AP VP simulator?

Does anyone know of a game/website that simulates video poker games but also gives cash back or free play? I feel like this would be good to simulate real results.

I suppose that some apps give $wagered, so i could just use that yo estimate cash back. But if anyone knows of anything please shout.

What is it you're trying to simulate, exactly?


I wrote a quick-and-easy simulation program for single-line 9/6 Jacks or Better, because I wanted to figure out the required BR for a specific play. ie: $1 denomination, play to $15,000 coin in....figure out how much money I would need (like 95% of the time I would need at least $3K, or whatever the result was).

It can easily be modified for other games [which I might actually modify right now], so that the user would have to do some "basic" inputs, like:

- Game type: JOB, DB, DDB, DW
- Pay table
- Frequency distribution [ie: # of combinations for each hand for 9/6 JOB can be found here: https://wizardofodds.com/games/video-poker/strategy/a-1-b-74-c-1-d-0-d-1-d-2-d-3-d-4-d-6-d-9-d-25-d-50-d-800/ ]
- Denomination: any denom
- Coin in: any amount of coin in
- Confidence level: if you enter 99%, then it tells you how much of a BR (session bankroll, not total bankroll) you'd need to be able to finish the play.


What is a total b**** and a half is figuring out how to do this simulation for multi-hand (triple play, five play, ten play, 50-play, 100-play, etc.) games, because you can't use the frequencies....but gotta do some real f****ed up coding.....although I am making a bit of progress, slowly but surely.




What is it, though, that you want to simulate with cash-back/free-play? Or are you just trying to figure out the return? ie: $25,000 coin-in on 9/6 JOB getting $1000 in FP, would be {[1-0.9954]*$25,000) + $1,000} = return (something like $880). The EV would be the $880/$25,000 ~ 3.5%. In other words, you'd be getting a 3.5% advantage over that $25,000 coin in.

That can be done with simple math, assuming you know the return of the game and know how much FP you'll get (or you have a good idea of how much FP you'll get).



OR are you trying to figure out how much of a total BR you need to be playing at "full kelly" (or half kelly or w/e)? Kelly doesn't work perfectly with video poker because you have high variance games (ie: RF, or AWAK in DDB or 4-deuces in DW).....but you can figure out your kelly-optimal bet by doing (EDGE / VARIANCE) * BANKROLL = WAGER. Divide both sides by bankroll, divide both sides by wager, [so you solve for BR], and voila! [PS: This isn't perfect because there are some other factors, like you aren't constantly playing at an advantage, but you have to do the entire amount of coin-in in order to realize your advantage...so if you stop at the $18,000 coin-in point but were aiming for $25,000 coin-in, you've really f****d yourself. This calculation doesn't take that into account.]



If you are looking for a game to actually play.....this is worthless. It can take a LONG TIME before you can be confident your results are close to expectation. Of course, it depends on your edge (0.5% vs 10%) and the variance (9/6 JOB or 9/6 DDB?), as well as where the edge comes from [if it comes from free-play/mail for doing coin in, low/no variance there. If your edge comes from hitting a premium hand, you're going to have extra high variance].

I sort of want the -feel- of playing a high expectation game. I want to get a very good estimate of my playing speed, hourly rate, variance... Not just on paper, but in time. If that makes sense.

Quote: RS

What is a total b****

you tell us

Quote: RS

and a half is figuring out how to do this simulation for multi-hand (triple play, five play, ten play, 50-play, 100-play, etc.) games, because you can't use the frequencies....

oh, you mean a pita

well, you can't use the frequencies but others can
this is easy stuff to do, in my opinion
either convolve the paytable (easy) or setup a Markov chain (more challenge)
simple math will follow
a little bit of code goes a long way

for 9/6 JOB VPforWinners shows the
# of game outcomes
job 9/6
10 play: 2,011
5 play: 406
3 play: 114
1 play: 10
i actually matched the listed probabilities in VPW using some simple code for the free program R
but have not done the 50 or 100 play (JUNEbe, we will see)
i mean, if silly me can did it accurately, should be simple stuff for the rest of the world, i would think so

of course, if the Wizard and/or JB has yet to do that, they should for multi-hand
more feathers in the hat

Quote: RS

What is it you're trying to simulate, exactly?

What is a total b**** and a half is figuring out how to do this simulation for multi-hand (triple play, five play, ten play, 50-play, 100-play, etc.) games, because you can't use the frequencies....but gotta do some real f****ed up coding.....although I am making a bit of progress, slowly but surely.

I am having a little problem understanding why you can't treat a 50-play game the same as 50 plays of a 1-play game in this case. The frequencies for each hand should be the same - just because all 50 hands had the same 5 cards dealt shouldn't change this. If you were to simulate, say, 130 million 1-play hands, each 5-card deal is expected to appear 50 times; does it matter if they appear all at once?

