March 7th, 2010 at 6:18:18 AM
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Quote:Warning: Simulations are showing this scheme to be extremely volatile, wiping out the bankroll very quickly due to the extreme size of the first bet. However, it has also shown a large number of really big wins.

I am messing with a scheme where I don't have a PL or DP bet at all.

I wait for a point to be set, and then place a huge bet Across all six numbers.

This bet can be Place bets, Buy Bets, or Puts w/odds, whatever provides best advantage at that table.

When any number hits, take them all down. (If you did Puts, you would have to leave the base, but that would only be $5 or $10, and it would still be in play, so not too bad.)

Wait for the next come out roll and point to be set, and then do the same thing.

Lots of down time with this system, although once you get up, you could turn it into a Regression system instead, where once the first big hit occurs, you take all the bets down to minimum place bets.

Playing this system, only one thing matters.

How often does a shooter roll a 7 right after they set the point?

Note: any junk numbers (2,3,11,12) don't count as a roll, so a roll sequence consisting of:

point 6 set, 2,2,3,7 counts as a 7 out following the point set.

Lots of simulation rolls using WinCraps show my break even point to be right around 4 bets paid for each first roll 7 out.

Here is my question. As I recall all the times I have spent at the craps tables, it does not seem to me that 1 out of every 4 times a point is set, that the shooter rolls a 7 right away. Oh sure, we have all seen it, and I know I have done it more than I care to admit, but 1 out of every 4 points that are set?

I would be interested in what all of your perceptions are. To your best recollection, based on the play you have seen at the tables, how often do you think a 7 is rolled right after the point is set?

Always borrow money from a pessimist; They don't expect to get paid back !
Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, don't matter!

March 7th, 2010 at 6:27:26 AM
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the same as if the past was unknown

one time in six, so you've had some bad luck

one time in six, so you've had some bad luck

"Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." .......... Mark Twain

March 7th, 2010 at 6:51:06 AM
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Quote:odiousgambitthe same as if the past was unknown

one time in six, so you've had some bad luck

I don't think it is bad luck, as it happens every time. The longest the scheme has worked is 250 hours (70 rolls/hr), but it always ends with the ratio of wins to 7 outs slightly less than 4 to 1

You are correct, a 7 rolls, 1 out of every 6 times, but I don't think that is the right number for me to expect though.

Since 2,3,11,or 12 don't resolve the bet in my favor, don't I need to discount those rolls.

The times I get a sequence of 6,2,7 for my scheme counts the same as if it were 6,7.

6 ways to make 4/10

8 ways to make 5/9

10 ways to make 6/8

so 24 ways for me to win big bet

6 ways to lose big bet

6 ways the bet is unresolved.

I think this means I have a 66.7% chance of winning the bet, right? (24/36)

And there is a 16.7% chance that the house will get another crack at my bet.

The unresolved bets mean my money is now at risk for another roll of the dice. So my 66.7% chance of a win is going to be reduced, but how do I calculate by how much?

Always borrow money from a pessimist; They don't expect to get paid back !
Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, don't matter!

March 7th, 2010 at 7:03:49 AM
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Quote:RaleighCrapsQuote:odiousgambitthe same as if the past was unknown

one time in six, so you've had some bad luck

I don't think it is bad luck, as it happens every time. The longest the scheme has worked is 250 hours (70 rolls/hr), but it always ends with the ratio of wins to 7 outs slightly less than 4 to 1

You are correct, a 7 rolls, 1 out of every 6 times, but I don't think that is the right number for me to expect though.

Since 2,3,11,or 12 don't resolve the bet in my favor, don't I need to discount those rolls.

The times I get a sequence of 6,2,7 for my scheme counts the same as if it were 6,7.

6 ways to make 4/10

8 ways to make 5/9

10 ways to make 6/8

so 24 ways for me to win big bet

6 ways to lose big bet

6 ways the bet is unresolved.

I think this means I have a 66.7% chance of winning the bet, right? (24/36)

And there is a 16.7% chance that the house will get another crack at my bet.

The unresolved bets mean my money is now at risk for another roll of the dice. So my 66.7% chance of a win is going to be reduced, but how do I calculate by how much?

On $32 across, your expected loss is $.333333 per every roll that you have the bet up. Just because you have a 66.7% change of winning the bet, the losses of throwing a seven completely wipes you out.

6 ways to make 4/10 - win 9/5 units

8 ways to make 5/9 - win 7/5 units

10 ways to make 6/8 - win 7/6 units.

