## Poll

2 votes (20%) | |||

4 votes (40%) | |||

No votes (0%) | |||

4 votes (40%) | |||

1 vote (10%) | |||

No votes (0%) | |||

1 vote (10%) | |||

1 vote (10%) | |||

4 votes (40%) | |||

2 votes (20%) |

**10 members have voted**

Quote:MentalThanks for the update on trigger frequencies.Quote:drrockI can also report that the trigger percentages for these games have been updated today at videopoker.com.

There are 11 different trigger points listed for the 13 game types. In general, it looks like the games with the higher variances have lower trigger points and vice versa. This would be expected due to the advantage of being able to choose which hands get a multiplier when there is a higher probability of more extreme winning dealt hands.link to original post

I have busy losing a boatload of money online today. However, I have some time to write some code now. I should be able to automate the search for cutoff values. I also figure you might want the weighted EVs for all deals above any given cutoff point. Some nomenclature: If I give a cutoff point as 50.0 or 40, I mean that dealt RFs and SFs are all accepted for multiplication. Here are the first 15 lines in my new output (printed here without commas). Are these the weighted EVs you need for your calculations or something else? You might also want the weighted EVs for all deals below the cutoff, but I think that can be derived from these numbers.

800.00000000 4 800.00000000

50.00000000 40 125.00000000

25.00000000 664 31.02409639

19.68085098 760 29.59126539

19.65957451 796 29.14209344

19.59574509 820 28.86268812

19.55319214 848 28.55529910

18.55319214 944 27.53813568

18.53191566 992 27.10235084

18.48936081 1316 24.98182746

18.46808434 1436 24.43750352

18.42553139 1468 24.30645236

18.40425491 1504 24.16517636

18.36170197 1600 23.81696789

9.00000000 5344 13.43621793

link to original post

I'm trying to understand how I would make use of the 3rd column here. As I said, I was thrilled with having your precise calculations for the first two columns, but since we never know for certain which rows will be in the "acceptance range" for a given multiplier, I am not understanding how some type of premultiplication per row is helpful. In Excel, after loading all the rows containing your first two columns, I just use a function called SUMIF that sums all the values in a given column given that numbers in the same rows of another column meet a particular criterion (or set of criteria). Using that and a number of SUMPRODUCT functions does the trick really quickly.

In my next to last sentence in the last paragraph, the criterion is whether or not the EV for a row is greater than or equal to whatever cutoff that I am currently testing.

However, I am still curious as to how you calculated the 3rd column as well as how you think it should be used. I see your explanation, "Some nomenclature: If I give a cutoff point as 50.0 or 40, I mean that dealt RFs and SFs are all accepted for multiplication," but don't see any reason to identify special cases for special handling. Almost certainly, all RFs, SFs, and probably quads will be automatically accepted, but if I have to make special formulas for some rows to do calculations, I see that as complicating my spreadsheet rather than simplifying it.

Plus I know exactly how to use the first two columns. At this point, I don't understand exactly how I could use the 3rd column.

I apologize that when I first read your post, I was quite fatigued and the prospect of determining 11 more sets of cutoffs didn't seem appealing to me. I live in the Midwest/Mideast region of the U.S. and I took advantage of a rare day in early February when the temperature reached into the 70s to get a rare round of golf in.

I do understand that there may be more news for Fortune X from videopoker.com in the near future. If it comes to pass, I will comment further.

Don't worry about it. The third column is just the weighted average of all EVs above whichever cutoff you are targeting. You are already calculating the one particular average that you need in your spreadsheet. No need to add extra data that is derived from other data you already have.Quote:drrockPlus I know exactly how to use the first two columns. At this point, I don't understand exactly how I could use the 3rd column.

link to original post

In C++, I precalculate all averages even though I only use one value at a time. However, I am able to reuse a saved result while iterating.

I am glad you could get out on the links.

