Volatility questions

My wife and I play DDB quite often at the only casino reasonably driveable for us. Unfortunately, they have pretty awful video poker, 7/5 DDB at the level we can afford to play. I know DDB is very volatile, and I wonder just how volatile it can be with this paytable both positive and negative. We probably play a combined total of 10,000 hands per session. One recent trip yielded a total of 8 4oaks between us. I've read that the average hands for a 4oak in this game is about 425 hands. We had a session a few weeks ago where we had 10 quads just in Queens, and 26 total quads between us.

What I'd like to know is how good and bad this game can get with quad output considering it to say 3 standard deviations? This should tell us what to expect as a range for 99.5% of all sessions with the extremely rare 4th standard deviation or more in that tiny remainder. I think this would be helpful to many of us who have to play on less than full pay machines because we are unable to get to AC or Vegas. In addition to the quad swing range, I'd also like to know what range of losses or gains fit into the 3 standard deviations. Thanks in advance to any and all math gurus who can answer these questions!

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silly

Sally

First I'd like to thank you for your considerate reply. I must confess I ran like a rabbit from math after High School. I studied English and Business Law. I don't understand binomial distribution at all. It seems to me that if the chance is 1 in 415, then that would be the average rate. But this doesn't seem to be the case. Let's see if I comprehend your work:

for the number of quads expected you established 23.52941176 as the mean right? Does that represent an average?
the best case at 4 SDs is 42.90942144 quads is that right?
3 SDs out to the positive is 38.06441902 quads
2 SDs is 33.2194166 quads
1 SD is 28.37441418 quads

Then the terrifying dark side computations say
-1 to -4 sd
18.68440935 quads
13.83940693 quads
8.994404507 quads
4.149402088 quads

So pretty much the worst case scenario would be to get 4 quads in 10,000 hands correct?

I didn't specify the session bankroll, but rather calculated our hands by timing our play and knowing our hands per hour output. If we assume we are able to fund 10,000 hands just for the sake of convenience here, I'm trying to understand the scale of how much you could lose or win out to the 4th SD. I appreciate your patience and nice work on these questions very much.

Quote: 4andaKicker

First I'd like to thank you for your considerate reply. I must confess I ran like a rabbit from math after High School. I studied English and Business Law. I don't understand binomial distribution at all.

binomial distribution is fairly simple to comprehend, sometimes a lot of math is involved so many use calculators.
Here is a link for some reading if anyone wants to learn more on the subject.
http://stattrek.com/probability-distributions/binomial.aspx?Tutorial=Stat

binomial is just about 2 possible outcomes, success and failure.

In 10 coin flips we expect 5 heads and 5 tails but we know we could have anywhere from 0 heads to 10 heads.
Not every possible outcome has the same probability.

Quote: 4andaKicker

It seems to me that if the chance is 1 in 415, then that would be the average rate. But this doesn't seem to be the case. Let's see if I comprehend your work:

for the number of quads expected you established 23.52941176 as the mean right? Does that represent an average? Yes, the mean is the average
the best case at 4 SDs is 42.90942144 quads is that right? Yes
3 SDs out to the positive is 38.06441902 quads
2 SDs is 33.2194166 quads
1 SD is 28.37441418 quads

Then the terrifying dark side computations say
-1 to -4 sd
18.68440935 quads
13.83940693 quads
8.994404507 quads
4.149402088 quads

So pretty much the worst case scenario would be to get 4 quads in 10,000 hands correct?

I answered in bold. You got it!

Quote: 4andaKicker

I didn't specify the session bankroll, but rather calculated our hands by timing our play and knowing our hands per hour output.

By knowing the number of hands, the game and the average $ bet, one can easily calculate a session bankroll and the corresponding bust rate (survival rate or Risk of Ruin)

Quote: 4andaKicker

If we assume we are able to fund 10,000 hands just for the sake of convenience here,
I'm trying to understand the scale of how much you could lose or win out to the 4th SD. I appreciate your patience and nice work on these questions very much.

My answer assumed you played 25cents VP.

-$536 mean
1SD = $811.71

-3SD -$2,971.13
-2SD -$2,159.42
-1SD -$1,347.71
mean -$536
1SD $275.71
2SD $1,087.42
3SD $1,899.13

SD being equal, you would have the same chance of being up $275 or being down $1347


VP game return of .95712
house edge = -0.04288 (1-0.95712)
mean (expected value) after 10,000 hands =
N*P (10,000*-0.04288)= -428.8 units

then *5 (for 5 max coin played)
-428.8 *5 = -$2144 units is our expected value in units after playing 10,000 hands.
(Yes, this assumes we have enough bankroll to last that long)

A unit can be any VP machine denomination. For examples: expected value
-536 $0.25
-214.4 $0.10
-2144 $1
-10720 $5

We now need to calculate the $SD (standard deviation, how far from the average)
so we can see what ranges we will be dealing with.

formula = SD * (square root # hands played)
SD for a max 5 coin bet = 32.4685 (from the paytable or a program)
* 100(the square root of 10,000) =
3,246.85 (this is per unit bet)
Multiply the above value by your VP denomination played

examples:
denomination, 1SD
$0.25 $811.7125
$0.10 $324.685
$1 $3,246.85
$5 $16,234.25


We now know the mean or average expected value and can add and subtract SDs to find our range.

example:
-$536 mean
3SD = plus or minus $2435.14 ($811.71*3) for 25cent VP

Sally

Thanks Sally for your great work. It is pretty sobering to see how poor the chance really is to come out on top. On the good side, we moved our son's birthday trip up to Sunday rather than this coming Tuesday, and my wonderful wife hit a Royal equivalent hand in Triple Double Bonus, (4 Aces w/kicker) and an astounding 5 more quad Aces (without kicker) in her play. I lost, but my wife won enough after all was done to place the victory in the third SD. Thanks again for explaining the math components here.