Losing $1 for every tier credit on a 99% machine :(

[David6]

$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

[Thomas_Robertson]

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, stu…@…com [vpFREE] <vpF…@…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

[Dunbar]

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 vpF…@…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] <vpF…@…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

[Harry_Porter]

99% DJ paytable in question is 1/1/4/6/8/10/25/50/100/1000 (var akin to ddb)

—In vpF…@…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)

[Dunbar]

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 vpF…@…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 vpF…@…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)

[Harry_Porter]

Haven’t look at details, but I’m going to guess that the 99.0% game has a much stronger variance (think 8/5 BP vs 9/6 DDB). Are you secure in your take on the relative loss potential?

—In vpF…@…com, <h_dunbar@…> wrote :

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 vpF…@…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 vpF…@…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)

[Dunbar]

Harry, how can you do worse on a game, when the only difference is that you get paid more for a royal flush? In the 99.0% and 98.6% versions of DJ you have the same probability of getting each kind of hand. And you collect the same payoff for all hands below a RF. But you collect 200 more units on a RF in the 99.0% game.

The bigger variance of the 99% DJ game might have an impact on losses if the big royal payoff caused strategy changes that lessened the chance of hitting other hands. But I used the hand frequencies for 98.6% DJ that I got from WizOfOdds, and simply changing the RF payout from 800 to 1000 produces a 99.0% game. So, if there are strategy changes, they can be safely ignored.

–Dunbar

—In vpF…@…com, <harry.porter@…> wrote :

Haven’t look at details, but I’m going to guess that the 99.0% game has a much stronger variance (think 8/5 BP vs 9/6 DDB). Are you secure in your take on the relative loss potential?

—In vpF…@…com, <h_dunbar@…> wrote :

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 vpF…@…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 vpF…@…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)

[Harry_Porter]

My mistake … I presumed the 98.6% paytable featured a more commonly found “1/2” for the first two paylines, vs “1/1” of the 99.0% variant.

—In vpF…@…com, <h_dunbar@…> wrote :

Harry, how can you do worse on a game, when the only difference is that you get paid more for a royal flush? In the 99.0% and 98.6% versions of DJ you have the same probability of getting each kind of hand. And you collect the same payoff for all hands below a RF. But you collect 200 more units on a RF in the 99.0% game.

The bigger variance of the 99% DJ game might have an impact on losses if the big royal payoff caused strategy changes that lessened the chance of hitting other hands. But I used the hand frequencies for 98.6% DJ that I got from WizOfOdds, and simply changing the RF payout from 800 to 1000 produces a 99.0% game. So, if there are strategy changes, they can be safely ignored.

–Dunbar

—In vpF…@…com, <harry.porter@…> wrote :

Haven’t look at details, but I’m going to guess that the 99.0% game has a much stronger variance (think 8/5 BP vs 9/6 DDB). Are you secure in your take on the relative loss potential?

—In vpF…@…com, <h_dunbar@…> wrote :

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 vpF…@…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 vpF…@…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)

[nightoftheiguana2000]

stut70 wrote: “$0.25 single line 99% Double Joker vp. …
On avg, how many hands are needed to approach the machine’s 99% return?”

At what point is 3SD only 1% of the EV?

0.01x0.01xhands = 3sqrt(29xhands)

hands = 29((3/.0001)^2) = 26,100,000,000

Lesson to learn: Nzero (see the FAQ)

vpFREE Bank_NO

vpFREE Bank_NO
Very early on, upon joining this board, iggy introduced a concept which he references in shorthand as “N0”.

View on www.west-point.org

Preview by Yahoo

[nightoftheiguana2000]

I suppose you could back off a bit on how you define “approach”.

You could say 3SD less than 1% of the total return, in other words in this case your 3SD range would be 98 to 100% return.

3sqrt(variance x hands) < 1% of hands

hands > variance x (3/1%)^2 > 2,610,000

still a big number, and 98% to 100% is still a big return range, and at 3SD you still have some outliers (about 0.27% of results)

[nightoftheiguana2000]

Average return is a useful metric, but you are unlikely to “get there” with a high variance game like video poker. On the plus side, it is basically a 50-50 proposition that you will either do better or worse than the average return, in other words, the average return is the center line. If you want a metric to “go after”, I would strongly suggest Nzero (variance/edge^2 hands).

[Dunbar]

—In vpF…@…com, <nightoftheiguana2000@…> wrote :

“Average return is a useful metric, but you are unlikely to “get there” with a high variance game like video poker. On the plus side, it is basically a 50-50 proposition that you will either do better or worse than the average return, in other words, the average return is the center line. If you want a metric to “go after”, I would strongly suggest Nzero (variance/edge^2 hands).”

