Fortune X Poker

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January 25th, 2023 at 12:57:50 PM permalink

Quote: rsactuary
Curious what the decision point is to take a multiplier lower than 12x? ie: where does it tilt from passing to taking the multiplier?
link to original post

The answer to that question can be found in my page on Fever X Poker.

Last edited by: Wizard on Jan 26, 2023

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

miplet

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January 25th, 2023 at 1:28:09 PM permalink

"four of eights" typo instead of "pair of eights" in the 8x section.

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gordonm888
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January 25th, 2023 at 1:52:41 PM permalink

Quote: rsactuary
Curious what the decision point is to take a multiplier lower than 12x? ie: where does it tilt from passing to taking the multiplier?
link to original post

I've been scratching my ahead about that as well.

Wizard says for 9/6 DDB
2X multiplier: Indifferent EV = 8.901087
3X multiplier: Indifferent EV = 4.889740
5X multiplier: Indifferent EV = 2.567263
8X multiplier: Indifferent EV = 1.411099

I have been looking at that and I think
2X multiplier: made Full House or better, which includes
'4 to a Royal Flush'
'3 of a kind: Aces'
3X multiplier: made Flush or better
5X multiplier: made Two Pair or better, which includes
'4 to a Straight Flush'
'1 pair: Aces'
8X multiplier: Indifferent EV = 1.411099 (TBD - haven't done this one yet.)

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

gordonm888
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January 25th, 2023 at 1:58:43 PM permalink

typo: replace the first "enough" with "not"

"I am now at the fourth stage of the multiplier progression, at 8x. Like the previous hand, a low pair is enough enough to keep the multiplier, so I declined it. "

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Johnzimbo

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January 25th, 2023 at 2:06:58 PM permalink

Is it Fever X or Fortune X?

Mental

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January 25th, 2023 at 5:39:02 PM permalink

Are you using a Markov chain to do the calculations?
It seems that there is a tension between extending the sequence to get a larger denominator in the RTP and getting the lowest loss per cycle back to the starting point. If you were playing for points, then you want to take the decisions that get the highest RTP. If you are approaching it from the point of view that you will play one bonus cycle and want to lose the least amount, you would get a different strategy and EV.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Wizard
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January 26th, 2023 at 5:36:39 AM permalink

Quote: Johnzimbo
Is it Fever X or Fortune X?
link to original post

Fever X.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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January 26th, 2023 at 5:40:58 AM permalink

Quote: Mental
Are you using a Markov chain to do the calculations?
It seems that there is a tension between extending the sequence to get a larger denominator in the RTP and getting the lowest loss per cycle back to the starting point. If you were playing for points, then you want to take the decisions that get the highest RTP. If you are approaching it from the point of view that you will play one bonus cycle and want to lose the least amount, you would get a different strategy and EV.
link to original post

Yes, I'm using a Markov Chain, at least I think you could call my method that. At every decision point, my goal for the player is to win the most money, deducuting subsequent wagers until a multiplier is accepted. I don't factor points into it.

The strategy for what cards to hold are the same as conventional video poker. This game just adds another decision, to accept a multiplier or play on.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Mental

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January 26th, 2023 at 5:56:44 AM permalink

Quote: Wizard
Quote: Mental
Are you using a Markov chain to do the calculations?
It seems that there is a tension between extending the sequence to get a larger denominator in the RTP and getting the lowest loss per cycle back to the starting point. If you were playing for points, then you want to take the decisions that get the highest RTP. If you are approaching it from the point of view that you will play one bonus cycle and want to lose the least amount, you would get a different strategy and EV.
link to original post

Yes, I'm using a Markov Chain, at least I think you could call my method that. At every decision point, my goal for the player is to win the most money, deducuting subsequent wagers until a multiplier is accepted. I don't factor points into it.

The strategy for what cards to hold are the same as conventional video poker. This game just adds another decision, to accept a multiplier or play on.
link to original post

When you play the pass line in craps in a vacuum, you are indifferent to taking odds. If you get rated on the odds bets the same as the pass line, you will take the max odds bets. You are getting something of value for the extra action. If you were taxed on your action, you would never take the max odds bets. So, whether you try to increase your action on a craps game depends on external factors not fundamental to the game of craps. The absolute EV of each cycle is the same, but the coin in from the odds bets increases the denominator and increases the RTP.

I would not be surprised to find that the differences in the RTP has to do with whether you are trying to get the best absolute return or whether you are trying to get the best (return / action) ratio.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

rsactuary

rsactuary

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January 26th, 2023 at 6:07:16 AM permalink

It is "Fortune X" on the website. I'm just really not a fan of this variant. Slows you down while you think about making your decision. I like to get in a groove and go.

Wizard
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January 26th, 2023 at 8:21:27 AM permalink

Quote: Mental
I would not be surprised to find that the differences in the RTP has to do with whether you are trying to get the best absolute return or whether you are trying to get the best (return / action) ratio.
link to original post

One could make a similar comparison to doubling down or splitting in blackjack or raising games like Ultimate Texas Hold 'em.

I assume the player has a gun to his head and forced to play until he gets and the completes a bonus feature, with the player's goal to lose as little as possible in the process.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Mental

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January 26th, 2023 at 10:12:13 AM permalink

Quote: Wizard
Quote: Mental
I would not be surprised to find that the differences in the RTP has to do with whether you are trying to get the best absolute return or whether you are trying to get the best (return / action) ratio.
link to original post

One could make a similar comparison to doubling down or splitting in blackjack or raising games like Ultimate Texas Hold 'em.

I assume the player has a gun to his head and forced to play until he gets and the completes a bonus feature, with the player's goal to lose as little as possible in the process.
link to original post

Not playing at all would be the strategy that loses as little as possible, right? I only gamble because the follow-on benefits plus the RTP total greater than 100%. Most players gamble for action. In either case, there is an extrinsic benefit to playing an extra round rather than terminating the sequence.

I don't have a membership at VP.com so I cannot see the rules. However, your OP seems to imply that forgoing the multiplier and playing for the next hand requires new coin in. If I am wrong on this , then my analysis is off base.

You are trying to calculate a RTP for the game. This is return on bets, not absolute win/loss. Let us assume I find a game left at 8X for the current hand. I get dealt a hand that is exactly neutral for taking the 8X or playing again for the 12X. This implies that the current hand is ++EV, and playing both hands has identical absolute ++EV. What is the RTP of the two choices? If the absolute payoff of the 8X choice is X bets, I would rather play two hands to accrue X bets worth of expected credits, not one. The 8X choice allows you to pocket exactly X bets for one bet. The RTP is X. The 12X choice allows you to pocket an expected X bets for two bets, so the RTP is X/2. The variance of the 8X choice is zero. The RTP of the 12X choice is quite high. I earn more bennies from the 12X choice and I don't really care about the extra variance. Note that I will never be throwing away an 8X trips or better, so I am not making painful choices to play the extra hand at a neutral EV.

Note that this neutral EV choice is very similar to playing e-craps where the odds bets are rated. I get extra variance and cashback, but no immediate gain in EV, by taking odds.

If VP.com is calculating relative RTP and you are optimizing absolute immediate gain, then you will use different strategies and get a different RTP.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Mental

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January 26th, 2023 at 10:54:47 AM permalink

I propose a new game called Fortune NSUD poker. The player plays ordinary 9/6 DDB for an ordinary 5-coin bet. One in 10 games randomly triggers a bonus feature. If you accept this feature, the next game will be a normal 5-coin game of NSUD. If you decline, the bonus is over. Mike always declines this feature since the game of NSUD is negative EV to the tune of of 0.27% (99.73% RTP). I always accept this feature and play the NSUD game.

After 990 games, I will have played an average of 900 games of DDB and 90 games of NSUD. Mike will have played 990 games of DDB.

We will disagree on the RTP of the game because we are making different strategy choices regarding the bonus game. We will both be playing negative EV games unless future benefits are factored in.

If you want to get the optimal RTP of the Fortune X game, you need to realize that the alternative to playing an extra slightly negative EV bonus game in some circumstances is to play a worse base game. The strategy cutoff point should not be based on neutral EV. I don't think it was a coincidence that the 8X/12X decision point lies right in the middle of the 3RF and High Pair flops.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Mental

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January 26th, 2023 at 11:21:23 AM permalink

If you use a neutral EV cutoff point for your bonus round, you are basically saying "Let me get back to playing the base game as quickly as I can. After all, the base game has an RTP of < 50%. I want the highest possible percentage of my games to be base games, not the lucrative bonus games. Let me terminate this bonus round."

I would go in a different direction.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Wizard
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January 26th, 2023 at 11:35:39 AM permalink

Quote: Mental
I don't have a membership at VP.com so I cannot see the rules. However, your OP seems to imply that forgoing the multiplier and playing for the next hand requires new coin in. If I am wrong on this , then my analysis is off base.

