Probability and simple dice game I am trying to learn some probability (self-learning) and having a hard time. Say I am building a casino and want to invent a game that is easy to understand, but statistically will give me money. My basic question is how one can design such a game.
I have an example.
Say you have a six-sided die and you have a game where you can bet on what the outcome of a roll is. You can double your money from guessing correctly. So, you would make a bet on a number and then roll. If you get it right, you double your money. If you get it wrong, you lose the bet. This is obviously a terrible game to play since there is only a 1/6 chance of guessing the correct number.
But, how would one calculate some kind of measure for how much money I can expect to make after, say. 1000 people play the game betting 1 dollar per round?
Is it as simple as saying that I have a 5/6 chance of wining each round, so I should expect to earn 5/6 dollars per round? So, if 1000 people play, then I should expect to earn $1000\cdot 5/6$ dollars? Do I need to somehow account for the money I pay out when someone wins the game? How can I do that?
If I change the game to be guessing odd or even, then the game is no longer in my favor since there is a 50% chance for the player to win a around. In that case I would expect to make 0 dollars from 100 people playing the game.
Is that correct?
 A: Expanded comment to answer:
In general, to calculate the net amount earned, you would subtract the expected losses from the expected winnings. Let's denote this as $E[W]-E[L]$ where $E$ denotes the expected value.
First, solving for $E[W]$ gets us
$$E[W] = P[W]*x_1 $$, where $x_1$ denotes the amount gained on a win.
Similarly, solving for $E[L]$ gives us:
$$E[L] = P[L]*x_2$$
Then, we simply apply the numbers to the given problem. For your first example, we have
$$E[W] = \frac{5}{6}*1$$
$$E[L] = \frac{1}{6}*1$$
$$E[W]-E[L] = \frac{5}{6}-\frac{1}{6}=\frac{4}{6} = \frac{2}{3}$$. This value denotes the amount you are expected to win per round of playing. Accordingly, finding the expected winnings for multiple rounds gives us $$\frac{2}{3} * \text{number of rounds}$$
Additionally, plugging in our numbers for the even/odd game gives:
$$E[W] = \frac{3}{6}*1$$
$$E[L] = \frac{3}{6}*1$$
$$E[W]-E[L] = \frac{3}{6}-\frac{3}{6}=\frac{0}{6} = 0$$, as you had calculated originally
If you want to make a simple yet profitable game, I'd recommend making a game that has so many combinations that it's hard to track the probabilities (such as blackjack, which, even with only one player and a dealer, gives over 6 million possible starting hands, assuming no card repetition (and that doesn't include the added combinations from "hitting"). Alternatively, choose a game like roulette, which is quite simple and has up to a 47.4% (almost 50-50) chance of winning if you only bet on one color. Dice are somewhat harder to make these types of games with since there are only 36  possible rolls of a pair of dice (6 per die), but if you add several rounds or rules to one game, you can complicate things significantly harder to calculate, while the game can be easy to understand.
