I did get a lot of Assassin's Vambraces in my last game, I mean like three or four of the darned things in the first 100 turns. Random number generator issues are notoriously hard to prove though, people forget that rolling 1, 1, 1 on three dice is not impossible, just unlikely. 1/216 is not even close to "never".

I would be curious to understand the mechanics for loot drops in Legendary Heroes, you do seem to get a lot of the same stuff sometimes. It's possible something is pre-seeded, or the same random number affects a lot of different results. It may just be that there is a relatively small list of "Uncommon armour", so if that was the loot drop you kept getting then chances of a duplicate or three are quite high.

This has been a problem for me for several versions now.  I'm still on 1.4, but I find that quite often in a game I will get 3-5 copies of a random item.  I don't know if that's because the pool of available items is so small, or if it's a seed issue with the RNG, but it's persistent enough from game to game that I'd like to see it get addressed.

gs

Agreed with Primal_Savage. I've noticed this too. Each play session is a different item, but I get multiple copies of it to the point where something doesn't feel right about the randomizer.

Major issue imo.  Landmarks and monsters also tend to be distributed in pairs.

Probability is confusing. If I have three dice, the chances of rolling a 1, a 1, and then a 2 are 1/216. But if I don't care about the order as long as I get a 1, a 1 and a 2 (so in other words 112 and 121 and 211 are also valid) then the chances treble to 3/216.

If there are 19 Uncommon armour drops and I have two drops, what are the chances both will be Assassin's Vambraces? 1/361. But if I have three drops, what are the chances at least two will be Assassin's Vambraces? 

19 / 19/ any

19/ any/ 19

any/ 19/ 19 are all valid.

In other words, (1/19 * 1/19 * 1/1) + (1/19 * 1/1 * 1/19) + (1/1 * 1/19 * 1/19). So my chances are actually 3/ 361 (assuming my maths is correct.).

If I have four drops, what are the chances at least two will be Assassin's Vambraces?

19 / 19/ any/ any

19/ any/ 19/ any

any/ 19/ 19 / any

any/ 19/ any/ 19

19/ any/ any/ 19

any/ any/ 19/ 19

So I reckon that's 6/361. Still doesn't seem very likely. But we've been looking for a specific item; what are the chances of any item showing up at least twice in four drops?

In other words it doesn't matter what the first drop is; what we're interested in is whether there will be a duplicate in the next three drops.

What are the chances there will be no duplicate in the next three drops? On the second drop there are 18 items it can be which are not duplicates, then there are 17 items it can be which are not duplicates, and finally there are 16 items which are not duplicates to any of the previous three items. (18/19)*(17/19)*(16/19)=4896/6859. So the chance that there will be a drop of two identical items is (6859 - 4896) / 6859, i.e. 1963 / 6859. Or to put it another way, about 29%. Doesn't seem so unlikely now, does it? In fact if I played two games my chance of getting two duplicate Uncommon armours in the first four drops in either game are now slightly better than half (1-(0.71*0.71), i.e. 51%.

I should put in a disclaimer that I am not a mathematician and I do make mistakes. However I'm fairly confident in my broad point that the chances of getting a duplicate in something over a certain number of drops are actually quite high if there were only 19 items to choose from to start with. I don't therefore think there is a problem with the random number generator. I think the solution (if it really bothers people that much) would be to add more items. If there were, say, 30 uncommon pieces of armour then I believe the chance of getting two duplicates over four drops would be about 10% rather than about 29%. Each of those items would need artwork and a description though, so quite a lot of work for something which is mainly an illusion.

I was basing my calculations on this. If I got four Uncommon armour drops, what are the chances two would be duplicates? The example given was actually any four out of twenty (when the sample is all uncommon items, i.e. 124), you can work that out but it's a lot harder. Intuitively I would actually expect four out of twenty to be more likely than two out of four, even with the increased sample size, but I'd have to work it out. Feel free to work out the maths for larger numbers!

But the point remains that you can get duplicates quite quickly even if the sample size is large. It's the birthday problem: from a sample of 365 days, how many "drops" do you have to get before you will probably have two people sharing the same birthday?

The answer is 23, which is far lower than most people guess.  https://en.wikipedia.org/wiki/Birthday_problem#Calculating_the_probability

You don't want a randomizer then, you want want a shuffle algorithm, which is entirely different.

I strongly disagree.  This would also greatly skew the rarity weighting factors in the items lists, since after each "find", the chance for every other type of item would be greatly boosted.  Personally, the idea that you'll only see certain items once in a blue moon is kind of appealing.

Besides, no one has actually confirmed a random number generator bug yet - it's merely suspected.

Given the relatively low number of items, I'm dubious whether there is a bug. This is exactly the same as the itunes shuffle "problem", and Apple came up with various solutions, including (for example) "without repetitions". "Without repetitions" is not random at all, but it may be more what people expect.

So two solutions spring to mind for legendary heroes:

1: add more items; 

2: add a world option for "no repetitions" for loot drops.

Erm. I can't remember maths I never had, like I said I'm not a mathematician. It looks a bit simple to me though. See;

http://math.stackexchange.com/questions/25876/probability-of-3-people-in-a-room-of-30-having-the-same-birthday

Given that one answer uses an approximation and the other refers to an academic paper, I don't think it's that simple a problem to solve.

One of the answers suggests that the chances of having three identical drops out of 100 drops from a universe of 365 is 70%, so your answer sounds low to me. If you actually understand what a Poisson approximation is then maybe you can use it to get the correct answer with our actual problem,  i.e. four identical drops out of 20 with a universe of 124 (or five identical drops out of 100 with a universe of 124).

Yes, I think that's what I was trying to get at. It's the point that in my example above, assuming my maths is correct the chance of getting two Assassin's Vambraces (i.e. one specified piece of armour) in four drops is only 6/361. However the chances of getting any two identical items in four drops is more like 29%.

With many drops and a specified number of duplicates, the problem gets very hard, but the principle remains that it's higher than people generally think. You are looking at the end result, e.g. four Assassin's Vambraces out of eight drops, but you would have found it equally odd if the game gave you four Weathered Shields (or whatever) out of eight drops. So it's the example where we're looking for any combination of four, we're not looking specifically for the example where we got four Assassin's Vambraces.

For five out of 100 you have the combination

1, 1, 1, 1, 1, (95 other numbers)

1, any, 1, 1, 1, 1 (95 other numbers) etc.

And then

any, 1, 1, 1, 1, 1, (94 other numbers)

etc.

"At least" is important, because if you calculate the chances of getting exactly four more drops of the first item in the next 99 drops you will get too low a probability. Getting 100 identical drops is a combination which matches "at least four matching the first drop".

It might be possible to write a computer program to do a brute force calculation of the combinations, assuming you could specify the problem correctly.

The web page which uses a Poisson approximation states that for three people out of 25 to share a birthday the chance is about 3%, but for three people out of 100 to share a birthday the chance is about 70%. What I take from this is that as you increase the number of drops the number of possible combinations increases massively; so if you increase the number of drops from 25 to 100, the chance of three duplicates is not quadruple, it goes up by 23 times.

If you apply this to the game, with a universe of 124 rather than 365, then I personally expect the chances of getting five drops to be quite high, perhaps even over 50% for 100 drops.

I've had a stab working out the Poisson approximation with a friend, assuming our maths is correct I believe the chance to get five duplicates out of 124 items with 100 drops  is approximately 27%.

The chance of getting four identical drops out of 124 items with 20 drops is approximately 0.25%. The chance passes 50% after about 77 drops.

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