Here are the rule screens.

There is a third screen on the game, but it seems to incorrectly apply to another game.

Yes, the player must put new coin in for the "next hand."

I analyze this game the way I analyze every game, as if a gun were pointed at the player's head and he were forced to play at least one time. The player's motive is to lose as little as possible. Since this is a 99%+ game, I don't think it makes much difference.

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"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Mental

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January 26th, 2023 at 11:47:07 AM permalink

Quote: Wizard
Yes, the player must put new coin in for the "next hand."

I analyze this game the way I analyze every game, as if a gun were pointed at the player's head and he were forced to play at least one time. The player's motive is to lose as little as possible. Since this is a 99%+ game, I don't think it makes much difference.

link to original post

Thanks for the screen shots of the rules.

I respectfully submit that trying 'to lose as little as possible' means favoring staying in the bonus games and avoiding the crushing -50.5% EV of the base games as much as possible.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Wizard
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January 26th, 2023 at 12:06:45 PM permalink

Quote: Mental
I respectfully submit that trying 'to lose as little as possible' means favoring staying in the bonus games and avoiding the crushing -50.5% EV of the base games as much as possible.
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I fully consider that in my analysis.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Mental

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January 26th, 2023 at 4:45:54 PM permalink

Quote: Wizard
Quote: Mental
I respectfully submit that trying 'to lose as little as possible' means favoring staying in the bonus games and avoiding the crushing -50.5% EV of the base games as much as possible.
link to original post

I fully consider that in my analysis.
link to original post

Maybe your cutoff is right, but you calculate the denominator wrong.

I did a Monte Carlo simulation of the game based on my understanding of the rules and using your cutoff numbers. I get a RTP of 99.0% for 100M hands. I will run a larger sim later.

I assume that the feature award 10.96% of the time when you are not already in a feature. The rules have a nonsense sentence regarding the probability.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Wizard
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January 26th, 2023 at 5:29:30 PM permalink

Quote: Mental

I did a Monte Carlo simulation of the game based on my understanding of the rules and using your cutoff numbers. I get a RTP of 99.0% for 100M hands. I will run a larger sim later.

Thank you for running the simulation. The mathematician at VideoPoker.com, who I'll call YH, gets 99.00% too.

To look at a possible point of the problem, consider the decision to accept or reject the 8X multiplier. I agree with YH on the probability of accepting the 2, 3, and 5x multipliers. In fact, here are both our probabilities of accepting at each level, assuming the player makes it this far.

Multipliers	YH	Wiz
2	0.003681	0.003681
3	0.024998	0.024998
5	0.028550	0.028550
8	0.198314	0.199515

You can see they accept the 8X multiplier more often than I do.

Here are some of my figures for the 8X level:

Indifference EV of base hand = 1.411099
Probability accept 8x = 0.211626
EV given multiplier accepted = 2.357209
EV given multiplier rejected = 0.622751
Total EV = 10.269053

Overall, I show winning the bonus is worth 9.920242 credits. To achieve the 99% RTP figure of YH, I'd need to get to 10.0256816.

Quote:
The rules have a nonsense sentence regarding the probability.

My rules or theirs?

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"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Mental

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January 26th, 2023 at 5:46:14 PM permalink

Quote: Wizard
Quote: Mental

I did a Monte Carlo simulation of the game based on my understanding of the rules and using your cutoff numbers. I get a RTP of 99.0% for 100M hands. I will run a larger sim later.

Thank you for running the simulation. The mathematician at VideoPoker.com, who I'll call YH, gets 99.00% too.

To look at a possible point of the problem, consider the decision to accept or reject the 8X multiplier. I agree with YH on the probability of accepting the 2, 3, and 5x multipliers. In fact, here are both our probabilities of accepting at each level, assuming the player makes it this far.

Multipliers YH Wiz
2 0.003681 0.003681
3 0.024998 0.024998
5 0.028550 0.028550
8 0.198314 0.199515

You can see they accept the 8X multiplier more often than I do.

Here are some of my figures for the 8X level:

Indifference EV of base hand = 1.411099
Probability accept 8x = 0.211626
EV given multiplier accepted = 2.357209
EV given multiplier rejected = 0.622751
Total EV = 10.269053

Quote:
The rules have a nonsense sentence regarding the probability.

My rules or theirs?

link to original post

link to original post

Their rules in the screenshot you provided: "The feature is awarded an average of 10.96% when active." Percentage of what?

I will try to pull out the decision percentages from my sims.

We cannot determine the overall RTP from simply knowing the absolute value of getting the bonus round. We also need to know the average coin in on each bonus round. I assume you derive the average length of the bonus round from your decision probabilities .If they are off, then your average length of the bonus round will be off.

A MC sim is not a good way to get the precise RTP, but it is so trivial to program that I am unlikely to have fubared it.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Wizard
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January 26th, 2023 at 6:07:38 PM permalink

Quote: Mental

Their rules in the screenshot you provided: "The feature is awarded an average of 10.96% when active." Percentage of what?

I assume they mean percentage of bets, when the previous hand did not reject a multiplier. In other words a clean hand, with no entitled multiplier.

Quote:

I will try to pull out the decision percentages from my sims.

We cannot determine the overall RTP from simply knowing the absolute value of getting the bonus round. We also need to know the average coin in on each bonus round. I assume you derive the average length of the bonus round from your decision probabilities .If they are off, then your average length of the bonus round will be off.

A MC sim is not a good way to get the precise RTP, but it is so trivial to program that I am unlikely to have fubared it.

Okay, I think I see where our point of departure is. Thank you for your help. Let me go back to the drawing board.

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"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Mental

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January 26th, 2023 at 7:12:38 PM permalink

Quote: Wizard

I calculated the cutoff percentages using your cutoff EVs. These do not agree with your numbers. I have not calculated my own cutoff values. I just used yours. Has YH shared their cutoff EV numbers with you?
Multipliers YH Wiz Mental
2 0.003681 0.003681 0.0036815
3 0.024998 0.024998 0.0250900
5 0.028550 0.028550 0.0293933
8 0.198314 0.199515 0.2116262

I am not even using weighted canonical hands. I am looping through all 2598960 deals to make the counting very simple. I was expecting to agree with your numbers.

link to original post

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

drrock

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January 27th, 2023 at 5:02:26 AM permalink

Without my own program, I get the exact same percentages for the Wizard�s cutoffs that Mental did. I used the strategy feature from Video Poker for Winners (VPFW). It has a table with Perfect Play Occurrences and Average EVs for various holds listed from high EVs to low. I did have to separate the open-ended 4-card straight flush draws from inside draws since VPFW doesn't and you would select to play the former at 5x while proceeding to 8x with the latter. Fortunately, the cutoffs in this case are not too near holds that depend on complex penalty card situations that put a hold on one side of the cutoff in one case and the other side in another.

I wonder whether a confidence interval for the mean of Mental�s simulation would allow for the possibility of a 98.9% EV instead of 99.0%. With direct calculation shown below, I get 98.8783% as the EV for long-term play using the Wizard�s cutoffs.

(Wizard, your blue table in your post conflicts with what you wrote. Under 8x, your table has the cell Wiz 0.199515; the prose below says your probability of accepting 8x is 0.211626, which agrees with Mental�s number in his blue table. Also, you say �they accept the 8X multiplier more often than I do,� when 0.198314 is LESS than both 0.199515 AND 0.211626, meaning YH accepts 8x less than you do, whichever of the two figures is correct.)

For purpose of this post (until the end), I�ll assume that the Wiz cutoffs are accurate � and that the figures in VPFW are accurate. The method below will still work if you eventually settle on more accurate figures. Just plug and chug.

1. Use a table of holds and their frequencies, ordered by their EVs. Use the cutoffs to determine the percentage or counts of holds above the cutoff and below the cutoff. Calculate the two average numbers of coins expected per hand above the cutoff and below the cutoff. Determine and keep the composite average coins expected for each multiplier. With 1x and 12x there is no cutoff, just a simple average. With 2x, 3x, 5x, and 8x, you get a weighted average of two averages.

2. Determine the probabilities of the next hand�s multiplier dependent on the current hand�s multiplier, starting from 1x. Work your way forward until you get to 12x.

3. For each of the 6 possible multipliers, set up an input/output equation to represent the interplay between the multipliers. These could correspond to rows in the transpose of a transition matrix of a Markov process if that is your line of thinking.

4. Replace one of those equations with a different equation that says the probabilities of the 6 multipliers must sum to one. (This is necessary because the 6 equations in #3 will not be independent.) The new tautological equation is that p1 + p2 + p3 + p5 + p8 + p12 = 1.

5. Solve that system of 6 equations for the probabilities. Multiply those probabilities by the expected coins calculated in #1; sum those products and divide by the 10 coins per hand investment to get the EV.

