## Poll

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Quote:onenickelmiracleAn 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.

I meant as coin-in. That is a good deal the way I said it. Plenty of hookers and blow for your money.Quote:waltomealI'll gladly take $714.28 for every $100 lost. But I'll acknowledge that $909.09 is even better.

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.

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

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

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.

Quote:mamatMedian 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.

$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.Quote:RSOh 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.

Thank you though RS.