Progressive EV

So, I was playing a dollar, 98.01% EV, Bonus poker with the progressive on the Royal over $12,000 (making it a 102.78% EV). However, there must be some mathematical analysis that takes into account the winner-take-all nature of the game. I’ve chatted with some other math guys in the past, but don’t think I’ve seen an analysis yet. Sure, on every pull before someone wins, the EV is 102.78%. However, it doesn’t stay at that level until I hit a Royal. Any words of wisdom on this?

Clarify what you mean by, "It doesn't stay at that level until I hit a Royal."

Keep in mind that the higher progressive should result in a strategy change. For example, with a progressive that high, if you are dealt a suited King-high straight flush, I think that's high enough that you discard the 9 to go for the Royal. There may be some other plays where you would go for the Royal in this case where otherwise you would not.

Yes, the probability of a Royal is non-decreasing until someone hits it because of the changed play. So, the point I'm making is such games have a stop play point (when the EV drops after a Royal hit). The probability of this end is a simple function of the current Royal probability and the number of players. However, I'm baffled on how to plan for the gambler's ruin and the bank one needs. This all gets more complex if there are 2 or more progressive outcomes.

Question: is it just the one machine that has the progressive, or is it a bank of them with a shared progressive?

I imagine that some of the more important decision points on strategy might be:

At what level of RF progressive pot do you draw to
- an Ace -Ten suited rather than an Ace suited?
- a King-Ten suited rather than 2 unsuited high cards, Ace or King highest?
- a Queen-Ten suited rather than 2 unsuited high cards, King highest?

- two royal cards as opposed to 3 SF cards? With say Js-Ts-8s-x-x or Js-Ts-9s-x-x?
- two royal cards as opposed to 4 to an outside straight with 1,2 or 3 high cards?
- two royal cards as opposed to 4 to an inside straight with 4 high cards?

- three to a royal flush as opposed to a high pair?

Quote: ThatDonGuy

Question: is it just the one machine that has the progressive, or is it a bank of them with a shared progressive?
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I believe at least two bars with 10-20 seats. At 3AM, I was the only player at a bar but the meter was still moving during pauses in my play.

Quote: GaryJKoehler

So, I was playing a dollar, 98.01% EV, Bonus poker with the progressive on the Royal over $12,000 (making it a 102.78% EV). However, there must be some mathematical analysis that takes into account the winner-take-all nature of the game. I’ve chatted with some other math guys in the past, but don’t think I’ve seen an analysis yet. Sure, on every pull before someone wins, the EV is 102.78%. However, it doesn’t stay at that level until I hit a Royal. Any words of wisdom on this?
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Maybe I am not getting the gist of your question.

The total EV of the session depends on the situation and your ability to stick with it until the jackpot hits. If you have a stand-alone jackpot and you or a team can hold it until the jackpot hits, the average EV per hand for the duration of the play is the EV you would get from an EV calculator for the next hand plus the contribution from meter rise. If a cycle is 40K hands and the meter is 0.5%, then you would on average hit at a jackpot level that is higher by $5*40K*0.005=$1000. This increases your EV by 0.5%.

There are minor second-order effects because the jackpot cycle and strategy may need to be adjusted. So, the EV rises monotonically until the jackpot is hit.

If you have a finite bankroll or time, your EV is the base EV plus (the meter enhancement reduced by the chance that you will actually hit) and recover the meter advance.

The winner take all nature has zero effect on the EV and certainly does not require a downward adjustment of EV. If you are playing a slot progressive with no strategy involved, then the EV is the EV of the next spin plus the meter rate with no adjustment needed for strategy changes.

In a competitive progressive situation, it is easier to think about in terms of the EV for the next hand/spin. If you stay until someone hits the jackpot, you will always be playing at an increasingly higher EV on each hand. However, if you are alone at first and then more and more competitors show up, then you will be playing a large fraction of your hands at a lower jackpot versus the latecomers. As a group, the the EV per hand will be the EV of the initial hand plus the meter rate. However, your average EV will be lower than that of the latecomers.

The same thing happens with linked slot progressives that hit proportional to the bet amount. If there is a number where the game is breakeven when you are playing by yourself, that number will not be accurate in the presence of increasing competition. You need to wait if your competition is expected to play faster than you when the progressive gets higher.

So, your ability to stay until the progressive hits does affect the average EV on your hands/spins, but the average will always be higher than the EV for that first single hand/spin.

