I am trying to prove this problem:
Suppose we have a game where the player draws $d$ items from a bag. Each can be one of $t$ types, all with the same chance. (Example: Draw 4 items from the chest, and each item can be a sword, a bow, a shield, or piece of armor, each with a 1/4 chance). Calculate the expected number of swords that the player will get. (In this case, 4 draws with a 1/4 chance each, so the result will simply be 1)
I came up with a rather convoluted function to calculate the chance of getting n count of that item, times that by n to get the expected outcome, and sum it up all the way up to d
$$\sum_{n=0}^d\left(\frac{1}{t}\right)^n \times \left(\frac{t-1}{t}\right)^{d-n} \times \frac{d!}{(d-n)!n!} \times n$$
Alternatively, we can also think of it as each draw giving us a $\frac{1}{t}$ chance of getting that item, so simply
$$ \frac{1}{t} \times d$$
Will get us that result. I tested it and both methods yields the same result. The problem comes when I try to prove the first method. I'm not a guru when it comes to sigma notations, so I'm wondering if there's a way to mathematically prove this instead of using logic
Your first expression can be rewritten by ignoring the zero when $n=0$ and taking out a factor of $\dfrac d t$ and cancelling an $n$ as $$\dfrac d t \sum_{n=1}^d \left(\frac{1}{t}\right)^{n-1} \left(\frac{t-1}{t}\right)^{d-n} \frac{(d-1)!}{(d-n)!\,(n-1)!}$$ and then letting $m=n-1$ as $$= \dfrac d t \sum_{m=0}^d \left(\frac{1}{t}\right)^{m} \left(\frac{t-1}{t}\right)^{(d-1)-m} \frac{(d-1)!}{((d-1)-m)!\,m!} \\ = \dfrac d t \left(\frac{1}{t}+ \frac{t-1}{t}\right)^{d-1} \\ =\dfrac d t $$