Below, I will start with step 1 for each of the 6 multipliers, calculating the expected coins won for each hand from 1x to 12x. I will note the transition probabilities of moving from a current multiplier to the next higher multiplier and note also the probabilities of moving back to 1x or 2x after accepting a multiplier or being forced to accept 12x.

Multiplier 1x.

Next hand�s multiplier: 89.04% 1x; 10.96% 2x.

Expected Coins = 5 x 0.9898078 = 4.949039.

Multiplier 2x.

0.36815% of hands are above cutoff of 8.901087 bets. Expected Coins = 2 x 84.363259 = 168.726518. Next hand�s multiplier: 0.36815% x 89.04% will be 1x; 0.36815% x 10.96% will be 2x.
99.63185% of hands are below cutoff of 8.901087 bets. Expected Coins = 4.655597. Next multiplier: 99.63185% of the time will be 3x.

Average expected coins for 2x: 0.36815% x 168.726518 + 99.63185% x 4.655597 = 5.259620.

Multiplier 3x.

2.50900% of hands are above cutoff of 4.889740 bets. Expected Coins = 3 x 37.581251 = 112.743753. Next hand�s multiplier: 2.50900% x 89.04% will be 1x; 2.50900% x 10.96% will be 2x.
97.49100% of hands are below cutoff of 4.889740 bets. Expected Coins = 4.109225. Next multiplier: 97.49100% of the time will be 5x.

Average expected coins for 3x: 2.50900% x 112.743753 + 97.49100% x 4.109225 = 6.834869.

Multiplier 5x.

2.93933% above cutoff of 2.567263 bets, EV 5 x 34.976559 = 174.882793; 97.06067% below, EV 4.039703. 2.93933% splits 89.04%/10.96% between next multipliers 1 and 2, while 97.06067% goes to 8x.

Composite Expected Coins: 9.061344.

Multiplier 8x.

21.16262% above cutoff of 1.411099 bets, EV 8 x 11.786043 = 94.288344; 78.83738% below, EV 3.113756. 21.16262% splits 89.04%/10.96% between multipliers 1 and 2, while 78.83738% goes to 12x.

Composite Expected Coins: 22.408686.

Multiplier 12x.

Next Multiplier: 89.04% 1x, 10.96% 2x.

Expected Coins: 12 x 4.949039 = 59.388467.

Call the probabilities for the multipliers p1, p2, p3, p5, p8, and p12. From above we know the relationships between all the probabilities.

E.g., you can get a hand with a 12x multiplier only if the previous hand had an 8x multiplier. A 12x multiplier follows an 8x multiplier 78.83738% of that time. So, we have p12 = 0.7883783 p8.

We also have p8 = 0.9706067 p5, p5 = 0.9749100 p3, p3 = 0.9963185 p2, and the two bigger equations

p1 = .8904 x (p1 + 0.0036815 p2 + 0.0250900 p3 + 0.0293933 p5 + 0.2116262 p8 + p12) and
p2 = .1096 x (p1 + 0.0036815 p2 + 0.0250900 p3 + 0.0293933 p5 + 0.2116262 p8 + p12)

It is important to remember Step 4 above if you want to solve for the 6 probabilities. Replace one of the equations with p1 + p2 + p3 + p5 + p8 + p12 = 1. If you are doing this by hand, you may want to replace one of the more complicated equations. If you have a computer solution, it doesn�t matter which equation is replaced.

You can verify that probabilities below satisfy those 6 equations. Supplying expected coins for each multiplier allows us to find the game EV.

1x: 63.5799%, 4.94904
2x: 7.8261%, 5.25962
3x: 7.7973%, 6.83487
5x: 7.6017%, 9.06134
8x: 7.3782%, 22.40869
12x: 5.8168%, 59.38847, totaling to 9.88783 coins. Dividing by 10 coins per hand yield 98.8783%.

This methodology does not guarantee optimality if the cutoffs are not chosen optimally. E.g., I can raise the EV slightly to 98.8784% by raising the 8x cutoff to something like 1.414304. But if YH or you found a better way to determine cutoffs, you can still use the method above to determine the EV.

Last edited by: drrock on Jan 27, 2023

Mental

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January 27th, 2023 at 5:14:34 AM permalink

Quote: drrock
This methodology does not guarantee optimality if the cutoffs are not chosen optimally. E.g., I can raise the EV slightly to 98.8784% by raising the 8x cutoff to something like 1.414304. But if YH or you found a better way to determine cutoffs, you can still use the method above to determine the EV.
link to original post

Nice work. It ill take some time to digest.

I was going to double check my cutoff percentages. With your corroboration, I think I will create a table with the cumulative cutoff probabilities versus EV instead. You need these for your method, and I can provide an independent calculation.

I emphasize that I have not worked out the cutoff values on my own. I will try to understand your method and see if it answers the questions I initially raised about optimizing for max gain until the end of the bonus versus max running RTP. The latter would lead to higher cutoffs (keeping the bonus round going longer).

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Mental

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January 27th, 2023 at 6:11:55 AM permalink

If you click the button you will see my cumulative list of hand counts for various cutoff EVs for 9/6 DDB. You can paste into a CSV file and open in a spreadsheet (using a comma delimiter). To convert to probabilities, divide by 2598960.

 
800.0000,4
400.0000,16
221.2766,52
160.0000,88
100.4255,196
50.0000,664
19.6809,760
19.6596,796
19.5957,820
19.5532,848
18.5532,944
18.5319,992
18.4894,1316
18.4681,1436
18.4255,1468
18.4043,1504
18.3617,1600
12.7169,1816
12.7114,4120
12.4894,5848
12.2729,5920
12.2673,6112
9.0000,9568
7.5310,16480
7.4570,21664
7.3830,22240
6.0000,27192
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0.3256,2555496
0.3256,2555640
0.3256,2555904
0.3256,2556000
0.3256,2556048
0.3256,2556072
0.3256,2556144
0.3256,2556360
0.3256,2556408
0.3256,2556576
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0.3255,2556708
0.3255,2556732
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0.3246,2578536
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0.3245,2578944
0.3245,2579088
0.3245,2579112
0.3245,2579208
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0.3245,2580336
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0.3244,2580960
0.3244,2581056
0.3244,2581176
0.3244,2581224
0.3244,2581440
0.3244,2581512
0.3244,2581752
0.3244,2581944
0.3244,2582088
0.3244,2582160
0.3244,2582232
0.3244,2582280
0.3244,2582520
0.3244,2582616
0.3244,2582664
0.3243,2582688
0.3243,2582760
0.3243,2583000
0.3243,2583120
0.3243,2583240
0.3243,2583336
0.3243,2583360
0.3243,2583432
0.3243,2583648
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0.3242,2585880
0.3242,2586048
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0.3242,2586696
0.3242,2586768
0.3241,2587080
0.3241,2587320
0.3241,2587368
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0.3241,2587584
0.3241,2587656
0.3241,2587824
0.3241,2587992
0.3241,2588328
0.3240,2588424
0.3240,2588592
0.3240,2588640
0.3240,2588760
0.3240,2589432
0.3240,2589576
0.3240,2589768
0.3240,2589984
0.3240,2590080
0.3240,2590176
0.3240,2590344
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0.3239,2590464
0.3239,2590656
0.3239,2590776
0.3239,2590800
0.3239,2590968
0.3239,2591160
0.3239,2591376
0.3239,2591448
0.3239,2591544
0.3239,2591688
0.3239,2591736
0.3239,2591808
0.3239,2591904
0.3239,2592000
0.3239,2592072
0.3239,2592096
0.3239,2592192
0.3238,2592240
0.3238,2592408
0.3238,2592432
0.3238,2592504
0.3238,2592552
0.3238,2592600
0.3238,2592816
0.3238,2592912
0.3238,2592960
0.3238,2593008
0.3238,2593176
0.3238,2593368
0.3238,2593392
0.3238,2593440
0.3238,2593632
0.3238,2593656
0.3238,2593776
0.3237,2593944
0.3237,2594232
0.3237,2594280
0.3237,2594448
0.3237,2594544
0.3237,2594568
0.3237,2594616
0.3237,2594760
0.3237,2594880
0.3237,2595120
0.3237,2595312
0.3236,2595456
0.3236,2595840
0.3236,2595936
0.3236,2595984
0.3235,2596152
0.3235,2596224
0.3235,2596560
0.3233,2596608
0.3233,2596824
0.3233,2596920
0.3232,2597040
0.3232,2597064
0.3232,2597304
0.3232,2597400
0.3232,2597544
0.3232,2597568
0.3232,2597760
0.3232,2597856
0.3231,2597952
0.3231,2597976
0.3231,2598072
0.3231,2598144
0.3231,2598576
0.3231,2598672
0.3231,2598696
0.3230,2598960

.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Wizard
Administrator

Threads: 1518
Posts: 27037

Joined: Oct 14, 2009

January 27th, 2023 at 6:16:24 AM permalink

Good post, drrock. It would take me hours to absorb all of it.