Quote: GaryJKoehler

So, I was playing a dollar, 98.01% EV, Bonus poker with the progressive on the Royal over $12,000 (making it a 102.78% EV). However, there must be some mathematical analysis that takes into account the winner-take-all nature of the game. I’ve chatted with some other math guys in the past, but don’t think I’ve seen an analysis yet. Sure, on every pull before someone wins, the EV is 102.78%. However, it doesn’t stay at that level until I hit a Royal. Any words of wisdom on this?
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Can we try a simple example to see if I can understand what your question is?

Suppose we have a machine that chooses among the digits 0-9. You pay one unit to guess the digit, and 100% of this amount goes to the meter before you guess. If you guess the digit correctly, you get paid the current contents of the meter and the meter resets to 0.

This is a winner-take-all progressive. If you play the game when the meter is 0, the EV is -0.9, right? The next-game EV gets better by 0.1 with every game until the progressive hits. However, the EV of the overall series of games is zero if you can play until it eventually hits.

The average EV per game keeps increasing as the meter rises. If the meter is at 9, the EV is (+9*0.1 + -1*0.9 = 0).

You can work out that the average EV of a series of games where you start with a meter M is ((M-10)/10 + 1).

This agrees with the average EV per game being 0 if you play starting with the meter at 0. The next-game EV is -1, and you add +1 for the 100% meter contribution.

Is any part of this example unclear? The math is not much different for your VP progressive. The meter contribution is lower, the cycle time is longer, and you have a base game EV to add in. However, you do the calculate the EVs the same way.

I guess I'm concerned about "However, the EV of the overall series of games is zero if you can play until it eventually hits." since it may be hit by someone else playing against the shared meter. And how would n competing players impact the gambler's ruin probabilities?

Quote: GaryJKoehler

I guess I'm concerned about "However, the EV of the overall series of games is zero if you can play until it eventually hits." since it may be hit by someone else playing against the shared meter. And how would n competing players impact the gambler's ruin probabilities?
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Now, you are asking about ROR, but your original post was asking about EV. The EV always increases until the jackpot is hit. ROR is trickier.

I assume you agree that my the EV of my hypothetical game is zero if started at zero and played until it hits. It stands to reason that the EV of the game would be zero for a group of players sharing the linked game equally or unequally. Whoever takes the first guess will have an EV of -0.9, the second will be -0.8, ..., the 10th player will be playing at an EV of zero, that is, breakeven. The jackpot will usually be hit before the meter gets to 11 units, but whoever plays the game with a meter of 11 or higher will be playing a very positive EV game until it hits.

It is easiest to talk about ROR for fair competition. If there are N players and each one takes their turn in an order that is decided randomly, this is fair competition. Your ROR for a single progressive chase actually decreases the more competition you have. Let us say that you only have 2 units in your bankroll and N=100. You will normally not even get a play. Let us say you are randomly chosen to play the 50th game after the meter reaches 11. You are a 194:1 dog to even get your turn to play because one of your competitors already hit the jackpot. You are over 7M:1 to get a second chance to play and the odds of you losing both of your units is over 8M:1.

Compare this to only having one competitor and you play the second, fourth, sixth, ... games. Your ROR works out to 65.61%. That is the probability that the jackpot is not hit before the fifth game.

I think your question is about the ROR over a series of identical progressive jackpot chases. My intuition is that your ROR increases with more competition because you have the variance of the underlying VP game plus the variance of the fraction of the jackpots that you hit out of all jackpots that are hit by anyone.

I cannot think of an easy formula for this ROR and the dependence on N. My thought is that it can be approximated by assuming that you and your competitors play and average of C games per attempt, where C is the jackpot cycle. After T jackpot attempts, you will accrue a total variance of T*V*(C-1), where V is the variance per hand excluding the jackpot. You will also experience a variance in how many jackpots you get relative to your fair share. For example, if you have three equal competitors, then you expect to hit 25 +/- 5 jackpots after T=100 attempts. Most of the variance will come from bad luck in getting your share if jackpots.

You can calculate the expected loss from 100 attempts, not including the jackpots. Any VP calculator will give you this loss rate based on the appropriate strategy. Then, you can add in just 15 jackpots in 100 attempts assuming you run very badly. (you have a 1.1% chance of hitting 15 or fewer jackpots.) If your base-game losses after 100 attempts minus 15 jackpots is greater than your starting bankroll, then you are ruined.