Standard Deviation in VP?

I'm trying to understand swing ranges in VP.

From the Wiz's VP Standard Deviation page:
https://wizardofodds.com/games/video-poker/appendix/3/

Variance:
Full pay deuces wild 25.834618
10/7 double bonus 28.255539
9/6 jacks or better 19.514676
8/5 Bonus poker 20.904082
9/6 DDB 41.985037 (Yike!!!!!! I hate to see what Triple double bonus is.)

Standard Deviation is the sq root of variance.
so for 9/6 JoB: sd = 4.42

to convert that to $:
sd = 4.42 * sqrt (# hands played) * amount wagered each hand

ie #1:
I play thru $750 in single line $1 9/6 JoB.
I lose $300 which is a 40% loss. (aka 60% return)

#hands = $750/5 = 150
sd = 4.42 * sqrt (150) x $5 = 4.42 x 12.25 x 5 = $270
thus anything +/- $270 is within 1 sd of normal.

so -$300 is just bad luck. (and if it was +$300, then it would have been great luck)


ie #2:
I play thru $750 in single line $5 9/6 JoB ($25 a hand).
I lose $300 which is a 40% loss. (aka 60% return)

#hands = $750/25 = 30
sd = 4.42 * sqrt (30) x $25 = 4.42 x 5.48 x 25 = $606
+/- $606 is within 1 sd of normal.

so here -$300 is within the norm?


ie #3 just for giggles (this is what I did for Diamond in a Day):
Play through $50k in single line $2 9/6 JoB ($10/hand).

# hands = 50k/10 = 5000
sd = 4.42 * sqrt (5000) x 10 = $3125

so theoretically DiaD will cost me $50k x (1- 99.5%) = $250
But losing up to $3125 is within the norm?

Your figures look correct. Don't forget the confidence that goes with each Standard Deviation though. 1SD is with 68% confidence. 2SD is with 95% confidence. 3SD is 99.7% confidence. Thus, if you really wanted to play something safe (RoR, etc) you'd want to look with maximum confidence. https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule

You have your EV, and this is essentially the center of your "bell curve" that is your Standard Deviations. So if your EV is -250, then on a graph from -10,000 to positive 10,000 you have a vertical line at -250 upward. From there you can draw a bell curve to your SD's. 250 plus or minus 3125 would be your 1st SD. The confidence that comes with this means "Every time you do this, this bell curve will capture the result 68% of the time."

Now if you draw your SD datapoints to 3SD, then you'd have your vertical line at -250 still but your plus or minus points would be -250 plus or minus 9375. Now when you draw your bell curve you can say "Every time I do this, this bell curve will capture the result 99.7% of the time."

So really 3SD is the realm of possibility, or "normal", as 99.7% of the time your results will fall within this bell curve. The further from 1SD to 3SD you are the more "unlucky" you can say you were (because you're farther and farther away from EV but still within the realm of possibilities). Does this mean you're absolutely cheated if it's outside of 3SD? Well, no not technically (don't forget that .3%), but if you have a decent sampling size I would absolutely start to raise my eye brows.

*do note I didn't fully double check your math on the 3125. Your other scenarios appeared correct so I took this as correct.

To cover 95% of results you should look within +/- 2 SD, but you have the right idea. Variance for 9/7 TDB is 98, FYI.

However thanks to the skew of VP (i.e. big variance due to payouts all sitting in the positive direction), this method will always over estimate your loss expectation because every losing hand only loses one unit. In your example 2, being -2 SD or more down is impossible but it should happen 2.5% of the time.

Quote: Romes

Your figures look correct. Don't forget the confidence that goes with each Standard Deviation though. 1SD is with 68% confidence. 2SD is with 95% confidence. 3SD is 99.7% confidence. Thus, if you really wanted to play something safe (RoR, etc) you'd want to look with maximum confidence. https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule

You have your EV, and this is essentially the center of your "bell curve" that is your Standard Deviations. So if your EV is -250, then on a graph from -10,000 to positive 10,000 you have a vertical line at -250 upward. From there you can draw a bell curve to your SD's. 250 plus or minus 3125 would be your 1st SD. The confidence that comes with this means "Every time you do this, this bell curve will capture the result 68% of the time."

