A bit of story
I've returned to Black Desert after a long time and now, as I've learned some math in my free time so... I'm wondering about some stuff from math point of view. Like how many Items and money (How much time) I have to waste to fully upgrade my gear.
There are websites that calculate this, but I'm more interested in math RN.
To the problem.
- An item has a chance to be upgraded, the chance to upgrade to level 1 is 70%, level 2—60% level 3—40% and from now on each level up to level 10 will divide the chance by two rounded to 2 decimal places. What's middle amount of upgrades a player has to do to get fully upgraded item.
My guess for this is
Let's say that $P(i)={P(1)=0.7,...,P(10)=0.0031}, i \in [1,10]$ so we use binomial distribution $(B(i)=(_x^n){P(i)}^x(1-p)^{n-x}$ and find at what $n$(amount of tries) we have the highest $B(i)$ and sum that up
Roughly I got 634 using excel
One more way to do this is use $1-(1-P(i))^n$ and find at what $n$ we have roughly 1 (same 4 digits after the comma) sum that up
And I got 13688 which is a LOT bigger than 634.
EDIT: This is wrong, thanks to David K we can say that this calculates amount of tries for probability to exceed 99.995
And one more way to do this is to find mathematical expectation which is roughly $M(i)=\frac{1}{P(i)}$ rounded down and summed up $\begin{align*} \sum_{i=1}^{10}\frac{1}{P(i)} \end{align*}$
This gives me 639 which looks a lot more truthful judging by first way.
- An item has a chance to be upgraded, the chance to upgrade to level 1 is 70%, level 2—60% level 3—40% and from now on each level up to level 10 will divide the chance by two rounded to 2 decimal places. If upgrade failed, you lose 1 level. What's middle amount of upgrades a player has to do to get fully upgraded item.
Absolutely no Idea how to solve this.
Questions
Why 2 way of solving this gives me such an obscure answer, am I missing something?
Did I understand binomial distribution right?
Is there any other way to solve 1st one?
Is there a way to use BINOM.INV (EXCEL Function) to solve the 1st one? I personally think that with something like BINOM.INV(n;P(i);0,9999) is right, but I really don't know if this is right.
What about minimum and maximum amount of tries?
How to solve the second one?
What about minimum and maximum amount of tries? (not necessary, I'll probably understand it myself, but I already spent so much time at this I'm getting frustrated)
For the first question, note that if $T_i$ denotes the random time upgrade from level $i-1$ to $i$, then $T_i$ is a geometric random variable with success probability $p_i$. A geometric random variable counts the number of trials to the first success and the mean of a geometric random variable with success parameter $p$ is given by $1/p$. If we denote the total time to upgrade as $T$, then \begin{align*} T = \sum_{i = 1}^{10} T_i \implies \mathbb{E}[T] = \sum_{i = 1}^{10}\mathbb{E}[T_i] = \sum_{i = 1}^{10}\frac{1}{p_i}, \end{align*} which is what you believed your answer to be.
For the second part, let $t_i$ denote the expected time upgrade to level $10$ starting from level $i$. Note that $t_i$'s satisfy the following relation for $i = 0, 1, 2, \dots, 9$: \begin{align*} t_i = p_i t_{i+1} + (1 - p_i)t_{i-1} + 1, \end{align*} with $t_{10} = 0$ and $t_{-1} = t_0$ (since you cannot go lower than $0$). You can solve this system of linear equations to obtain $t_0$, which is your required answer.