Average Number of Points Hit Per Shooter in Craps
Thread Rating:
We all know the shooter can't keep hitting points forever. Sooner or later he eventually 7's out.
But, what do you guys think the average number of points a shooter is able to hit before rolling the seven and passing the dice?
Quote: Wizard
I believe that's rolls per shooter. Since rolls per passline decision is 3.376, that makes the average decisions per shooter about 2.5. Of those decisions, 27.1% are point winners, so the average number of points made by the shooter is about 0.68.
In other words, if you make even one point, you're having a better-than-average roll. Not very intuitive, is it?
Yes, you really do only have a bit better than 1 in 4 chance of establishing and making a point. A big chunk of winning passline bets happens from natural 7/11 rolls: 22.2% to be precise. When you add them in you get 49.3%, the overall chance of winning a line bet.
The Wiz linked to a page where he shows the math to come up with the 8.53 number.
Scroll down about 1/3 the page - just below the average rolls per hour charts.
It's 8.53 THROWS per shooter.
Quote: JimmyMacDoes anyone know the average number of points hit by a shooter before they 7 out?
A better question would be the math for:
The probability of a shooter making 0 points,
1 point,2 points, 3 points etc.
Let x be the answer. The probability of eventually making the next point is (6/24)*(1/3)+(8/24)*(2/3)+(10/24)*(5/11) = 201/495 = 0.460606.
The expected number of points made is 0.460606/(1-0.460606) = 201/294 = 0.683673469.
Quote: guido111A better question would be the math for:
The probability of a shooter making 0 points,
1 point,2 points, 3 points etc.
I had run a sim of 8,522,945 million dice rolls. Here are those results:
1,000,000/shooters
682796/total points wins
| count | points wins | % | or more | or less |
|---|---|---|---|---|
| 594110 | 0 | 0.5941100000 | ||
| 241267 | 1 | 0.2412670000 | 0.4058900000 | 0.8353770000 |
| 97983 | 2 | 0.0979830000 | 0.1646230000 | 0.9333600000 |
| 39547 | 3 | 0.0395470000 | 0.0666400000 | 0.9729070000 |
| 16136 | 4 | 0.0161360000 | 0.0270930000 | 0.9890430000 |
| 6535 | 5 | 0.0065350000 | 0.0109570000 | 0.9955780000 |
| 2596 | 6 | 0.0025960000 | 0.0044220000 | 0.9981740000 |
| 1067 | 7 | 0.0010670000 | 0.0018260000 | 0.9992410000 |
| 425 | 8 | 0.0004250000 | 0.0007590000 | 0.9996660000 |
| 188 | 9 | 0.0001880000 | 0.0003340000 | 0.9998540000 |
| 91 | 10 | 0.0000910000 | 0.0001460000 | 0.9999450000 |
| 55 | 11+ | 0.0000550000 | 0.0000550000 | 1.0000000000 |
The math would be a challenge
Quote: guido111A better question would be the math for:
The probability of a shooter making 0 points,
1 point,2 points, 3 points etc.
from the WoO site:
59.39% chance of not hitting a point
40.61% chance of hitting a point (or more)
edit:https://wizardofodds.com/ask-the-wizard/craps-probability/
Quote: MathExtremistso the average number of points made by the shooter is about 0.68.
In other words, if you make even one point, you're having a better-than-average roll. Not very intuitive, is it?
Just goes to show that an "average" can be a useless number.
Knowing the standard deviation would give a better understanding to the "average", and that still may show a number that is still useless.
Quote: WizardCan you give me the URL where I say that?
you do not exactly say that.
https://wizardofodds.com/ask-the-wizard/craps-probability/
" So the probability of making a point, given that a point was established is, (5/12)*(5/11)+(4/12)*(4/10)+(3/12)*(3/9) = 40.61%"
So,
59.39% chance of not hitting a point
40.61% chance of hitting a point (or more)
Quote: guido111
https://wizardofodds.com/ask-the-wizard/craps-probability/
" So the probability of making a point, given that a point was established is, (5/12)*(5/11)+(4/12)*(4/10)+(3/12)*(3/9) = 40.61%"
can one use The Wizard's above formula to calculate exactly the probability of a shooter getting only 1 point before a 7out? How about 2 points or more?
Quote: MathExtremistI believe that's rolls per shooter. Since rolls per passline decision is 3.376, that makes the average decisions per shooter about 2.5. Of those decisions, 27.1% are point winners, so the average number of points made by the shooter is about 0.68.
0.68 is a small average for point winners
Quote: MathExtremistIn other words, if you make even one point, you're having a better-than-average roll. Not very intuitive, is it?
