To find the exp(x^2) of the bonus, recall variance = exp(x^2) - (e(x))^2. Not that we directly care, but the overall return of the game is exp(base game) + exp(bonus) = 0.593301 + 0.308198 = 0.901498. The expected win from the bonus on any given spin is prob(bonus)*(average bonus) = 0.021644 * 14.23921236 = 0.308198. It would have been nice to just multiply that by the probability of winning the bonus of 0.021644, but you can't. However, the player doesn't always win the bonus. In the bonus there are 12 doubled free games. Next, let's do the variance of the bonus, given that the player won the bonus in the first place. Let's start with the variance of the base game.Īs a reminder, the variance equals exp(x^2) - (e(x))^2. In this case of this game, var(entire game) = var(base game) + var(bonus) + 2*cov(base game,bonus) To begin, recall var(x + y) = var(x) + var(y) + 2*cov(x,y) Let's use the 3-10-5 pick-8 pay table as an example. As a reminder, it plays like conventional spot keno, except if the last ball drawn contributes to a win, the player gets 12 free games with a 2x multiplier. I had to dust off my college statistics books to help me with that one. Somebody asked me about the variance of Cleopatra Keno.
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