I wrote a nested loop to search for optimum EV cutoffs among a range of possible choices. My table lists the multiplier, the optimum cutoff EV for that multiplier, and the range of values I searched to find that optimum. The range refers to the line numbers in my dealt EV list with the first line being zero. A cutoff EV of 9.00 means I am accepting multipliers for all dealt full houses and any hands with higher EV. For the 2x multiplier in DDB, I searched the 19th to 29th entries, and found the optimum cutoff on line 23 (N = 22). I am searching 20,000 combinations of the four cutoff values. This takes just a few seconds.

Mult EV { L < N < H }

2 9.00000 { 18 < 22 < 28 }

3 5.36633 { 22 < 27 < 32 }

5 3.53191 { 28 < 33 < 38 }

8 1.42183 { 70 < 80 < 90 }

Optimum EV = 25.23842 / 25.49155 = 99.00698

Double Double Bonus trigger frequency = 0.1100

Mult EV { L < N < H }

2 9.00000 { 10 < 14 < 17 }

3 6.00000 { 12 < 15 < 19 }

5 2.59574 { 19 < 23 < 28 }

8 1.43293 { 53 < 57 < 63 }

Optimum EV = 24.33395 / 24.43180 = 99.59947

Jacks or Better trigger frequency = 0.1165

If you iteratively optimize each cutoff level separately, you may not find the global minimum. My method is very quick and finds the global optimum.

If anyone disagrees with my EVs for this particular set of cutoffs that I believe are optimum, we might need to compare notes on the numerical values for each multiplier.

Also, if anyone has the trigger frequency for other VP games/paytables, I can generate the optimum cutoffs with very little effort.

Quote:Mental

Mult EV { L < N < H }

2 9.00000 { 18 < 22 < 28 }

3 5.36633 { 22 < 27 < 32 }

5 3.53191 { 28 < 33 < 38 }

8 1.42183 { 70 < 80 < 90 }

Optimum EV = 25.23842 / 25.49155 = 99.00698

Double Double Bonus trigger frequency = 0.1100Mult EV { L < N < H }

2 9.00000 { 10 < 14 < 17 }

3 6.00000 { 12 < 15 < 19 }

5 2.59574 { 19 < 23 < 28 }

8 1.43293 { 53 < 57 < 63 }

Optimum EV = 24.33395 / 24.43180 = 99.59947

Jacks or Better trigger frequency = 0.1165

If anyone disagrees with my EVs for this particular set of cutoffs that I believe are optimum, we might need to compare notes on the numerical values for each multiplier.

Also, if anyone has the trigger frequency for other VP games/paytables, I can generate the optimum cutoffs with very little effort.

link to original post

Good work!

I can confirm that I get exactly the same EVs, and effectively the same cutoffs as you, after inserting the new trigger points and increasing the precision of the dealt hand EVs.

If you want to check the next few decimals, I can extend those EVs to 99.00698028% for 9-6 DDB and 99.59947112% for 9-6 Jacks. At least that is what I got. I am not certain whether the data used is totally accurate to that many places, but just thought that might give us extra assurance if your figures match with mine even more closely.

I think that I am the only poster who has spoken about iteration. So, I guess I need to clarify that each iteration yields new values for all 4 cutoffs simultaneously on its way to maximizing game EV, along with other information in addition to the game EV, such as the percentage of visits to each state, and the relative future values of each state, including the details as to how those values are calculated. And generally convergence occurs essentially instantaneously after 2 or 3 iterations. I can put in numbers that are quite a bit off and still get convergence in 4 or 5 iterations.Quote:MentalIf you iteratively optimize each cutoff level separately, you may not find the global minimum. My method is very quick and finds the global optimum.

The search method is fine, and with today's computers, it is a quick way to get answers. Alternatively, I have used direct matrix calculation using sums and averages from those raw dealt hand counts and EVs that you provided with no search involved. I use only a 9 x 9 matrix; so even the inversion is essentially instantaneous in Excel. Certainly a search is valuable in verifying whether the technique is accurate. Here are the optimal cutoffs that the Markov Chain analysis provided (they are all consistent with the endpoints of the gaps that you identified in your code boxes).

