

Then wait we shall. I will keep my statistics to myself.

(Although if you meant getting a specific leggo, you've been sneaky and solved the problem yourself in your second sentence there... not as satisfying as a percentage or a p-value perhaps, but it's the correct answer)

I'm not a maths expert, but I calculate it as this.

There are currently 194 legos in the game, including the new ones.

45 of these are voids, so discount those, and 14 of the others can't be summoned from any shards.

This leaves 135 possible legos to summon from ancient or sacred shards.

If you pull from sacred with a 6% lego chance and a 135-1 chance of getting and specific lego that would make it 2250-1 chance of pulling a lyssandra from any single sacred shard.

But that doesn't factor in the mercy system.

If you calculate on ancient shards with a 0.5% chance than it's 27000-1 chance.

So if someone had opened 6750 sacred shards (or 3375 on a double chance event) then it would be quite probable that they had 3 of her.

But if you want to win against RNG and get 3 lyssandra's from just 3 sacred shards, then the chances are 11390625000-1 of winning.

Ah, so if everyone on the planet played raid, half of one person would be able to get 3 consecutive of the same legendary. That's quite comforting to know haha :P

I think what may be missed in the above;

How many (y) shards does it take to pull x,  (meaningless)

and what are the odds of pulling x given y opportunities, are two very different questions.

it is impossible to say (predict) with accuracy how many shards one must open to recieve a given champ.

it is 100% predictable to say what the odds are of receiving a given champ on the next shard.

probabilities are predictable given a number of opportunities.  So I can never tell you with 100% accuracy how many times you will need to flip a nickel to get a head.  It may, might, could, possibly, etc, be 1, 2, 3, or more.  But given 1, 2, 3, or more flips, I can tell you the odds of getting 1, 2, or 3 heads.

It is always 50/50 on flipping a coin.  If there are 99 heads in a row  then the probabilities of tails increases.  

I don't mean to speak for @Angwil but this is incorrect and is the gambler's fallacy. Each coinflip is statistically independent of one another. 

It is necessary to differ between judging the overall sum of events (= coin flips) and the single event. 

Dthorne is correct, that the next SINGLE coin flip is 50%, regardless of the number you flipped before.

The probability of achiving a special event (=tail) within a certain number of flips can be calculated....but it is only a probability and not a certainty.

You state the odds are ALWAYS 50/50.

Then you state the odds change depending on previous results. 

Do you not see the incompatibility of your own statements?

Each result is independent of previous results.  Even if you flipped 100 heads, the odds on the next flip are 50/50.

This is sampling with replacement.  There is also a thing called sampling without replacement, where the odds can change based on previous results.  E.g.  If I put 10 numbered slips in a hat, numbered 1-10, and I pull out the 9, what are the odds that the next slip I pull out will be the 7?

it depends.  If I tore up and tossed out the 9, then the odds are 1 in 9 for any other number (0 for repeating the 9).  But if I replaced the 9, the odds of any number remain 1 in 10, including repeating the 9.

if I open a shard and pull duchess, she is not removed from the possibilities.  So the odds of repeating that result are exactly the same as they were before the first pull.  This is sampling with replacement.

Think of a deck of cards.  No matter the game if you see the ace of spades played,  you know the next card dealt cannot be the ace of spades.  The odds of dealing any of the remaining cards have change.  Sampling without replacement.

But if someone asks you to pull a card, any, card, look and put it back.  Then the odds of pulling the same card or any other have not changed.  Sampling with replacement.