We’ve killed rag 19 times and haven’t seen a weapon.
Heck. I’ve lost a half dozen 80%+ chance hands on the river in a Holdem session. It totally can and does happen all the time.
Drop rates are calculated over thousands if not millions of kills. Streaks and variance are very likely so 100 kills with nothing isn’t even odd.
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He’s not saying it’s due to happen, he’s using math to show the probability that a streak of this kind would happen.
It’s more likely to flip a coin 100 times and get tails every time than what he is claiming happened. Yes, it is possible and yes every flip is 50%, but 100 in a row being tails is ridiculous to the point of impossibility.
Edit:, My comment about coins is for the OP and his 100 runs with a 33% chance on Reed, not the MC example.
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this is why gambling is the devil, people don’t grasp this concept.
like flipping a quarter. if it lands on heads 10 times in a row, it doesn’t have a higher chance to land on tails the next time.
I lost 6 pairs of BIS pants to sub 60s I was carrying, I refused to be in a group with any sub 60s after that and haven’t seen them drop ever again.
Well. It wouldn’t be less likely because a coin is 50:50. Coins aren’t three sided. Every single coin flip is a universe unto itself independent of every flip before and after. You can have 100 flips be all tails. It would mean you are extremely unlucky, but it can happen. Gambler’s fallacy. Doubling up strategies fail because of this. Variance will kill you without an infinite bankroll due to this.
I’ve only killed Jed twice, he dropped it both times, i won it the 2nd time and have never needed to do UBRS since, i also have burst of knowledge to pair it with, i got that the first time i killed that boss.
Exactly. There is never a guarantee it will be heads. Just the same 50% chance. This is why people go home broke. Variance.
my pally took 124 runs just to SEE HOJ the first time
my warrior on 7th , 18th , 31st , and 44th seen hoj.
I said more likely to flip a coin 100 times…which is correct.
I don’t see where the argument is at this point. We both agree that it’s 50% every time, but I’m talking about a set of events, 100 coin flips. And the OP and other poster are talking about a set of events, 100 Jed runs with no Reed.
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Which is entirely possible as each event is the same 33% independent of every single event before and after. You are just as likely to see it after one run or 1000 runs. Same 33%. You are never guaranteed to see it 33 times in 100 runs. It could be 100 times or 0. Just as likely. If a million people run it, it will drop 330,000 times, but one person may have seen 0 and someone else 1000.
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You are never ever guaranteed to see it in your entire life even if you run it a billion times. This is absolutely true.
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You are guaranteed to see it 33% of the time if you run it an infinite number of times though 
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Yep. Since no one can ever run it infinitely, you can be the one unlucky schmuck who never sees it drop.
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You really don’t understand the concept of variance. You would be a terrible gambler.
Google “Variance in gambling” the first article is a great description of why your argument is extremely faulty. Farming drops is basically gambling hence why it is so addictive.
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One more time and I give up.
We’re talking about the probability of a particular set of results from a sequence of events. Yes it is just as likely to get heads every other time in 100 as it is to get all tails. All outcomes of a set of 100 coin flips are just as likely or unlikely as any other. But I don’t care about the other possibilities. I’m talking about this one particular possibility of 100 tails in a row. Or 100 runs of Jed without a Reed.
I’m not suggesting that doubling down on black over and over is a good strat. I’m saying that it is close enough to impossible for black to get hit 100 times in a row as to make it meaningless. That doesn’t mean I’m going to bet red on the 100th time thinking , “This time it will hit”.
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We’re NOT talking about a single drop. We’re talking about a particular seqence of drops.
That’s where your confusion is.
For a coin flip, your odds of getting a head is 0.5.
For two coin flips, your chance of getting two heads is 0.5 * 0.5 = 0.5² = 0.25.
The reason for that is branching. There are four possible outcomes of two coin flips:
| flip 1 | flip 2 |
|---|---|
| heads | heads |
| heads | tails |
| tails | heads |
| tails | tails |
Since the probability of getting heads is the same as the probability of getting tails, each sequence has the same probability as the others: 0.25. Note that there are two ways to get one of each, but only one way to get a pair of tails.
If you add another flip, the number of sequences doubles again.
| flip 1 | flip 2 | flip 3 |
|---|---|---|
| heads | heads | heads |
| heads | heads | tails |
| heads | tails | heads |
| heads | tails | tails |
| tails | heads | heads |
| tails | heads | tails |
| tails | tails | heads |
| tails | tails | tails |
So the probability of one sequence is 1 in 8 or 0.125.
Note that there are multiple ways of getting 2 tails and one heads (for example), but only one way to get all tails.
Also note that this doesn’t predict what the next flip will be. Only that you’ll only get a particular sequence 1 in 8 times you run the experiment.
You can express this in general with Bernoulli trials, but for the simple case of “a streak of N events of probability P” it simplifies down to a probability of P ^ N.
The probability of NOT getting a Briarwood Reed from Jed is 1 – 0.3 = 0.7.
The probability of not getting a Briarwood reed for three Jed kills in a row is 0.7 ^ 3 = 0.343. Meaning you should expect that to happen about one third of the time.
The probability of not getting a Briarwood Reed after 10 kills is 0.7 ^ 10 = 0.028. Meaning you should expect this streak to happen about one in 35 sets of 10 kills.
Again, if you’re on kill 6 without a reed and want to know what the probability of the next kill dropping one is, it’s still 0.3. You don’t yet know if the current set of ten kills is a streak without any reeds.
And predicting kill 11 after a streak is the same way. There’s still a 0.3 probability of seeing the reed.
If you kill Jed 10 times without seeing a reed and THEN ask “what’s the probability of the next 5 kills ALSO having no reed”… it’s still 0.7 ^ 5 = 0.168. The past streak isn’t part of the calculation.
But if you ask “what’s the probability of seeing a streak of 15 with no reed” that is calculated as 0.7 ^ 15 = 0.0047 (one in about 211). Unsurprisingly, this is the same as saying “what’s the probability of seeing a streak of 10 followed by a streak of 5” which is (0.7 ^ 10) × (0.7 ^ 5) = (0.7 ^ 15) = 0.0047.
I hope that makes sense. Statistics is often quite difficult to explain.
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Red and black opens up a whole can of worms as people always forget 0 and 00 which really screw your odds in the favor of the house.
Just saying as a long time poker, craps and blackjack player that 100 losses with relatively attainable odds comes as no surprise to me at all. You seem shocked someone could lose a 33% chance 100 times in a row. That is not only possible, but virtually unavoidable based on how independent events work.
You can’t take anything as a set of events. It is always 33% whether it is your first kill or millionth kill. Still 33%. Nothing can change that. Repetition is no guarantee of eventual success.
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