Mult DDB Jacks

2 8.973937 8.544021

3 4.916917 4.750044

5 2.576148 2.549935

8 1.413936 1.421897

These are NOT vulture indifference points, but do answer the question at what point would the expected value (current and future) be equal whether accepting the current multiplier or rejecting it and waiting for at least one more multiplier, assuming continuous play at optimal expected value. The values in the Mental code boxes are in fact all the next higher point after these equilibrium points that is possible with a dealt video poker hand; so this is another point of agreement.

Quote:drrock

I can confirm that I get exactly the same EVs, and effectively the same cutoffs as you, after inserting the new trigger points and increasing the precision of the dealt hand EVs.

If you want to check the next few decimals, I can extend those EVs to 99.00698028% for 9-6 DDB and 99.59947112% for 9-6 Jacks. At least that is what I got. I am not certain whether the data used is totally accurate to that many places, but just thought that might give us extra assurance if your figures match with mine even more closely.

I think that I am the only poster who has spoken about iteration. So, I guess I need to clarify that each iteration yields new values for all 4 cutoffs simultaneously on its way to maximizing game EV, along with other information in addition to the game EV, such as the percentage of visits to each state, and the relative future values of each state, including the details as to how those values are calculated. And generally convergence occurs essentially instantaneously after 2 or 3 iterations. I can put in numbers that are quite a bit off and still get convergence in 4 or 5 iterations.Quote:MentalIf you iteratively optimize each cutoff level separately, you may not find the global minimum. My method is very quick and finds the global optimum.

The search method is fine, and with today's computers, it is a quick way to get answers. Alternatively, I have used direct matrix calculation using sums and averages from those raw dealt hand counts and EVs that you provided with no search involved. I use only a 9 x 9 matrix; so even the inversion is essentially instantaneous in Excel. Certainly a search is valuable in verifying whether the technique is accurate. Here are the optimal cutoffs that the Markov Chain analysis provided (they are all consistent with the endpoints of the gaps that you identified in your code boxes).

Mult DDB Jacks

2 8.973937 8.544021

3 4.916917 4.750044

5 2.576148 2.549935

8 1.413936 1.421897

These are NOT vulture indifference points, but do answer the question at what point would the expected value (current and future) be equal whether accepting the current multiplier or rejecting it and waiting for at least one more multiplier, assuming continuous play at optimal expected value. The values in the Mental code boxes are in fact all the next higher point after these equilibrium points that is possible with a dealt video poker hand; so this is another point of agreement.

