Average Joe Betting System

[tango]

I just finished reading a book titled: Easy Blackjack for the Average Joe. In this book the author explains how he has evaluated 300 billion hands of blackjack and extracted from that evaluation the percentage of hands lost and won in a row. He used that information to develop a betting system and he then compared that system to 8 other established betting systems. Long story short: His betting system came out with a slightly positive result and I was wondering how that could have happened in a negative-expectation game over that many hands. I'm looking for some clarification. How did the author make this work? He provided the math in tables and it seems to line up with his findings of groups of wins and losses in a row. Anyone read this book that can shed some light?

[ihate17]

Originally Posted by tango

I just finished reading a book titled: Easy Blackjack for the Average Joe. In this book the author explains how he has evaluated 300 billion hands of blackjack and extracted from that evaluation the percentage of hands lost and won in a row. He used that information to develop a betting system and he then compared that system to 8 other established betting systems. Long story short: His betting system came out with a slightly positive result and I was wondering how that could have happened in a negative-expectation game over that many hands. I'm looking for some clarification. How did the author make this work? He provided the math in tables and it seems to line up with his findings of groups of wins and losses in a row. Anyone read this book that can shed some light?

I just did 400 billion hands on my fingers and toes and your writer is full of it. Oh and no charge for the info.

ihate17

[Sonny]

Originally Posted by tango

In this book the author explains how he has evaluated 300 billion hands of blackjack and extracted from that evaluation the percentage of hands lost and won in a row.

You don’t need to analyze 300 billion hands to find out the number of wins and losses in a row. That’s just simple math:

Wins in a row = 0.43^Number of wins
Losses in a row = 0.49^Number of losses

If you include pushes then the numbers change slightly but the losses will always outnumber the wins. That’s the reason progression systems don’t work in the long run.

Originally Posted by tango

How did the author make this work?

Easy – he lied. He probably “adjusted” the numbers to get the results he wanted. I haven’t ready the book but I can assure you that his system does not show a positive expectation for 300 billion hands. There’s some “funny math” going on there.

-Sonny-

[sagefr0g]

Originally Posted by Sonny

You don’t need to analyze 300 billion hands to find out the number of wins and losses in a row. That’s just simple math:

Wins in a row = 0.43^Number of wins
Losses in a row = 0.49^Number of losses

...-

what is this? same as
_______X
0.43

where x=#wins?

that gives you some number of wins in a row?

[Sonny]

Originally Posted by sagefr0g

what is this? same as
_______X
0.43

where x=#wins?

that gives you some number of wins in a row?

It's the exponential function. Winning two hands in a row would be 0.43^2 (aka 0.43², 0.43 squared, or to the power of 2). Winning three hands in a row would be 0.43^3 (0.43³, 0.43 cubed, or to the power of three).

-Sonny-

[sagefr0g]

Originally Posted by Sonny

It's the exponential function. Winning two hands in a row would be 0.43^2 (aka 0.43², 0.43 squared, or to the power of 2). Winning three hands in a row would be 0.43^3 (0.43³, 0.43 cubed, or to the power of three).

-Sonny-

ok that's what i thought you was saying.
so but anyway i don't understand it.
is that some sort of probability math?
tells you the likelyhood of an event not when it's likely to happen?

[Sonny]

Originally Posted by sagefr0g

is that some sort of probability math?

In this case, the probabilities are independent so you can just multiply them together. Here’s a quick rundown:

One win = 0.43
Two wins = 0.43 * 0.43 (same as 0.43^2)
Three wins = 0.43 * 0.43 * 0.43 (same as 0.43^3)
Four wins = 0.43 * 0.43 * 0.43 * 0.43 (same as 0.43^4)
etc.

-Sonny-

[sagefr0g]

Originally Posted by Sonny

In this case, the probabilities are independent so you can just multiply them together. Here’s a quick rundown:

One win = 0.43
Two wins = 0.43 * 0.43 (same as 0.43^2)
Three wins = 0.43 * 0.43 * 0.43 (same as 0.43^3)
Four wins = 0.43 * 0.43 * 0.43 * 0.43 (same as 0.43^4)
etc.

-Sonny-

got it. so it does tell you how likely just not when.
we wanna know when!
knowing that a'int to likely i guess.
probably just could know that circa 63%, 90%, 95% stuff that cancels it's self out sorta thing. standard deviation?
such is life

[Sonny]

Originally Posted by sagefr0g

got it. so it does tell you how likely just not when.
we wanna know when!

There's a formula that gives the average number of hands played before getting ruined, but I don't have it with me. I'll look it up when I get home. That will give you a general idea of how long you might last, but you could easily hit that streak at any time.

-Sonny-

[sagefr0g]

Originally Posted by Sonny

There's a formula that gives the average number of hands played before getting ruined, but I don't have it with me. I'll look it up when I get home. That will give you a general idea of how long you might last, but you could easily hit that streak at any time.

-Sonny-

that would be interesting. i don't think i'll do it lol. but i'd like to see it.
that wouldn't be probability of reaching goal with time constraints or something like that would it?