So in this particular game, a drop rate of a particular item is 1/384. There are 3 unique items with this drop rate.
I would assume this would mean the chances of receiving any drop is 3/384 or 1/128.
This is where it gets tricky. If I wanted to calculate the odds of something dropping I get confused as to what I would do. Say, I repeated this process 100 times, meaning that I've hit this chance of 1/128 100 times, would the probability of me getting this item be 111/128 (symbolizing 100 attempts)? Or is this a gambler's fallacy and my odds are still 1/128?
Let's say I got two items back to back, so I luckily hit this 1/128 twice. Would my odds be (1/128)^2 or would it be different?
Like let's say I got a drop after 2 kills. then another drop 3 kills later, would my odds of that happening be (2/128)*(3/128)?
>>8137578
>I would assume this would mean the chances of receiving any drop is 3/384 or 1/128.
Depends if the odds are independent - i.e. does one of the items dropping impact the odds of the other two items dropping from the same kill.
>Or is this a gambler's fallacy and my odds are still 1/128?
Correct. If you want to calculate the aggregate chance over the 100 attempts that at least one drop happened, however, I believe you can take the inverse probability 127/128 of nothing happening and raise that fraction to the power of 100 attempts, resulting in approximately 45% chance of no drops over that period, which means approximately 55% chance of at least one drop over that period. Calculating the odds of more detailed outcomes would add a lot of complexity.
>Let's say I got two items back to back, so I luckily hit this 1/128 twice. Would my odds be (1/128)^2 or would it be different?
Believe so but I'm rusty.
>Like let's say I got a drop after 2 kills. then another drop 3 kills later, would my odds of that happening be (2/128)*(3/128)?
Again, I'm rusty, but I suppose using the above technique, (1 - (127/128)^3) * (1 - (127/128)^2) = 0.7%. Someone else can probably correct that.
>>8137578
>Say, I repeated this process 100 times, meaning that I've hit this chance of 1/128 100 times, would the probability of me getting this item
36,932305226788006779354614874355%
>>8137578
If you wan't to find the odds of getting at least one drop in 100 attempts you start by calculating the odds of not getting anything after 100 attempts and subtract that from one.
1 - (127/128)^100 = 54%
>>8137603
Didn't he ask for the exact probability of getting it once first - "meaning that I've hit this chance of 1/128 100 times, would the probability of me getting this item be 111/128 (symbolizing 100 attempts)? Or is this a gambler's fallacy and my odds are still 1/128?" and only there for more than one - "Let's say I got two items back to back, so I luckily hit this 1/128 twice. Would my odds be (1/128)^2 or would it be different?"
>>8137608
>Didn't he ask for the exact probability of getting it once first
111/128 is a significantly worse attempted estimate for the odds of exactly one drop than the odds of at least one drop, so it's not too unfair to presume he was asking the odds of at least one drop.
Op Here. There are no consecutive odds in the game. And this was misworded, I meant that I haven't hit the drop table for a drop, I've just killed this boss 100 times symbolizing I had done it 100 times. Is my 101st kill odds of getting a drop equal to my first kill odds of getting a drop?
>>8137616
yea, if you flip a quarter 100 times and get tails, the probability that the 101th would be heads is still 1/2
>>8137616
The 101'st attempt should be the same as the 1'st attempt. If you wanted to calculate the probability of the 101'st attempt being your first success, refer to the geometric distribution
(p)(1-p)^(k-1)
where p is chance of success on any given attempt and k is number of trials, in this case, 101.
>>8137578
In your example, say you're trying to find the odds of receiving at least 1 drop in 100 attempts, refer to the binomial distribution and subtract the odds of receiving no drops from the sum of all probabilities, 1.
