## Poll

5 votes (29.41%) | |||

12 votes (70.58%) |

**17 members have voted**

also please explain how i might be incorrect.

Quote:GalatraxHello I was wondering if my math is correct on this and if this strategy I made really gives you an edge... here is how you do it.... bet 10$ on field, and bet 5$ on Any seven/Big Red... I calculated it out and that should give you a 61.11% chance of winning and only a 38.89% chance of losing each roll which theoretically gives you a 22.22% advantage OVER casino every roll! correct?? is this a viable way to play if so I call this strategy BETTING FOR MONACO

You propose to combine two wagers, each with a negative expectation, and come out with a +22% advantage every single roll? Perhaps a better name for this strategy is "BETTING FOR SOMALIA"

Quote:GalatraxMy intention in replying is not to argue but just to point out that statistically i do not see how i am incorrect. the math adds up. please explain the only way to lose is if you roll a 5 6 or 8 any other number you win....statistically i dont see the flaw

There are 36 ways the dice can land. Map out how much you would win or lose for each of the 36 possible outcomes and add them up. Do you still get +22%? There are a lot of ways to make 5,6 and 8. You lose $15 for each of them.

Package requires a single-toss resolution.

Package has a return from the Field numbers 2, 3, 4, 9, 10, 11, 12 and the number 7

Field will pay double on 2 and triple on 12 and single on other winners plus return the Field bet

Big Red will pay quadruple plus return the Red bet

Package Expected Value is (1/36)(10 + 20) + (14/36)(10 + 10) + (1/36)(10 + 30) + (6/36)(5 + 20) = 13.888889

Percent disadvantage is (13.888889 - 15)(100) / (15) = - 7.407407 %

2 1/36 = 2.78%

3 2/36 = 5.56%

4 3/36 = 8.33%

5 4/36 = 11.11%

6 5/36 = 13.89%

7 6/36 = 16.67%

8 5/36 = 13.89%

9 4/36 = 11.11%

10 3/36 = 8.33%

11 2/36 = 5.56%

12 1/36 = 2.78%

nope it doesnt like i said 11.11 +13.89+13.89 =38.89% to lose =61.11%win

Quote:pwcrabbProposed bets are $10 Field and $5 Big Red.

Package requires a single-toss resolution.

Package has a return from the Field numbers 2, 3, 4, 9, 10, 11, 12 and the number 7

Field will pay double on 2 and triple on 12 and single on other winners plus return the Field bet

Big Red will pay quadruple plus return the Red bet

Package Expected Value is (1/36)(10 + 20) + (14/36)(10 + 10) + (1/36)(10 + 30) + (6/36)(5 + 20) = 13.888889

Percent disadvantage is (13.888889 - 15)(100) / (15) = - 7.407407 %

please explain this

Quote:Galatraxmy question is does this method give advantage over casino if not please explain why not...

Please do this: Map out how much you would win or lose for each of the 36 possible outcomes when you throw two dice, and add them up. It will be clear whether you have an advantage or not.

hint: You never win both of your bets, but you lose, and lose often, both bets.

583 lose

554 win

being that i have a 22% advantage at rolling a winning number this makes me a winner....

For rolling a 3, with how much money do you walk away? What is the probability of that outcome? Multiply these two numbers.

Do analogous analyses for each of the nine remaining possible dice outcomes.

Add the results of each of these eleven analyses.

Compare that sum with your initial investment of $15.

_______________________________________________________

Simply adding probabilities is mathematically absurd. Your prior results have no analytical meaning.

37.8

51.65

171.7

65.55

51.65

37.8

79.5

547.35 WINNING NUMBERS

583.35 5.6.8 LOSING NUMBERS

HOUSE EDGE OF LIKE 1. SOMETHING ADVANTAGE

Quote:Galatrax472.36W 583L I APOLIGIZE THANK YOU FOR HELPING ME SEE THE LIGHT IDK I GUESS I JUST GOT EXCITED BEFORE I TOTALLY TESTED IT OUT THIS WAY

No problems. Check out the article on dice probabilities on the "Wizard of Odds" site for a good tutorial.

For the 2: (1/36) ($10 + $20) = $0.833333

For the 3: (2/36) ($10 + $10) = $1.111111

For the 4: (3/36) ($10 + $10) = $1.166667

For the 5: (4/36) ( $0 ) = $0.000000

For the 7: (6/36) ($5 + $20) = $4.166667

et cetera

There are 36 ways the dice can land, with each of them being equally likely. Here they are presented with the result of your bet for each possible roll.

