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[Implemented] Blue Shards


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Well, I don't know what game you play, kitkat, but I got no blue shard in 80 encounters. Neither did many of my friends/teammates in an hour span. This is not how grinding should be. It's absurd.

 

UPDATE: Took 110 wild encounters to get my first Blue Shard. My thief Quag is PP maxed (40), so I took 2 trips to the PC and... counted.

 

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Edited by OldKeith
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Well, I don't know what game you play, kitkat, but I got no blue shard in 80 encounters. Neither did many of my friends/teammates in an hour span. This is not how grinding should be. It's absurd.

 

UPDATE: Took 110 wild encounters to get my first Blue Shard. My thief Quag is PP maxed (40), so I took 2 trips to the PC and... counted.

 

Blue shards find me

[spoiler][spoiler][spoiler] kitkat [/spoiler][/spoiler][/spoiler]

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You just had bad luck

 

No. The encounter rate for clamperl is 30%, and I believe blue shard has 5% chance of appearing, giving an expected ~67 encounters between each blue shard. If I'm being generous, you will have about 80 encounters every hour underwater, due to the slow speed and the time it takes to use thief. Thus you are expected to spend two and a half hours on getting three blue shards, which is enough for a single move. Oh, and need I remind you that every single elemental punch (perhaps the most common tutor moves in the game) requires three blue shards? For a pokemon with two elemental punches you are expected to spend five hours to complete half of it's moveset. No bad luck here, just bad gameplay.

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No. The encounter rate for clamperl is 30%, and I believe blue shard has 5% chance of appearing

 

You're right on the encounter rate, but the held item rate is doubled - which is a rather significant difference.

 

 

I'd also like to note:

 

Yellow shard species have a collective 65% encounter rate, they hold the yellow shard at 5%.

 

The suggested 25% would be a ridiculous buff, the Green shard species currently has a 25% hold rate and they have a 5% encounter rate.

 

The rates are designed to scale in accordance to encounter rates.

 

 

I believe that the problem here is not the rate of the blue shards, but the focus put on them.

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You're right on the encounter rate, but the held item rate is doubled - which is a rather significant difference.

 

Okay my bad, divide the expected time by 2.

 

I believe that the problem here is not the rate of the blue shards, but the focus put on them.

 

Well, there was a suggestion to solve this, which was appearently ignored by everybody:

https://forums.pokemmo.eu/index.php?/topic/50427-redistributing-e-punches/

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Okay my bad, divide the expected time by 2.

 

Probability works differently to that; you're not adding the previous 10% onto the next one, and then eventually you hit an 100% - you're rolling a 10% each and every time.

Eventually however, you have rolled so many times that it's improbable not to have hit it.

 

Which is not the same as a x2 modifier.

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Probability works differently to that; you're not adding the previous 10% onto the next one, and then eventually you hit an 100% - you're rolling a 10% each and every time.

Eventually however, you have rolled so many times that it's improbable not to have hit it.

 

This scales differently to just a x2 modifier.

 

Well, I was talking about the expected value, not the probability (i.e. it does scale up to 2x modifier). In this case, the probability of encountering a clamperl with a blue shard is (3/10) * (1/10) = (3/100), so you are expected to get 3 blue shards per 100 encounters, or 1 blue shard per ~33 encounters. By the previous assumption that you will encounter 80 pokemon per hours if you thief them all, you are expected to spend 100/80 hours, or 75 minutes, to find three blue shards. This happens to be exactly half of the previously estimated time (two and a half hours.) Assuming you want more than just three shards, you will probably be hunting for a while, and as time increases, the average time of collecting shards will converge towards the expected value.

 

By the way, I'm doing a master's degree in mathematics (albeit not statistics), but I'm sorry if I was not clear on the terminology in my first post.

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Well, I was talking about the expected value, not the probability (i.e. it does scale up to 2x modifier). In this case, the probability of encountering a clamperl with a blue shard is (3/10) * (1/10) = (3/100), so you are expected to get 3 blue shards per 100 encounters, or 1 blue shard per ~33 encounters. By the previous assumption that you will encounter 80 pokemon per hours if you thief them all, you are expected to spend 100/80 hours, or 75 minutes, to find three blue shards. This happens to be exactly half of the previously estimated time (two and a half hours.) Assuming you want more than just three shards, you will probably be hunting for a while, and as time increases, the average time of collecting shards will converge towards the expected value.