Quote:
(Wizard, your blue table in your post conflicts with what you wrote. Under 8x, your table has the cell Wiz 0.199515; the prose below says your probability of accepting 8x is 0.211626, which agrees with Mental�s number in his blue table. Also, you say �they accept the 8X multiplier more often than I do,� when 0.198314 is LESS than both 0.199515 AND 0.211626, meaning YH accepts 8x less than you do, whichever of the two figures is correct.)

Yes, I my prose below the table was wrong.

Before I go on, I will say that if I assume exactly a 9% probability of triggering the feature when not already in the feature as well as define the RTP as the ratio of money won to all money bet, then I match the RTP of YH. Maybe it's just a coincidence. I also don't know if the 10.96% probability is outside the feature or over all bets.

However, for this post, I'd like to focus on just the 8X decision.

The value of getting to the 12X multiplier is 12*0.989808 - 2 = 9.877694018.

We want to find ev, such that the player is the player is indifferent to accepting or rejecting the 8x multiplier.

reject 8x = accept 8x
9.877694018 + ev = 8*ev
9.877694018 = 7*ev
ev = 9.877694018/7 = 1.41109914

In a custom program I wrote, I show the probability of reaching 1.411099145 is 0.211626.

Assuming 0.211626 is reached, I show the average return is 2.357209.
Assuming 0.211626 is not reached, I show the average return is 0.622751.

Note that 0.211626*2.357209 + 0.211626*0.622751 = 0.989808.

I show the total value of reaching the 8X multiplier is thus:

0.211626 * 8 * 2.357209 + (1-0.211626) * (0.622751 + 9.877694) - 2 = 10.269053

Does anyone disagree with that?

.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

drrock

Threads: 3
Posts: 99

Joined: Mar 6, 2012

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January 27th, 2023 at 6:45:04 AM permalink

Thanks, Mental. I actually used a transition matrix and modeled this problem as a Markov process, but put together the play-by-play arithmetic calculations shown above so that it could be more easily digested, checked, and replicated. If you have a detailed table of the frequencies by each exact EV, using the transition matrix, you can calculate what Gary Koehler calls the biases of being in each state.

Then to calculate the cutoffs, if z is the EV of a hold without a multiplier, your decisions are to accept the multiplier if:

2 z2 + V(1) > z2 + V(3),
3 z3 + V(1) > z3 + V(5),
5 z5 + V(1) > z5 + V(8),
8 z8 + V(1) > z8 + V(12)

Your decisions are then to accept the multiplier

2x when z2 > V(3) - V(1),
3x when z3 > [V(5) - V(1)] / 2
5x when z5 > [V(8) - V(1)] / 4, and
8x when z8 > [V(12) - V(1)] / 7

Unfortunately, you cannot calculate these V's before you go through the calculations outlined earlier, so you need to first have some starting point for the cutoffs, and then calculating them iteratively, and hoping for convergence.

You can get starting values by just equating what would happen on the next couple hands depending upon what decision you would make. E.g., if y is the EV of the game with no multiplier, solve for z8 thru z2 backwards, by looking at the next two hands, and for simplicity assuming that the next hand you don't get a multiplier. You could make other assumptions, but these are just starting values, so there is a trade-off between simplicity and accuracy

8 z8 + y > z8 + 12y or z8 > 11y / 7

Analysis of this would allow you to figure out what the average payout would be when you have an 8x multiplier. You can use this in the next equation setting y8 to that average payout.

5 z5 + y > z5 + y8 or z5 > (y8 - y) / 4

Similarly,

3 z3 + y > z3 + y4 or z3 > (y4 - y) / 2

and

2 z2 + y > z2 + y2 or z2 > (y2 - y)

If you wanted to have y take into account the possibility that the next hand would have a 2x multiplier 10.96% of the time, you could go back and rework those equations, but I think it would not appreciably shorten the original iterative solution that we are seeking.

Good luck and happy calculating!

Mental

Threads: 16
Posts: 1549

Joined: Dec 10, 2018

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January 27th, 2023 at 7:27:26 AM permalink

Quote: Wizard
Good post, drrock. It would take me hours to absorb all of it.

Quote:
(Wizard, your blue table in your post conflicts with what you wrote. Under 8x, your table has the cell Wiz 0.199515; the prose below says your probability of accepting 8x is 0.211626, which agrees with Mental�s number in his blue table. Also, you say �they accept the 8X multiplier more often than I do,� when 0.198314 is LESS than both 0.199515 AND 0.211626, meaning YH accepts 8x less than you do, whichever of the two figures is correct.)

Yes, I my prose below the table was wrong.

Before I go on, I will say that if I assume exactly a 9% probability of triggering the feature when not already in the feature as well as define the RTP as the ratio of money won to all money bet, then I match the RTP of YH. Maybe it's just a coincidence. I also don't know if the 10.96% probability is outside the feature or over all bets.

However, for this post, I'd like to focus on just the 8X decision.

The value of getting to the 12X multiplier is 12*0.989808 - 2 = 9.877694018.

We want to find ev, such that the player is the player is indifferent to accepting or rejecting the 8x multiplier.

reject 8x = accept 8x
9.877694018 + ev = 8*ev
9.877694018 = 7*ev
ev = 9.877694018/7 = 1.41109914

In a custom program I wrote, I show the probability of reaching 1.411099145 is 0.211626.

Assuming 0.211626 is reached, I show the average return is 2.357209.
Assuming 0.211626 is not reached, I show the average return is 0.622751.

Note that 0.211626*2.357209 + 0.211626*0.622751 = 0.989808.

I show the total value of reaching the 8X multiplier is thus:

0.211626 * 8 * 2.357209 + (1-0.211626) * (0.622751 + 9.877694) - 2 = 10.269053

Does anyone disagree with that? .
link to original post

I am happy that we settled the issue of the probabilities for your cutoff values. However, YH must be using different cutoff values if the probabilities are right. You must be aware that 1.446068 is a critical EV value. All of the JJ,QQ,KK holds have this EV, so setting the cutoff above or below this value is a big deal. When I look at the table, I cannot see any cutoff that would yield YH's probabilities. Maybe a transcription error?

I still disagree that you can optimize the cutoffs in a vacuum without considering how the cutoff affects the fraction of hands spent in the bonus round. I think the drrock's Markoff model addresses this aspect of the problem.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

drrock

Threads: 3
Posts: 99

Joined: Mar 6, 2012

Thanked by

January 27th, 2023 at 9:47:17 AM permalink

Quote: Mental
If you click the button you will see my cumulative list of hand counts for various cutoff EVs for 9/6 DDB. You can paste into a CSV file and open in a spreadsheet (using a comma delimiter). To convert to probabilities, divide by 2598960.

link to original post

Very cool! Yes, this is exactly the type of table necessary to carry out the analysis to the finish. I particularly like having the hand counts rather than probabilities, so that you can deal with any possible rounding error and also perform checks.

E.g., I had the same concern about YH's 8x cutoff that you noted in a later post. It seems the only way that probability of acceptance could occur would be if the cutoff point was exactly equal to the EV for a pair of Js, Qs, or Ks AND there was some arbitrary acceptance for some of the holds while moving to 12x for some others.

Because some of the frequencies of hands are in the hundreds of thousands (like those with a high pair), I would prefer to have three more decimal places in the EVs for the holds so that I could be more certain that rounding error doesn't creep into my final EV calculations.

However, I would think this table is perfect for checking out alternative cutoffs by trial and error even if we could not figure out the exact analytical method for determining them.

gordonm888
Administrator

Threads: 61
Posts: 5359

Joined: Feb 18, 2015

January 27th, 2023 at 11:17:08 AM permalink

This is turning into a really outstanding thread, especially thanks to mental and drrock (a member I have not been familiar with.) I really love the table mental contributed, thanks. And drrock's explanation of the analytic approach.

One game aspect I want to ask about again, for clarity. When you are "in the bonus feature" and deciding to accept or reject the multipliers, you only wager 5 coins in for each hand and are not paying the 5 coin fee (per ahnd) for the feature, correct? because when you are deciding between an 8x multiplier or a 12x multiplier you do not have a 10.9% chance of starting a new bonus feature sequence on that hand, correct?

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

drrock

Threads: 3
Posts: 99

Joined: Mar 6, 2012

January 27th, 2023 at 11:29:18 AM permalink

Quote: Wizard
Good post, drrock. It would take me hours to absorb all of it.

Yes, I my prose below the table was wrong.

Before I go on, I will say that if I assume exactly a 9% probability of triggering the feature when not already in the feature as well as define the RTP as the ratio of money won to all money bet, then I match the RTP of YH. Maybe it's just a coincidence. I also don't know if the 10.96% probability is outside the feature or over all bets.

However, for this post, I'd like to focus on just the 8X decision.

The value of getting to the 12X multiplier is 12*0.989808 - 2 = 9.877694018.

