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?
Average Joe Betting System
- Thread starter tango
- Start date
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.
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-
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.
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 said:
what is this? same as
_______X
0.43
where x=#wins?
that gives you some number of wins in a row?
_______X
0.43
where x=#wins?
that gives you some number of wins in a row?
-Sonny-
sagefr0g said:
got it. so it does tell you how likely just not when.
we wanna know when!
we wanna know when!
-Sonny-
21st century blackjack
I've seen a lot of stuf lately about Twenty-first Century Blackjack. Does anyone know what the actual betting strategy is? I have seen positive reviews but I don't think I want to slog through hundreds of pages of drivel and anecdotes to get to the 3 pages of meat. Thanks!
And yes, I know... nothing actually works other than counting. I like the research. By the way, if anyone knows where there is a large results chart anywhere, I'd appreciate that as well. Thanks again, everyone!
Carpe Este
I've seen a lot of stuf lately about Twenty-first Century Blackjack. Does anyone know what the actual betting strategy is? I have seen positive reviews but I don't think I want to slog through hundreds of pages of drivel and anecdotes to get to the 3 pages of meat. Thanks!
And yes, I know... nothing actually works other than counting. I like the research. By the way, if anyone knows where there is a large results chart anywhere, I'd appreciate that as well. Thanks again, everyone!
Carpe Este
The results of Thomason's strategy can be found at the link below:
"Player 3: This player uses a popular strategy found in the book 21st Century Blackjack by Walter Thomason. Without going into details, this is basically a 2-3-4-5 positive progression, reset on shuffle, stop after four consecutive losses. But any progression will do as they are all the same. "
http://www.blackjack-scams.com/html/prog__systems.html
-Sonny-
"Player 3: This player uses a popular strategy found in the book 21st Century Blackjack by Walter Thomason. Without going into details, this is basically a 2-3-4-5 positive progression, reset on shuffle, stop after four consecutive losses. But any progression will do as they are all the same. "
http://www.blackjack-scams.com/html/prog__systems.html
-Sonny-
thanks!
I am 50-something, and used to count in my youth. I calculated percentages to a 4-deck shoe. When they went to 6 and 8 decks, I was done. I remember playing at the Dunes, in a split-level section at the rear, single-deck 1-on1 and positively crushing them regularly. Now they are standing 3-deep at $15 minimum tables with those damn continuous-shuffle machines. What a gyp!
Thnks for the link!
I am 50-something, and used to count in my youth. I calculated percentages to a 4-deck shoe. When they went to 6 and 8 decks, I was done. I remember playing at the Dunes, in a split-level section at the rear, single-deck 1-on1 and positively crushing them regularly. Now they are standing 3-deep at $15 minimum tables with those damn continuous-shuffle machines. What a gyp!
Thnks for the link!
Thanks for the responses to my wins/losses in a row question. Since I have you guys here I have one other question for you: If my choices of table types are either a six-deck game with H17 and surrender available or a six-deck game with S17 but no surrender, which game should I choose? Which is more valuable? The dealer having to stand on 17 or having the option to surrender those ugly 15's and 16's when I can?
Go for the S17 game. From the Basic Strategy Engine on this site you can compare the house edges and basic strategies:
6D S17 DAS = 0.44% house edge
6D H17 DAS Late Surrender = 0.58% house edge
The S17 game is better for a basic strategy player.
-Sonny-
6D S17 DAS = 0.44% house edge
6D H17 DAS Late Surrender = 0.58% house edge
The S17 game is better for a basic strategy player.
-Sonny-
Yes; if you're interested, you should take a course on probability and statistics from your school or from a community college - it will be very helpful.
For instance, the probability of getting three wins in a row is 0.43^3, but the probability of getting exactly three wins in four hands is different, because you have the possibility of:
WWWL
WWLW
WLWW
LWWW
In this case, the probability is higher than 0.43^3, because there 4 ways to get three wins. The probability then becomes 4*0.43^3*0.49^1. This is an easy example because you can count it easily.
Now, when you consider the probability of winning 501 or more hands out of 1000, you need statistics (or a lot of patience to count the combinations). Statistics is basically a way of counting up the combinations really fast. There's things like binomial distribution, normal distribution, and other things that show up over and over again, both in gambling fields as well as non-gambling fields.
For instance, the probability of getting three wins in a row is 0.43^3, but the probability of getting exactly three wins in four hands is different, because you have the possibility of:
WWWL
WWLW
WLWW
LWWW
In this case, the probability is higher than 0.43^3, because there 4 ways to get three wins. The probability then becomes 4*0.43^3*0.49^1. This is an easy example because you can count it easily.
