Reducing variance by not always making the perfect move

For somebody like me, who it hurts to lose $200 (I'm playing nickel FPDW), I'm wondering if there is a way to play "safer", while still getting back 100.7% or so.

I know, I shouldn't play if I can't handle the ups and downs, but I'm thinking that this makes sense, because wouldn't it be better to have a better chance of making slightly less money, than having a lesser chance of making slightly more money? Maybe it ends up being a wash, which is why I made this thread, but I'm just wondering if there is a strategy out there. One that I already do is I never hold a K high 2-to-a-royal.

Also another thing I thought of, is if you were between denominations ($.25 too low, $1 too high), you could always play the $1 with a revised strategy, because let's say you totally butchered the strategy, and now you're only getting back 100.70%, well that's better than getting back 100.76% betting 4x less right? The question is, how much does it lower variance.

This makes me think of the St. Petersburg coin flip paradox, which shows that just because the EV is infinitely positive, it's not a good bet in reality. So sometimes variance is very important.

This is the topic of Risk Aversion (RA) play. The idea is - exactly what you want - to give up some part of EV for reduced variance. With reduced variance your risk of ruin (RoR) will be lower.
This will require a new strategy, based on your bankroll. You don't maximize EV by all decisions, you maximize CEV (maximum log-utility). This is well known in Blackjack, and should be applicable to VP as well.

In principle, in a second stage, once you play on RA strategy, you can trade the lower variance into higher betsize maintaining your original RoR (as you had used by max-EV strategy). Since betsize will be higher, the total EV per decision might be higher.

OK, I deleted some of my OP, because I realized I've been doing something wrong, I think. To calculate the effect of mistakes on my overall payback, I've been multiplying my accuracy (in money, not hands) by my edge of .0076. I think what I should be doing is multiplying it by 100.76. The way I was doing it, you could play only 50% accurate and still make a profit. With my corrected math, it looks like even being 99.24% accurate isn't good enough to be +EV. Wow, looks like those little mistakes can add up then. I'm about 99.8% accurate, which takes 26% away from my edge! 100.76 * .998 = 100.56%, then .2 * .76 = 26%. Correct me if I'm not doing it right. But ya it looks like I'll be doing every little penalty card correctly then.

You're not doing it right.

Quote: MangoJ

You're not doing it right.

No? Then how do you do it? Why wouldn't you take the payback % of game * the % you get correct (via WinPoker)? Again, this is the % correct in money, not hands; not keeping 4 to a royal is much worse than not keeping KT suited.

I think, your calculation of the penalty is correct.
What I don't quite get is what makes you think your "exceptions" decrease variance.

Quote: weaselman

I think, your calculation of the penalty is correct.
What I don't quite get is what makes you think your "exceptions" decrease variance.

I only listed one exception, but if you're wondering how exceptions decrease variance, well let me explain, because they absolutely do.

First let me say that variance has nothing to do with payback %. If you played a game that every time you bet $1, you lost $1, 100% of the time, that game would have no variance (0 or 1, not sure which). With that said, if I'm looking at the hand of As Ad Kd Qd Jd, which would you hold? The 4 to a royal obviously, right? Well you have a 1 in 47 chance of hitting that, so variance will be way higher on that play than if you held the pair of aces, which would only have a variance of the pair, 3 oak, full house, and 4 oak. Those hands are more likely than 1 in 47 chance, and they also pay less than the royal. Understand? Now I'm not saying to do that play, but if you did plays like all the time and ignored the EV, you'd significantly cut down on variance.

Now, the reason why I'm not going to make these deviations, is I originally was only multiplying the cost of the mistakes by my advantage, not the overall payback. I have no idea how I made that dumb mistake, but you could get 99% of your plays wrong, and you'd still have a +EV game according to my foolish math. So ya, it looks like with the corrected math, you will lose quite a bit of EV if you start making deviations. And if the deviations are really close and don't cost you EV, well obviously they won't affect variance much if they're that insignificant. Not sure what the coefficient would be for the cost of deviation vs reduction in variance, but if it's perfectly related, well then deviations are probably a bad idea. However there are some that you could do if you felt like it and it was really close, such as if you never wanted to keep KTs.

