Poll

5 votes (29.41%)
12 votes (70.58%)

17 members have voted

Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 4:19:28 PM permalink
Hello 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
also please explain how i might be incorrect.
Ayecarumba
Ayecarumba
  • Threads: 236
  • Posts: 6763
Joined: Nov 17, 2009
Thanked by
Galatrax
February 9th, 2017 at 4:24:17 PM permalink
Quote: Galatrax

Hello 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"
Simplicity is the ultimate sophistication - Leonardo da Vinci
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 4:26:27 PM permalink
My 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
MB
MB
  • Threads: 8
  • Posts: 86
Joined: Aug 20, 2015
Thanked by
Galatraxteliot
February 9th, 2017 at 4:41:50 PM permalink
You lose $15 on 5, 6, or 8. You win $10 on 7. You win $15 on 2/12. You win $5 on everything else. Do the math.
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 4:44:46 PM permalink
when 12 pays 3 times you win 25 most casinos ive seen pay double 2 and triple 12
MB
MB
  • Threads: 8
  • Posts: 86
Joined: Aug 20, 2015
February 9th, 2017 at 4:53:11 PM permalink
It doesn't change the conclusion.
Ayecarumba
Ayecarumba
  • Threads: 236
  • Posts: 6763
Joined: Nov 17, 2009
Thanked by
SanchoPanza
February 9th, 2017 at 4:55:16 PM permalink
Quote: Galatrax

My 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.
Simplicity is the ultimate sophistication - Leonardo da Vinci
pwcrabb
pwcrabb
  • Threads: 4
  • Posts: 185
Joined: May 15, 2010
Thanked by
Galatrax
February 9th, 2017 at 4:56:30 PM permalink
Proposed 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 %
"I suppose I was mad. Every great genius is mad upon the subject in which he is greatest. The unsuccessful madman is disgraced and called a lunatic." Fitz-James O'Brien, The Diamond Lens (1858)
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 4:57:07 PM permalink
Outcome 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
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 5:01:52 PM permalink
please explain this lol im confused
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 5:03:03 PM permalink
Quote: pwcrabb

Proposed 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
pwcrabb
pwcrabb
  • Threads: 4
  • Posts: 185
Joined: May 15, 2010
Thanked by
Galatrax
February 9th, 2017 at 5:06:49 PM permalink
Galatrax fails to recognize that probability is necessary but not sufficient for analyzing his question.
"I suppose I was mad. Every great genius is mad upon the subject in which he is greatest. The unsuccessful madman is disgraced and called a lunatic." Fitz-James O'Brien, The Diamond Lens (1858)
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 5:09:12 PM permalink
my question is does this method give advantage over casino if not please explain why not...
Ayecarumba
Ayecarumba
  • Threads: 236
  • Posts: 6763
Joined: Nov 17, 2009
Thanked by
Galatrax
February 9th, 2017 at 5:13:42 PM permalink
Quote: Galatrax

my 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.
Simplicity is the ultimate sophistication - Leonardo da Vinci
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 5:27:56 PM permalink
yes i lose 38.89% 583$ of time statistically...but i win 61.11 percent of time 250 for 7 69$ 12,5 41.7$ 2,195 for all other numbers
583 lose
554 win
being that i have a 22% advantage at rolling a winning number this makes me a winner....
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 5:41:04 PM permalink
472.36W 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
pwcrabb
pwcrabb
  • Threads: 4
  • Posts: 185
Joined: May 15, 2010
Thanked by
Galatrax
February 9th, 2017 at 5:46:37 PM permalink
For rolling a 2, with how much money do you walk away? What is the probability of that outcome? Multiply these two numbers.

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.
"I suppose I was mad. Every great genius is mad upon the subject in which he is greatest. The unsuccessful madman is disgraced and called a lunatic." Fitz-James O'Brien, The Diamond Lens (1858)
Galatrax
Galatrax
  • Threads: 2
  • Posts: 14
Joined: Feb 9, 2017
February 9th, 2017 at 5:59:35 PM permalink
51.7
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
Ayecarumba
Ayecarumba
  • Threads: 236
  • Posts: 6763
Joined: Nov 17, 2009
February 9th, 2017 at 6:04:49 PM permalink
Quote: Galatrax

472.36W 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.
Simplicity is the ultimate sophistication - Leonardo da Vinci
pwcrabb
pwcrabb
  • Threads: 4
  • Posts: 185
Joined: May 15, 2010
February 9th, 2017 at 6:11:16 PM permalink
No, Galatrax

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
"I suppose I was mad. Every great genius is mad upon the subject in which he is greatest. The unsuccessful madman is disgraced and called a lunatic." Fitz-James O'Brien, The Diamond Lens (1858)
Doc
Doc
  • Threads: 46
  • Posts: 7287
Joined: Feb 27, 2010
February 9th, 2017 at 6:35:33 PM permalink
O.K., I'll type it all out for you. Strategy: $10 field bet (pays $20 on 2, $30 on 12, and $10 on 3, 4, 9, 10, or 11) and $5 Big Red (pays $20 on 7). Non-sevens lose the $5 Big Red bet, and 5, 6, 7, and 8 lose the field bet.

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.
Last edited by: Doc on Feb 9, 2017
DeMango
DeMango
  • Threads: 36
  • Posts: 2958
Joined: Feb 2, 2010
February 9th, 2017 at 9:45:42 PM permalink
What scares me is two people voted for the proposition! Who is that masked bandit?
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
FleaStiff
FleaStiff
  • Threads: 265
  • Posts: 14484
Joined: Oct 19, 2009
February 9th, 2017 at 9:55:54 PM permalink
22 percent advantage.... ain't heard of any casinos going bankrupt and ripping out their craps tables.
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 2967
Joined: Jun 17, 2011
February 10th, 2017 at 2:59:38 AM permalink
I think one problem might be confusing probability of winning versus making money in the long term. Perhaps a simple case might be easier.

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.
sodawater
sodawater
  • Threads: 64
  • Posts: 3321
Joined: May 14, 2012
Thanked by
RisingDough
February 10th, 2017 at 3:07:06 AM permalink
I 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%.
DeMango
DeMango
  • Threads: 36
  • Posts: 2958
Joined: Feb 2, 2010
February 10th, 2017 at 7:22:51 AM permalink
Quote: sodawater

I 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!
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
Joeman
Joeman
  • Threads: 36
  • Posts: 2422
Joined: Feb 21, 2014
February 10th, 2017 at 8:28:25 AM permalink
Quote: Galatrax

Outcome 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

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:


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%.
"Dealer has 'rock'... Pay 'paper!'"
Joeman
Joeman
  • Threads: 36
  • Posts: 2422
Joined: Feb 21, 2014
February 10th, 2017 at 8:32:36 AM permalink
Alternatively, you could just do a weighted average of the edge for each bet:

((10 * 2.78%) + (5 * 16.67%)) / 15 = 7.41%
"Dealer has 'rock'... Pay 'paper!'"
  • Jump to: