how many favor the initial steep regression betting?

[spiker]

freak things? sounds like a normal cold table. shoulda played the don'ts

[heavy]

In answer to your original question - I'm in favor of an early steep regression - after one or two hits - or three no-hit tosses. No point in leaving action out there that isn't earning its keep . . .

[Dave73]

freak things? sounds like a normal cold table. shoulda played the don'ts

shoulda coulda woulda.... its so easy to say after the fact. In a random game you NEVER know what is coming.

Im usually a fan of regression. IF im playing a place number ill keep it up for 2-3 rolls max. Quite recently at one of the OK Indian casinos a guy next to me was playing max bets on all numbers, I had the inside. We hit a few numbers and I said " Time to take em down" He gave me a look like I was crazy...two craps numbers then a seven showed.

One in the hand is worth two in the bush.

[AtGame7]

Just returned from a 7day casino trip & ended up a loser!!!!! Had to dig out of a VERY BIG hole my entire stay. I know that the math states that most rolls will never exceed four rolls before the devil shows. With that thought in mind I figured that it was best to load up moderately heavy & then regress after 1 or 2 hits!!! I have NEVER seen so many 3 & 5 roll hands in my entire life! Not to mention the large assortment of (PSO's) that were thrown into the mix. Maybe it was just one of those freak things that occurs from time to time!! how many of you (DI'S) favor the (ISR) betting over the flat betting? (of course this is betting on yourself only ?)

If you don't mind a noobie question here, I'm curious about the bolded part above. Odds on the 7 showing are 6-1 which means probability of once in 7 rolls, right? How do you come up with the math saying four rolls?

Thanks for any light you can shed on this for me.

[wudged]

To determine the probability of an event occurring within n trials, you use the formula

pn = 1 - (1 - p) ^ n

where pn is the probability in n trials and p is the probability of 1 trial.

So if you substitute in 1/2 for pn (50/50 chance of it occurring) and 1/6 for p (chance of a 7 rolling) then solve for n, you get 3.8 rolls on average for at least one occurrence of 7.

To look at it another way, (1 - p) ^ n is going to calculate the chances of the event not occurring n times in a row. For example, not rolling a 7 two times in a row would be (1 - 1/6) ^ 2 or 5/6 * 5/6. From there you subtract that result from 1 and that's the chance a 7 rolled at least once.

[Mad Professor]

For those interested in delving into Regression-betting at a level of detail never thought possible (including optimizing betting-ratios, regression trigger-points, and bankroll-doubling/re-doubling formulas, etc.); I would invite you to carefully read the following series:

Regression Avoids Depression - P-1

Regression Avoids Depression - P-2

Regression Avoids Depression - P-3

Regression Avoids Depression - P-4

Regression Avoids Depression - P-5

Regression Avoids Depression - P-6

Regression Avoids Depression - P-7

Regression Avoids Depression - P-8

Regression Avoids Depression - P-9

Regression Avoids Depression - P-10

Regression Avoids Depression - P-11

Regression Avoids Depression - P-12

Regression Avoids Depression - P-14

Regression Avoids Depression - P-15

Regression Avoids Depression - P-16

Regression Avoids Depression - P-17

Regression Avoids Depression - P-18

Regression Avoids Depression - P-19

Regression Avoids Depression - P-20

Regression Avoids Depression - P-21



MP

[AtGame7]

To determine the probability of an event occurring within n trials, you use the formula

pn = 1 - (1 - p) ^ n

where pn is the probability in n trials and p is the probability of 1 trial.

So if you substitute in 1/2 for pn (50/50 chance of it occurring) and 1/6 for p (chance of a 7 rolling) then solve for n, you get 3.8 rolls on average for at least one occurrence of 7.

To look at it another way, (1 - p) ^ n is going to calculate the chances of the event not occurring n times in a row. For example, not rolling a 7 two times in a row would be (1 - 1/6) ^ 2 or 5/6 * 5/6. From there you subtract that result from 1 and that's the chance a 7 rolled at least once.

I graduated high school (barely), but did pretty well in English so I would appreciate it if you could translate that to my native language please.

Thanks in advance

[wudged]

We know the chance of a 7 rolling is 1/6. So the chance of 7 not rolling is 5/6.

To figure out the chance of a 7 rolling twice in a row, you would multiply 1/6 * 1/6. So to make 2 rolls in a row without a 7 is 5/6 * 5/6.

What this formula does is tell you the chance of rolling at least one 7 in X number of rolls. If we want to find the "break-even" point (when you have a 50/50 chance of having rolled at least one 7 versus not rolling any 7s) then the formula tells you it takes 3.8 rolls. So half the time you will roll at least one 7 before 3.8 rolls, and half the first 7 will not occur until after 3.8 rolls.

Since you obviously can't have 3.8 rolls in a real life situation, more than 50% of the time you will see a 7 show up before 4 rolls of the dice.

[AtGame7]

Thank you, I appreciate the explanation.

[Mad Professor]

Since you obviously can't have 3.8 rolls in a real life situation, more than 50% of the time you will see a 7 show up before 4 rolls of the dice.

Which brings up the inevitable question about the 4-Rolls-No-7 wager, and whether it is a viable bet in the hands of a skilled-shooter:

4-Rolls-No-7……Worthwhile Bet or Waste of Money?


MP

[Buy The Four]

I feel much more comfortable upping my bet than I do starting my hand with big action. I'd rather put money on the passline for a 100, 200, 400...then leave if I lose three in a row. You win five bets at the 100 dollar level and you leave the casino with 500.00.

204 across or any other across action for 3 hits will yield minimal returns. It is much more advantageous to flat bet or place a larger bet on a signature number for one hit and down. If you throw fives and nines....why waste (134 using MP 204) when you can place the 5 and 9 for 100 each (less action, but not much)...collect 140 and take both down. Do that four times and you have 600+ win....don't make the game complicated!

[Mad Professor]

That's a good point, BuyTheFour, but it's also important to note that the MP $204-Across Regression is NOT designed for skilled-shooters where you already KNOW their most-dominant numbers.

Instead, the MP-$204 is designed for either DI group-shoot situations, or for when you encounter previously-unknown-to-you dice-influencers, where again, you don't know ahead of time what their most-dominant highest-advantage numbers are.

Like you suggest, if you already know a given shooter's highest-advantage numbers; then obviously you would want to put the lion's share of your betting-exposure on those, and not use such a widely-cast betting-net.

Maybe when H has a chance to post some more of the GAC weekend in-casino toss-stats, we can run the MP-$204 against them to see how they fair (even if the roll-stats most of the half-dozen or so Fire-bet hands from that weekend have gone 'missing in action').

MP

[Buy The Four]

Hey MP

Thanks for the reply. I hope you didn't think I was "calling" you out or the MP-204. I just want to let people know they should bet to their advantage. I know you have articles on that subject.