How often does the 7 out come right after the point is set?

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?

the same as if the past was unknown

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

Quote: odiousgambit

the 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?

Quote: RaleighCraps

Quote: odiousgambit

the 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.

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.

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: RaleighCraps

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.

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

It's early, your math mileage may vary,

--Dorothy

I played $320 across on the Wizard's craps game and got 7 on the first roll three times in a row.

Quote: Headlock

I 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!

Quote: DorothyGale

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: RaleighCraps

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.

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.

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

If you place $320 across the HA is 2.29%. You could probably reduce that some with buy and put bets. I wonder if the HA could be reduced by placing more or less on certain numbers?

Quote: DorothyGale

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: RaleighCraps

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.

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

It's early, your math mileage may vary,

--Dorothy

This is correct. It's really much simpler to derive, however. It's not different than figuring the probability for one place bet on the six. There are only eleven combinations that resolve the bet, so 5/11 = .4545 = p(win) and 6/11 = .5454 = p(lose).

Here, you don't care about six combinations, so you have 30 ways to resolve one of the bets, 24 winners and 6 losers, so 6/30 = .2. This means one out of five times, on average of course, the shooter will seven out before rolling any point number.

Suppose you place $32 across and take everything down after one hit.
The ratios of the frequency of points are 5 (6/8) to 4 (5/9) to 3 (4/10), so we weight the results:

5 * 7 = $35
4 * 7 = $28
3 * 9 = $27
---
$90 / 12 = $7.50

Now, we weight the win-loss results:

.2 * -$32 = -$6.4
.8 * +$7.5 = +$6.0
-----
-$.40

To figure the HA, we have to also weight the amounts bet, because when you win on e of those bets, the others are not at risk. So,

5 * 6 = 30
7 * 5 = 35
3 * 32 = 96
---
161 / 15 = 10.73

-.4 / 10.73 = -.0373

This looks right, as it's a mixture of the HAs on the different place bets. Of course, if you buy the outside numbers under favorable table rules, it will be lower, but you still have a negative expectation, as we all expected, right?

I put these numbers into my calculation program:

bet outcome ways
6 +7 25
5 +7 20
5 +9 15
32 -32 15

The expectation is -.40, standard deviation $15.42.
For 60 bets, expectation is -$24, standard deviation $119.46, probability of breaking even or better is about .42.

Cheers,
Alan Shank

Quote: RaleighCraps

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.

You are missing the other outcomes where you still lose. Note that these probabilities do not give you the HA which is covered in other posts.

You have established that there are 30 rolls that affect you - 24 are points and 6 sevens. I'll carry this out to hitting 4 points which you figure is about your break even point.

7 out before hitting any point: 6 / 30 = 20%
1 point then 7 out : 24 / 30 * 6 / 30 = 16%
2 points then 7 out : (24 / 30)^2 * 6 / 30 = 12.8%
3 points then 7 out : (24 / 30)^3 * 6 / 30 = 10.2%
4 points then 7 out : (24 / 30)^4 * 6 / 30 = 8.2%

Add these up and you see that 67.2% (just over 2/3) of the time you lose or break even. (And that is assuming that you haven't started pressing yet.)

If you consider that hitting two points gets back about half of your total bet, then almost half of the time, you lose more than half of your bet.

I look at relative probablities like this as a kind of poor man's way to illustrate the variance.

Quote: odiousgambit

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

No quibble, no problem. It is good to be questioned, to make sure I am not missing something.

The answer to the question as you phrased it is, "The shooter 7 outs 100% of the time after they set the point."

My question, as phrased, is intended for a one bet scenario.
Shooter sets a point (this has no real bearing, it is just a convenient trigger point. I could do my bet at any time.)
For my casino, I buy the 4,5,9,10 for $90. I place the 6,8 for $90.
When any one of my numbers hits, all the bets come down.
If a 7 comes out first, I lose.

So, I couldn't care less when a shooter 7 outs, as long as they do it AFTER they rolled any one of these 4,5,6,8,9,or 10.
( I am also working on a bet pattern using PUTS of varying amounts, so that I will win the same amount, no matter what number is rolled. That will remove the variance I have now which is dependent on which number rolled when I won.)


