Thanks, Harry. I looked for it at WizOfOdds but couldn't find it. Should've checked vpFREE2.
That 99% pay table makes the chance of losing $2K a tiny bit smaller. The difference between 99.0% and 98.6% amounts to an $80 difference in EV at the end of $20,000 of coin-in. So it's not going to have much impact on one's chance of losing $2000.
--Dunbar
---In vpFREE@yahoogroups.com, <harry.porter@...> wrote :
---In vpFREE@yahoogroups.com, <harry.porter@...> wrote :
99% DJ paytable in question is 1/1/4/6/8/10/25/50/100/1000 (var akin to ddb)
---In vpFREE@yahoogroups.com, <h_dunbar@...> wrote :
---In vpFREE@yahoogroups.com, <h_dunbar@...> wrote :
Assuming that...
1. "99% Double Joker" is the 98.6% pay table, and
2. no errors were made in play...
then losing $2000 in $20K coin-in at a 25c game is way worse than 2 standard deviations. It's about a 0.04% event, 1 in 2500.
If you assume that errors costing 0.2% of EV were made, the chance of losing $2K is still only about 1 in 1600. And even if mistakes totaling 0.6% of EV were made (turning it into a 98% game), the chance of losing $2K is still 1 in 670, or about 3 standard deviations.
Like you, I can't answer the original poster's question without more specific parameters.
--Dunbar
(Calcs were done using Dunbar's Risk Analyzer for Video Poker 2.0)
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