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)
---In vpFREE@yahoogroups.com, <007kzq@...> wrote :
Your question, particularly your use of the word "approach," is too vague to answer. You can determine how many standard deviations from expected value your result was, and I'll guess it was about 2, but, since you already knew you had bad luck, I don't think that tells you much.
On Tue, Dec 6, 2016 at 11:18 AM, stut70@... [vpFREE] <vpFREE@yahoogroups.com> wrote:$0.25 single line 99% Double Joker vp.
four $5k coin-in days = $20k coin-in.
$5k coin-in = 500 tc
It cost me $2000 for 2000 tc. :(
$2k lost/$20k coin-in = 10% loss = 90% return
$20k coin-in on a .25 machine = 16k hands.
On avg, how many hands are needed to approach the machine's 99% return?
thx
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Posted by: h_dunbar@hotmail.com
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