vpFREE2 Forums

significance of N0

#1

the significance of N0 is that if your goal is to win, you have to
play N0 hands (variance/(er-1+comp)^2) to get even an 84% chance of
winning

for example, for full pay deuces wild:

at one hand your chances of net winning are 17%
at ten hands your chances of net winning are 40%
at 100 hands your chances of net winning are 38%
at 1000 hands your chances of net winning are 35%
at 5000 hands your chances of net winning are 44%
at 10000 hands your chances of net winning are 48%

at N0 (450,000) hands your chances of net winning are 84%
at 4N0 (1.8 million) hands your chances of net winning are 98%
at 9N0 (4 million) hands your chances of net winning are 99.9%

you can calculate N0 for any game that you know the variance and er
N0 = variance/(er-1+comp)^2 hands
while you're at it, you can also calculate:
kelly bankroll (~11%ror) = variance/(er-1+comp) bets
sharpe ratio = (er-1+comp)/sqrt(variance) (higher number is better
bet)

#2

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:

the significance of N0 is that if your goal is to win, you have to
play N0 hands (variance/(er-1+comp)^2) to get even an 84% chance of
winning

at N0 (450,000) hands your chances of net winning are 84%
at 4N0 (1.8 million) hands your chances of net winning are 98%
at 9N0 (4 million) hands your chances of net winning are 99.9%

you can calculate N0 for any game that you know the variance and er
N0 = variance/(er-1+comp)^2 hands

What you have done appears similar to a table I created several years
ago, but I never knew if it was valid. The table has "total games
played" on one axis and variance on the other, and cell values
showing the +/- extremes around the expected return at a specific
confidence level. I used the standard CLT formula.

I believe the table is valid (and possibly useful to some
players) because the return and standard deviation of a game can both
be expressed as a fraction, multiple, or percentage of a betting
unit. The return of a game is usually written as a percentage of a
betting unit, but a 101% return can also be written as 1.01 betting
units. The standard deviation is usually expressed as a multiple or
fraction of a betting unit, but it can also be written as a
percentage. For example, JorB has a Variance of 19 (squared betting
units), or a standard deviation of 4.36 betting units, or 436% of the
expected return.

To give an example using the table, if a game has a variance of
20 (close to 9/6 JorB) and I play 1 million games, the table
entry is 0.74%. This game returns 99.54%, and lets assume comps
of .46% bring the return to 100.00%. This means a player can expect
their actual return (with perfect play) to be in the range of 100.00%
+/- .74% (or between 99.26% and 100.74%), with 95% confidence.

My original reason for building the table was to identify (for any
given return + comps and game variance) the number of games I must
play perfectly to be guaranteed (with 95% confidence) of not losing
any money. That way I can decide what the return must be, and how
many games I must play, in order not to be an overall loser. But the
table might also be useful to people concerned about the relationship
between the variance and total return of a game, although that wasn't
my original reason for making it.

#3

yup, it's basically the same thing

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:
>
> the significance of N0 is that if your goal is to win, you have to
> play N0 hands (variance/(er-1+comp)^2) to get even an 84% chance

of

> winning
>
> at N0 (450,000) hands your chances of net winning are 84%
> at 4N0 (1.8 million) hands your chances of net winning are 98%
> at 9N0 (4 million) hands your chances of net winning are 99.9%
>
> you can calculate N0 for any game that you know the variance and

er

> N0 = variance/(er-1+comp)^2 hands
>>>>>>>>>>>>>>>>>>>>>>>>

What you have done appears similar to a table I created several

years

ago, but I never knew if it was valid. The table has "total games
played" on one axis and variance on the other, and cell values
showing the +/- extremes around the expected return at a specific
confidence level. I used the standard CLT formula.

I believe the table is valid (and possibly useful to some
players) because the return and standard deviation of a game can

both

be expressed as a fraction, multiple, or percentage of a betting
unit. The return of a game is usually written as a percentage of a
betting unit, but a 101% return can also be written as 1.01 betting
units. The standard deviation is usually expressed as a multiple

or

fraction of a betting unit, but it can also be written as a
percentage. For example, JorB has a Variance of 19 (squared

betting

units), or a standard deviation of 4.36 betting units, or 436% of

the

expected return.

To give an example using the table, if a game has a variance of
20 (close to 9/6 JorB) and I play 1 million games, the table
entry is 0.74%. This game returns 99.54%, and lets assume comps
of .46% bring the return to 100.00%. This means a player can

expect

their actual return (with perfect play) to be in the range of

100.00%

+/- .74% (or between 99.26% and 100.74%), with 95% confidence.

My original reason for building the table was to identify (for any
given return + comps and game variance) the number of games I must
play perfectly to be guaranteed (with 95% confidence) of not losing
any money. That way I can decide what the return must be, and how
many games I must play, in order not to be an overall loser. But

the

table might also be useful to people concerned about the

relationship

between the variance and total return of a game, although that

wasn't

ยทยทยท

--- In vpFREE@yahoogroups.com, "brumar_lv" <brumar_lv@y...> wrote:

my original reason for making it.

#4

the derivation of N0 (which does rely on clt) is pretty
straightforward:

the -2sd curve is:

(er-1) x hands - 2sqrt(variance x hands)

for full pay deuces wild it is:

.00762 x hands - 2sqrt(25.83462 x hands)

this can be ploted using a spreadsheet

to find the minimum take the first derivative and solve for zero:

(er-1) - 1/sqrt(variance x hands) = 0
hands = variance/(er-1)^2
which is also the formula for N0 which is the x-intercept of the -1sd
curve

to find the loss at the minimum, plug N0 into the -2sd formula:

(er-1) x variance/(er-1)^2 - 2sqrt(variance x variance/(er-1)^2))
variance/(er-1) - 2variance/(er-1) = -variance/(er-1)
which is also the formula for the kelly bankroll

so, for the -2sd curve (2.28% of results are worse assuming clt) the
highest loss is a kelly bankroll (variance/(er-1) bets) and it occurs
at N0 (variance/(er-1)^2) hands, for full pay deuces wild 5-coin
quarters this would be $4,238 at 444,930 hands