There are two problems with your calculations:

First, any "mistake" you do will not lead to the immediate loss of your wager. If you play 99% of your hands right, this doesn't mean that you get 1% less in EV. Wrong played hands still have a non-zero EV, they are just not optimal.

Second, although you label your play deviations "mistakes", they aren't. Mistakes are randomly distributed over all possible hands. Your play deviations however are linked to specific hands, which in turn is linked to specific payouts. Hence you have a strong correlation between payouts and deviations. Just multiplying some average numbers will not do the job.

You need a table of your play deviations, their EV reduction due to your deviations, and the relative frequency they are occuring.

Quote: SilentBob420BMFJ

First let me say that variance has nothing to do with payback %.

I stopped reading here (sorry). I know what variance is.

Quote:

The 4 to a royal obviously, right? Well you have a 1 in 47 chance of hitting that, so variance will be way higher on that play than if you held the pair of aces, which would only have a variance of the pair, 3 oak, full house, and 4 oak.

This is where I resumed. 5 in 47 actually, we are talking about deuces wild, aren't we? So you prefer losing with certainty to winning with probability? Is that it?

Quote:

Now I'm not saying to do that play, but if you did plays like all the time and ignored the EV, you'd significantly cut down on variance.

Well. If you were to just surrender your bet without playing, that would give you the variance of zero ... :)

Quote: MangoJ

There are two problems with your calculations:

First, any "mistake" you do will not lead to the immediate loss of your wager. If you play 99% of your hands right, this doesn't mean that you get 1% less in EV. Wrong played hands still have a non-zero EV, they are just not optimal.

Second, although you label your play deviations "mistakes", they aren't. Mistakes are randomly distributed over all possible hands. Your play deviations however are linked to specific hands, which in turn is linked to specific payouts. Hence you have a strong correlation between payouts and deviations. Just multiplying some average numbers will not do the job.

You need a table of your play deviations, their EV reduction due to your deviations, and the relative frequency they are occuring.

I've been waiting for somebody to ask this, even though I stated it many times (odd how that works). I'll say it again: When I say 99%, I am NOT talking about % of HANDS correct, I'm talking in MONEY, which absolutely does mean exactly what you said it doesn't; 1% in mistakes will cost you 1% in EV. For instance, if there is a play that has an EV of 10, and instead of doing that play, you keep doing the one below it for 9.9, that will cost you 1% of your EV. This is what I mean by "mistakes". What WinPoker does is it takes the EV you chose of each hand, adds it up, and divides it by what the best EV is, and there you have your "mistake %". To say that getting 1 out of every 100 hands wrong means you'll have 1% less EV would be completely incorrect, obviously.

I agree with your last sentence. And I'll add that you'd also need their effect on variance, unless it's perfectly correlated, to which you'd just multiply it by whatever number that is.

Quote: weaselman

I stopped reading here (sorry). I know what variance is.


This is where I resumed. 5 in 47 actually, we are talking about deuces wild, aren't we? So you prefer losing with certainty to winning with probability? Is that it?



Well. If you were to just surrender your bet without playing, that would give you the variance of zero ... :)

5 in 47 is your chance of getting a natural royal OR a deuces royal, and to answer your question, no. Read my reply to Mango.

Your question is very similar to Surrendering 88 vs a 10-Value in Stand Soft 17 Blackjack. Basic Strategy says to split them and increase the variance on a presumed -0.23EV x2 wager. Surrender offers a zero variance -0.50 EV. This situation happens, and Surrender though not correct will lower the variance overall, with little penalty to the game (usually -.0041 vs. -.004 EV).

The third dimension to your OP is how often does the situation occur? that fraction times the increased loss is your new EV. In a 100.7% return game (EV is +.007) one would not want to compromise more than a total of 0.002 EV (the new EV would be +.005 or 100.5%).

N&B