I am trying to build a variation off of the anything can happen in the short time frame.
We know that if we play 1 million rolls, we will lose, and it will be whatever the HA is.
However, we play for an hour, or 4 hours, and can win.
I am trying to see what can happen if you are only in play for very short durations. The idea being, I have no bet in play when a majority of the 7 outs occur.
If the average shooter rolls 8 times before they 7 out, and I am in play for only 1 or 2 rolls, then perhaps I should miss 75% of the 7 outs that occur. But I pay dearly to get this 'advantage'. I risk losing $540, and I only win $105 to $176. So I need to win about 4x to every loss, which, ironically, is about the same number of place bets you need to hit to break even if you just place the minimum across the board each time a shooter is rolling.
But, deciding those bets takes a large number of rolls, and we know the more rolls we play, the more the number distribution will fall exactly as expected, which means we lose.
I am in play a very, very small number of rolls, so my chance for variation from standard should be much higher.

Here is my analogy.
Flip a coin 4x. I will bet you that you do not get 2 Heads and 2 Tails. I will bet the distribution is 75%-25%, or even 100%-0%.
Now say you are going to flip the coin 10,000 times. I WILL NOT make the same bet, that the distribution will be 75%-25%. With that many rolls, it is bound to be much closer to the 50%-50%.

I am not touting this system by any means. I am experimenting with this thought process to see what might happen. So far the results have been interesting. I have had many times where my bankroll has lasted over 250 hours, which is way longer than anything else I have tried, including just making $100 DP bets, with no odds. The swings are wild, and there have been many games where I lost my 2k bank, within 1 hour. BUT, I have had a few games where I am up over 10K after 40 hours of play.
One other issue with this scheme is, I suspect my avg bet and comp rating would take a hit, since I am not playing a majority of the points.

Quote: seattledice

7 out before hitting any point: 6 / 30 = 20%
1 point then 7 out : 24 / 30 * 6 / 30 = 16%
2 points then 7 out : (24 / 30)^2 * 6 / 30 = 12.8%
3 points then 7 out : (24 / 30)^3 * 6 / 30 = 10.2%
4 points then 7 out : (24 / 30)^4 * 6 / 30 = 8.2%

Add these up and you see that 67.2% (just over 2/3) of the time you lose or break even. (And that is assuming that you haven't started pressing yet.)

If you consider that hitting two points gets back about half of your total bet, then almost half of the time, you lose more than half of your bet.

I see what you are saying. Since I was taking the bet down after the one roll decision, I was treating each bet as a new occurrence. I think you are saying whether the second roll needed is the next roll, or 30 rolls from now, is irrelevant. It is still the 'second' roll I need. Since dice have no memory it does not matter when that second roll occurs.
I can understand it as you, Dorothy, and Alan have explained it, but when I try and correlate it with dice sessions I have had, it 'seems' like the math is not right.
I almost always place across, after the point is set. And yes, I do lose all of my bets sometimes with no hit at all. More often though, I hit one or two numbers, and then the 7. And of course, there are times where 8 or 9 numbers get hit. This new scheme would make me a winner in all but that first situation.

Quote: RaleighCraps

Quote: seattledice

7 out before hitting any point: 6 / 30 = 20%
1 point then 7 out : 24 / 30 * 6 / 30 = 16%
2 points then 7 out : (24 / 30)^2 * 6 / 30 = 12.8%
3 points then 7 out : (24 / 30)^3 * 6 / 30 = 10.2%
4 points then 7 out : (24 / 30)^4 * 6 / 30 = 8.2%

Add these up and you see that 67.2% (just over 2/3) of the time you lose or break even. (And that is assuming that you haven't started pressing yet.)

If you consider that hitting two points gets back about half of your total bet, then almost half of the time, you lose more than half of your bet.

