Romes
Joined: Jul 22, 2014
• Posts: 5445
August 21st, 2014 at 1:24:40 PM permalink
UPDATE
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:

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.
Playing it correctly means you've already won.
Romes
Joined: Jul 22, 2014
• Posts: 5445
August 21st, 2014 at 3:12:06 PM permalink
Okay, I did the follow up probabilities and I'm getting a POSITIVE EV. So, what did I do wrong???

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
Playing it correctly means you've already won.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 3852
Thanks for this post from:
August 21st, 2014 at 5:04:00 PM permalink
Where are you getting +1 for a seven on the second throw?
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.
bahdbwoy
Joined: Aug 23, 2013
• Posts: 163
August 21st, 2014 at 5:31:24 PM permalink
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 3852
August 21st, 2014 at 7:56:03 PM permalink
Here's what I get:

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
Romes
Joined: Jul 22, 2014
• Posts: 5445
August 21st, 2014 at 8:23:42 PM permalink
Quote: ThatDonGuy

Where 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.
Playing it correctly means you've already won.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 3852
August 21st, 2014 at 8:59:28 PM permalink
Quote: Romes

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.

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".
Romes
Joined: Jul 22, 2014
• Posts: 5445
Thanks for this post from:
August 21st, 2014 at 9:48:50 PM permalink
Quote: bahdbwoy

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.
Playing it correctly means you've already won.
Romes
Joined: Jul 22, 2014
• Posts: 5445
August 21st, 2014 at 9:54:38 PM permalink
Quote: ThatDonGuy

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

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.
Playing it correctly means you've already won.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
August 21st, 2014 at 10:12:26 PM permalink
Quote: Romes

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.

you did not take into account every possible state the game can be in.
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
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