link to original post

Mult EV { L < N < H }

2 9.00000 { 18 < 22 < 23 }

3 5.36633 { 22 < 27 < 28 }

5 3.53191 { 28 < 33 < 34 }

8 1.42183 { 70 < 80 < 81 }

Optimum EV = 25.23842 / 25.49155 = 99.00698

Double Double Bonus trigger frequency = 0.1100 search 1980

Mx Cutoff N Accept Visit Yield Cost Yield Mx Yield 1x

1 0.1000 100 1.0000000 8.09091 8.00845 16.18182 0.000000 8.008445

2 9.0000 22 0.0036815 1.00000 1.05192 2.00000 0.124232 0.927692

3 5.3663 27 0.0250900 0.99632 1.36194 1.99264 0.563666 0.798275

5 3.5319 33 0.0293933 0.97132 1.76029 1.94264 0.998592 0.761703

8 1.4218 80 0.2103518 0.94277 4.21338 1.88554 3.748819 0.464559

12 0.1000 100 1.0000000 0.74446 8.84243 1.48891 8.842433 0.000000

Optimum EV = 25.238415613826 / 25.491552087638 = 99.006978967222

Mult EV { L < N < H }

2 9.00000 { 10 < 14 < 17 }

3 6.00000 { 12 < 15 < 19 }

5 2.59574 { 19 < 23 < 28 }

8 1.43293 { 53 < 57 < 63 }

Optimum EV = 24.33395 / 24.43180 = 99.59947

Jacks or Better trigger frequency = 0.1165 search 4410

Mx Cutoff N Accept Visit Yield Cost Yield Mx Yield 1x

1 0.1000 100 1.0000000 7.58369 7.54910 15.16738 0.000000 7.549102

2 9.0000 14 0.0020562 1.00000 1.02307 2.00000 0.055255 0.967811

3 6.0000 15 0.0039616 0.99794 1.07135 1.99589 0.116939 0.954413

5 2.5957 23 0.0769323 0.99399 2.06454 1.98798 1.343857 0.720685

8 1.4329 57 0.2122726 0.91752 3.99237 1.83504 3.518896 0.473474

12 0.1000 100 1.0000000 0.72276 8.63351 1.44551 8.633514 0.000000

Optimum EV = 24.333946511014 / 24.431802907803 = 99.599471241816

I have added more digits of precision. I am using 64-bit precision numbers. At some point, 32-bit calculations will diverge from those results. In my table, Yield means expected winnings in units of 5-coin bets. I set N=100 as a placeholder for 1x and 12x, but these Ns are not needed nor used. The last two columns should sum to the Yield column. Yield Mx is the winnings from multiplied hands. Yield 1x is the winnings from unmultiplied hands

Yesterday, I noticed that videopoker.com added some custom pay schedules to their selection of Fortune X games. I believe a Gold membership is required to view and play these. Right now, you need Gold to play Fortune X at all, but that will likely change when they introduce other new games for the Gold members.Quote:drrockI do understand that there may be more news for Fortune X from videopoker.com in the near future. If it comes to pass, I will comment further.

There are 55 variations among the 13 game types; so, 42 new pay schedules. The mean trigger point for all choices is 10.9555%, so that may be where the original 10.96% number came from.

The increase in EV for games over their base counterparts is quite tiny but positive, with a median of 0.020% and the middle 80% of differences being between 0.005% and 0.044%.

I will be glad to run the new pay tables as a cross check with you. I looked at the NSUD deuces wild game, and it seems that the indifference EV has to change and the trigger levels have to be much higher for the math to work out.Quote:drrockYesterday, I noticed that videopoker.com added some custom pay schedules to their selection of Fortune X games. I believe a Gold membership is required to view and play these. Right now, you need Gold to play Fortune X at all, but that will likely change when they introduce other new games for the Gold members.Quote:drrockI do understand that there may be more news for Fortune X from videopoker.com in the near future. If it comes to pass, I will comment further.

There are 55 variations among the 13 game types; so, 42 new pay schedules. The mean trigger point for all choices is 10.9555%, so that may be where the original 10.96% number came from.

The increase in EV for games over their base counterparts is quite tiny but positive, with a median of 0.020% and the middle 80% of differences being between 0.005% and 0.044%.

link to original post

https://www.playperfectllc.com/fortunex

I confirm your values for the two games.

Thanks, Gary. It is nice to see the formal Markov matrices.Quote:GaryJKoehlerGreat work Mental and Drrock. I just got time and did a simple non-discounted, discrete time, Markov Decision analysis that can be found at

https://www.playperfectllc.com/fortunex

I confirm your values for the two games.

link to original post

This Fortune X game is trickier because you have to split the deals into high variance deals and low variance deals, then multiply the high-variance deals by the square of the right multiplier. I couldn't figure out an easy way to do this split in my existing code. When I calculate the variance of a normal game, I just use the frequencies of all the different hands. In Fortune X, the high-value hands types get higher multipliers, on average. Therefore, I cannot just use the frequencies of all the different final hands.

Since I don't want to muck with my kernel code, I just added code to estimate the variance in the MC simulation that I had already created.

The variance for DDB seems to be around 571 to 581 bets^2 (using 10-coin bets). The variance expressed in terms of 5-coin bets is around 2300 bets^2. Yikes!

The variance for JB seems to be around 270 bets^2 (using 10-coin bets).

