"DC7" Question
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The flaw is mathematically solved for on Page 3.
I looked 5 pages back and didn't see anything about this, so I hope I'm not repeating a question but here goes... I'm mostly an advantage player in a casino, but one of my fun games I love to drink and relax at is craps. Even so, the AP in me can't help but try to play "better odds" bets. That being said I don't always play the 'best' odds bets, but I am primarily a Don't Pass player (and shooter - my lord I'm really good at crapping out) and Don't Come (with odds) player.
Anyways I came across this youtube video for a don't side strategy he called DC7:
https://www.youtube.com/watch?v=oc_fkoi6TGU
Basically, the goal is to get an even money bet on any number against the 7.
Here's an example:
1) $5 pass, $5 don't pass (i.e. Switzerland) - Only lose $5 on come out 12.
2) Point Established
Bet $5 DC, and put 2x Odds on the original Don't Pass bet. One of 4 things can happen:
- If shooter 7's out, you still win $5. Go back to 1)
- Shooter points you lose $5 + 2x odds, but win $5 pass bet and have your DC move to the number for even bet against a 7. Go back to 1)
- Shooter throws 2, 3, or 11. Odds wise, the 3 cancels out the 11, and the 2 is a 1/36 advantage, but on the come out you're 1/36 disadvantage to 12, so cancels? Stay at 2)
- Shooter throws any other point number. Go to 3)
3) Any other point thrown
Pick up the DC odds and you essentially have an even money bet with the new number vs 7.
My Question
Your original pass/don't pass will be a wash at this point, and you're left with an even money bet on a point number vs 7. For example, on a 10 you're 2-1 likely to win the bet and you're getting paid 1-1, so it's a good bet. The scare of getting to this point (7 when you're on the DC) is removed because that wins your original Don't Pass and odds (only laying odds when you're on the DC).
While I haven't put crazy effort in to solving this, I have given it a couple passes and feel I'm missing something because this seems like a great way to get even money bets against a 7. Lose 12 come out, but win 2 any DC bet. Is the probability of the shooter going back to back on a number to point that overcoming vs getting favorable even money bets against a 7? It seems as though every step of the way you're the favorite (minus the 12 come out 2 DC canceling each other). A 7 wins for you every step of the way and doesn't hurt you on the come out or DC bet. Again, you're then left with a positive edge bet looking for the most likely number, 7.
All possible outcomes… Assuming Laying Odds so first throw 7 out wins 1 unit:
Outcome --------------------- Effect (Numeric) ------------------------- Probability ---------------------- EV
Come out 12 ------------------- -1 ---------------------------------------- .027778 ------------------- -.027778
First Throw 7 Out ------------- +1 ---------------------------------------- .166667 ------------------- .166667
First Throw 11 ----------------- -1 ---------------------------------------- .055556 ------------------- -.055556
First Throw 3 ------------------ +1 ---------------------------------------- .055556 ------------------- .055556
First Throw 2 ------------------ +1 ---------------------------------------- .027778 ------------------- .027778
TOTAL ---------------------------------------------------------------------------------------------------- .166667
First Throw Point 4/10 -------- -4 ---------------------------------------- .083333 ------------------- -.333333
First Throw Point 5/9 --------- -3 ---------------------------------------- .111111 ------------------- -.333333
First Throw Point 6/8 -------- -2.4 --------------------------------------- .138889 ------------------- -.333333
TOTAL First Point 4/10 ---------------------------------------------------------------------------------- -.166667
TOTAL First Point 5/9 ----------------------------------------------------------------------------------- -.166667
TOTAL First Point 6/8 ----------------------------------------------------------------------------------- -.166667
*** "First Throw Point" means the first throw AFTER the point has been established. I.E. technically the second throw of the shooters turn.
*Note, for the “first throw point 4/10” the result is -.166667, but getting a 4/10 vs 7 which has a positive EV (2 to 1) of .333333
*Note, for the “first throw point 5/9” the result is -.166667, but getting a 5/9 vs 7 which has a positive EV (3 to 2) of .222222
*Note, for the “first throw point 6/8” the result is -.166667, but getting a 6/8 vs 7 which has a positive EV (6 to 5) of .090909
Calculating this in with our previous EV (-.166667):
When 4/10 is the point = .333333 – .166667 = .166667
When 5/9 is the point = .222222 – .166667 = .055556
When 6/8 is the point = .090909 – 1.66667 = -.075756
TOTAL EV FOR ENTIRE "DC7 SYSTEM" = +.146467
Here's what I get:
I will assume a bet of 6 so you can get full odds
Original pass: +6
Original don't pass: -6
Don't come: -6
2x odds on don't: +6 if it's 4/10, +8 if it's 5/9, or +10 if it's 6/8
Your total will be 0, +2, or +4 on a 6-unit bet, which is 0, +1/3, or +2/3 on 1 unit.
