DW44 Bankroll Requirements

[PeaceFr0g]

Hi,I'm new here and tried a search, but didn't turn up anything on this (hard to believe on a site dedicated to VP, so I probably did a poor job searching).  There's a deuces wild (44 pay table) that returns 99.96% at my casino. It's my understanding that the variance for this machine is 25.5.  I know a little about standard deviation, but do not know what this 25.5 number represents.  Is it coins per hour?  Coins per 100 hands?  Etc. Basically, I'm just looking to find out what a proper bankroll should be for playing max 5 coins per spin.  Also, this is a 50 play machine and I play 4 hands.  Shouldn't this reduce variance or smooth it out sooner?  I've already played this machine some and if I had to go by guess or gut feel, I'd say bankroll should be around $40k-$60k in order to be assured of getting to around 5% of expectation.  But I really have no idea and would appreciate some more solid numbers.  Thanks. Peace

[shadowman]

Variance is a measure of volatility. It is the square of the standard deviation. It's useful for comparative purposes. A value of 25.5 is sort of middle of the road for VP. However, all VP is volatile compared to BJ for example. 


 
Multi-play variance is a tricky subject. If you are comparing quarter 4-play to dollar than it would be less volatile. However, compared to playing single line quarter it is more volatile.

[PeaceFr0g]

Variance is a measure of volatility. It is the square of the standard deviation. It's useful for comparative purposes. A value of 25.5 is sort of middle of the road for VP. However, all VP is volatile compared to BJ for example. 


 
Multi-play variance is a tricky subject. If you are comparing quarter 4-play to dollar than it would be less volatile. However, compared to playing single line quarter it is more volatile.Thank you shadowman.  So would you have any idea what size bankroll I would need (in dollars, playing 4 hands on a $1 machine x 5 max. So $20 per spin) to assure reaching expectation before going broke?  Was my $50k figure close?  Not close?  I'd really appreciate some idea even if it's just an approximation.  Thanks. 

[PeaceFr0g]

Edit: Obviously, I'm assuming perfect play on a 99.96% pay table. 

[shadowman]

Since this is a negative game, although only slightly, it requires an infinite bankroll to avoid bankruptcy. A game must be positive to compute a bankroll.
 
However, I assume there's some promotions, cashback or bounceback available at this casino. If they total close to .5% then you have a good play.
 
I'm not an expert on bankroll calculations but assuming you could play 500 hands/hour with the .5% advantage, the game is worth $50/hour overall. I would guess that the bankroll requirement for this would be around $100K to have less than a 2% chance of going bankrupt. But, that is just a guess.
 
Maybe our resident excel expert, new2vp, will give you something better.

[PeaceFr0g]

Thanks again.  I probably didn't ask my question carefully enough.  I realize that no bankroll is enough for a -EV game.  There are comp points, that would make this game profitable, but only slightly (definitely not profitable after paying taxes on the signers).  The real reason I'm playing is for comp "status" (a whale card, which gives you many perks).  This requires that I run $1M through the machine.  Theoretically, this should only cost me $400 ($1M - 99.96%) -lol.  But obviously, it can wind up costing me quite a bit more than that in real dollars.  I'm more than willing to pay $400 for the bennies this comp status would give me.  However, it would not be worth +$6k.  So I'm wondering what the worst case scenario (dollar-wise) would be through 50k spins if I were to run say, 2.5 standard deviations on the wrong side of the mean.  Again, I plan on being done after 50k spins whether up or down. 

[New2vp]

EV and Standard Deviation are in the same units and that unit could be thought of as bets.  With the standard APDW, the game your are referring to, 99.96% means that the average return for each bet is 0.9996 bets.  The standard deviation is 5.0496 bets and is an indicator that allows one to estimate how far from the average one is likely to get after several plays.  Since variance is the square of standard deviation, the 25.5 number that you refer is in somewhat awkward units of "bets squared."If you are going to answer your question for multi-play as opposed to single play, you also need to know the covariance of the game because:Variance(4-play) = Variance(1-play) + 3 x CovarianceThe covariance for most games are not readily available but can be inferred if you have VPFW (and some patience).  I'll just say that, for this game, it is 3.0492 in the same units as the variance you quoted.For 4-play, that takes your variance up to 34.6466 bets squared.  For 50,000 plays, down 2.5 standard deviations would be sqrt(50000) x (2.5) x SqRT(34.6466) = 3290.45 bets x $20/bet or $65,809.

[shadowman]

And, if you're 2.5 SDs over the EV you will have made a nice profit. Those risk-reward issues are definitely important.
 
For those who are wondering why the multi-play increases the variance, it is because of the dealt hand issue. For example, you get dealt 4 deuces about once in every 55K hands. Since peacefrog is only planning on playing 50K hands that makes it quite possible they would miss getting them. The cost is $4000 vs. being one short on single line cost $1000.
 
In addition, the odds of a dealt RF is also diminished by the fewer dealt hands. Of course, the dealt deuces and royals are all part of the .9996 EV so not having as good a chance at them increases the risk.

[Eduardo]

Thanks, new2vp and shadowman. That's exactly what I was going to say, if I undertstood any of it.

[faygo]

Thanks, new2vp and shadowman. That's exactly what I was going to say, if I undertstood any of it.
 
Unlike Eduardo, I have a through understanding of Mathematics as witnessed by one of my answers to a test paper.