We want to find ev, such that the player is the player is indifferent to accepting or rejecting the 8x multiplier.

reject 8x = accept 8x
9.877694018 + ev = 8*ev
9.877694018 = 7*ev
ev = 9.877694018/7 = 1.41109914

In a custom program I wrote, I show the probability of reaching 1.411099145 is 0.211626.

Assuming 0.211626 is reached, I show the average return is 2.357209.
Assuming 0.211626 is not reached, I show the average return is 0.622751.

Note that 0.211626*2.357209 + 0.211626*0.622751 = 0.989808.

I show the total value of reaching the 8X multiplier is thus:

0.211626 * 8 * 2.357209 + (1-0.211626) * (0.622751 + 9.877694) - 2 = 10.269053

Does anyone disagree with that?
link to original post

Thanks Wiz for the comments.

But, yes, unfortunately I have to disagree with the way you calculated this cutoff, even though it gives you an oh-so-close to correct answer. On the left hand side of the equation, you are considering not only the current hand, but also the next hand. On the right hand side of the equation, you are only considering the current hand. You are subtracting 2 bets from the LHS. Perhaps the RHS has one bet implicitly subtracted from it. A way of getting closer to the answer might be to subtract another house edge from the RHS. This would increase that cutoff to a higher number that is closer to what I got as an initial cutoff. This may fix that number and may be close enough for government work as they say, but you will never know when the differences will be significant enough to torpedo your calculations.

An alternative approach follows. I have already discussed a way to shortcut through these calculations in an earlier post. What follows is really to try and convince you that abandoning the static approach considering a single bonus cycle while adopting the difference in values of the two different streams of future hands depending upon the decision to accept the multiplier is the way to go.

To even up the two sides of the equation, consider what will happen next hand if you accept 8x this time. The next hand on the right hand side will be a 1x hand 89.04% of the time and a 2x time 10.96% of the time.

And the 2nd hand after the current hand will also have a different distribution of multipliers depending upon your decision in the current hand. The same issue will occur on the 3rd, 4th, 5th, etc. hands until the differences die out 15 or 20 hands in the future.

Implicitly, the left hand side of your equation has the house edge for 20 coins in it (2 hands x 10 coins) while the right hand side has the house edge only for 10 coins. So, this is sort of an apples and oranges comparison.

A problem with trying to think of this problem as a single bonus cycle analysis is that there are two different forces working simultaneously that affect the length of the bonus.

If you start with the first hand of a bonus cycle when the 2x multiplier appears (using the probabilities for your cutoffs), we see that the bonus cycle could end on that hand or end on any one of the next 4 hands with probabilities:

1 hand: 0.36815%
2 hands: (1- 0.36815%) x 2.50900%
3 hands: (1- 0.36815%) x (1 - 2.50900%) x 2.93933%
4 hands: (1- 0.36815%) x (1 - 2.50900%) x (1 - 2.93933%) x 21.16262%
5 hands: (1- 0.36815%) x (1 - 2.50900%) x (1 - 2.93933%) x (1 - 21.16262%)

Although it may seem simple enough to try to consider only these possibilities in forming our opinion of the game and determine the cutoffs, we also must consider that it takes from 0 to infinity hands for any bonus to start because we have to include the 1x hands when considering the variable length of a cycle. So before the cycle above can start, we will have a string of 1x hands with these probabilities:

0 hands: 10.96%
1 hand: (1 - 10.96%) x 10.96%
2 hands: (1 - 10.96%)^2 x 10.96%
3 hands: (1 - 10.96%)^3 x 10.96%
4 hands: (1 - 10.96%)^4 x 10.96%, and so on.

Now, I'm not saying that we can't use a couple geometric series to calculate all the equalities necessary for the various indifferent cutoffs, but the complications of the varying starts and finishes of any one bonus cycle will make this comparison different.

I won't repeat my previous post where I showed how a cutoff could be calculated, but I will say that the calculation that you have already done will be fine for a starting estimate, and in fact will practically be quite close to the final cutoff. Additionally, since the list of EVs possible is discrete with some large empty spaces in it, you may get the effect of optimality even if your cutoffs are off by a fraction. As long as a candidate cutoff is inside the same interval as the cutoff from a more accurate analysis, the effects will be just the same as if you had the "truer" cutoff.

The other way of thinking of this is to imagine not only what occurs on the current and next hands but also to think of what will occur on the next 15 or so hands depending upon whether you choose to accept the current multiplier or wait at least one more hand until the next one.

I'll illustrate by showing what will occur on the next 5 hands, depending upon whether you accept an 8x or wait until the next hand for the 12 x. In order to determine the cutoff, you will need to know the expected value that will occur on all the future hands to compare with the difference between 8 x EV of current hand minus EV of current hand (in coins). As I have said earlier, until iteration is completed you may not have the EVs for 2x thru 8x, but you can make an estimate, which will likely give you reasonably close calculations for candidate cutoffs.

Multiplier	1x	2x	3x	5x	8x	12x
EV coins	4.94904	5.25962	6.83487	9.06134	22.40869	59.38847

Hand #	1x	2x	3x	5x	8x	12x	EV
1	0.89040	0.10960	0	0	0	0	4.9831
2	0.79317	0.09763	0.10920	0	0	0	5.1853
3	0.70900	0.08727	0.09727	0.10646	0	0	5.5974
4	0.63654	0.07835	0.08695	0.09483	0.10333	0	7.3314
5	0.59081	0.07272	0.07806	0.08477	0.09204	0.08159	11.5163
6	0.62015	0.07633	0.07246	0.07610	0.08228	0.07268	10.8157

At around Hand 18 or so, these probabilities will settle down to an equilibrium distribution with future hands having nearly identical probabilities and expectations.

If you reject 8x, you will get 12x and the next 6 hands will look like this:

Hand #	1x	2x	3x	5x	8x	12x	EV
1	0	0	0	0	0	1	59.38847
2	0.89040	0.10960	0	0	0	0	4.9831
3	0.79317	0.09763	0.10920	0	0	0	5.1853
4	0.70900	0.08727	0.09727	0.10646	0	0	5.5974
5	0.63654	0.07835	0.08695	0.09483	0.10333	0	7.3314
6	0.59081	0.07272	0.07806	0.08477	0.09204	0.08159	11.5163

At around Hand 18 or so, these probabilities will also settle down to an equilibrium distribution with future hands having nearly identical probabilities and expectations. Even though hands 2 thru infinity here are the same as hands 1 thru infinity in the next table, taking the staggered difference of hands 2 thru infinity in the two tables will NOT be zero. And this difference is important in considering whether or not to accept the 8x multiplier.

Last edited by: drrock on Jan 27, 2023

drrock

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January 27th, 2023 at 11:48:44 AM permalink

Quote: gordonm888
This is turning into a really outstanding thread, especially thanks to mental and drrock (a member I have not been familiar with.) I really love the table mental contributed, thanks. And drrock's explanation of the analytic approach.

One game aspect I want to ask about again, for clarity. When you are "in the bonus feature" and deciding to accept or reject the multipliers, you only wager 5 coins in for each hand and are not paying the 5 coin fee (per ahnd) for the feature, correct? because when you are deciding between an 8x multiplier or a 12x multiplier you do not have a 10.9% chance of starting a new bonus feature sequence on that hand, correct?
link to original post

No, it is a 10-coin wager per hand on each and every hand; so you are losing more than half on most hands. But the 12x hands occur fairly regularly, and you clean up on those. Here is what the game experience will be, in expected coins:

Multiplier	Wager	Exp. Payout	G/L in %	% @ this multiplier
1x	10	4.9490	-50.51%	63.580%
2x	10	5.2596	-47.40%	7.826%
3x	10	6.8349	-31.65%	7.797%
5x	10	9.0613	- 9.39%	7.602%
8x	10	22.4087	124.09%	7.378%
12x	10	59.3885	493.88%	5.817%
Avg	10	9.8878	- 1.12%	Sum = 100%

The expected payouts and probabilities are based on my calculations, which haven't been verified, and the Wizard's cutoffs, so you could expect some differences. Plus, of course, no games play exactly as expected.

Last edited by: drrock on Jan 27, 2023

gordonm888
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January 27th, 2023 at 11:49:27 AM permalink

Quote: Wizard
Good post, drrock. It would take me hours to absorb all of it.

Quote:
(Wizard, your blue table in your post conflicts with what you wrote. Under 8x, your table has the cell Wiz 0.199515; the prose below says your probability of accepting 8x is 0.211626, which agrees with Mental�s number in his blue table. Also, you say �they accept the 8X multiplier more often than I do,� when 0.198314 is LESS than both 0.199515 AND 0.211626, meaning YH accepts 8x less than you do, whichever of the two figures is correct.)

Yes, I my prose below the table was wrong.