Now if you draw your SD datapoints to 3SD, then you'd have your vertical line at -250 still but your plus or minus points would be -250 plus or minus 9375. Now when you draw your bell curve you can say "Every time I do this, this bell curve will capture the result 99.7% of the time."

So really 3SD is the realm of possibility, or "normal", as 99.7% of the time your results will fall within this bell curve. The further from 1SD to 3SD you are the more "unlucky" you can say you were (because you're farther and farther away from EV but still within the realm of possibilities). Does this mean you're absolutely cheated if it's outside of 3SD? Well, no not technically (don't forget that .3%), but if you have a decent sampling size I would absolutely start to raise my eye brows.

*do note I didn't fully double check your math on the 3125. Your other scenarios appeared correct so I took this as correct.

ah.. a practical use of SD is to estimate how much $ to bring to avoid risk of ruin. thx.

+/- 99.7% of theoretical is normal?!
ahh.. it's within the realm of possibility.

would you use +/- 1sd as being within norms? and anything greater than that as being really unlucky/lucky?

I might consider 1.5-2 SD if I wanted to feel relatively safe. After all, 2SD is 95% confidence. You said it pretty correctly though, 1SD is almost 70% of the outcomes, so that's not a poor assumption, and anything outside of it is in the realm of possibilities, but the further out the unluckier you got... up to 3SD. Outside 3SD you need to look at other conditions such as cheating.

When my business partner and I did a bigger VP play we took 1.5 SD, and I'm glad we did... We were down over 1SD about 60% of the way through, YIKES... then variance balanced out and I hit a royal (my first) putting us up a bit, only to finish down a hair (SPOT ON TO OUR EV) by the end. The variance in VP can be wild, especially depending on the game.

Quote: Romes

I might consider 1.5-2 SD if I wanted to feel relatively safe. After all, 2SD is 95% confidence. You said it pretty correctly though, 1SD is almost 70% of the outcomes, so that's not a poor assumption, and anything outside of it is in the realm of possibilities, but the further out the unluckier you got... up to 3SD. Outside 3SD you need to look at other conditions such as cheating.

When my business partner and I did a bigger VP play we took 1.5 SD, and I'm glad we did... We were down over 1SD about 60% of the way through, YIKES... then variance balanced out and I hit a royal (my first) putting us up a bit, only to finish down a hair (SPOT ON TO OUR EV) by the end. The variance in VP can be wild, especially depending on the game.

This is not true for practical VP plays. The idea that 68% of the results will be within 1 SD of the mean and 95% will be within 2 SD's is only true for the normal or Gaussian distribution.

The central limit theorem states that overall results will eventually converge to normal but it doesn't state how fast it will converge. For a lot of gambling games like blackjack or baccarat this isn't a problem because they're essentially binomial which converges to normal very quickly (a few hundred hands), but video poker takes millions of hands to converge because its distribution is so heavily skewed by jackpots like the royal flush.

If you simulate a typical session of 96JOB (a few thousand hands) you can see for yourself that the results aren't bell shaped or Gaussian. It will be bimodal with 2 humps corresponding to days you hit a royal flush and days you didn't. You can't use normal approximation for a distribution that is not bell-shaped.

Sorry if it sounds like I'm picking on you but a lot of people have posted this mistake on the forum. Normal approximation isn't a practical solution for video poker, slots, or any game with a large jackpot unless you intend to play millions of hands. EV and SD aren't sufficient statistics for these games and you need to run a simulation.

Quote: Tortoise

...For a lot of gambling games like blackjack or baccarat this isn't a problem because they're essentially binomial which converges to normal very quickly (a few hundred hands)...

Oh reeeeaaallly? That's why we're about 15k hands in and no where near our EV? =) To eliminate blackjack variance one needs a minimum of about 50k-100k hands. I do agree video poker might be a bit more give the frequency skew of the royal. Just take a look at the page the OP posted. Look at the variances of VP. When doing SD's in blackjack it's 1.1*AvgBet... 1.1 as opposed to numbers 4x-10x that.