As you know, NOTI, it’s not really a 50-50 proposition until you’ve played a whole lot of hands. There have to be some extra losses to make up for the bigger wins you get when you hit a Royal. For example, using the OP’s example of 16,000 hands of 99% Double Joker, you’ll do worse than the expected 99% return about 59% of the time. After 40,000 hands it’s about 54-46. After 100,000 hands, it’s still about 53-47.

–Dunbar

[Harv_Cohen]

good morning. When playing 10 play, if you draw to a four card royal, what are the odds of making three royals ? Thank you.

···

Sent from Windows Mail

From: vpFREE3355 vpfree3…@…com [vpFREE]
Sent: ‎Saturday‎, ‎December‎ ‎10‎, ‎2016 ‎11‎:‎13‎ ‎AM
To: vpFREE3355 vpfree3…@…com [vpFREE]

Average return is a useful metric, but you are unlikely to “get there” with a high variance game like video poker. On the plus side, it is basically a 50-50 proposition that you will either do better or worse than the average return, in other words, the average return is the center line. If you want a metric to “go after”, I would strongly suggest Nzero (variance/edge^2 hands).

[Dunbar]

The odds of making 3 royals from a draw to a 4-card-royal (10-play game with no wild cards) are 1006 to 1.

Here are the odds for each number of RF’s

RFs
Chance
One in…
0
0.80649
1.2
1
0.17532
5.7
2
0.01715
58
3
0.00099
1006
4
3.78E-05
26437
5
9.87E-07
1013419
6
1.79E-08
55940708
7
2.22E-10
4503226987
8
1.81E-12
552395843687
9
8.75E-15
114345939643109
10
1.90E-17
52599132235830100

The chance of hitting 2 or more RF’s is 1 in 55. That’s a good number to remember if you’re concerned about getting a W-2G in a 25c game.

–Dunbar

—In vpF…@…com, <hjclasvegas6969@…> wrote :

good morning. When playing 10 play, if you draw to a four card royal, what are the odds of making three royals ? Thank you.

[Harv_Cohen]

Thank you, a fellow gambler was correct, a gambling ex math teacher was wrong as I suspected ! lol.

···

Sent from Windows Mail

From: vpFREE3355 vpfree3…@…com [vpFREE]
Sent: ‎Sunday‎, ‎December‎ ‎11‎, ‎2016 ‎6‎:‎55‎ ‎AM
To: vpFREE3355 vpfree3…@…com [vpFREE]

The odds of making 3 royals from a draw to a 4-card-royal (10-play game with no wild cards) are 1006 to 1.

Here are the odds for each number of RF’s

RFs
Chance
One in…
0
0.80649
1.2
1
0.17532
5.7
2
0.01715
58
3
0.00099
1006
4
3.78E-05
26437
5
9.87E-07
1013419
6
1.79E-08
55940708
7
2.22E-10
4503226987
8
1.81E-12
552395843687
9
8.75E-15
114345939643109
10
1.90E-17
52599132235830100

The chance of hitting 2 or more RF’s is 1 in 55. That’s a good number to remember if you’re concerned about getting a W-2G in a 25c game.

–Dunbar

—In vpF…@…com, <hjclasvegas6969@…> wrote :

good morning. When playing 10 play, if you draw to a four card royal, what are the odds of making three royals ? Thank you.

[steven_wojciechowski]

So you do get a 1099 if you hit multi royals playing quarters, Well that sucks I always wondered about that… since they are all really separate games. I still hope it happens some day…

···

---

Dunbar wrote:

The odds of making 3 royals from a draw to a 4-card-royal (10-play game with no wild cards) are 1006 to 1.

—In vpF…@…com, wrote :

good morning. When playing 10 play, if you draw to a four card royal, what are the odds of making three royals ? Thank you.

[kcace1024]

You get a W-2G not a 1099. If you win a drawing you will receive a 1099.

[David6]

can you use loses to offset the 1099?

if not, any way to mitigate that 1099 winnings?

[Vegasvpplayer]

If the 1099 win was based on gambling activity you can. You have to maintain good records to document all your wins and losses. If you are using tax preparation software to do your taxes DO NOT enter the Form 1099 in the place requested. You have to include it under gambling winnings (along with any W-2G’s you have). Later you can list gambling loses up to the amount you claim as winnings. If your software allows, it’s probably a good idea to state gambling winnings include all W-2g’s and gambling related 1099’s.

···

Sent from my iPhone

On Dec 12, 2016, at 8:38 AM, stu…@…com [vpFREE] <vpF…@…com> wrote:

can you use loses to offset the 1099?

if not, any way to mitigate that 1099 winnings?