Hello all,
in column A, I have 6 items with drop chances. But, a drop drops 3 items, based on their percentage.
In cell B, I have how many items I need.
I want in cell C to calculate, how many tries its going to take me to get an X amount of item 1 for example.
To provide an example:
Item 1 has a drop chance of 16%. A single drop drops 3 items however (Item 1-Item5, based on percentage)
I need 5 items 1.
How many tries is it statistically going to take me to get the 5 items 1?
Cheers
Last edited by metal.stuff; 06-17-2016 at 05:28 AM.
Wouldn't it be:
=B1/(3*A1)
If item one is at 16.6667% and you get three shots at it per drop, then each drop gets you a 50% chance (16.67*3) of landing an item 1. You would therefore need 10 drops, on average, to get 5 of item 1 (5/50%). Does that logic fit your scenario?
Thanks for the reply, but Im afraid that is not the case. 3 shots at a 16% drop does not equal (3*16%). I found that out here http://wowwiki.wikia.com/wiki/Chance...t_desired_drop
And then, if someone is really good at math here, i would need to extend the question. Imagine you get a chest which drops 3 out of the 6 items. The percentage changes, because once you know the first of the 3 drops, the other two have different percentages, (because first item is 1/6, second is 1/5, third one is 1/4). Following this logic, when will i get 5 pieces of item 1.
Hope someone knows math here better then me
Last edited by metal.stuff; 06-17-2016 at 09:57 AM.
Interesting link, but I'm not entirely sure that formula is measuring the same thing you're looking for. It seems to me that the formula in your link is measuring the odds of getting ONE in X number of tries, whereas my formula (and your request?) appear to be looking at the expected return from a given number of tries. If you think of coin flipping, the two questions: "What are the odds of getting at least one 'Heads' in two flips?" and "How many 'heads' can I expect, on average, from two flips?" ...are completely different questions.
"What are the odds of getting at least one 'Heads' in two flips?" - is a question for the formula in your link, which correctly returns 75%
"How many 'heads' can I expect, on average, from two flips?" - is an expected value question which returns an answer of 1.
Are you certain that the formula in the link is the one that best applies to your scenario?
To follow up with your example, the odds of getting at least one Item1 in a drop would be 1-.84^3=40.7%, but your expected value from a drop would be .48 Item1s (Expected value is higher because you may get 2 item ones amongst the 3 dropped items).
If the game is designed such that you can't get two of the same item in a drop, then we would need to combine the two formulas and go with:
=B1/(1-(1-A1)^3)
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