I think the main problem is, you are assuming your odds bets are paying off at even money (how else do you come to the conclusion that if you 7 out, you win $5 regardless of the point on which you placed the odds bet) - but since you're laying, they're either 1-2, 2-3, or 5-6, depending on the point.
This assumes the base bet is $6 (that way, you get true odds regardless of the point)
On the initial roll:
1/36 of the time, it is 12, and you lose $6
11/36 of the time, it is 2, 3, 7, or 11, and you break even
1/12 of the time, the initial roll is 4; you make a $6 DC bet and $12 odds against the 4
In this situation:
1/6 of the time, the next roll is 7; you lose your $6 DC bet, but gain $6 on your odds ($12 @ 1-2), so you break even
1/12 of the time, the next roll is 4; you lose your $12 odds, and your DC bet has a point of 4
In that case, the ER of the DC bet is 1/3 x (-6) + 2/3 x (+6) = +2
1/12 of the time, the next roll is 10; you take down your odds, and your DC bet has a point of 4
In that case, the ER of the DC bet is 1/3 x (-6) + 2/3 x (+6) = +2
2/9 of the time, the next roll is 5 or 9; you take down your odds, and your DC bet has a point of 5 or 9
In that case, the ER of the DC bet is 2/5 x (-6) + 3/5 x (+6) = +6/5
5/18 of the time, the next roll is 6 or 8; you take down your odds, and your DC bet has a point of 6 or 8
In that case, the ER of the DC bet is 5/11 x (-6) + 6/11 x (+6) = +6/11
The total EV for the initial roll of 4 up to this point is:
1/6 x 0 + 1/12 x (-10) + 1/12 x 2 + 2/9 x 6/5 + 5/18 x 6/11 = -41/165
Call this number P.
5/6 of the time, your second roll is something other than 2, 3, 11, or 12; the ER is P.
1/6 of the time, your second roll is 2, 3, 11, or 12; the ER is 1/2 x 6 + 1/3 x (-6) + 1/2 x = 0 = +1,
and your odds are still in play; you make another DC bet, so you are back to where you were before the
second roll, but with a gain of 1
The total EV is 5/6 x P + 1/6 x 5/6 x (P + 1) + 1/6 x 1/6 x 5/6 x (P + 2) + ...
= 5/6 x P x (1 + 1/6 + (1/6)2 + ...) + 5/6 x 1/6 x (1 + 1/6 x 2 + (1/6)2 x 3 + ...)
= 5/6 x P x (1 / (1 - 1/6)) + 5/6 x 1/6 x (1 + 1/6 + (1/6)2 + ...)2
= P + 5/36 x (1 / (1 - 1/6))2
= P + 5/36 x 36/25
= P + 1/5 = 33/165 - 41/165 = -8/165
1/12 of the time, the initial roll is 10; the EV is also -8/165
1/9 of the time, the initial roll is 5; you make a $6 DC bet and $12 odds against the 5
In this situation:
1/6 of the time, the next roll is 7; you lose your $6 DC bet, but gain $8 on your odds ($12 @ 2-3), so the ER is 2
1/9 of the time, the next roll is 5; you lose your $12 odds, and your DC bet has a point of 5
In that case, the ER of the DC bet is 2/5 x (-6) + 3/5 x (+6) = +6/5
1/9 of the time, the next roll is 9; you take down your odds, and your DC bet has a point of 4
In that case, the ER of the DC bet is 2/5 x (-6) + 3/5 x (+6) = +6/5
1/6 of the time, the next roll is 4 or 10; you take down your odds, and your DC bet has a point of 4 or 10
In that case, the ER of the DC bet is 1/3 x (-6) + 2/3 x (+6) = +2
5/18 of the time, the next roll is 6 or 8; you take down your odds, and your DC bet has a point of 6 or 8
In that case, the ER of the DC bet is 5/11 x (-6) + 6/11 x (+6) = +6/11
The total EV for the initial roll of 5 up to this point is:
1/6 x 2 + 1/9 x (-54/5) + 1/9 x 6/5 + 1/6 x 2 + 5/18 x 6/11 = -41/165
Call this number P.
5/6 of the time, your second roll is something other than 2, 3, 11, or 12; the ER is P.