Before I go on, I will say that if I assume exactly a 9% probability of triggering the feature when not already in the feature as well as define the RTP as the ratio of money won to all money bet, then I match the RTP of YH. Maybe it's just a coincidence. I also don't know if the 10.96% probability is outside the feature or over all bets.

However, for this post, I'd like to focus on just the 8X decision.

The value of getting to the 12X multiplier is 12*0.989808 - 2 = 9.877694018.

We want to find ev, such that the player is the player is indifferent to accepting or rejecting the 8x multiplier.

reject 8x = accept 8x
9.877694018 + ev = 8*ev
9.877694018 = 7*ev
ev = 9.877694018/7 = 1.41109914

In a custom program I wrote, I show the probability of reaching 1.411099145 is 0.211626.

Assuming 0.211626 is reached, I show the average return is 2.357209.
Assuming 0.211626 is not reached, I show the average return is 0.622751.

Note that 0.211626*2.357209 + 0.211626*0.622751 = 0.989808.

I show the total value of reaching the 8X multiplier is thus:

0.211626 * 8 * 2.357209 + (1-0.211626) * (0.622751 + 9.877694) - 2 = 10.269053

Does anyone disagree with that?
link to original post

Just noting a typo in the line that I bolded. That line should read "0.211626*2.357209 + (1-0.211626)*0.622751 = 0.989808" The results are unchanged.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Mental

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January 27th, 2023 at 12:19:15 PM permalink

You can put a text table in a code block and it will be uniformly spaced as would on a xterm. Multiple spaces will not be collapsed. Here is a short snippet. I am trying to reproduce your numbers in the MC sim. They look correct, but i need to do a longer sim to get the values to closer to the analytical results.


 1     -nan  0.92177  PX:   63.5959  AveX:    4.9511
 2  0.00363  0.00028  PX:    7.8228  AveX:    5.2909
 3  0.02524  0.00197  PX:    7.7944  AveX:    6.8180
 5  0.02919  0.00222  PX:    7.5977  AveX:    8.9405
 8  0.21188  0.01563  PX:    7.3759  AveX:   22.3139
12     -nan  0.05813  PX:    5.8132  AveX:    4.9294
aveMult: 1.7619316

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Mental

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January 27th, 2023 at 12:40:59 PM permalink

Quote: drrock
Quote: gordonm888
This is turning into a really outstanding thread, especially thanks to mental and drrock (a member I have not been familiar with.) I really love the table mental contributed, thanks. And drrock's explanation of the analytic approach.

One game aspect I want to ask about again, for clarity. When you are "in the bonus feature" and deciding to accept or reject the multipliers, you only wager 5 coins in for each hand and are not paying the 5 coin fee (per ahnd) for the feature, correct? because when you are deciding between an 8x multiplier or a 12x multiplier you do not have a 10.9% chance of starting a new bonus feature sequence on that hand, correct?
link to original post

No, it is a 10-coin wager per hand on each and every hand; so you are losing more than half on most hands. But the 12x hands occur fairly regularly, and you clean up on those. Here is what the game experience will be, in expected coins:

Multiplier Wager Exp. Payout G/L in % % @ this multiplier

1x 10 4.9490 -50.51% 63.580%

2x 10 5.2596 -47.40% 7.826%

3x 10 6.8349 -31.65% 7.797%

5x 10 9.0613 - 9.39% 7.602%

8x 10 22.4087 124.09% 7.378%

12x 10 59.3885 493.88% 5.817%

Avg 10 9.8878 - 1.12% Sum = 100%

The expected payouts and probabilities are based on my calculations, which haven't been verified, and the Wizard's cutoffs, so you could expect some differences. Plus, of course, no games play exactly as expected.
link to original post

Thanks for clarifying the cost of games in the bonus round. I had wondered about this, but I am using 10 coins for all games in the MC sims.

The need for a simultaneous equation solution is implicit in your equation: p1 + p2 + p3 + p5 + p8 + p12 = 1. If you tweak any probability, all the others must change. Your math is much better than all my handwaving arguments.

I get an average multiplier of only 1.76. Recall that the average multiplier in UX has to be around 2x to make the final EV similar to the base game. the reason 1.76 works here is that all the 2x, 3x, 5x, and 8xs multiply especially good starting hands. In UX, they multiply random starting hands.

Thanks for the really interesting math. My quibble is that you switched to coin EVs when the thread started with EV in units of 5-coin bets. With a 10-coin game, I guess this is clearer because the 'bet unit' could be defined as 5 or 10 coins.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

drrock

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January 27th, 2023 at 1:02:25 PM permalink

Quote: Mental
My quibble is that you switched to coin EVs when the thread started with EV in units of 5-coin bets. With a 10-coin game, I guess this is clearer because the 'bet unit' could be defined as 5 or 10 coins.
link to original post

Yep, I do it precisely for the reason that you stated, primarily because I can get myself confused if I don't do it this way. I know the wizard likes to use bets. Sorry for any ambiguity. Perhaps, I could have included a statement as to why I was proceeding in the manner that I did. Also, I had already done analysis on this game before the Wizard started this thread. The game has been on videopoker.com for a while now.

Mental

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January 27th, 2023 at 1:35:14 PM permalink

Quote: drrock
Quote: Mental
My quibble is that you switched to coin EVs when the thread started with EV in units of 5-coin bets. With a 10-coin game, I guess this is clearer because the 'bet unit' could be defined as 5 or 10 coins.
link to original post

Yep, I do it precisely for the reason that you stated, primarily because I can get myself confused if I don't do it this way. I know the wizard likes to use bets. Sorry for any ambiguity. Perhaps, I could have included a statement as to why I was proceeding in the manner that I did. Also, I had already done analysis on this game before the Wizard started this thread. The game has been on videopoker.com for a while now.
link to original post

I thought you had an amazing grasp of the issues. It didn't make sense that you just stumbled onto the problem yesterday.

Quote: drrock
1x: 63.5799%, 4.94904
2x: 7.8261%, 5.25962
3x: 7.7973%, 6.83487
5x: 7.6017%, 9.06134
8x: 7.3782%, 22.40869
12x: 5.8168%, 59.38847

I was able to reproduce the above table from your long post in my MC simulation:

trials:   100M
 1 A:    -nan  xC: 0.92176  xP:  63.5896  xAve:   4.9483
 2 A: 0.00367  xC: 0.00029  xP:   7.8239  xAve:   5.2513
 3 A: 0.02506  xC: 0.00195  xP:   7.7951  xAve:   6.8506
 5 A: 0.02943  xC: 0.00224  xP:   7.5998  xAve:   9.0427
 8 A: 0.21157  xC: 0.01561  xP:   7.3761  xAve:  22.3831
12 A:    -nan  xC: 0.05816  xP:   5.8155  xAve:  59.3376
aveMult: 1.7620904
FORTUNE X EV: -1.194567%

A is the acceptance ratio, xC is the fraction collected at each multiplier, xP and xAve are supposedly the same as the definitions in your table.

92% of the time, you are collecting at a 1x multiplier. This includes times that you are at 8x, but declining to use the multiplier and collecting a modest starting hand at a 1x multiplier.

Note that this run comes in well below 99% RTP. I need a billion games before this converges reasonably well. And I am not drawing to the flops or it would take longer to converge.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

gordonm888
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January 27th, 2023 at 2:32:48 PM permalink

Sorry to slow up this conversation, but I am not following. If the feature is triggered at a 10.9% frequency when you are not in the feature, and at a 0% frequency when you are in the feature, then the overall trigger frequency of the feature is much less than 10.9%.

Using Wizard's criteria and Mental's EV table, I calculate that once you have triggered the feature your average number of rounds in the feature are 4.648904721 hands: lets call it approx 4.649 hands.

At a 10.9% probability of triggering the feature, it requires an average of approx. 9.174 hands at a 1X multiplier to trigger the feature. So, the average feature cycle length is 9.174+4.649 = 13.823 hands and thus, the average wager invested in a feature is 138.23 units.

My main point is that it may be optimal for a player to choose a multiplier with hands that have a slightly lower EV than Wizard has calculated because that will shorten the average feature length, i.e., you would have a slightly lower EV per feature but you would slightly increase the frequency for triggering features if you terminate the feature more quickly. This is probably only relevant at the 8x decision point where there are categories of hands that are slightly below Wizard's threshold values.

BTW, I calculate that the 13,823 hands of the average feature cycle break down as follows:
at 1x
12.823 hands that pay 4.949 for a return = 63.4621
at 2x
0.003681 hands that pay 168.7266 for a return = 0.62108
at 3x
0.024998 hands that pay 112.7436 for a return = 2.81836
at 5x
0.02825 hands that pay 174.8825 for a return = 4.9404
at 8x
0.204868 hands that pay 94.2896 for a return = 19.3169
at 12x
0.738199 hands that pay 59.388 for a return = 43.84016

Total return from 138.23 units wagered =134.999 which is a total return of only 97.66%. So, I probably have made an arithmetic error, but I hope you see the methodology.

