I have a question about a progressive machine at a local casino I go to.

Let's say you have a progressive VP machine with the best pay table being 9/5 Double Double Bonus for 97.87% with perfect play for the base jackpot. Say this is a quarter machine so you play 1.25 dollars a hand. (The actual machine is a ten play and the progressive seems to get very big at times, with 10 separate progressives for each hand).

Anyway, my question is: How many pennies need to be added to the jackpot per hand to make this a negative EV machine for the casino in the long run if someone was using perfect play? Now I understand its better to wait for the jackpot to become large before you play, but this is just a curiosity to try and understand progressives better.

Two possibilities:

Your expected loss per hand is 1.25*(1-.9787)=2.66 cents. If the jackpot goes up by three cents per missed royal hand, that is bad for the casino.

On the other hand, I"m thinking the jackpot only increases conditional on not hitting a royal in the previous hand. In that case, your EV conditional on no royal is less than 97.87%, maybe 2% less as an approximation, so the answer would be about 1.25*(1-.9587)=5.16 cents.

This machine adds 3 cents per hand so which one is it or is there some other story?

Huh?

This sounds similar to an argument that multi-line slots increases hold because a jackpot win on one line precludes a jackpot win on the other lines. Where did you get 2%?

2% is about the EV that comes from a Royal Flush in most video poker games.

Your EV conditional on no royal for a given hand is more complicated. There are some reasons why it might be more than the EV of the game minus 2%, like holding a pair or three of a kind eliminates any chance of a royal but may not be bad for EV. There are also lots of hands with very low EV that eliminate your chances of getting a royal, like an inside straight draw (DDB specific).

Your idea is approximately correct if a single unit always gains 3 cents per hand. But are you positive it truly gains 3 cents per hand? Even when the machine is +EV? The machine could be set to not constantly increment the jackpot. Is it networked at all? If it shares the progressive with any other unit, then it is definitely -EV. Even if none of my questions are true, they still may offer the machine since errors still allow the casino to turn a small profit.

I also say approximately correct because the average EV machine should be calculated like this (I'm ignoring strategy changes for now because the math becomes even more messy):

(0.978729) + (0.978729 + 3/40365/125)*(40364/40365) + (0.978729 + 2*3/40365/125)*(40364/40365)^2 + (0.978729 + 3*3/40365/125)*(40364/40365)^3 + ...

Then this infinite sum is divided by:

(1 + (40364/40365) + (40364/40365)^2 + (40364/40365)^3 + ...)

Doing it this way weighs the lower payout states heavier because you play hand 1 100% of the time but hand 22,000 (which pays 99.18%) only is played 57.98% of the time since we'll hit a royal the other 42.02% of the time.

I did this for a 3c increment for ~220,000 terms (my spreadsheet was not happy about it) and I get an average return 100.21%. I got 100.00% when I did 141k terms.

I was playing it yesterday when the average jackpot was 2.3 K, which is getting close to 101% and that was when I noted the three cents per hand. Its a bank of 8 machines all feeding the same jackpot. I was playing at 4 AM, so no one else was around at the time I observed the three cent increase. And that was definitely 3 cents on each of the 10 jackpots.

If I may speculate on why it exists I think there are a few reasons. The first is that there are a bunch of different types of games which feed the same jackpot. You can play jacks or better 8/5 for 97.2, deuces wild with a payout less than 97 and joker poker at something like 96 base payout before the jackpot. The DDB payout schedule was a shocker when I put it in. The second reason I think this might be legit is that you have to play ten hands for a 12.5 dollar per hand investment. There was also a cheaper option to play 3 or 5 hands at the machine with smaller progressives.

I understand why you didn't include strategy deviations, but by the time the jackpots are at 2.3K, the royal goes from 1/40000 to 1/30000, so that make a significant difference.

One way which I thought of similar to your math was just to calculate the expected jackpot at any particular hand. The probability you are i hands away from the jackpot hit is just

(1/40000)*(1-1/40000)^(i-1)
and the jackpot is .03i+1000

so you can just take an infinite sum and see the expected jackpot would be 2200. Even money jackpot would be about 1950, so you could frequently expect a positive EV situation, depending on how many people play it at different times.