1/6 of the time, your second roll is 2, 3, 11, or 12; the ER is 1/2 x 6 + 1/3 x (-6) + 1/2 x = 0 = +1,
and your odds are still in play; you make another DC bet, so you are back to where you were before the
second roll, but with a gain of 1
Again, the total EV is P + 1/5 = -8/165
1/9 of the time, the initial roll is 9; the EV is also -8/165
5/36 of the time, the initial roll is 6; you make a $6 DC bet and $12 odds against the 6
In this situation:
1/6 of the time, the next roll is 7; you lose your $6 DC bet, but gain $10 on your odds ($12 @ 5-6), so the ER is 4
5/36 of the time, the next roll is 6; you lose your $12 odds, and your DC bet has a point of 6
In that case, the ER of the DC bet is 5/11 x (-6) + 6/11 x (+6) = +6/11
5/36 of the time, the next roll is 8; you take down your odds, and your DC bet has a point of 8
In that case, the ER of the DC bet is 5/11 x (-6) + 6/11 x (+6) = +6/11
1/6 of the time, the next roll is 4 or 10; you take down your odds, and your DC bet has a point of 4 or 10
In that case, the ER of the DC bet is 1/3 x (-6) + 2/3 x (+6) = +2
2/9 of the time, the next roll is 5 or 9; you take down your odds, and your DC bet has a point of 5 or 9
In that case, the ER of the DC bet is 2/5 x (-6) + 3/5 x (+6) = +6/5
The total EV for the initial roll of 6 up to this point is:
1/6 x 4 + 5/36 x (-126/11) + 5/36 x 6/11 + 1/6 x 2 + 2/9 x 6/5 = -41/165
Call this number P.
You know the drill...the total EV is P + 1/5 = -8/165
5/36 of the time, the initial roll is 8; the EV is also -8/165
The final result:
5/36 of the time, the initial roll is 2, 3, 7, or 11; the EV is 5/36 x 0 = 0
1/36 of the time, the initial roll is 12; the EV is 1/36 x (-6) = -1/6
5/6 of the time, the EV is 5/6 x (-8/165) = -4/99
The total EV is -1/6 - 4/99 = -41/198.
Using a base bet of 1, it would be 1/6 of that, or -41/1188, which is 3.45%
That 12 (or 2 if you're in Reno/Tahoe) on the initial roll is a killer
Quote: ThatDonGuyWhere are you getting +1 for a seven on the second throw?
On the second throw the 6 pass and 6 don't pass are a wash. I'm laying odds on the don't pass bet that winnings equal 12 (2x base unit), thus, lose 6 on the DC, lose 6 on the Pass, but win 6 + 12 on the Don't Pass + Odds... Which makes you 1 unit, or in this case, 6 ahead.
I'm sorry I caused confusion when I said 2x Odds, as I could see why that would lead to confusion. I'm laying whatever odds I need to win 2 units if a 7 is rolled immediately. For the 4, 5, 9, and 10 it would be odds to win 2 units. So if 6 is the base unit, and a 9 was rolled, then I would lay 18, to win 12 (2x base bet). As shown in my calculations above, this results in losing 3 units for the 5 or 9 on a back to back point number, 4 units lost on 4 or 10, and 2.4 units lost on the 6 or 8.
I apologize for the mixed wording, but it shouldn't change my math at all.
Quote: RomesI'm sorry I caused confusion when I said 2x Odds, as I could see why that would lead to confusion. I'm laying whatever odds I need to win 2 units if a 7 is rolled immediately. For the 4, 5, 9, and 10 it would be odds to win 2 units. So if 6 is the base unit, and a 9 was rolled, then I would lay 18, to win 12 (2x base bet). As shown in my calculations above, this results in losing 3 units for the 5 or 9 on a back to back point number, 4 units lost on 4 or 10, and 2.4 units lost on the 6 or 8.
I apologize for the mixed wording, but it shouldn't change my math at all.
I am looking at the statement:
"Note, for the “first throw point 4/10” the result is -.166667, but getting a 4/10 vs 7 which has a positive EV (2 to 1) of .333333"
If the first throw is a 4, then you make a Don't Come bet of 1 and put 4 odds on your Don't Pass against the 4.
There is a 1/12 chance that the next roll is 4; you lose the 4 you bet on odds, and the ER of the Don't Come is now 1/3 x (-1) + 2/3 x (+1) = 1/3, for a total ER of (-4 + 1/3) = -11/3, and an EV of 1/12 x (-11/3) = -11/36 = -0.305556.
There is a 1/6 chance that the next roll is 7; you lose the 1 you bet on Don't Come, but gain 2 on odds (4 @ 1-2), so the EV here is 1/6 x 1 = 0.166667.
When these are combined, that's still -5/36 = -0.138889.
Yes, I realize I haven't taken the other rolls into account, but I have no idea how you get "When 4/10 is the point = .333333 – .166667 = .166667".
Quote: bahdbwoyenjoy
http://wtwii.wordpress.com/2006/08/24/debunking-a-craps-system/
1) They're laying whatever they need to win a single unit, not 2x units, which instead of breaking even makes the 7 on the 2nd roll a winner .166667 of the time.
2) The first 'minor flaw' is that you lose the 12 on a come out. But it fails to address the EV of winning the 2 on the 2nd roll from the DC. These have the same odds of winning and losing. If you're going to point out the minor loss, you should probably point out the minor win.