6 ways to make 7 - lose 6.4 units.

(6 * 9/5 + 8 * 7/5 + 10 * 7 / 6 - 6 * 6.4)/36 = -.066667 units. If a unit is $20 dollars, you would be betting $132 across with an expected loss on every roll of $1.333.

You are at risk for this much each time that the dice is rolled and the dice has no memory.

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You want the truth! You can't handle the truth!

March 7th, 2010 at 7:12:43 AM
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If you looked at it more like 66.7% you win, 16.7% you lose, and 16.7% you TIE, then you should be able to see that it's still 66.7% you win on the next roll...The 2/3 chance doesn't get reduced, anymore than a tie in blackjack "exposes" you to a greater risk on the next hand.

March 7th, 2010 at 7:18:21 AM
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Quote:RaleighCraps

How often does a shooter roll a 7 right after they set the point?

Note: any junk numbers (2,3,11,12) don't count as a roll, so a roll sequence consisting of:

point 6 set, 2,2,3,7 counts as a 7 out following the point set.

Let A be the event of rolling a 2,3,11,or 12. Then p(A) = 6/36 = 1/6.

Let B be the event of rolling a 7. Then p(B) = 1/6.

What you want is some number of A's followed by a B. That is, you want

sum(n = 0 to infinity) p(A)^n*p(B) =

(1/6)^0*(1/6) + (1/6)^1*(1/6) + (1/6)^2*(1/6) + ... = 1/5 = 20%.

This is a geometric series, so the sum is trivial.

That is, 20% of the time you will roll 7 before hitting a number other than 2,3,11 or 12. This is not the intuitively expected number of 1/6. It is a bit more frequent.

Quote:RaleighCrapsAs I recall all the times I have spent at the craps tables, it does not seem to me that 1 out of every 4 times a point is set, that the shooter rolls a 7 right away.

1/5 is more often than 1/6, but it is not 1/4.

It's early, your math mileage may vary,

--Dorothy

"Who would have thought a good little girl like you could destroy my beautiful wickedness!"

March 7th, 2010 at 8:25:17 AM
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I played $320 across on the Wizard's craps game and got 7 on the first roll three times in a row.

March 7th, 2010 at 9:21:01 AM
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Quote:HeadlockI played $320 across on the Wizard's craps game and got 7 on the first roll three times in a row.

Aha! Absolute proof that placing bets on the numbers increases the likelihood of a seven being rolled! Now, I know not to do that. Thanks!

March 7th, 2010 at 9:25:30 AM
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Quote:DorothyGaleQuote:RaleighCraps

How often does a shooter roll a 7 right after they set the point?

Note: any junk numbers (2,3,11,12) don't count as a roll, so a roll sequence consisting of:

point 6 set, 2,2,3,7 counts as a 7 out following the point set.

Let A be the event of rolling a 2,3,11,or 12. Then p(A) = 6/36 = 1/6.

Let B be the event of rolling a 7. Then p(B) = 1/6.

What you want is some number of A's followed by a B. That is, you want

sum(n = 0 to infinity) p(A)^n*p(B) =

(1/6)^0*(1/6) + (1/6)^1*(1/6) + (1/6)^2*(1/6) + ... = 1/5 = 20%.

This is a geometric series, so the sum is trivial.

That is, 20% of the time you will roll 7 before hitting a number other than 2,3,11 or 12. This is not the intuitively expected number of 1/6. It is a bit more frequent.Quote:RaleighCrapsAs I recall all the times I have spent at the craps tables, it does not seem to me that 1 out of every 4 times a point is set, that the shooter rolls a 7 right away.

1/5 is more often than 1/6, but it is not 1/4.

It's early, your math mileage may vary,

--Dorothy

Dorothy,

Thanks for providing that math. I felt it would be less than 1/6, but I didn't know how to get to that number through mathematics.

So my break even point is one 7 out out of every 4 decisions, and my expected losing rolls is 1 out of every 5. Seems like the math is in my favor here. I'm sure I am missing something though.

Always borrow money from a pessimist; They don't expect to get paid back !
Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, don't matter!

March 7th, 2010 at 9:39:21 AM
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your question should be "How often does a shooter roll a 7-out after they set the point? " not "How often does a shooter roll a 7 *right after* they set the point? "

assuming I now understand

just a quibble, sorry

assuming I now understand

just a quibble, sorry

"Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." .......... Mark Twain