This was mainly a curiosity and something that I wanted to think about to see how frequently I could expect the jackpot to be in positive territory. I'm never going to play it below a 2k jackpot.

I think I see what you're doing. I think you're confusing the % the royal contributes to the payout with the % frequency, the latter being much smaller. I don't understand why you want to know your EV conditional on no royal -- it's a contract wager -- by the time you know the condition, you're already in the game, and it doesn't change the EV of the game sequence. Just like hitting a point of 4 in craps changes the EV of that wager, it doesn't change the EV of the sequence that got you there.

I'm really just looking for the number of pennies added per hand that makes the machine long-run +EV. You can ignore my rambling if it doesn't help answer the question.

How often you'll see the jackpot in +EV territory if it adds 3 cents a hand?

Depending on how people play the machine: 35 to 40% of the time.

The Royal needs to be about 7800 credits ($1950) to be +EV before taxes. The variance is 100 bets per hand for the single hand game...lol Standard DDB has a variance of 40 bets per hand. 10-play makes this significantly worse, making the variance per single line played almost double this value. When the machine is +EV, then you have a variance of about 2000 betting units a deal in 10 play! Thats really sick.

Variance is a little bit overrated as a measure of risk in these situations. The entire distribution of outcomes matters. If the variance comes from really good states of the world it is less important than if it comes from really bad states of the world. I understand that strategy deviations increase the downside a little bit, but most of the variance increase is from good states getting better.

How do you do the variance calculation for a 10 line game anyway?

I understand its risky, but I've played riskier than this before with less EV so it seems like it has potential.

This is a good point, you can just think that youre playing a ~96% DDB machine (the approximate return without the royal) and you're hoping to hit the royal.

Multiline variance calculations are confusing, you have to consider the variance of your deal distribution and weight that by the hands you play (since you get the same deal in every hand of multiplay). Here are the two best descriptions I have ever found on the subject, but the exact numbers on DDB aren't published, and I don't even know how to the exact math over the actual multiplay variance calculation, so I just estimated for DDB.

http://wizardofodds.com/games/video-poker/appendix/3/

http://www.jazbo.com/videopoker/nplay.html

Something I was wondering about was how the fact that there are 10 separate jackpots affects the probability of a +EV situation. It seems to me, if the expected jackpot is +EV at any given point, then more largely independent jackpots will raise the probability of a +EV situation compared to a one jackpot machine, while the opposite would be true if the expected jackpot was -EV.

In my sample size of two observations 4 days apart, the total jackpot was very close to the expectation of 2.2 K average both times (always in the 2.15-2.35 K range), even though one time most of the jackpots were large while the other time most were small and a few were enormous.

So while the probability of one jackpot being above 1.95 is 35-40%, the probability that the average of 10 is above may be well above 50%. Does this seem reasonable?

I'm not sure of that, because there will be times where 4 to a Royal draws will be plentiful and help knock out jackpots more frequently, and of course there will that 1 in 649,740 chance like this, which will knock out all the jackpots. (I wish I didn't understand the concept of -EV and was playing max lines there dammit!)

I thought of another way of thinking about this problem. Lets say you start when the jackpot is at an even money 1944 jackpot. If you play until you hit the jackpot, what is your edge? Without the additional run-up of the jackpot your return in expectation is 100%, but with the additional 3 cents per hand the edge is over 2%. The calculation is exactly like the one you did previously except starting at 100% instead of 97.8%. This assumes you capture the entire bank of machines.

I'm not sure how to do this calculation for the 10 different progressive jackpots. I think the edge might be lower but I'm not sure yet.

Let me make sure I understand the machine correctly. This is a 10-handed game with each hand having a separate progressive for the RF? You keep saying when you hit "the jackpot". The only way to hit "the jackpot" is getting dealt the royal. Otherwise you'll generally be getting 1 of the 10 jackpots for all other royals. If the machine is like this, also it's unlikely all 10 jackpots will be equal and all above 1944 for that matter. The math on doing this will be much more difficult if they are unequal as well.

Yes you understand the machine correctly.

Am I correct in thinking that the math for the strategy for this machine is the same as the strategy for the average jackpot?

I understand that picking a stopping point if you own the entire bank is extremely complicated and I have no idea how to do it besides some sort of simulation.