123BroRick123
Posted by 123BroRick123
Oct 20, 2018

Equation?

I'd like to know the mathematical equation for finding out the best percentage of these 4 bad bets. These bets are very bad in the long run, so I'm only going to raise high the bets, and bet ONE TIME ONLY and leave.
1. a Bet that Wins 83.3 per cent needing 6 wins to recoup (30/36)
2. a Bet that Wins 91.6 per cent needing 13 wins to recoup (33/36)
3. a bet that Wins 94.4 per cent needing 20 wins to recoup (34/36)
4. a Bet that Wins 97.2 per cent needing 39 wins to recoup (35/36)
No wise cracks, please. Just math concerning the winning percentages with the wins needed to recoup. Thank you. Bro. Rick
For ONE ROLL ONLY.

Comments

OnceDear
OnceDear Oct 21, 2018

Please clarify your question ( which would have got more attention as a forum post )

By 'wins 83.3 per cent' Do you mean has 83.3% probability of winning?

How many units do you win when you win for each of these options? Assuming your bet is one unit?



For the mathematical equation, you need to calculate the Expected Value of each wager. The one with the most positive EV is, on average, the 'best' bet.



EV= (Probability of winning x Profit of winning) + (Probability of losing x 'Profit' of losing)



If you potentially lose all of your 1 unit stake, this becomes



EV= (Probability of winning x Profit of winning) - (Probability of losing x 1)



EV= (Probability of winning x Profit of winning) - (1-Probability of winning)



So, consider a fair coin toss.

EV= (0.5 x1)-(1-0.5) =0

I.e It's a neutral bet and long term you will win or lose 0 units per game on average.



Now consider a wager on rolling a 6 with a single die, where the house gives you 4 units if you win ( on top of returning your stake)

Probability of a win is 1/6=0.166667: Profit of winning is 4

EV=(0.1666667 x 4)-(1-0.166667)

EV=-0.1666667

Normally expressed as a percentage: You'll lose 16.67% of the total amount you wager, on average. That's worse than if the gave you fair odds which is 5 units for a win.

123BroRick123
123BroRick123 Oct 21, 2018

Please clarify your question ( which would have got more attention as a forum post )



By 'wins 83.3 per cent' Do you mean has 83.3% probability of winning?



How many units do you win when you win for each of these options? Assuming your bet is one unit?



Thank you for your wonderful answer. I know these are terrible bets in the long run, but for ONE ROLL ONLY they might be valid.

Number 1 is your typical Iron Cross. Number 2 is with an added Lay4 bet. Number 3 is with an added Lay4 Bet and an added Hard4 bet. Number 4 is Anything But A Hard4 (or Hard6 or Hard 8 or Hard10.

1. 83.3 percent chance of winning 7.5 and losing 41

2. 91.6 percent chance of winning 7.4 and losing 93

3. 94.4 percent chance of winning 7.6 and losing 153

4. 97.2 percent chance of winning 1.7 and losing 97



Thank you very much on your input.

Gratitude opens doors to abundance.

Bro. Rick

OnceDear
OnceDear Oct 21, 2018

1. 83.3% chance of winning 7.5 and 16.7% chance of losing 41

1. 83.3% chance of winning 7.5/41 and 16.7% chance of losing 1

EV=(0.833 x 7.5/41 - 0.167)

EV=-0.01462195121 or -1.462%. Not bad if your pay-table is correct.



2. 91.6% chance of winning 7.4 and 8.4% chance of losing 93

2. 91.6% chance of winning 7.4/93 and 8.4% chance of losing 1

EV=(0.916 x 7.4/93 - 0.084)

EV= -0.01111 or -1.1111%. A bit better than option 1



3. 94.4% chance of winning 7.6 and 5.6% chance of losing 153

3. 94.4% chance of winning 7.6/153 and 5.6% chance of losing 1

EV= (0.944 x 7.6/153 - 0.056)

EV=-0.009108 or -0.91%. Better still



4. 97.2% chance of winning 1.7 and 2.8% chance of losing 97

4. 97.2% chance of winning 1.7/97 and 2.8% chance of losing 1

EV=(0.972 x 1.7/97 -0.028)

EV=-0.0109649 or -1.09%





So. Assuming your pay-table is correct, none of the bets are too bad. Option 3 would be best. However, I'm not familiar with craps and have taken your probabilities at face value. Can all those wagers definitely be resolved with one roll?

