Insufficient information. However, provide game variance and permit the assumption that game results are "near-normally" distributed for values central to the mean, and it's possible to calculate the answer you are looking for (... which I would leave as an exercise for NOTI, whose much more facile with the math).
Variance is key here for a simple reason ... you're looking for results that exceed the EV by a given margin. The high the game variance, the greater the proportion of results that will exceed that margin.
---In vpFREE@yahoogroups.com, <mickeycrimm@...> wrote :
Variance is key here for a simple reason ... you're looking for results that exceed the EV by a given margin. The high the game variance, the greater the proportion of results that will exceed that margin.
---In vpFREE@yahoogroups.com, <mickeycrimm@...> wrote :
This is some math I don't know how to do. In a 98.0% video poker game what is the chance of breaking even in a one million hand sample space? Thanks for any help.
Mickey
Mickey
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