Question of the Day - 09 December 2019

[Editor's Note: The reason that you've never been able to get an adequate answer to this question is, perhaps, you didn't have an expert of the caliber of Michael "Wizard of Odds" Shackleford to ask. We do! Here's what he has to say. But be forewarned, this isn't easy stuff.]

To determine a random outcome, slot machines, video poker, video keno, and all electronic games of chance first will choose a random integer between 0 and some very large number, typically (2^32)-1, which equals 4,294,967,295. This number will then be mapped to a range the game really needs. For example, a double-zero roulette game might divide that random number by 38 and then map that remainder to a table of all 38 possible outcomes in roulette.* 

That said, there are two ways I can think of to determine a win point for must-hit-by jackpots. One is to use the same method to pick a random number between the minimum and maximum value and then wait for the jackpot to cross that point. 

Another method is to go spin by spin. Let's say the meter is going to increase by x and the total between the jackpot, before the increase, and the must-hit-by point is y.  The probability of the jackpot hitting should be x/y.  For example, let's say the jackpot is at $723, it must hit by $1,000, and it is due to go up by $0.03.  Then the probability the jackpot will hit is 0.03/(1000-723) = 0.000108, or 1 in 9233. 

They probably do the second method, as it's not good for game security to have a pre-destined jackpot point. The first method doesn't seem very kosher to me as a game programmer. However, the net result is the same: the jackpot hitting at any point with equal probability.

I must add a warning that there is a great deal of evidence that AGS slot machines, with a must-hit-by feature, do not follow this method, but almost always delay triggering the jackpot until it's nearly at the maximum point. 

The short answer to your question is one can start with a random integer and transform it into whatever random result the game is looking to achieve. I learned all about how to do this in one of my actuary exams on numerical analysis. 

*  Perfectionists may remark that 38 does not divide into (2^32)-1 evenly.  I believe there is a gaming regulation to ensure every outcome is equally likely by setting the maximum value of the original random number so that the desired outcomes divides it evenly.  In the case of double-zero roulette, 38 does divide into 4,294,967,290 evenly.  Since the random range starts with 0, they could achieve the proper effect by tossing out an original random number if it was equal or greater to 4,294,967,290 and draw another one.