The value of a progressive play is maximized if the jackpot is assumed
to be at the point at which the payback is 100%, including meter
movement. This assumes that there's no chance of quitting the play,
getting kicked off, or running out of money and there will be no
competition and no other reason that the player won't have it all to
himself until he hits it. It also assumes there is no cost to the
additional time spent by playing more conservatively and that no
players on the team will get paid, make any mistakes, steal, etc.
Take 9/6 Jacks or Better with a 1% meter. It doesn't matter how high
it is. A 5 x $1 non-progressive breaks even at around $4872 and pays
back 99% at something like $2900. Playing as if the meter were fixed
at $2900 maximizes the value of the play.
>I'm lost. Would anyone care to dumb this down for me? Thanks.
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>---In vpFREE@yahoogroups.com, <harry.porter@...> wrote :
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> Thanks for the push ... as you note, the calcs for a few examples are readily worked out.
>
> My "gut sense" failed me on this one. Yes, a strategy based on a RF value that yields an ER = (100% - meter advance rate) optimizes the net payback.
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>[Non-text portions of this message have been removed]
Posted by: 007 <007@embarqmail.com>
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