- how long are his sessions, and how many pulls per session?

- what is the payback percentage of the slot machine

- what is the machine's standard deviation (I imagine this would be the most difficult to figure out, since you would need to determine the frequency that each payout level hits)


In any event, if he plays 'x' hrs per session, at 'y' pulls per hour, on a 'z' percent payback machine with a 'sd' standard deviation per pull, the odds of this event occuring is (I think):


72*(1- normdist({10xy*(1-z)}/{sd*sqrt(10xy)}))


To break it down:


EV/month = -10xy*(1-z)

SD/month = sd*sqrt(10xy)


Num +SD's needed to overcome house edge =

(-EV/month) / (SD/month)


Prob of this many +SD's or better =

1 - normsdist{(-EV/month) / (SD/month)}


Prob of overcoming this many SD's 72 months in a row = 72 * (prob of this many +SD's or better)


The main problem is in finding the SD/pull of the machine. Other than that I think it is fairly straightforward. Please correct me if I am way off, or if there is a much easier way to do this.


-- Homer