 

By the way, I'm doing a master's degree in mathematics (albeit not statistics), but I'm sorry if I was not clear on the terminology in my first post.

 

 

Hmm, not so sure about that.

 

By that logic you're directly implying that each 10 encounters you will have 3 clamperls, and for each 10 clamperl you encounter 1 of them will have a blue shard.

 

The problem is that probability works a bit differently. 

 

For example. a 30% encounter rate obviously means that you have a 30% chance of encountering a clamperl.

Which means that the chance of not encountering clamperl is 70%.

 

 

 

The opposite or complement of an event A is the event [not A] (that is, the event of A not occurring); its probability is given by P(not A) = 1 − P(A). As an example, the chance of not rolling a six on a six-sided die is1 – (chance of rolling a six) 2533e9c427bbaf1ddab37b798c3e8e0b.png

 

On top of that, If two events, A and B are independent then the joint probability is eb53a4b561f5b841683829dc4f9df598.png

for example, if two coins are flipped the chance of both being heads is 9a2da4fb1306ffce7f5ba28bc4f3d733.png.

 

So let's take 3 encounters. If you want to calculate what the chance is of encountering at least 1 clamperl in 3 encounters it's easier to start from the chance of not encountering a clamperl (you either get 3 times clamperl, 2 times clamperl, 1 time clamperl or no clamperl, so you can see why somebody would start from not getting a clamperl and doing 1 minus that chance so then you will have every instance of getting at least 1 clamperl).   Calculating the chance of not encountering a clamperl in 3 encounters is by that logic (7/10) x (7/10) x (7/10) = 34,3%. So by the above formula: 1 minus that chance is the chance of getting at least 1 clamperl in 3 encounters which is 1 - 34,3% = 65,7%

 

Applying that to the logic: The chance of getting at least 1 blue shard in 100 encounters.

Chance of getting a blue shard in 1 encounter is like you said (1/10) x (3/10) = 3/100

But that doesn't mean you get 3 blue shards in 100 encounters.

It goes like this:

The chance of not getting a blue shard in 1 encounter is 97/100.

The chance of not getting a blue shard in 2 encounters is (97/100) x (97/100)

etc.

So the chance of not getting a blue shard in 100 encounters is (97/100) x (97/100) x (97/100) x...... etc 100 times in total or in short (97/100)^100 = 0,04755 or 4,755 %

That means that the chance of getting at least 1 blue shard in 100 encounters is 1-0,04755 = 95,245%

You have to interpret 3/100 not as 3 blue shards in 100 encounters, but merely a 3% chance of getting a blue shard in 1 encounter. There is no certainty in probability unless the probability becomes 1.

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You guys are both right, except you're just evaluating different things.
One is the expected value (Google it if you don't know what it means)
One is the probability over 100 encounters

Expected value is probably more reliable

 

Thank you for clarifying, although it is above me how my last post could have possibly been misinterpreted.

 

The reason why I'm using the expected value is that we are working with a binomial distribution here. As the sample size n increases, a binomial distribution with expected value x will converge towards a normal distribution with mean value x, i.e. it is safe to assume that we have a normal distribution around the expected value. While it is true that this approximation is dependent on n to be large, we will generally assume that players will encounter houndreds of pokemon, so the approximation will be very accurate. So, why does this matter? If we know anything about the normal distribution, it's that the probability of the value being close to the mean value is very high. I did nowhere say that you are going to find three blue shards per 100 enounters, but the probability of finding close to three blue shards is very high. I'm not going to bother with finding my statistic tables just to make an example, but the underlying point here as that if you are expected to spend 75 minutes on hunting for three blue shards (in other words, the expected value for the amount of time it takes to find three blue shards is 75 minutes), it is a very high probability that it will take you close to 75 minutes - and in the long run, the average time you have spent on hunting for 3 blue shards will be very close to 75 minutes. (If I had an actual example, I would show the probability of finding three blue shards within 65 to 85 minutes, and I would expect it to be somewhere around 75% to 95%.)

EDIT: Just read the changelog, I think this suggestion can be put on hold now.

Edited by PandaJJ
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  • 1 year later...
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