123BroRick123
123BroRick123 Oct 21, 2018

So. Assuming your pay-table is correct, none of the bets are too bad. Option 3 would be best. However, I'm not familiar with craps and have taken your probabilities at face value. Can all those wagers definitely be resolved with one roll?



Yes. Here is the breakdown on my "ANYTHING BUT EASY 4". Numbers: 3-Hard4-5-6-8-9-10-11 wins $7. Number 7 wins $6. Numbers 2 & 3 win $14 (double on 2 & 3 in the Field).



Bets: Put $10 on the 5, $12 on the 6, $12 on the 8, $140 on the Lay4, and $20 on the Hard4.

140 on Lay4 = $67 -$20 (on the Field) -$41 (on the 5/6/8/Field) = $6

20 on Hard4 =$140 (7x) +$7 (on the Field) -$140 (on the Lay4) = $7



36 ROLL EXPECTATION:

2 = 14 x1 = 14

3 = 7 x2 = 14

Easy4 = -140 -20 +7 = -153x2 = -306

Hard4 = 7 x3 = 21

5 = 7 x4 = 28

6 = 7 x5 = 35

7 = 6 x6 = 36

8 = 7 x5 = 35

9 = 7 x4 = 28

10 = 7 x3 = 21

11 = 7 x2 = 14

12 = 14 x1 = 14



260 -306 = -46





LOSES 2 OUT OF 36 !

Win percent: 34/36 = 94.4 !

Put up 201

Average Win = 260/34 = 7.647

Loss = -153

Have to Win 20x



A few years ago I went to a $50,000 Craps Tournament in Vegas. You had to roll an 11 when playing to get an invitation. I made it and they gave everyone at the table about $5,000 in chips and play was to be one hour long. I put up most of the chips according to this method, and the shooter threw a 3-1, an easy 4!!! Ouch.



Thank you for your explanation and the math involved. I was par excellent in algebra but needed help in this. Thank you. Bro. Rick

BaccaratKid
BaccaratKid Oct 22, 2018

Just take everything you have or can get your hands on and put it all on one hand of Baccarat on Banker. Why? Because the Banker will win more hands than the Player. Take the 5% beat on the commission charged to all Banker wins and be happy you won. Your Martingale fantasy you are toying with above will only lead to misery as you face that huge final bet, even while knowing the outcome in your heart of hearts. It's all fun and games until that moment arrives. Free drinks. Maybe even a suite and meals galore, depending on your average bet. Then.......BAM!! They get it all back WITH INTEREST and you are left facing that uphill climb that never quite reaches your starting point. Worse than smoking crack because your health doesn't deteriorate so you are left to deal with the mental anguish with mind and body intact and the relief of death nowhere in sight! lol

OnceDear
OnceDear Oct 22, 2018

Quote:

Just take everything you have or can get your hands on and put it all on one hand of Baccarat on Banker.



What's the fun in that. If you lose, you are destroyed. If you win, what do you do for an en-core? bet it all again? Losing everything might be the best option.



Marty is a fun way to lose money. It doesn't increase the house edge, so no harm done. $:o)

odiousgambit
odiousgambit Oct 23, 2018

With 36 in the denominator looks like Craps.



I think you'll need to clarify some things,



* you do mean "return" of 83.3% etc?

* what do you mean by number of wins 'to recoup'?

* how do you recoup if you bet one time only?

OnceDear
OnceDear Oct 23, 2018

HI ODG,

He did sort of clarify and it is indeed craps, which is not something I know much about. He mentions bets like Iron Cross, so I'm not sure his percentages are correct for any one roll wager.

odiousgambit
odiousgambit Oct 23, 2018

stupidly I left a comment without noting there were already comments, sorry



the iron cross is a combination of bets, a bet on the 5, 6, 8, and the field ... Wizard breaks it down on a per roll basis, as an HE of "(5/22)×2.778% + (5/22)×1.111% + (6/22)×0.463% + (6/22)×0.463% = 1.136%." Note that the final EV is the sum of all the other EVs, which as others have long pointed out is probably the best way to look at combination bets ... they never cancel each other out.



