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onenickelmiracle
onenickelmiracle
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December 29th, 2016 at 11:36:14 PM permalink
This is so irritating to me, the phrase, "about the same" or "in the same range". You'll see a question about slots, are some machines better than others, are end machines looser, etc. Then the expert will say "slot directors have told me most of the machines they place are all in the same range". That's a nondenial denial. An 86% machine will return $714.28 for every hundred lost, and an 89% machine $909.09.These numbers dont seem the same to me and describing them as being in the same range doesn't mean anything. A $500 loss is not about the same as a $400 loss.
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waltomeal
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December 30th, 2016 at 12:10:49 AM permalink
Quote: onenickelmiracle

An 86% machine will return $714.28 for every hundred lost, and an 89% machine $909.09.


I'll gladly take $714.28 for every $100 lost. But I'll acknowledge that $909.09 is even better.
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onenickelmiracle
onenickelmiracle
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December 30th, 2016 at 12:29:07 AM permalink
Quote: waltomeal

I'll gladly take $714.28 for every $100 lost. But I'll acknowledge that $909.09 is even better.

I meant as coin-in. That is a good deal the way I said it. Plenty of hookers and blow for your money.
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RS
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December 30th, 2016 at 2:55:19 AM permalink
86% of 714+100 is not 714
89% of 909+100 is not 909


The problem is with the way you came up with the numbers. Although innacurate it's close enough. 86% and 89%, IMO, are in the same range. What's different is that the expected loss is 27% higher for the 86% game than the 89% game.

It'd be like saying 90/9/6 JOB (99.9916% IIRC) is in the same range as NSUD (99.728%). I think we'd all agree on that. Yes? If you do the math the way you did it above, you'd come up with $1.19M coin in for the JOB game and $36,700 for NSUD.


You can do lots of stuff with math.....problem is figuring out how to do it honestly.
onenickelmiracle
onenickelmiracle
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December 30th, 2016 at 3:17:38 AM permalink
I had thought if you divided a bankroll by the hold, it gives the average coin-in losing it all. Is this not true, or are you saying it is a rule of thumb?
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mamat
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December 30th, 2016 at 3:43:31 AM permalink
Quote: onenickelmiracle

I had thought if you divided a bankroll by the hold, it gives the average coin-in losing it all. Is this not true, or are you saying it is a rule of thumb?

Median coin-in losing it all...will be a less than "average coin-in", maybe a LOT less...

...unless bankroll is relatively large (probably on the order of NO, but not sure exactly where).
RS
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onenickelmiracle
December 30th, 2016 at 3:45:29 AM permalink
Oh I see what you did. 714 coin in returned 614. And 909 coin in returned 809.


The problem is you're comparing the amount of coin in you can do on each machine to say "they aren't in the same range". Instead, determine the return or expected loss for the same amount of coin in. Do $1,000 coin in and you get a return of $860 and $890, or a loss of $140 and $110 respectively. IMO, it doesn't make sense to compare the amount of coin in you can do for an expected loss of $X. See my earlier example of 90/9/6 JOB and NSUD.

It's a play on numbers.
RS
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December 30th, 2016 at 3:47:27 AM permalink
Quote: mamat

Median coin-in losing it all...will be a less than "average coin-in", maybe a LOT less...

...unless bankroll is relatively large (probably on the order of NO, but not sure exactly where).



But it is the mean average. And I think people use the word "average" when referring to the mean average. Otherwise, they'll say the median or the mode average, if that's what they're referring to.
onenickelmiracle
onenickelmiracle
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December 30th, 2016 at 4:06:57 AM permalink
Quote: RS

Oh I see what you did. 714 coin in returned 614. And 909 coin in returned 809.


The problem is you're comparing the amount of coin in you can do on each machine to say "they aren't in the same range". Instead, determine the return or expected loss for the same amount of coin in. Do $1,000 coin in and you get a return of $860 and $890, or a loss of $140 and $110 respectively. IMO, it doesn't make sense to compare the amount of coin in you can do for an expected loss of $X. See my earlier example of 90/9/6 JOB and NSUD.

It's a play on numbers.

$3500 coin-in, $490 versus $385. Your $1000 example makes it seem similar, but doesn't seem so similar anymore. You trade me $110 for $140 and I'll trade you $385 for $490. The same range is so vague, it implies some equality, implies it doesn't matter where you play, results will be about the same. Like saying McDonald's hamburgers and Wendy's have the same quality or in the same quality range.

Thank you though RS.
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onenickelmiracle
onenickelmiracle
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December 30th, 2016 at 4:26:11 AM permalink
Then you have to consider the truthfulness of the range. Maybe they dont count 30 of 2000 that are 90% or maybe have 40 that are 92-93% and they don't know it. Can a slot be set wrong like a VP machine and they don't know it? Yeah it happens. Then the 84s out there. They'll keep adding those 84% and keep adding these 84%, will still tell the experts 86-89%. Next thing you know half the floor is 84% and on YouTube you'll hear 86-89%.
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onenickelmiracle
onenickelmiracle
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December 30th, 2016 at 4:47:28 AM permalink
I've talked to a few slot directors and they have assured me if you hardly use their services, you'll hardly know you're being ripped off.
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MathExtremist
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RS
December 30th, 2016 at 9:34:48 AM permalink
I completely misinterpreted the premise of this thread.

"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
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