EDIT: I realized that I was actually calculating variance based on 5-coin bets. My corrected variance estimates based on 10-coin bets are:

DDB: 143.0 bets^2

JB: 68.7 bets^2

If by "game" variance, you mean single hand AND using equilibrium long-run probabilities, there really is not too much involved in that but a lot of tedium. For DDB, you can look at the Wizard's page and it shows a line-by-line breakdown showing multiplier, base win, and probability. Unfortunately many of the high-valued hands have very low probabilities, but if you had his unrounded probabilities instead of what is in the table, it would be a simple matter of E[x^2] - [E(x)]^2.Quote:GaryJKoehlerAnyone up for working on the game variance?

link to original post

He does show Return on each line, and in general, it has more significant digits, so we can get more precision by using Probability = Return/(Multiplier*Base Win/2). His probabilities in that table sum to 0.999997. The probabilities derived from the formula above sum to 0.99999784. Both of them show an expected value of 0.990070, quite close to the figure he shows at the bottom of his table 0.990069.

Using the derived probabilities and not worrying too much that they don't sum to unity, I got an average second moment of 144.2964 and a variance of 143.3161 using squared 10-coin bets. If you want that in 5-coin bets squared, you can multiply by 4. Personally, I prefer to take square roots and just deal with standard deviations in terms of bets. Here, that would translate to 11.9715 bets. This could be compared to a regular 9-7 Triple Double Bonus game which has a standard deviation of 9.9137 bets. To me the comparison of 9.91 to 11.97 is more meaningful than a comparison of 98 to 143, when using variances.

But the "game" that is at videopoker.com is not single line. It is either Triple Play, Five Play, or Ten Play. I am not certain that Fortune X will be available anywhere soon at single line only. Any variance calculations of the actual game would involve convolutions and covariance. And of course the variances would be different depending upon which option and which base game was selected.

When I started this post with the qualifier "equilibrium long-run probabilities," I was avoiding the idea that in the short run the hands are not independent so even the single-hand game would require some rather involved covariance calculations unless we are dealing far enough in the future so that equilibrium can be considered.

This isn't quite right, but doesn't seem to be taught in many statistics classes. Try this example. Consider a simple game with no house edge. It costs $1 to play with results of $0, 50% of the time; $1, 40%; and $6 ,10% of the time. Some quick calculations show the Expected Value as $1 and the Variance as 3 squared dollars (whatever squared dollars mean). Now consider a situation in which there is a multiplier of 11x: results of $0, 50% of the time; $11, 40%; and $66 ,10% of the time. Again quick calculations show an expected value of $11 and a variance of 3*11^2 = 363 squared dollars.Quote:MentalIf you played UX with no strategy changes from the base game, the variance expressed in 5-coin bets^2 would just be the variance of the base game multiplied by the weighted average of the squares of the multipliers.

Now consider a game that costs double the first game ($2) and returns the 1x situation 90% of the time and the 11x situation 10% of the time. So the Expected Value of the new game is 90% x 1 + 10% x 11 = $2, again a game with no house edge. Using the rule "the weighted average of the squares of the multipliers," we should get a variance of 3 x (90% x 1^2 + 10% x 11^2) = 39.

But if we work through the new game possibilities, we get $0, 50%; $1, 36%, $6, 9%; $11, 4%; and $66, 1%. If you find the variance here, you should get an answer of 48 instead of 39. So what is the missing piece? Below is the formula for the variance of a random variable that is a mixture of two other random variables:

p1V(x1)+p2V(x2)+p1p2(Ex1-Ex2)^2

The last term has to be put in because if the two random variables are further apart, the variance is larger. Even easier to see intuitively would be to imagine that both multiplier situations in and of themselves had zero variance. If in the 1x situation, you always got $1 and the 11x situation you always got $11, each of them would have 0 variance; however, once you found out that 90% of the time you had 1x and 10% of the time you had 11x, the difference in the means would guarantee that you had more than zero variance.

In our example, plugging into that formula yields

0.9(3) + 0.1(363) + (0.9)(0.1)*(1 - 11)^2 = 2.7 + 36.3 + 9 = 48

In single hand Ultimate X, it would have more terms, but the concept would be the same. You would have terms for all the differences of each pair of multipliers squared.