If you really want to do, say, 3-hand, then what you do is, you determine the perfect strategy draw for each of the 134,459 "unique" hands (e.g. a 10-high straight flush in spades is the same as one in diamonds, and 4s 4h 4d 6s 6h is the same as 4h 4d 4c 6c 6d, but not 4h 4d 4c 6c 6s as in this case one card in the pair has a suit that is not in the three as well), calculate the frequencies of each result for each hand, and then multiply each one by the number of actual hands of that type that can be dealt (e.g. there are 4 10-high straight flushes, and 12 hands of three 4s and two 6s where both of the 6s have suits that are also in the 4s).

As for taking cash back into account, it depends on how it works. If it's a straight percentage of your coin-in, regardless of wins, then it should be a simple case of adding the rebate value to each result of the paytable; for example, if it's a 0.1% return, then a RF is now worth 800.001, and a losing hand is worth 0.001. If your app only takes integers, then multiply everything, including the denomination, by 1000 (i.e. you are now playing a 1000-credit game that pays 1 for a loss, 1001 for a high pair, 2001 for two pair, and so on, up to 50,001 for a straight flush and 800,001 for a royal flush).

Quote: ThatDonGuy

I am having a little problem understanding why you can't treat a 50-play game the same as 50 plays of a 1-play game in this case.

for example 9/6 JOB
3 single line hands are way different from 1
3 hand play
so i guess the hands are not all independent

the chance to gain 0 in 3 hands of 9/6 JOB = 16.711263%
from this part dist (in units)
-3: 16.226625%
-2: 19.151648%
-1: 19.072747%
0: 16.711263%

one round of 3 play i gets (so does VPW)
-3: 26.26%
-2: 15.51%
-1: 15.88%
0: 13.72%

more JUNEbe next week

Doesn't "video poker for winners" do this kind of analysis?

No idea what Sally is saying...

Quote:

I am having a little problem understanding why you can't treat a 50-play game the same as 50 plays of a 1-play game in this case. The frequencies for each hand should be the same - just because all 50 hands had the same 5 cards dealt shouldn't change this. If you were to simulate, say, 130 million 1-play hands, each 5-card deal is expected to appear 50 times; does it matter if they appear all at once?

If you care about the game's volatility, it certainly does matter when they appear. If I'm only looking at the EV, then it doesn't matter.

Quote:

If you really want to do, say, 3-hand, then what you do is, you determine the perfect strategy draw for each of the 134,459 "unique" hands (e.g. a 10-high straight flush in spades is the same as one in diamonds, and 4s 4h 4d 6s 6h is the same as 4h 4d 4c 6c 6d, but not 4h 4d 4c 6c 6s as in this case one card in the pair has a suit that is not in the three as well), calculate the frequencies of each result for each hand, and then multiply each one by the number of actual hands of that type that can be dealt (e.g. there are 4 10-high straight flushes, and 12 hands of three 4s and two 6s where both of the 6s have suits that are also in the 4s).

That's what JB said, too. Is this using wizard's "my VP methodology" page? Do you cycle through each hold and each possible dealt cards...then record the best returning hold?

Quote: RS

That's what JB said, too. Is this using wizard's "my VP methodology" page? Do you cycle through each hold and each possible dealt cards...then record the best returning hold?

I came across the method on my own when trying to develop a VP strategy Android app, but it's the same - how many ways can there be of doing it?

As for "cycling through each hold and each possible dealt cards", the answer is, "sort of" - you do have to check each hold separately, of course, but there are shortcuts to having to cycle through each possible set of drawn cards, especially when you consider that there are 1,370,754 different deals when you discard all five cards.
Here's the "quick version":
First, for each set of 4 cards, you determine the sum of the 48 possible values for the fifth card. When you want to determine the value for discarding a particular card from a deal, you take the value for the set of 4 held cards, and subtract the value for the 5 cards dealt, since the discarded card is included in that 4-card set but can't be drawn again; divide by 47 to get the expected value for discarding that card.
Next, for each set of 3 cards, you determine the sum of the 49 possible values for the fourth card, but then divide by 2 as every set appears twice (e.g. for set 1,2,3, you add the values for sets 1,2,3,4, 1,2,3,5, ..., 1,2,3,52, but, for example, 5-card set 1,2,3,10,20 is in both 1,2,3,10 and 1,2,3,20, so divide the total by 2). To determine the value for 2 discards, take the value for the set of 3 cards held, subtract the values for the 4-card sets including each discard, but now it turns out that you subtracted the 5-card set that included both discards one too many times, so add that back.
Do something similar for the 2-card sets (but divide the sum by 3 instead of 2, and it's "the 2 minus the sum of the 3s plus the sum of the 4s minus the 5"), then the 1-card set, then the "set" of discarding all five cards.
Whichever of the 32 discard options (including holding all five) has the highest expected value is what you do.