Last edited by: gordonm888 on Jan 27, 2023

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Mental

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January 27th, 2023 at 3:59:58 PM permalink

Quote: gordonm888
Sorry to slow up this conversation, but I am not following. If the feature is triggered at a 10.9% frequency when you are not in the feature, and at a 0% frequency when you are in the feature, then the overall trigger frequency of the feature is much less than 10.9%.

Using Wizard's criteria and Mental's EV table, I calculate that once you have triggered the feature your average number of rounds in the feature are 4.648904721 hands: lets call it approx 4.649 hands.

At a 10.9% probability of triggering the feature, it requires an average of approx. 9.174 hands at a 1X multiplier to trigger the feature. So, the average feature cycle length is 9.174+4.649 = 13.823 hands and thus, the average wager invested in a feature is 138.23 units.

My main point is that it may be optimal for a player to choose a multiplier with hands that have a slightly lower EV than Wizard has calculated because that will shorten the average feature length, i.e., you would have a slightly lower EV per feature but you would slightly increase the frequency for triggering features if you terminate the feature more quickly. This is probably only relevant at the 8x decision point where there are categories of hands that are slightly below Wizard's threshold values.

BTW, I calculate that the 13,823 hands of the average feature cycle break down as follows:
at 1x
12.823 hands that pay 4.949 for a return = 63.4621
at 2x
0.003681 hands that pay 168.7266 for a return = 0.62108
at 3x
0.024998 hands that pay 112.7436 for a return = 2.81836
at 5x
0.02825 hands that pay 174.8825 for a return = 4.9404
at 8x
0.204868 hands that pay 94.2896 for a return = 19.3169
at 12x
0.738199 hands that pay 59.388 for a return = 43.84016

Total return from 138.23 units wagered =134.999 which is a total return of only 97.66%. So, I probably have made an arithmetic error, but I hope you see the methodology.
link to original post

It would be nice if the VP.com rules were very clear on this point. Instead, the pertinent sentence is a very unclear. The bonus round is so long that observation of the game would quickly clarify the rule. I am not a member of the VP.com site.

It would be nice to confirm that the game can go directly from 12x to 2x on the next hand. If there was one dead hand after every bonus round, our models would be far off the mark.

I have argued the opposite of you point. I would prefer to stay in a neutral EV situation in the bonus round rather than move into a -EV situation in the base game. I would use a higher cutoff and have a lower acceptance rate for the current multiplier. Of course, only the 8x has close decision points.

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

drrock

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January 27th, 2023 at 9:32:54 PM permalink

Quote: Mental
Quote: drrock
Quote: Mental
My quibble is that you switched to coin EVs when the thread started with EV in units of 5-coin bets. With a 10-coin game, I guess this is clearer because the 'bet unit' could be defined as 5 or 10 coins.
link to original post

Yep, I do it precisely for the reason that you stated, primarily because I can get myself confused if I don't do it this way. I know the wizard likes to use bets. Sorry for any ambiguity. Perhaps, I could have included a statement as to why I was proceeding in the manner that I did. Also, I had already done analysis on this game before the Wizard started this thread. The game has been on videopoker.com for a while now.
link to original post
I thought you had an amazing grasp of the issues. It didn't make sense that you just stumbled onto the problem yesterday.

Quote: drrock
1x: 63.5799%, 4.94904
2x: 7.8261%, 5.25962
3x: 7.7973%, 6.83487
5x: 7.6017%, 9.06134
8x: 7.3782%, 22.40869
12x: 5.8168%, 59.38847

I was able to reproduce the above table from your long post in my MC simulation:
trials:   100M
 1 A:    -nan  xC: 0.92176  xP:  63.5896  xAve:   4.9483
 2 A: 0.00367  xC: 0.00029  xP:   7.8239  xAve:   5.2513
 3 A: 0.02506  xC: 0.00195  xP:   7.7951  xAve:   6.8506
 5 A: 0.02943  xC: 0.00224  xP:   7.5998  xAve:   9.0427
 8 A: 0.21157  xC: 0.01561  xP:   7.3761  xAve:  22.3831
12 A:    -nan  xC: 0.05816  xP:   5.8155  xAve:  59.3376
aveMult: 1.7620904
FORTUNE X EV: -1.194567%
A is the acceptance ratio, xC is the fraction collected at each multiplier, xP and xAve are supposedly the same as the definitions in your table.

92% of the time, you are collecting at a 1x multiplier. This includes times that you are at 8x, but declining to use the multiplier and collecting a modest starting hand at a 1x multiplier.

Note that this run comes in well below 99% RTP. I need a billion games before this converges reasonably well. And I am not drawing to the flops or it would take longer to converge.
link to original post

Wow, many of your simulation numbers are quite close to what I calculated directly, accurate in many places to the 3rd and 4th significant digit. That looks like verification to me that the calculations hang together fairly well.

drrock

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January 27th, 2023 at 11:17:20 PM permalink

There are more questions and statements in this thread that I would like to address, but I haven't slept much lately and have to get up early Saturday a.m. Since I have to brief now (I know, I am not good at that), I won't get to everything for a day or two.

However, as I was looking at trying to respond to some of Gordon's comments, I realized something about this table that the Wizard put together.

Multipliers	YH	Wiz	Accept %
2	0.003681	0.003681	0.0036815
3	0.024998	0.024998	0.0250900
5	0.028550	0.028550	0.0293933
8	0.198314	0.199515	0.2103518

My own analysis actually ties exactly to YH's numbers. For my posts, in explaining an alternate methodology I stated I was using the Wizard's cutoffs, even though I had a different cutoff for 8x, which I mentioned at the end of my first post.

At first glance, only YH's 2x probability ties exactly with the probabilities that I cited in my first post. But YH's column (and the Wizard's column) are representing something different than the column that I added titled "Accept %". All of the numbers in this last column except the 8x number were in my first post.

The first two columns are identifying the probability of each of the first 4 multipliers occurring in any given bonus round. Only one of the multipliers can occur in any bonus round. The last column shows the probability that a particular multiplier will be accepted. The first two columns refer to any entire bonus round from start to finish. The last column refers ONLY to hand in a bonus round, that being when the particular multiplier is being considered.

For 2x, the first hand in any bonus round, the numbers are the same. If the probability in this first hand in the bonus round of taking the 2x multiplier is 0.36815%, the probability of 2x occurring in any given bonus round is in fact 0.36815%, which in this case rounds to 0.3681%, tying to YH's number that is shown to 4 decimal places.

That leaves 99.63185% of the time when we will continue the bonus round to the 2nd hand. If we get to the 2nd hand, my number of 2.50900% can be multiplied by 99.63185% and yields YH's number for 3x of 2.4998%. This leaves 99.63185% - 2.49977% = 97.13209% chance that we move to the 5x multiplier. Sorry about jumping back and forth between 4 and 5 decimal places, but my numbers were to 5 while YH's were to 4. In doing these calculations, I was actually using the total precision allowed by Excel and rounding to 4 or 5 to be consistent with what is in each column.

When we consider the 5x multiplier, 2.93933% of the hands call for acceptance, and 2.93933% x 97.13209% is YH's number, 2.8550%. This leaves 94.27705% of the time that we move on to the 8x multiplier.

So, for 8x, 21.03518% of the hands qualified for acceptance (this is different than the Wizard's number because YH and I had a little different cutoff than the Wizard put forth; in my first post, at the very end, I stated that by using a slightly higher cutoff, I could increase the EV, although only by 0.0001%). Getting back to the main point of THIS post, 21.03518% x 94.27705% is equal to 19.8314% which again is identical to YH's number.

Mental and I didn't like this number, because we could see that 19.8314% was not consistent with any possible cutoff number if we were considering ONLY the hand when 8x was under consideration. I can now see that the Wizard understood this particular number better than we did. Again, it is NOT the acceptance percentage during a hand in which the 8x appears. Rather it is the percentage of time in any one bonus cycle that the 8x multiplier will be applied.

Now, given that I can see my numbers tie exactly to YH's numbers AND that Mental's simulation numbers are so close to what I calculated, I am becoming more confident that the actual EV is closer to what I stated, a shade UNDER 99% rather than 99.01%. However, when the numbers are that close and the calculations are so involved, I suppose that rounding at the wrong moment could be the difference.

When I have more time, I will see if I can reconcile more of the calculations and respond to other points in these posts.

Mental

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January 28th, 2023 at 4:33:56 AM permalink

Quote: drrock
Now, given that I can see my numbers tie exactly to YH's numbers AND that Mental's simulation numbers are so close to what I calculated, I am becoming more confident that the actual EV is closer to what I stated, a shade UNDER 99% rather than 99.01%. However, when the numbers are that close and the calculations are so involved, I suppose that rounding at the wrong moment could be the difference.
link to original post

I get much larger errors than this if I use a 32-bit float instead of a 64-bit double to sum the EV for a VP game. If you add many numbers, the rounding errors will do a random walk away from the more precise value.