3) Their 2nd 'minor flaw' doesn't seem to make any sense (unless I'm really missing something)
Quote:
Another Minor Flaw: 11 on the Come-out
If an 11 is rolled on the Come-Out roll, the Don’t Come bet is lost but the original bet and odds bets are not affected. This amounts to another small ‘leak’ in the system which we glossed over in the strategy. It’s small, happens only once in 18 sequences and only loses the Don’t Come portion. Whether or not this case is covered in descriptions of the strategy varies. (I left it out of my original description by accident.) Like the previous flaw, it doesn't break the system, but it does reduce the expected outcome, which remains positive until the next flaw is accounted for.
If an 11 is rolled on the '2nd' roll (for simplicity) then the DC is lost, and nothing else is effected. However if a 3 is rolled the DC is won, and nothing else is effected. These two should effectively 'cancel' each other out, once again leaving an advantage from the 2 being rolled since 12 is a push and 7 is a 1 unit winner. Assuming I'm missing something here (please explain if I am) the author still abides this is a +EV system 'for now.'
4) I accounted for the 'major' flaw to every aspect/number in my breakdown on my second post. Basically, the author is saying you lose so much when the point is rolled back to back that it doesn't make up for the 7 out Don't Pass win. He mentions that people fail to realize the point will be rolled a little more than 1/9th of the time. Again, a difference here is I'm laying odds to win 2 units on a 7 out, but this is not true...
You have, on average, a little more than a 1/9 (.1111) chance of having the point number rolled = .0128. This, again on average, will result in the loss of -3.0667 units (4+3+2.4 / 3). You win 1 unit when the 7 is rolled however, so that's 6/36 = .16667. Again this will result in the win of 1 unit.
So the probabilities (weighted with units) are:
(.0128 * -3.1333) + (.16667 * 1) = -.0401 + .16667 = +.1266, a positive EV.
The only time it changes is when your Don't Pass odds result overall in a push. Then this has a negative expectation (-.0401 + 0 = -.0401).
Conclusion
So again, I'm not sure what I'm missing (I do believe every system is flawed and not a positive expectation) but I am not satisfied with the link provided as reasonable doubt.
Quote: ThatDonGuyI am looking at the statement:
"Note, for the “first throw point 4/10” the result is -.166667, but getting a 4/10 vs 7 which has a positive EV (2 to 1) of .333333"
If the first throw is a 4, then you make a Don't Come
Again, verbiage is mine enemy today... What this is signifying is AFTER the point is made, I'm calculating in their "first throw" (really the second overall since the first made the point) if they were to hit the point back to back.
This is another thing the article listed doesn't touch on. When you "lose" on a back to back point, you still end up with a +EV bet! Say the point is 4, you lay your odds and place your DC, and they role 4 again. Doh, you lose the Don't Pass + odds, but the DC travels to the 4. So you're a 4 vs 7 at even money, this is a +EV bet and should be accounted for.
you did not take into account every possible state the game can be in.Quote: RomesSo again, I'm not sure what I'm missing (I do believe every system is flawed and not a positive expectation) but I am not satisfied with the link provided as reasonable doubt.
A very common error produced by many
(I even catch myself doing it. I bet Don never does. The strong and silent type never make mistakes in my opinion)
This is a state problem that can be solved quickly and easily using Matrix Algebra
The video, and that guy talking is really rude with his vulgar and offensive language, says to lose only 2 DC bets
per shooter
I simulated this both ways before doing any math
green/red the shooter
A fast sim shows a player edge = -0.98%
done
Conclusion
Romes, stop calling others rude and offensive names (your other thread Romes)
typical male gambler, yes my opinion
Sally
Quote: ThatDonGuyThere is a 1/12 chance that the next roll is 4; you lose the 4 you bet on odds, and the ER of the Don't Come is now 1/3 x (-1) + 2/3 x (+1) = 1/3, for a total ER of (-4 + 1/3) = -11/3, and an EV of 1/12 x (-11/3) = -11/36 = -0.305556.
There is a 1/6 chance that the next roll is 7; you lose the 1 you bet on Don't Come, but gain 2 on odds (4 @ 1-2), so the EV here is 1/6 x 1 = 0.166667.
When these are combined, that's still -5/36 = -0.138889.
Yes, I realize I haven't taken the other rolls into account, but I have no idea how you get "When 4/10 is the point = .333333 – .166667 = .166667".
I'm not sure where you're getting 1/3 x (-1) + 2/3 x (1) = 1/3 and why it's being combined with -4 + 1/3...
If a 7 is thrown on the 2nd throw, this results in a win of 1 unit. Thus 1/6 (.1667 * 1) = .1667.
If the 4 point is made, the result is a loss of 4 units, thus 1/12 (.0833 * -4) = -.3333
Forgetting the other numbers, this means the Don't Pass bet vs 7 on the 'second' roll has an ER = -.1667. THIS is where the article fails to address what happens next. The DC moves to the 4 (since it was just pointed). This is a bet you're a 2-1 favorite in (i.e. 66% vs 33%). Thus, on average you'll win 1/3 (.3333) of the time and you'll win 1 unit.