AgreedQuote:MentalThis Fortune X game is trickier because you have to split the deals into high variance deals and low variance deals,

I'm not certain what you are doing here, but one of us is off by a factor of 4 somewhere. This is why I generally like using coins rather than 5-coin bets when bets are actually 10 coins each. I'm too tired to want to check right now, but I did notice that my number of 143 is right around 1/4 of the number that you saw in your simulations. If the numbers you used from your simulations were in terms of 5-coin bets, then you should have got a number that was about 4 times the number that I reported.Quote:MentalThe variance for DDB seems to be around 571 to 581 bets^2 (using 10-coin bets). The variance expressed in terms of 5-coin bets is around 2300 bets^2. Yikes!

I looked back at my code and realized that I was actually calculating the 5-coin variance. Just divide my variance estimates by four and we agree now.Quote:drrockIf by "game" variance, you mean single hand AND using equilibrium long-run probabilities, there really is not too much involved in that but a lot of tedium. For DDB, you can look at the Wizard's page and it shows a line-by-line breakdown showing multiplier, base win, and probability. Unfortunately many of the high-valued hands have very low probabilities, but if you had his unrounded probabilities instead of what is in the table, it would be a simple matter of E[x^2] - [E(x)]^2.Quote:GaryJKoehlerAnyone up for working on the game variance?

link to original post

He does show Return on each line, and in general, it has more significant digits, so we can get more precision by using Probability = Return/(Multiplier*Base Win/2). His probabilities in that table sum to 0.999997. The probabilities derived from the formula above sum to 0.99999784. Both of them show an expected value of 0.990070, quite close to the figure he shows at the bottom of his table 0.990069.

Using the derived probabilities and not worrying too much that they don't sum to unity, I got an average second moment of 144.2964 and a variance of 143.3161 using squared 10-coin bets. If you want that in 5-coin bets squared, you can multiply by 4. Personally, I prefer to take square roots and just deal with standard deviations in terms of bets. Here, that would translate to 11.9715 bets. This could be compared to a regular 9-7 Triple Double Bonus game which has a standard deviation of 9.9137 bets. To me the comparison of 9.91 to 11.97 is more meaningful than a comparison of 98 to 143, when using variances.

But the "game" that is at videopoker.com is not single line. It is either Triple Play, Five Play, or Ten Play. I am not certain that Fortune X will be available anywhere soon at single line only. Any variance calculations of the actual game would involve convolutions and covariance. And of course the variances would be different depending upon which option and which base game was selected.

When I started this post with the qualifier "equilibrium long-run probabilities," I was avoiding the idea that in the short run the hands are not independent so even the single-hand game would require some rather involved covariance calculations unless we are dealing far enough in the future so that equilibrium can be considered.

link to original post

My corrected variance estimates based on 10-coin bets are:

DDB: 143.1 bets^2

JB: 68.7 bets^2

I only looked at the Wizard FX site briefly when this thread started. I did not realize (or forgot) he had probabilities for individual payoffs.

There is some value in doing calculations by a completely independent method from the Wizard just as a check.

Also, I could get the multi-line variance rather easily using the MC sim.

JOB:

RTP: 99.5646 VAR: 67.91 1-play

RTP: 99.5566 VAR: 89.89 3-play

RTP: 99.5578 VAR: 109.78 5-Play

RTP: 99.5687 VAR: 168.61 10-Play

DDB:

RTP: 98.9149 VAR: 143.09 1-play

RTP: 98.9173 VAR: 188.03 3-Play

RTP: 98.9411 VAR: 233.66 5-Play

RTP: 98.9599 VAR: 345.60 10-Play

One billion starting hands for each sim, so 10B final hands for the 10-play sim. That sim took 9 minutes to run.

It took me two minutes to code this extra multi-play mode once the code to calculate 1-play variance was in place.

