Could I get a sample paytable with at least $10$ combos for a $4$ reel slot machine with $6$ symbols on each reel with a house edge of $1 \%$?
Pay table is the combinations in which you win for example "Apple-Apple-Apple-Apple" gives you 50x gold, "Any-Any-Any-Apple" 2x gold and so on. House edge is the funds that the house doesn't give back to the players and keeps as profit instead.
My math sucks.
On
You have $6^4=1296$ possible outcomes. You clearly are intended to assume all the choices come up equally. To have exactly $1\%$ house advantage, after each combination comes up once, you need to pay out $0.99\cdot 1296=1283.04$, which we can round to $1283$ The simple way is to pick $10$ combinations and find $10$ numbers that add to $1283$ and assign one payoff to each combination. An approach more like a real machine is to make a small prize, say $2$, for Apple, Apple, Any, Any. That is $36$ combinations, so makes $72$ of your total payout. There are lots of answers, but this should show you how to get to one.
OK, this solution assumes that I understand what you want (which I'm not too sure of). If all 6 values on each reel are unique (i.e. they don't repeat), then there's only $1$ way to get each of $10$ combinations, and obviously the probability of it is $\frac{1}{6^4}$. If you play $n$ times, the probability to get this specific combination $k$ times is just Binomial (assuming each sample is independent of the previous ones): $\binom{n}{k}(\frac{1}{6^4})^k (1-\frac{1}{6^4})^{n-k}$, so you expect that you'll get this combination on the average $\frac{n}{6^4}$. If the payoff of combination $j$ is $\pi_j$ and the cost of playing once is $r$, the expected payoff from combination $j$ would be $$ R(j)=\frac{n(\pi_j-r)}{6^4} $$ Does this answer your question?