6/8 Regression-Progression Strategy

Hi Everyone,

I need some help calculating the odds of 5 consecutive point-then 7s before either a 6 or 8 is rolled.
Ah, but there is the kicker, let's assume the following data:

6 or 8 comes up 30.28% instead of the random 27.78%
7s come up 14.83% instead of the random 16.67%
We only care about the throws after the point is established. Disregard the Pass Line bets.

The bets will be a 6 and 8 Progression that will always win money on every session until the 5 point-7s occur.
Like all Progressions, it is impossible to alter bets to overcome the house odds, but on average, there will be a bunch of smaller wins and then 1 large loss. In this case, the large loss comes when 5 point-7's in a row occur. (I understand this is inconsequential to figuring out the odds, but I thought I would mention it in case some one was wondering.)

So I got to thinking, what are the odds that 5 point-7 outs in a row will occur given the slight edge mentioned above. Additionally, will that edge be enough to overcome the house edge over the long term?

Thank you in advance for your help!

Quote: SteveStew

So I got to thinking, what are the odds that 5 point-7 outs in a row will occur given the slight edge mentioned above. Additionally, will that edge be enough to overcome the house edge over the long term?

Let's start with this: the house edge will never be overcome. Your odds of winning will always be less than proportionate to what you're getting from a win.

That said, I'm not sure I can answer your question. The chance of five consecutive points sevening out is 7.39%. If a seven, yo, or craps on the come-out roll will save you, the chance of a seven-out on five consecutive rounds is 0.973%. If a six or eight on any roll will also save you, it's 0.00242%. If a six or eight will save you on any roll, but a win/loss on the come-out roll can't, it's 0.0183%.

I must stress, however, that as small as these numbers seem, if any other scenario is a win, I guarantee you that your losses in these scenarios will be large enough to outpace the wins you'll get.

Hi 24Bingo!

Thank you very much! I was speaking of 5 Seven-outs only after the point is established and a 6 or 8 can save me. So is that 1.83% or .0183%? I understand that is for random, correct? Do you have any idea how I can figure out the odds if the 6 or 8 are thrown 30.28% to save me and the 7 thrown 14.83% to lose?

Thank you!

I'm sorry, I missed the fact that you think you have the dice in your favor. That will absolutely overcome the house edge, but a progression's not the way to take advantage, at all. I'll get to how you should take advantage, but first, the progression, assuming fair dice.

I guess I need to know, exactly under what outcomes...

A: You get paid (or at least get back to even) and start the progression over.
B: You don't get to start over, but you're not any closer to going bust.
C: You're one step closer to going bust.

.0183% - and that does mean .0183% - is if A is any 6 or 8, or any point hitting; B is any round ending on the come-out roll either way; C is any seven-out; and the dice are fair.

The problem with your question is that it sounds as if not any 7 will finish you. At any point, with your odds, you have a 67.94% chance of just hitting a 6 or 8 before you hit a seven (as opposed to the normal 62.5%), but if it makes a significant difference whether these numbers hit on a come-out roll or not, or what the point is when they hit, the math gets harder.

However, I should stress that if you think this is the case, I recommend you simply place the 6 and 8, rather than bothering with a progression, since at 7:12, you'll get about a 7.6% yield. I somewhat doubt you have it, but if you do, I recommend you do this betting an eighteenth of what, all told, you're willing to risk on each at a time (a ninth at a time total), always rounding down and ensuring each individual bet (not the sum) is divisible by six (and, again, round down to the nearest multiple of six, never up). When you don't have the dice, try betting a teeny bit on the pass line just to stay at the table - if you can find a table with a small enough minimum compared to your bet, it shouldn't get you into too much trouble, especially since a losing streak means you get the dice quicker.

Hi 24Bingo,

Great response. I agree with you, a progression/regression would not be the best way to take advantage of favorable dice.

First, let me be more specific about your outcome question. The only outcome that will create a loss for the progression is when a 7 is thrown before either a 6 or 8 only after a point has been established and for 5 consecutive hands. In other words, once a 6 or 8 is hit the regression-progression starts over.

Second, let me give you an example of the Regression-Progression:

1. Establish a point. It does not matter what number the point is or whether the point is made for this discussion.
2. Place $X on 6 and $X on 8.
3. If a 6 or 8 is thrown before a 7, regress both bets to $.4X each. Regression-Progression starts over at Step 1 on next hand.
4. If a 7 is thrown: next hand Step 2 will progress to $3X each on the 6 and 8.
5. If 2 consecutive hands throw a 7 before a 6 or 8 is thrown, Step 2 goes to 9X; 3 consecutive = 27X; 4 consecutive = 81X; 5 consecutive = you stop and lose 244X.

So you can see, you will always make money on every hand until the 5 consecutive point-7s occur at which point you lose large.

Thirdly, I understand every Regression-Progression will lose in the long run with random games, that is not my point. My point was to give you an idea of the Progression in order to help describe the probability I am seeking; that is, what is the probability that 5 consecutive hands will throw a 7 before either a 6 or 8, after a point has been established given the following probabilities: 7 occurs 14.83% and 6 or 8 occur 30.28% of the time.

Fourth, can you please tell me where I am going wrong with my calculations. I started by adding 7 and 6/8 probabilities: 14.83 and 30.28 =45.11. Then 30.28/45.11 to get a 67.12% probability of a 6 or 8 before a 7. Inversely, 32.88% probability of a 7 before a 6 or 8. Next, I simply took 7 probability .3288 to the 5th power to get .00384 which is .384% which is 3.84 times per 1000 decisions. You came up with .0183% which is .183 per 1000 decisions. I am more than 20 times off.

Lastly, thank you so very much for taking the time to have this discussion.

Cheers,
SteveStew

Hi 24Bingo,

I am sorry, I am new to this Forum and did not notice there was a Math section. I posted the probability question there and ChesterDog answered. Thank you so much for your posts!

Great site, Wizard!

Cheers,
Steve Stew