Having a Problem with Gambling Theory And Other Topics

[KeyToTheMint]

My problem is playing .50 cent-dollar limits and using a sample
size of 50,000+ hands I find out:

win rate= 4.6 big bets/ 100 hands with
standard deviation = 14.77

Using Mason's formula in the chapter, how much do you need?
I compute I need only 106.705 big bets in my bankroll to assure a win. Needless to say this cannot be right.

In practice there was a subset of hands, nearly 7000 hands in
that sample, where I not only didn't make a profit but was
down 224 big bets. What gives?

Any comments welcome.

[BluffTHIS!]

You need to reread it to thoroughly to understand the concepts of standard deviation and variance.

[ QUOTE ]
I compute I need only 106.705 big bets in my bankroll to assure a win.

[/ QUOTE ]

NOTHING assures a win, especially if you don't play well. And even if you are a good player, the formula only predicts what bankroll level will give you the BEST CHANCE of surviving down swings a winning player might encounter and avoid what is termed 'risk of ruin'.

If you had Bill Gates' bankroll and play bad enough you would lose it, only it would take you a very long time, with only death preventing that outcome at those limits.

[maurile]

[ QUOTE ]
to assure a win.

[/ QUOTE ]
This part isn't right.

What risk of ruin are you using? And what unit is your standard deviation measured in? (Big bets per 100 hands?)

[KeyToTheMint]

Sorry if I wasn't clear. My standard deviation is 14.77 per
100 hands and the term "assure a win" comes directly from
the book. In a footnote: Mason writes the risk of ruin in
his charts are approximately 1.1 percent.

My question is using a win rate of 4.6/100 big bets with
a standard deviation of 14.77 big bets/100 is 107 Big
bets enough to "assure a win"? In other words using those
specific numbers and masons formula, is the answer indeed
107? Furthermore, I know ituitively this is wrong and
already encountered a sequence where i lost 224 big bets,
more than twice what I needed. However, I was up over 800
big bets at the time so I didn't go broke and theory wasn't
proved wrong in practice. The point is though, if I took
just a 107 big bet bankroll and started playing at the
beginning of the 7000 hand run i would have tapped out
long before I even got to 7000 hands. Is this just attributed
to 1.1 percent risk of ruin or is my 107 big bet bankroll
computed wrong on my part?

[maurile]

[ QUOTE ]
My question is using a win rate of 4.6/100 big bets with a standard deviation of 14.77 big bets/100 is 107 Big bets enough to "assure a win"?

[/ QUOTE ]
Yes. The formula is:

Bankroll Requirement = (-1)*ln(RiskOfRuin)*(SD^2)/(WinRate*2)

In your case, (-1)*ln(0.011)*(14.77^2)/(4.6*2) = 106.9

So if you can bear a 1.1% chance of going broke, your bankroll requirement is 107 big bets.

[ QUOTE ]
Furthermore, I know ituitively this is wrong and already encountered a sequence where i lost 224 big bets, more than twice what I needed. However, I was up over 800 big bets at the time so I didn't go broke and theory wasn't proved wrong in practice. The point is though, if I took just a 107 big bet bankroll and started playing at the beginning of the 7000 hand run i would have tapped out long before I even got to 7000 hands. Is this just attributed to 1.1 percent risk of ruin or is my 107 big bet bankroll computed wrong on my part?

[/ QUOTE ]
You computed your bankroll requirement correctly.

The formula isn't telling you that you only have a 1.1% chance of ever experiencing a 107-BB downswing. (The chance of that is much higher.) It's telling you that you only have a 1.1% chance of experiencing a 107-BB downswing starting at any pre-defined point.

In other words, if you plot your bankroll along the y-axis of a graph and your total number of hands played along the x-axis, you will get a sort of squiggly, zig-zaggy pattern that generally slopes upward and to the right.

Let A = the chance that, sometime after hand #1, your bankroll will ever be 107 BB less than it was at hand #1.

Let B = the chance that, sometime after hand #50, your bankroll will ever be 107 BB less than it was at hand #50.

If A = 1.1%, and B = 1.1%, then (A or B) has to be greater than 1.1%.

You could assign a different variable like that to every hand you play. The chance that you will suffer a 107-BB downswing from any predetermined point is the same: A = B = C = D = . . . = X = Y = Z = 1.1%.

But the chance that you'll ever suffer a 107-BB downswing from any point is much larger: (A or B or C or D or . . . or X or Y or Z) > 1.1%.

The formula Mason gives takes into account that a 107-BB downswing is not likely to bust you . . . unless you happen to experience it right off the bat (before increasing your bankroll at all), which is unlikely.

[KeyToTheMint]

Thanks Maurile, it makes sense to me now.