I get a headache trying to figure out what he is getting at

odiousgambit
odiousgambit Oct 23, 2018
123BroRick123
Posted by 123BroRick123
Oct 13, 2018

PROGRESSION

Try this progression on your WinCraps:

On Pass Line, no Odds or on Don't Pass, no odds.

1-2-3-5-7-9-11

On a minimum $5 table, Bet

$5-$10-$15-$25-$35-$45-$55

Follow these rules:

Always Start with $5.

After every Win, go Down one step. DO NOT GO UP ONE STEP!

After every Loss, go UP one step. DO NOT GO DOWN ONE STEP!

STOP at after every Bet on $1, Start again with $5.

STOP at after every Bet on $11, Start again with $5.

I'm Wining over $80 an hour on my WinCraps with this method

Comments

OnceDear
OnceDear Oct 13, 2018

Take it to the real world and report back from your mansion when you have made your fortune.....

... Or accept that you've just discovered an amusing way to lose money, like the rest of us dreamers.





Before you criticize, try this system on WinCraps. Say 40 games of 120 Rolls. (4,800 Rolls), then report back.

Zcore13
Zcore13 Oct 13, 2018

Why not bump your minimum to $50 and bring home $800 an hour??



Yes, you're right. Again, before you criticize, try this system on WinCraps. Say 40 games of 120 Rolls. (4,800 Rolls), then report back.



They don't have craps in Arizona, but when I get to Vegas I might Bet 10x this progression.

OnceDear
OnceDear Oct 13, 2018

I don't believe the above comment is a quote of ZCore': I believe that the OP edited it with his reply instead of adding his reply as a new comment. Surely unintentional if he did.



Anyhow... 123. Try this scheme with real money if ... 1) You want to have some fun and 2) You are happy to lose your bankroll.



If you have a profit, then come back and rave about it... But that won't make it a winning system: Even progressive bettors win some sessions.

odiousgambit
odiousgambit Oct 14, 2018

If you're using Wincraps, make a program so the betting is automatic and learn to use the Hyper-speed function. I'll eat my hat if that doesn't wise you up fast.

123BroRick123
Posted by 123BroRick123
Sep 18, 2018

IRON CROSS

IRON CROSS VARIAtIONS

SORRY ABOUT THE ORIGINAL NUMBERS, HERE ARE THE CORRECTIONS

$10 on 5, $12 on 6, $12 on 8, $7 on Field
Every number but 7 wins $7 Wins 30 times, Loses 6 times Leave up 1 roll, no more than 2 rolls
or
$10 on 5, $12 on 6, $12 on 8, $7 on Field, $100 on either Lay4 or Lay10
Every number but 4 wins $7 Wins 33 times, Loses 3 times Leave up 2 rolls, no more than 3 rolls
or
$10 on 5, $12 on 6, $12 on 8, $7 on Field, $146 on either Lay4 or Lay10, #21.00 on Hard4
Every number but an easy4 wins $7 Wins 34 times, Loses 3 times Leave up 3 rolls, no more than 4 rolls

Bro. Rick

If you are Betting 5 times this or more, then Buy the 5.

Comments

mustangsally
mustangsally Sep 18, 2018

$5 on 5, $6 on 6, $6 on 8, $7 on Field

Every number but 7 wins $7.00 Wins 30 times


you actually believe that Enron Accounting poppycock?



a 5 rolls and

you

call losing the $7 Field

and winning $7 on Place 5

a

WIN!

exactly what is the win? (comps I think knot!)



ha

poor try!



Sally

odiousgambit
odiousgambit Sep 19, 2018

welcome to the forum, rick, you can learn a lot here but as you can see it will be rough sailing - this is not a place for posting silly ideas and looking for cheerleaders



I have no idea if you really think the iron cross or its variations ever made money for anybody, and I don't know if you are any good at math [dude, you gotta weigh outcomes with probability or it's laughable], but you should know to put something like that up here where people know better than to think the IC is a good bet and also generally have some competence in math .... well, you'd better have a thick skin or you might as well skip on over to some other forum

GWAE
GWAE Sep 25, 2018

So you will have 30 wins of $7 for 210

And

6 loses of 41 for $246



Sounds like a good plan to me.