Roll # | Dice Outcome | Bet Outcome |
---|---|---|

1 | 1&1 = 2 | 20-5 = 15 |

2 | 1&2 = 3 | 10-5 = 5 |

3 | 2&1 = 3 | 10-5 = 5 |

4 | 1&3 = 4 | 10-5 = 5 |

5 | 2&2 = 4 | 10-5 = 5 |

6 | 3&1 = 4 | 10-5 = 5 |

7 | 1&4 = 5 | 0-15 = -15 |

8 | 2&3 = 5 | 0-15 = -15 |

9 | 3&2 = 5 | 0-15 = -15 |

10 | 4&1 = 5 | 0-15 = -15 |

11 | 1&5 = 6 | 0-15 = -15 |

12 | 2&4 = 6 | 0-15 = -15 |

13 | 3&3 = 6 | 0-15 = -15 |

14 | 4&2 = 6 | 0-15 = -15 |

15 | 5&1 = 6 | 0-15 = -15 |

16 | 1&6 = 7 | 20-10 = 10 |

17 | 2&5 = 7 | 20-10 = 10 |

18 | 3&4 = 7 | 20-10 = 10 |

19 | 4&3 = 7 | 20-10 = 10 |

20 | 5&2 = 7 | 20-10 = 10 |

21 | 6&1 = 7 | 20-10 = 10 |

22 | 2&6 = 8 | 0-15 = -15 |

23 | 3&5 = 8 | 0-15 = -15 |

24 | 4&4 = 8 | 0-15 = -15 |

25 | 5&3 = 8 | 0-15 = -15 |

26 | 6&2 = 8 | 0-15 = -15 |

27 | 3&6 = 9 | 10-5 = 5 |

28 | 4&5 = 9 | 10-5 = 5 |

29 | 5&4 = 9 | 10-5 = 5 |

30 | 6&3 = 9 | 10-5 = 5 |

31 | 4&6 = 10 | 10-5 = 5 |

32 | 5&5 = 10 | 10-5 = 5 |

33 | 6&4 = 10 | 10-5 = 5 |

34 | 5&6 = 11 | 10-5 = 5 |

35 | 6&5 = 11 | 10-5 = 5 |

36 | 6&6 = 12 | 30-5 = 25 |

Total = | -35 |

On average, for each 36 rolls of the dice, you will lose $35 with this strategy. Yes, you would win net money on 22 of these 36 rolls and lose money on just 14 of them, but you would lose more money than you would win.

Somebody please let me know if I have screwed this up somewhere.

Assume single zero (as I'm in the UK!).

Roulette - 1-24

Suppose you put $5 on 1-12 and $5 on 13-24 . Then it seems simple; you gain $5 if any of these numbers come up but lose $10 is 25-36 or 0. Your view might be that you win 24 times but only lose 13 times, so have an advantage over the casino. However you're forgetting that you lose more when you lose.

The maths looks at the long term - that is eventually all 37 numbers will [nearly] come up equally.

Win 24 * $5

Lose 13 * $10

Total bet $370

Total profit/loss -$10

House Edge = 10/370 = 2.70%

Now use the same logic for all the rolls of the dice and see what you come up with.

$10 on the field

$5 hop the sevens (saves money vs big red!)

$5 each hop the 5s, 6s, and 8s.

Win probability = 100%.

Advantage over the casino = 100%.

Quote:sodawaterI can modify this system to have 100 percent chance of winning:

$10 on the field

$5 hop the sevens (saves money vs big red!)

$5 each hop the 5s, 6s, and 8s.

Win probability = 100%.

Advantage over the casino = 100%.

I like it! That's 11 hops for $55. The 9 rolls you win $10! Such a deal!

You're almost there. Now, you just have to multiply the percentages in your table by the return for each combination and sum up the values:Quote:GalatraxOutcome Probability

2 1/36 = 2.78%

3 2/36 = 5.56%

4 3/36 = 8.33%

5 4/36 = 11.11%

6 5/36 = 13.89%

7 6/36 = 16.67%

8 5/36 = 13.89%

9 4/36 = 11.11%

10 3/36 = 8.33%

11 2/36 = 5.56%

12 1/36 = 2.78%

nope it doesnt like i said 11.11 +13.89+13.89 =38.89% to lose =61.11%win

Roll | Combinations | Hit % | Return | Value |
---|---|---|---|---|

2 | 1 | 2.78% | 15 | 0.4167 |

3 | 2 | 5.56% | 5 | 0.2778 |

4 | 3 | 8.33% | 5 | 0.4167 |

5 | 4 | 11.11% | -15 | -1.6667 |

6 | 5 | 13.89% | -15 | -2.0833 |

7 | 6 | 16.67% | 10 | 1.6667 |

8 | 5 | 13.89% | -15 | -2.0833 |

9 | 4 | 11.11% | 5 | 0.5556 |

10 | 3 | 8.33% | 5 | 0.4167 |

11 | 2 | 5.56% | 5 | 0.2778 |

12 | 1 | 2.78% | 25 | 0.4167 |

Totals | 36 | 100 | -- | -1.1111 |

So, for every $15 ($5 + $10) you wager, you get $13.89 back, for a house edge of 7.41%.