So now taking the ENTIRE system in to account (including this traveling DC on a point), you have -.1667 + .3333 = +1.667, as shown in my math above. The numbers change slightly for the other point numbers, but are also shown to have a positive expectation, except for 6 and 8. Even so the total sum of these Expected Values is positive, which is my Final Total EV for the "DC7" system.
Choose the "no risk dont come" option in the betting system menu.
Just like the article says, the major flaw is when you roll the point immediately after establishing it while your DP odds are active.
This betting style relies heavily on hedges; breaking one of the WoO's gambling commandments.
Quote: Romes1) They're laying whatever they need to win a single unit, not 2x units, which instead of breaking even makes the 7 on the 2nd roll a winner .166667 of the time.
2) The first 'minor flaw' is that you lose the 12 on a come out. But it fails to address the EV of winning the 2 on the 2nd roll from the DC. These have the same odds of winning and losing. If you're going to point out the minor loss, you should probably point out the minor win.
3) Their 2nd 'minor flaw' doesn't seem to make any sense (unless I'm really missing something)
If an 11 is rolled on the '2nd' roll (for simplicity) then the DC is lost, and nothing else is effected. However if a 3 is rolled the DC is won, and nothing else is effected. These two should effectively 'cancel' each other out, once again leaving an advantage from the 2 being rolled since 12 is a push and 7 is a 1 unit winner. Assuming I'm missing something here (please explain if I am) the author still abides this is a +EV system 'for now.'
4) I accounted for the 'major' flaw to every aspect/number in my breakdown on my second post. Basically, the author is saying you lose so much when the point is rolled back to back that it doesn't make up for the 7 out Don't Pass win. He mentions that people fail to realize the point will be rolled a little more than 1/9th of the time. Again, a difference here is I'm laying odds to win 2 units on a 7 out, but this is not true...
You have, on average, a little more than a 1/9 (.1111) chance of having the point number rolled = .0128. This, again on average, will result in the loss of -3.0667 units (4+3+2.4 / 3). You win 1 unit when the 7 is rolled however, so that's 6/36 = .16667. Again this will result in the win of 1 unit.
So the probabilities (weighted with units) are:
(.0128 * -3.1333) + (.16667 * 1) = -.0401 + .16667 = +.1266, a positive EV.
The only time it changes is when your Don't Pass odds result overall in a push. Then this has a negative expectation (-.0401 + 0 = -.0401).
Conclusion
So again, I'm not sure what I'm missing (I do believe every system is flawed and not a positive expectation) but I am not satisfied with the link provided as reasonable doubt.
others I'm sure can and maybe will explain this better but in layman terms..
flaw 1 & 2 he is just showing the DC is not getting up there for free. but even if they were the only flaws you would still have a huge advantage.
flaw 3 is the issue.. you odds bet in the long run is going to win $$ just as much $$ as it loses
- 12 odds on 6
- 6x10 win = 60
- 5x12 lose = 60
when you repeat point immediately you lose your DP odds so that would be 1 count in the 5x12 loss above..
when you 7out immediately though you lose your DC and win your DP odds.. you are "paying for" so you get the "free DC" with your DP odds winning so instead of winning 10 on one of the 6x10 counts above you win 5 ($5 table ex) which means you dont break even anymore and always lose
Quote: RomesIf a 7 is thrown on the 2nd throw, this results in a win of 1 unit. Thus 1/6 (.1667 * 1) = .1667.
If the 4 point is made, the result is a loss of 4 units, thus 1/12 (.0833 * -4) = -.3333
Forgetting the other numbers, this means the Don't Pass bet vs 7 on the 'second' roll has an ER = -.1667. THIS is where the article fails to address what happens next. The DC moves to the 4 (since it was just pointed). This is a bet you're a 2-1 favorite in (i.e. 66% vs 33%). Thus, on average you'll win 1/3 (.3333) of the time and you'll win 1 unit.
So now taking the ENTIRE system in to account (including this traveling DC on a point), you have -.1667 + .3333 = +1.667, as shown in my math above.
There's your error.
There's a 1/12 chance that your second roll is a 4, so the EV of the Don't Come bet is .3333 x 1/12 = .027778, not .3333.
If you're multiplying the 4 that you lose with the odds bet by 1/12 (which is correct), you have to multiply the 1/3 you are now expected to win with the Don't Come bet by the same 1/12. (You'll do it again with the probabilities for 5, 6, 8, 9, and 10 on the second roll - but you have to take into account the odds losses.
Quote: bahdbwoyothers I'm sure can and maybe will explain this better but in layman terms..
flaw 1 & 2 he is just showing the DC is not getting up there for free. but even if they were the only flaws you would still have a huge advantage.
flaw 3 is the issue.. you odds bet in the long run is going to win $$ just as much $$ as it loses
- 12 odds on 6
- 6x10 win = 60
- 5x12 lose = 60
when you repeat point immediately you lose your DP odds so that would be 1 count in the 5x12 loss above..
when you 7out immediately though you lose your DC and win your DP odds.. you are "paying for" so you get the "free DC" with your DP odds winning so instead of winning 10 on one of the 6x10 counts above you win 5 ($5 table ex) which means you dont break even anymore and always lose
When the shooter 7 out immediately after the point you win 1 'unit', and when you point immediately, I ran the probabilities (weighted with 4, 3, and 2.4 unit bets). The main flaw in the article previously mentioned is that it's assuming you're laying odds to "break even" if the shooter 7's out immediately after the point is established. I'm not sure I understand what you're 6x10 and 5x12 means?