I see what you are saying. Since I was taking the bet down after the one roll decision, I was treating each bet as a new occurrence. I think you are saying whether the second roll needed is the next roll, or 30 rolls from now, is irrelevant. It is still the 'second' roll I need. Since dice have no memory it does not matter when that second roll occurs.
I can understand it as you, Dorothy, and Alan have explained it, but when I try and correlate it with dice sessions I have had, it 'seems' like the math is not right.
I almost always place across, after the point is set. And yes, I do lose all of my bets sometimes with no hit at all. More often though, I hit one or two numbers, and then the 7. And of course, there are times where 8 or 9 numbers get hit. This new scheme would make me a winner in all but that first situation.

My numbers were for leaving the bets up until you are at least ahead or seven out, which is different than taking them down after one hit. It's true that if you take all the place bets down after one hit, then 80% of the time you will win, but the wins you get will be not quite enough to make up for the 100% loss that happens 20% of the time.

I read your post and skimmed all of the responses, it seems like you were asking for general feelings.

In my experience, and as a person who believes in tables going up and down, I would do one of a few things:

1. Wait for a couple of rolls to go before you throw down anything.
2. Throw down a small amount, then when you're feeling it press the fuck out of the table.

We all know this is high risk high reward, you can't get around that. All you can do is have the best time. I guarantee you, if they seven out before you throw down big money you'll feel amazing. If you throw down big money just as the table gets hot, you'll feel amazing. Just try and block out the times you throw down money and the table sevens out...

Quote: Headlock

I played $320 across on the Wizard's craps game and got 7 on the first roll three times in a row.

I don't get it. You had 24 ways to win and only six ways to lose. The software must be faulty.

Quote: ahiromu

I read your post and skimmed all of the responses, it seems like you were asking for general feelings.

In my experience, and as a person who believes in tables going up and down, I would do one of a few things:

1. Wait for a couple of rolls to go before you throw down anything.
2. Throw down a small amount, then when you're feeling it press the fuck out of the table.

We all know this is high risk high reward, you can't get around that. All you can do is have the best time. I guarantee you, if they seven out before you throw down big money you'll feel amazing. If you throw down big money just as the table gets hot, you'll feel amazing. Just try and block out the times you throw down money and the table sevens out...

You are correct. I was asking for your general feelings as to how often a 7 out occurs right after the point is set. I then wanted to compare that to what the math says the frequency should be.

So the math seems to indicate that a 7 out before any place number has been rolled will occur 20% of the time, or 1 time for every 5 points that are set.
Maybe I have just been on lucky tables, but I don't think I have seen that sort of frequency on my tables. I would be losing my $52+ across, or $130+ across without a single payout. I know it happens, I remember the gut punch feeling I get each time it happens, but my gut says it has not been punched 1 out of every 5 points.

Quote: goatcabin

Suppose you place $32 across and take everything down after one hit.
The ratios of the frequency of points are 5 (6/8) to 4 (5/9) to 3 (4/10), so we weight the results:

5 * 7 = $35
4 * 7 = $28
3 * 9 = $27
---
$90 / 12 = $7.50

Now, we weight the win-loss results:

.2 * -$32 = -$6.4
.8 * +$7.5 = +$6.0
-----
-$.40

To figure the HA, we have to also weight the amounts bet, because when you win on e of those bets, the others are not at risk. So,

5 * 6 = 30
7 * 5 = 35
3 * 32 = 96
---
161 / 15 = 10.73

-.4 / 10.73 = -.0373

This looks right, as it's a mixture of the HAs on the different place bets. Of course, if you buy the outside numbers under favorable table rules, it will be lower, but you still have a negative expectation, as we all expected, right?

I put these numbers into my calculation program:

bet outcome ways
6 +7 25
5 +7 20
5 +9 15
32 -32 15

The expectation is -.40, standard deviation $15.42.
For 60 bets, expectation is -$24, standard deviation $119.46, probability of breaking even or better is about .42.

Cheers,
Alan Shank

Alan,
Thank you for breaking this down for me. I printed this so I can study it, and I understand how you made the calculations. I am going to recalc them using the favorable Buy bets I get in my game. I need to keep working with this until I can create these calcs on my own. It is still too fuzzy to me to do it on my own, but at least I understand it now when you present it.

Quote: odiousgambit

the same as if the past was unknown

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

That's been the trend every time I've played craps. A hot hand is anyone who can reach double digits.