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A bit of an oddball question.

I'm playing a game where I have a certain number of items, $n$ (in this case 9), and a certain action upgrades each item by one with a chance of $\rho$ (in this case, 33%). I'm trying to get all of the items to a target number of upgrades, $u$. Each item $i\in 0\ldots n$ also already has a certain number of upgrades, $u_i$.

I'm trying to calculate the expected number of upgrade actions $m$ (and the standard deviation from that expectation) that it should take to get them all to that threshold. Ideally, if possible, I'm also curious about finding the actual probability curve, too.

This started with me playing with Poisson statistics in my head. I know that the Poisson distribution is:

$$P(m) = \frac{\mu^m}{m!}e^{-\mu}$$

where $\mu$ is the average. I also know that, for any given item:

$$\mu_i=\frac{u-u_i}{\rho}$$

But the Poisson distribution has a key problem. It's nonzero for all numbers that could not actually reach $u$ upgrades. So I'm not sure if this is actually the correct approach at all.

...and I'm genuinely not sure where to go from here, when talking about all items considered together. Can anyone offer a pointer?

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    This looks like a generalization of the Coupon Collector’s problem in which the target number for each “coupon” can be different. – 2017-02-27
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    @amd I believe that's not the case, because with the coupon collector you are getting a single, random upgrade per action. If I understand correctly, here you are upgrading *all* items (each with an independent probability p) simultaneously every time you take the action. – 2017-02-27
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    @Anonymous That's correct - each causes an upgrade to apply to all items with independent probabilities. – 2017-02-27

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Once you have the probability curve for each item on its own, i.e. for each item $i$ you have the probability $P_i(a)$ that you will reach the desired "upgrade threshold" for that item with $a$ actions, you can obtain the probability that $a$ actions will suffice to upgrade all your items to the desired threshold as the product of all $P_i(a)$ (i.e. $\Pi_{i=1}^n P_i(a)=P_1(a)\cdot P_2(a)\cdot ... \cdot P_n(a)~$).

Note that in general this product might look really ugly and hard to interpret unless you accept some simplifications (yielding a looser bound, that tells you e.g. that with $10$ actions you have a probability at least $60\%$, when in fact it might be closer to $70\%$). Then again, if you have a computer crunching the numbers for you, it will just spit out the right answer with very little effort on your part.

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    ....huh, is it really that simple? Time to check... – 2017-02-27
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    If the individual upgrade probabilities are independent, then it really is that simple. – 2017-02-27