Quote: ThatDonGuyThere's your error.
There's a 1/12 chance that your second roll is a 4, so the EV of the Don't Come bet is .3333 x 1/12 = .027778, not .3333.
If you're multiplying the 4 that you lose with the odds bet by 1/12 (which is correct), you have to multiply the 1/3 you are now expected to win with the Don't Come bet by the same 1/12. (You'll do it again with the probabilities for 5, 6, 8, 9, and 10 on the second roll - but you have to take into account the odds losses.
Hmmm, this seems like it could be heading the right direction, I'm just struggling to connect the dots logically. To me, I'm breaking this down step by step as to not miss anything, but I'm not sure why you'd multiply your DC by 1/12. While it's sitting on the DC, it's not correlated to ANY number. Only AFTER the 4 is re-thrown can we take the DC's EV in to account for a POINT number. This loses the DP odds (all calculated on page 1), and now the DC bet is assigned odds, for a point number, because it's assigned a point number. While it's sitting on the DC it has a positive expectation due to the 2 and 7 being a winner (again on page 1). Any other point number would carry it's own separate EV which can only be established after it's rolled. So when the 4 is rolled, the DC is calculated to be a 1/3 favorite for 1 unit, so it's EV is +.3333. Again, I'm not sure why you'd want to multiply this by 1/12?
To me, when it's on the DC, at that 'state' you don't care if a point number is rolled for a direct outcome. The only direct outcomes are from 2, 3, 7, 11, and the actual point (which is being taken in to account with the 4 units multiplied by the 1/12 odds). The only time we lose those odds (on a point 4) is when a 4 is thrown. So we're happy unless the point is thrown, at which time the state of the DC becomes it's own +.3333 EV bet.
I apologize if I'm just 'not getting it' but perhaps I've confused myself a bit by hashing out these numbers over and over.
Quote: RomesHmmm, this seems like it could be heading the right direction, I'm just struggling to connect the dots logically. To me, I'm breaking this down step by step as to not miss anything, but I'm not sure why you'd multiply your DC by 1/12. While it's sitting on the DC, it's not correlated to ANY number.
Only AFTER the 4 is re-thrown can we take the DC's EV in to account for a POINT number. This loses the DP odds (all calculated on page 1), and now the DC bet is assigned odds, for a point number, because it's assigned a point number. While it's sitting on the DC it has a positive expectation due to the 2 and 7 being a winner (again on page 1). Any other point number would carry it's own separate EV which can only be established after it's rolled. So when the 4 is rolled, the DC is calculated to be a 1/3 favorite for 1 unit, so it's EV is +.3333. Again, I'm not sure why you'd want to multiply this by 1/12?
Because that bet doesn't always have a .3333 EV - it only gets that way if the point is 4 (or 10), which only happens 1/12 (well, 1/12 for the 4, and 1/12 for the 10) of the time. That's why I multiplied by 1/12, just as we both multiplied the -4 for the odds bet loss by 1/12.
You have to take all of the probabilities into account - and then make sure they add up to 1.
Quote: ThatDonGuyBecause that bet doesn't always have a .3333 EV - it only gets that way if the point is 4 (or 10), which only happens 1/12 (well, 1/12 for the 4, and 1/12 for the 10) of the time. That's why I multiplied by 1/12, just as we both multiplied the -4 for the odds bet loss by 1/12.
You have to take all of the probabilities into account - and then make sure they add up to 1.
I thought that's what I did when calculating the numbers individually...?
Quote: Romes
*Note, for the “first throw point 4/10” the result is -.166667, but getting a 4/10 vs 7 which has a positive EV (2 to 1) of .333333
*Note, for the “first throw point 5/9” the result is -.166667, but getting a 5/9 vs 7 which has a positive EV (3 to 2) of .222222
*Note, for the “first throw point 6/8” the result is -.166667, but getting a 6/8 vs 7 which has a positive EV (6 to 5) of .090909
Just play one side.
You sure you're an AP?
Quote: RS1/36 rolls in which you have a do and don't wager, a 12 is rolled, you lose the do and push on the don't, for a net loss.
Just play one side.
You sure you're an AP?
Is this a serious post? Did you even attempt to read my 2nd post breaking down all of the EV's? Did you see how on the DC you're "ahead" by the 2 on all "craps" outcomes?
Quote: RomesIs this a serious post? Did you even attempt to read my 2nd post breaking down all of the EV's? Did you see how on the DC you're "ahead" by the 2 on all "craps" outcomes?
