Slime Slayer: From Zero To Black Steel King - Ch. 2 - First drop

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The chance of something within 181 days was about 83.8%.
No no, thats not how gacha works. No matter how many times you try, the percentage won't change.

Unless the number of slime is already decided like, out of 100 slime's there only 1 win slime, then the chance will go up on each fail.
 
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No no, thats not how gacha works. No matter how many times you try, the percentage won't change.

Unless the number of slime is already decided like, out of 100 slime's there only 1 win slime, then the chance will go up on each fail.
They are talking about the probability over 181 attempts rather than each individual attempt, but I think their math isn't exactly right. The odds of failing 181 times in a row is 16.22%, and 1 - .1622 would be 83.78%, but this isn't the chance of the next attempt being a success. That only works with the base probability.

The probability of 1 success over 181 attempts would be calculated with (n!/(n-x)!x!)(p^x)(q^n-x) where n is the number of attempts, x is the number of successes, p is the chance of success and q is the chance of failure.

This would be 181•.01•.1638=.2965, or 29.65%

This is still only the chance of 1 success in 181. It doesn't tell us the chance that the 181st one is a success, just that there has been 1 success until this point.

The chance that specifically the 181st is a success would be 1% like you said.
 
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FYI, this is going to be a very slow burn if things continue at this pace. The MC doesn't do much of anything until pretty far along.
 
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No no, thats not how gacha works. No matter how many times you try, the percentage won't change.

Unless the number of slime is already decided like, out of 100 slime's there only 1 win slime, then the chance will go up on each fail.
I said
within 181 days
Not “on the 181st day”.
I think their math isn't exactly right.
It's right to one hundredth of a percent. (Few people want more precision than that.)
The probability of 1 success over 181 attempts
At this point, you confuse one success with exactly one success, whereas I didn't refer to one as such at all. I wrote of
something within 181 days
The main character has no reason to prefer exactly one drop. That would be like asking “Is there a doctor in the house?” and hoping that there weren't two.

The logical complement of something within 181 days is nothing within 181 days. The odds of nothing within 181 days are .99^181, making the odds of something 1 - .99^181.
 
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They are talking about the probability over 181 attempts rather than each individual attempt, but I think their math isn't exactly right. The odds of failing 181 times in a row is 16.22%, and 1 - .1622 would be 83.78%, but this isn't the chance of the next attempt being a success. That only works with the base probability.

The probability of 1 success over 181 attempts would be calculated with (n!/(n-x)!x!)(p^x)(q^n-x) where n is the number of attempts, x is the number of successes, p is the chance of success and q is the chance of failure.

This would be 181•.01•.1638=.2965, or 29.65%

This is still only the chance of 1 success in 181. It doesn't tell us the chance that the 181st one is a success, just that there has been 1 success until this point.

The chance that specifically the 181st is a success would be 1% like you said.
welp, conditional probability is hard
 
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Yeah ... that's the childhood friend's fault for getting mad if MC actually gets a love interest other than her. Should have been honest. I wouldn't blame him for being mad.
 

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