Also, I don't know about others, but I play about 10x-15x as many hands playing VP as I do playing blackjack (pending BJ dealer, etc). So while the variance is substantially less in BJ I'm going to smooth the variance out in VP 10x-15x faster.

Overall, I want to clarify that I am agreeing with Tortoise, and his explanation should make sense as to why. Thanks for bringing something I hadn't really considered to lite, I learned something =D. In terms of "how much money should I take" to a casino for a trip or play though, I'd still think 2SD of Gaussian distribution would cover your plays. It won't be an exact "humpy" chart as Tortoise described, but you should catch an approximation of results near what you want under it. It's a quick and easy way to go "My EV is X and my SD is Y, so I'll take 2Y with me today..." The argument here being "I do plan on playing millions of hands of VP... eventually."

Quote: Romes

In terms of "how much money should I take" to a casino for a trip or play though, I'd still think 2SD of Gaussian distribution would cover your plays. It won't be an exact "humpy" chart as Tortoise described, but you should catch an approximation of results near what you want under it. It's a quick and easy way to go "My EV is X and my SD is Y, so I'll take 2Y with me today..." The argument here being "I do plan on playing millions of hands of VP... eventually."

if 1sd = $3125, how much to bring for 2SD?

Quote: 100xOdds

if 1sd = $3125, how much to bring for 2SD?

$6250 =).... During our big VP play our EV was about -$300, with 1SD being about +-2100... we took $3k, about 1.5 SD.

Quote: Romes

$6250 =).... During our big VP play our EV was about -$300, with 1SD being about +-2100... we took $3k, about 1.5 SD.

ahh..
when I did DiaD on 9/6 JoB, I brought $4k, which apparently is ~1.3SD.

again, apparently I'm not a degen gambler. I had an adequate safety cushion to gamble thru $50k in a day.
I just didn't know it :)

How much more variance is in multi-hand video poker?
What i mean is when you get 1 hand -for example 88kj2 and you chose to keep 88 and then it goes and gives you 100 times 3 new cards besides those 8s...each for money

I found this to be an interesting, albeit old, discussion so just chiming in.

I certainly don’t fancy myself a statistician, but I do like to be an informed gambler so I find these topics very interesting.

I ran an example that I think may be helpful for those semi-serious players who care enough to know the stats but complacent enough to gamble on the strip (I say that in jest, as I do think there are many benefits to staying on the strip and that’s where I always stay when I go). I used “real world” numbers found on the strip from the last time we were in Vegas which was a few months ago.

6/5 Bonus Poker (shitty pay out but in my experiences by far the most common on the strip), playing $0.50 (ie $2.50/push), 10,000 hands which I would say would be a fair estimate for a moderate to heavy player who spends a weekend in Vegas. I used 4.57 for the SD which is technically the SD of 8/5 Bonus but I didn’t know 6/5 off hand.

-10k hands equals $25k coin in
-EV of (-782.50)
-using the posted formula above (sd * sqrt hands played * denomination) would yield: SD 1,142.50

Shifting the bell curve to peak at the EV would give the following:
+/- 1 SD (-$1,925 to +$360)
+/- 2 SD (-$3,067.50 to +$1,502.50)
+/- 3 SD (-$4,210 to +$2,645)

Now as previously noted VP is actually not normally distributed, so how useful these estimations are I guess is up for debate. From a statistician or even “professional gamblers” view point they probably aren’t particularly helpful. But at least in my opinion I think for any casual, or even semi-casual player it can be useful to shed some light on what can be expected and as previously noted I think for the vast vast majority of players it can be used as an approximation for bankroll management purposes.

Personally I would look at the 2 SD mark and focus on my average anticipate loss per 10k hands played/$25k coin in would be (-$782.50) and 95% of the time I would be expected to end up between (-$3,067.50) to + $1,520. I would personally also be fairly confident setting my bankroll at $3,000 if I planned to play the amount above.

Again, the pure statisticians are probably cringing but for the rest of us I think it suffices.

Also, I think using the SD amounts really highlights the importance of finding better pay tables. Playing a full pay 8/5 Bonus machine would result in that “negative EV bell shift” from -782.50 to -$207.50 and that bulk of anticipated outcome values found for 1SD to -1,350 to +$935 which just seems so much better.