2, 3, 7, 11 are a net push on the don't come and come bets.
Quote: RomesI thought that's what I did when calculating the numbers individually...?
You're multiplying the odds bet loss by 1/12 (for 4 & 10), 1/9 (for 5 & 9), or 5/36 (for 6 & 8), but you're recalculating the EV for the Don't Come bet without multiplying by those numbers as well. Probability doesn't work that way.
The EV for a Don't Come bet the minute it is placed is -3/220. While the EV once a point of 4 is established is 1/3, you still have to multiply by the 1/12 probability that the point became 4 in the first place. Otherwise, the EV when you roll a 7 when you place the bet is -1, not -1/6.
Quote: RomesI looked 5 pages back and didn't see anything about this, so I hope I'm not repeating a question but here goes... I'm mostly an advantage player in a casino, but one of my fun games I love to drink and relax at is craps.
If you drink and gamble, you're a fool.
Quote: RS2, 3, 7, 11 are a net push on the don't come and come bets.
You're either not very intelligent or trolling. Either way please stop posting in my thread. You are not contributing what-so-ever to the conversation.
Quote: Sonny44If you drink and gamble, you're a fool.
It's something I get for entertainment value and I play the same drunk or sober. I understand that I'm playing a losing game when playing craps, and regardless of drinking I've always still remembered this. Having an AP mind I'll refuse to put out any more money than is for 'entertainment' when I'm at a disadvantage. I don't think it's a 'foolish' thing to do to have fun from time to time (the only time I drink and play is on vacations after counting for 5-6 hours to blow off steam). I personally enjoy it a lot, and if you don't that's fine, but please don't call me a fool for enjoying my vacation.
Quote: ThatDonGuyYou're multiplying the odds bet loss by 1/12 (for 4 & 10), 1/9 (for 5 & 9), or 5/36 (for 6 & 8), but you're recalculating the EV for the Don't Come bet without multiplying by those numbers as well. Probability doesn't work that way.
The EV for a Don't Come bet the minute it is placed is -3/220. While the EV once a point of 4 is established is 1/3, you still have to multiply by the 1/12 probability that the point became 4 in the first place. Otherwise, the EV when you roll a 7 when you place the bet is -1, not -1/6.
SOLVED MATHEMATICALLY
I kept confusing numbers, so I went get back to basics and figured out why you're right, and the system is flawed. Here's all possible rolls. Let's assume we're at the situation where the point is 4, odds on the Don't Pass are laid (4 units), and we have 1 unit on the DC. The next possible outcome is contained by the following rolls:
2 = (1/36) = .0278
3 = (2/36) = .0556
4 = (3/36) = .0833
5 = (4/36) = .1111
6 = (5/36) = .1389
7 = (6/36) = .1667
8 = (5/36) = .1389
9 = (4/36) = .1111
10 = (3/36) = .0833
11 = (2/36) = .0556
12 = (1/36) = .0278
SUM = (36/36) = 1
---------------------------------------------------------------------------------
Now, let's weight the results with our winnings and losses (second number is parenthesis)
2 = (1/36)(1) = .0278
3 = (2/36)(1) = .0556
4 = (3/36)(1)(1/3) + (3/36)(-4) = .027777778 + -.33333333 = -.3056
5 = (4/36)(1)(1/4.5) = .0247
6 = (5/36)(1)(1/11) = .0126
7 = (6/36)(1) = .1667
8 = (5/36)(1)(1/11) = .0126
9 = (4/36)(1)(1/4.5) = .0247
10 = (3/36)(1)(1/3) = 0278
11 = (2/36)(-1) = -.0556
12 = (1/36)(0) = .0000
SUM = -0.0087
*After checking the other numbers below... they ALL come out to be -0.0087
5 = (4/36)(1)(1/4.5) + (4/36)(-3) = .024691358 + -.3333333 = -.3086
6 = (5/36)(1)(1/11) + (5/36)(-2.4) = .012626263 + -.3333333 = -.3207
8 = (5/36)(1)(1/11) + (5/36)(-2.4) = .012626263 + -.3333333 = -.3207
9 = (4/36)(1)(1/4.5) + (4/36)(-3) = .024691358 + -.3333333 = -.3086
10 = (3/36)(1)(1/3) + (3/36)(-4) = 027777778 + -.33333333 = -.3056
Notice as mentioned by ThatDonGuy ::hero::, the 4 (for example) actually has 2 outcomes (losing the Odds, and getting a 1 to 1 no 4 bet). In each outcome you need to take in to account the probability of actually rolling a 4. This confused me earlier, but when put this way makes perfect sense. Basically, I was over valuing the bet after the DC traveled to the number before.
Thank you to everyone who contributed to the thread. I knew the system had to be flawed, but couldn't see it on my own. I appreciate the help!
Quote: odiousgambitI hope you really knew all along there was a flaw, just had to find it.