EV: 98.980782% double
EV: 99.833816% float

Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.

Wizard
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January 28th, 2023 at 6:18:24 AM permalink

Quote: gordonm888
One game aspect I want to ask about again, for clarity. When you are "in the bonus feature" and deciding to accept or reject the multipliers, you only wager 5 coins in for each hand and are not paying the 5 coin fee (per ahnd) for the feature, correct? because when you are deciding between an 8x multiplier or a 12x multiplier you do not have a 10.9% chance of starting a new bonus feature sequence on that hand, correct?
link to original post

That is NOT correct. You must pay the 5-coin fee when in the feature to get the multiplier.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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January 28th, 2023 at 6:23:37 AM permalink

Quote: gordonm888

Just noting a typo in the line that I bolded. That line should read "0.211626*2.357209 + (1-0.211626)*0.622751 = 0.989808" The results are unchanged.
link to original post

You're right; thank you for the correction. It was a typographical error here and didn't affect any of my actual math.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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January 28th, 2023 at 6:25:01 AM permalink

I ran a simulation, based on my cutoff EV's posted earlier. Here are the results.

Multiplier	Hands	Probability	Average win	Exp Win
1	37,912,665,236,113,600	0.921736	0.933257	0.430108
2	11,844,716,482,335	0.000288	16.887036	0.004863
3	80,466,574,413,720	0.001956	7.516150	0.022056
5	91,895,255,960,475	0.002234	6.994974	0.039070
8	642,215,069,457,060	0.015614	2.357100	0.147211
12	2,392,720,463,072,770	0.058172	0.989648	0.345419
Total	41,131,807,315,499,900	1.000000		0.988727

The total hands should end in 500,000. Blame Microsoft for only so-many digits of precision in Excel.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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January 28th, 2023 at 6:45:27 AM permalink

I redid my pure math model and get the following, based on the 10.96% feature probability.

Feature Probability = 10.96%

Event	Weight	Probability	Average win	Multiplier	Exp win
Game not in feature	0.890400	0.635799	0.989808	1	0.314660
2X multiplier accepted	0.000403	0.000288	16.872652	2	0.004861
2X multiplier rejected	0.109197	0.077973	0.931120	1	0.036301
3X multiplier accepted	0.002740	0.001956	7.516250	3	0.022057
3X multiplier rejected	0.106457	0.076017	0.821845	1	0.031237
5X multiplier accepted	0.003129	0.002234	6.995312	5	0.039075
5X multiplier rejected	0.103328	0.073782	0.807941	1	0.029806
8X multiplier accepted	0.021867	0.015614	2.357209	8	0.147224
8X multiplier rejected	0.081461	0.058168	0.622751	1	0.018112
12X multiplier	0.081461	0.058168	0.989808	12	0.345451
Total	1.400442	1.000000			0.988783

This agrees very closely to my simulation.

So, a rather crude spreadsheet YH gave me shows a feature probability of 11%. I questioned him on the difference with the 10.96% on the game screen and he said something to the effect of 10.96% being an average over multiple different games.

If we use 11% as the feature probability, my pure-math model becomes...

Feature Probability = 11.00%

Event	Weight	Probability	Average win	Multiplier	Exp win
Game not in feature	0.890000	0.634851	0.989808	1	0.314190
2X multiplier accepted	0.000405	0.000289	16.872652	2	0.004874
2X multiplier rejected	0.109595	0.078176	0.931120	1	0.036396
3X multiplier accepted	0.002750	0.001961	7.516250	3	0.022114
3X multiplier rejected	0.106845	0.076214	0.821845	1	0.031318
5X multiplier accepted	0.003141	0.002240	6.995312	5	0.039177
5X multiplier rejected	0.103705	0.073974	0.807941	1	0.029883
8X multiplier accepted	0.021947	0.015655	2.357209	8	0.147607
8X multiplier rejected	0.081758	0.058319	0.622751	1	0.018159
12X multiplier	0.081758	0.058319	0.989808	12	0.346350
Total	1.401903	1.000000			0.990069

The figure YH gives for the return is 0.99006980.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

drrock

Threads: 3
Posts: 99

Joined: Mar 6, 2012

January 28th, 2023 at 10:39:17 AM permalink

Quote: gordonm888
Sorry to slow up this conversation, but I am not following. If the feature is triggered at a 10.9% frequency when you are not in the feature, and at a 0% frequency when you are in the feature, then the overall trigger frequency of the feature is much less than 10.9%.

Using Wizard's criteria and Mental's EV table, I calculate that once you have triggered the feature your average number of rounds in the feature are 4.648904721 hands: lets call it approx 4.649 hands.

At a 10.9% probability of triggering the feature, it requires an average of approx. 9.174 hands at a 1X multiplier to trigger the feature. So, the average feature cycle length is 9.174+4.649 = 13.823 hands and thus, the average wager invested in a feature is 138.23 units.

The average time to get to trigger the feature is a straightforward formula: 1/0.1096 = 9.124088 hands. If you hadn't truncated the feature percentage, you would have got the same answer. However, the expected number of hands at the 1x multiplier is one FEWER than that as the last hand will actually have the multiplier. So, any process going forward from here should use 8.124088 hands at 1x BEFORE the bonus round is triggered.

When I use the Wizard's cutoffs, I get 4.653665528 hands on average in the bonus section, but I don't use that number directly in calculating the EV from there. But your post, along with rereading and grasping the method that the Wizard was using triggered a perhaps simpler to understand calculation of the EV with the Wizard's numbers.

With 8.124088 hands at 1x, prior to a bonus round, we should collect an expected 40.2064258 coins. From there, I can construct the following table.

Multiplier	# Bns Hands	Prob	Cycle Length	Exp Coins
2x	1	0.36815%	9.12408759	208.9329437
3x	2	2.49977%	10.12408759	157.6057758
5x	3	2.85503%	11.12408759	223.8540412
8x	4	19.95149%	12.12408759	147.2992944
12x	5	74.32556%	13.12408759	115.5131735

Average			12.77775312	126.3442852

Dividing the last entry in the last row by the next-to-last entry in the last row yields an average of 9.88783271 coins or an EV of 98.8783%, the same number that I got for the Wizard's cutoffs in my first post. And, it's not totally surprising since I used the same numbers as in that post.

How did I get the next to last column? I added the possible lengths of the bonus round (1 thru 5) to the average length of the string of 1x's that occur prior to a bonus round.

How did I get the last column? I started with the 40.2064258 coins earned on average during the string of 1x's prior to the bonus round. Then for the 2x row, I added 168.7265179 coins, the average coins earned when we accept the 2x multiplier. This number is in my first post. For the 3x row, I included the 40.2064258 coins, added in 4.6555974 coins that we got when the 2x multiplier was active but we declined it and then added in the 112.7437526 coins that we expect when the current hand is above the 3x cutoff. And on subsequent rows, I start with 40.2064258 coins, add in the expected 1x amount hand by hand for all the previous multipliers that we declined and then add in the expectation for the multiplier that is accepted.

So, even though I disagreed (very slightly) with the Wizard's 8x cutoff, we can still get nearly all the EV that I got even though I matched YH's cutoffs. The jury is still out on what was missed to reconcile the two competing (though close) EVs of 98.8784% and 99,01%. Even if it is slightly imprecise, the Wizard's cutoff methods get quite close. Rather than switching to a more exotic and difficult method, another alternative that could be employed if you had a table like Mental provided is to increase your cutoff to exclude 1 more level (or decrease your cutoff to include 1 more level), one at a time, and see what effect this has on the EV. Simply select the cutoff that gives you the best EV after getting numbers that should be reasonably close to the precise cutoff.

This is perhaps not elegant or aesthetically pleasing, but I would think it would be an effective check to see if you are close or have actually reached at least a local maximum if not the overall maximum.

Quote: gordonm888
My main point is that it may be optimal for a player to choose a multiplier with hands that have a slightly lower EV than Wizard has calculated because that will shorten the average feature length

It is really difficult, for me at least, to reason exactly where the cutoff should be without chugging through the math. Basically, on a great hand you want to take advantage of a current multiplier; however, if you have even higher multipliers following you want to wait if your current hand is too low. Yes, if you end the bonus round more quickly, you will get to the next bonus round more quickly, which is why in my post re: cutoffs, I use the V notation.

In this type of game, the expected value of the next 25 or so hands is always greater when you take advantage of the opportunity to move to a higher multiplier vs. the next 25 or so hands when you decide to cash in and start over. The question that the cutoff answers is whether the CURRENT hand's EV is large enough to offset the difference described in the previous sentence. If you are wondering about hands 26 through infinity, the difference in the EVs in the two instances will be so negligible that you can almost certainly ignore them.

drrock