Now, when you consider the probability of winning 501 or more hands out of 1000, you need statistics (or a lot of patience to count the combinations). Statistics is basically a way of counting up the combinations really fast. There's things like binomial distribution, normal distribution, and other things that show up over and over again, both in gambling fields as well as non-gambling fields.
i was gonna do that after i retired just on a lark. thing was i figured there would be information overload. things i wasn't really interested in.
i wish i could find a book with it all simplified and directed at gambling problems.
just guessing if it's five hands maybe the probabilities a little higher sort of thing?
so just a quick question does statistics give you just as accurate of an answer as you'd get if you went and did all those combinations?
oh and if not to much trouble. what's the differance in a binomial distribution and a normal distribution. i think i know what a normal distribution is. only thing binomial i remmeber is the binormial equation which i think i forgot lol.
i wish i could find a book with it all simplified and directed at gambling problems.
just guessing if it's five hands maybe the probabilities a little higher sort of thing?
so just a quick question does statistics give you just as accurate of an answer as you'd get if you went and did all those combinations?
oh and if not to much trouble. what's the differance in a binomial distribution and a normal distribution. i think i know what a normal distribution is. only thing binomial i remmeber is the binormial equation which i think i forgot lol.
OK, here's the formula I was talking about before:
RoR = (F^T-F^BR)/(F^T-1)
Expected number of plays before ruin = (BR/(L-W))-(T/(L-W))*(1-F^BR)/(1-F^T)
Where:
W = Probability of a win
L = Probability of a loss
F = L/W
BR = Your bankroll
T = Your bankroll + Opponent’s bankroll
For example:
W = 60%
L = 40%
F = 0.6/0.4 = 0.667
BR = 10 units
T = 10 + 50 = 60 units
RoR = (0.67^60-0.67^10)/(0.67^60-1) = 1.73%
Avg. Plays = (10/-0.2)-(60/-0.2)*(1-0.667^10)/(1-0.667^60) = 244.8
So you have a 1.73% chance of losing your 10 units, a 98.27% chance of winning your opponents 50 units and an average of 245 plays before you go broke (in the few cases when you do).
-Sonny-
RoR = (F^T-F^BR)/(F^T-1)
Expected number of plays before ruin = (BR/(L-W))-(T/(L-W))*(1-F^BR)/(1-F^T)
Where:
W = Probability of a win
L = Probability of a loss
F = L/W
BR = Your bankroll
T = Your bankroll + Opponent’s bankroll
For example:
W = 60%
L = 40%
F = 0.6/0.4 = 0.667
BR = 10 units
T = 10 + 50 = 60 units
RoR = (0.67^60-0.67^10)/(0.67^60-1) = 1.73%
Avg. Plays = (10/-0.2)-(60/-0.2)*(1-0.667^10)/(1-0.667^60) = 244.8
So you have a 1.73% chance of losing your 10 units, a 98.27% chance of winning your opponents 50 units and an average of 245 plays before you go broke (in the few cases when you do).
-Sonny-
thanks Sonny. i don't really understand the logic of all that. usually when somebody starts using exponents to real world problems i get lost. lol
so but taking it for granted that all that does make sense i noticed that like in your example you've got your opponent with only 50 units. so i guess the question is to ask if it's reasonable to dice up an opponents really bankroll (say for instance a casino) where you say ok i just want 50 units of your big fat bankroll and the hell with the rest. lol. so you do your math as above and take your shot. somehow i don't think it's entirely reasonable because it looks to me like yeah you got a 98.27% chance of winning that 50 units but what's not said is maybe that could take you God only knows how many hands maybe way more than the average 245 plays before you go broke sort of thing? or am i just not getting it? it's interesting. i know that CVCX has calculators that'll tell you ok if you play 245 hands say for some game and you risk say 10 units then it'll come back and give you the probability of maybe winning 50 units or what ever. doesn't quite seem like what your giving above is the same thing or not?
so but taking it for granted that all that does make sense i noticed that like in your example you've got your opponent with only 50 units. so i guess the question is to ask if it's reasonable to dice up an opponents really bankroll (say for instance a casino) where you say ok i just want 50 units of your big fat bankroll and the hell with the rest. lol. so you do your math as above and take your shot. somehow i don't think it's entirely reasonable because it looks to me like yeah you got a 98.27% chance of winning that 50 units but what's not said is maybe that could take you God only knows how many hands maybe way more than the average 245 plays before you go broke sort of thing? or am i just not getting it? it's interesting. i know that CVCX has calculators that'll tell you ok if you play 245 hands say for some game and you risk say 10 units then it'll come back and give you the probability of maybe winning 50 units or what ever. doesn't quite seem like what your giving above is the same thing or not?