I said in my OP and in subsequent posts that I knew there was and that I was having trouble proving it mathematically. =)
lets say you get 6 as the point for a comeout 11 times in a session..
probability says you will win 6 and lose 5 correct on your dp odds?
6 x $10 (2units) = 60 win
5 x $12 (2units) = 60 loss
net= 0 / 0 units
if none are pso's then you are all even at the end correct? assuming for now no 12/2/3/11/7 on a comeout for or dc initial roll, correct?
but lets say 1 time you pso which means you lose a dc comeout so now you have the following on your odds:
1 x 5 (1unit) =$ 5 win from pso (dont odds win 10 and pays for dc loss of $5 so net $5)
5 x $10 (2units) = 50 win
5 x $12 (2units) = 60 loss
net = -5 / 1 unit
Thanks to all for the math lessons and helping Romes and others along who may have 'believed' in this system.
Quote: jkluv7This was one of the funnest posts I've read in a long time.
Thanks to all for the math lessons and helping Romes and others along who may have 'believed' in this system.
If you actually read the OP, or even 3 posts above your own, you'd note that no one ever claimed to 'believe' the system at any point in the way. I was simply running the math on it because I wanted too, and I was missing something small because I knew the 'system' did not have a positive EV.
Over on my thread, i had a similar idea to yours. Except I suggested playing the don't pass, and then a don't come on every roll.
The difference that i added, was that instead of taking 1x odds, or taking 2x odds. I was taking a certain amount depending on what point came up. I think it effected the overall EV, but idk what that exact effect was. I have been playing craps on a computer, so they will let you bet any amount of odds, and pay you with dollars AND cents.
You're supposed to put down
12 to win 10 on 6/8
15 to win 10 on 5/9
20 to win 10 on 4/10
I'm putting down
10 to win 8.33 on 6/8
10 to win 6.66 on 5/9
10 to win 5.00 on 4/10
Any way that you can calculate the EV on this "system"? Just curious as to how bad it is.
As a starter (only because others in this thread missed it) I didn't have this idea for the "DC7" system, nor did I ever think it "worked" out to some positive expectation. The reason for this thread was because I was doing the math on it (for fun) and knew I was missing something because I was in fact getting a positive expectation. So I was just trying to find where my error was basically.
Next, I don't dare claim to be a craps mathematician like some of the other posters (that I look up to as though they are) on here (ThatDonGuy, Sally, etc, etc, etc). Since you have a much different 'system' than this other system I came upon and was checking for fun, I would recommend you create a new thread, detail it with exactly how you play, and then ask any math questions in relation to it.
A big thing to include in that would be your base bet (or table min). If you're betting $10 Don't Pass, then $10 odds, that's technically not even 1x as you couldn't' expect to even win 1 'unit' from your odds (which most live casinos won't even let you do). So my semi-educated short cut guess would be that your system is 'barely' better than the standard -1.41% you get for the DP/DC. Again, this is nothing more than an educated guess, but I'd say your 'system' is a little over -1.00% EV... Something like -1.20%.
My bet on the dont pass was $5 and the bet on the dont come was $5.
Quote: uwf9990And yeah, I know that there is no system that has positive EV. I was just curious if this would be better/worse than the strategy that you suggested. Because if it's worse, then i might try to use your strategy when i go to the casino next time instead of the one that i came up with.
If you load up enough odds, the DP + DC is the best possible way to bet the game, from a math standpoint (you can get the HE down to like .08%). With your $5 DP/DC bets and always $10 odds, I'd say you're just under the 2x odds ratio, so probably something like -1.00%, or just under, for your EV. =)
Does this sound right?
1) comeout: doey-don't
2) point established: lay single or double odds, make DC (all the flat bets are in the same units)
3) if DC resolved on 1st roll, replace
if DC goes to a point: remove DP odds
4) if DC on point loses, replace DP odds, make another DC
if DC on point wins (i.e. 7 out), back to beginning
He also said he would quit making DC bets if the shooter knocked two off.
Did I miss s.t.?
Quote: goatcabinI watched this video, but like most of these demonstrations of a betting system, it's hard to put it into a specific set of rules. It would be most effective if he started out with the rules, then demonstrated with rolls.
Does this sound right?
1) comeout: doey-don't
2) point established: lay single or double odds, make DC (all the flat bets are in the same units)
3) if DC resolved on 1st roll, replace
if DC goes to a point: remove DP odds
4) if DC on point loses, replace DP odds, make another DC
if DC on point wins (i.e. 7 out), back to beginning
He also said he would quit making DC bets if the shooter knocked two off.
Did I miss s.t.?
That seems correct, I believe. This was a thread from a little while ago. The math was figured out and in on the previous pages if you're interested =). Overall the flaw I was looking for in the 'system' was over valuing the new DC number, and under estimating how often the shooter would go "set point hit point" on 2 rolls.
Him quitting on a shooter whom knocks him off twice is just superstition about a 'hot' shooter.