DDB

Pi 0.63479140564632 0.078457364742804 0.078168526105095 0.076207274969926 0.073967291996158 0.058408136539694

EV 0.99006980284488

Var 143.3832424

JOB

Pi 0.6208048596121 0.081860516292936 0.081692194118271 0.081368563577346 0.075108691907174 0.059165174492176

EV 0.99599471006083

Var 66.27868599

The values seem to agree with the simulation.

It looks right, but I am curious how you did the calculation. Are you using the Wizard's tables or are you getting the variance of the unique deals/draws and starting from there?Quote:GaryJKoehlerAssuming I didn't screw-up, I just cycled through all hands for the 6 states using the optimal break points for the multiplier decisions and using the final steady-state probabilities (Pi below).

DDB

Pi 0.63479140564632 0.078457364742804 0.078168526105095 0.076207274969926 0.073967291996158 0.058408136539694

EV 0.99006980284488

Var 143.3832424

JOB

Pi 0.6208048596121 0.081860516292936 0.081692194118271 0.081368563577346 0.075108691907174 0.059165174492176

EV 0.99599471006083

Var 66.27868599

The values seem to agree with the simulation.

link to original post

I can easily calculate the variance of each draw relative to the EV. Can this be used to quickly calculate the multi-play variance?

As for multi-hand games, I didn’t compute the variances for them, but the wiz showed the values are linear in the number of lines of play, as I recall.

Sorry, Gary, but I confused you with another poster who does not have the raw outcomes for each hand. I should have been remembered from your article on "Ultimate X Poker Bonus Streak" that you have a detailed hand calculator.Quote:GaryJKoehlerFor each of the 134,459 unique hands, I have the probability of the hand, the number of cases for the outcomes of the optimal holds and the sum of outcomes and the sum of the square of outcomes. So I just loop through the list 6 times, one for each state adjusting values by the appropriate multiplier and weighting the probabilities by the steady state probabilities.

As for multi-hand games, I didn’t compute the variances for them, but the wiz showed the values are linear in the number of lines of play, as I recall.

link to original post

Variances for N-play VP games is not linear N due to the covariance of drawing to the same hold. It would be linear if every draw was to a different random dealt hand.

DDB

Pi 0.63479140564632 0.078457364742804 0.078168526105095 0.076207274969926 0.073967291996158 0.058408136539694

EV 0.99006980284488

Var 142.5663773

JOB

Pi 0.6208048596121 0.081860516292936 0.081692194118271 0.081368563577346 0.075108691907174 0.059165174492176

EV 0.99599471006083

Var 65.45201477

Needless to say, none of the streamers came out as winners, and most didn't even last the full 60 minutes. The pay tables used at Yaamava are challenging to say the least (7/5 DDB is atrocious), 7/5 Bonus is the best there and adds two one hundredths of one percent to the pay table. The commenters are saying that it is more volatile than Hot Roll, which is already volatile. Essentially you are getting a 1/8-1/10 chance of getting the multiplier, and once you get it 95% of the time you will roll to the 8X level (so three dead spins) and 70% of the time you will roll all the way to the 12x multiplier (so four dead spins). The rate of the multiplier happening needs to go up so that people can feel more comfortable playing it, or the added bet increment should go down to two like double super times pay. I gave it a whirl at five play quarters, and lost $430 in about thirty minutes.

Quote:calwatchYou can see the game being played here:

Needless to say, none of the streamers came out as winners, and most didn't even last the full 60 minutes. The pay tables used at Yaamava are challenging to say the least (7/5 DDB is atrocious), 7/5 Bonus is the best there and adds two one hundredths of one percent to the pay table. The commenters are saying that it is more volatile than Hot Roll, which is already volatile. Essentially you are getting a 1/8-1/10 chance of getting the multiplier, and once you get it 95% of the time you will roll to the 8X level (so three dead spins) and 70% of the time you will roll all the way to the 12x multiplier (so four dead spins). The rate of the multiplier happening needs to go up so that people can feel more comfortable playing it, or the added bet increment should go down to two like double super times pay. I gave it a whirl at five play quarters, and lost $430 in about thirty minutes.

link to original post

What is the variance of double super times pay?