Yes, you really do only have a bit better than 1 in 4 chance of establishing and making a point. A big chunk of winning passline bets happens from natural 7/11 rolls: 22.2% to be precise. When you add them in you get 49.3%, the overall chance of winning a line bet.
So, when the pass line wins it wins about 45% of the time on the come out roll.
22.2%/49.3%
Would it be fair to say then that if I go to a table and wait for a roller to hit 1 number and then play the don't pass that I would have a little better shot of winning? (not counting the come out roll where I am at a disadvantage because of the 7)
I know if you run scenario's with a million rolls, the numbers probably tell you that will not have any advantage because the dice have no memory.
But, in real life when I go to a table, I may play long enough for the dice to go around the table about 3 times.
I'm just looking for an edge for the 2 hours I'm at the table.
This site is a great forum. I really appreciate all your advice.
You just answered your own question.Quote: JimmyMacWould it be fair to say then that if I go to a table and wait for a roller to hit 1 number and then play the don't pass that I would have a little better shot of winning? (not counting the come out roll where I am at a disadvantage because of the 7)
I know if you run scenario's with a million rolls, the numbers probably tell you that will not have any advantage because the dice have no memory.
On the other hand, there are plenty of gamblers that do just that - wait for a shooter to make a point - to 'qualify' him as a good shooter.
Quote: JimmyMacSo, if I'm reading this right.. if I was to walk up to a full crap's table, I could use that typically the shooter will not hit his point?
The shooter has a 40.61% on average of hitting any point.
Specifically,
4/10 33.33% (3/9)
5/9 40% (4/10)
6/8 45.5% (5/11)
So that makes on average a 59.39% chance of not hitting any point.
Quote: JimmyMacIf he hits one of his points he is doing average.
As stated above, the average is .68 so he is "above" average
Quote: JimmyMacFor him to even hit two points before crapping out and passing the dice would be above average.
A shooter "7outs" when he does NOT hit his point.
The term is NOT "crapping out". That ONLY applies on a comeout roll of a 2,3 and 12. The same shooter still keeps the dice after "crapping out".
Once a shooter has hit a point 1 or more times he IS already above average.
40.6% of all shooters hit 1 or more points.
16.1% of all shooters hit 2 or more points
6.66% of shooters hit 3 or more points
and so on by the table on page 1.
Quote: ChesterDog
(Wizard, I'm sure you meant to write: (6/24)*(1/3)+(8/24)*(2/5)+(10/24)*(5/11).)
Thanks, you're absolutely right. I just corrected my original post. I'm going through the flu, so my mind is not thinking straight.
Based on the above:
59.39% shooter will not his his point.
40.6% shooter will hit 1 or more points
16.1% shooter will hit 2 or more points
6.66% shooter will hit 3 or more points
If I take the opposite:
40.61% shooter will his his point
59.4% shooter will not hit 1 or more points
83.9% shooter will not hit 2 or more points
93.34% shooter will not hit 3 or more points
Therefore, if I choose to play the Don't Side of the table and I sit out until a shooter establishes a point and hits it and then join the game by placing my money on the Don't Pass (get past the come out roll) and the shooter establishes a 2nd point then my odds of winning go from 59.4% to 83.9%.
If I could hold out for a shooter to hit 2 points and then join the game, get past the come out roll and establish a point, I'd be at a 93.34% advantage of winning.
Do I have this correct?
Quote: JimmyMacOkay, so then what you have described then tells me that if I am playing the don't pass then my odds of winning the next established point increase every time a player hits his point.
...
Do I have this correct?
No -- that's the Gambler's Fallacy. In truth, the odds don't change over time. If the shooter has just made 10 passes in a row, his chance of making the next one are still 49.3%.
The only time aggregate probabilities matter are at the beginning of your group of events. Once you're in the middle of your set of trials, what happened in the past doesn't matter at all.
Edit, sorry, got interrupted before posting. Interesting how the examples are so similar...
If you are just as likely to have 10 passes in a row as you are to have 1 pass, why would the house offer 999-1 odds against it happening?
Quote: JimmyMacOkay, so then what you have described then tells me that if I am playing the don't pass then my odds of winning the next established point increase every time a player hits his point.
Based on the above:
59.39% shooter will not his his point.
40.6% shooter will hit 1 or more points
16.1% shooter will hit 2 or more points
6.66% shooter will hit 3 or more points
If I take the opposite:
40.61% shooter will his his point
59.4% shooter will not hit 1 or more points
83.9% shooter will not hit 2 or more points
93.34% shooter will not hit 3 or more points
Therefore, if I choose to play the Don't Side of the table and I sit out until a shooter establishes a point and hits it and then join the game by placing my money on the Don't Pass (get past the come out roll) and the shooter establishes a 2nd point then my odds of winning go from 59.4% to 83.9%.
If I could hold out for a shooter to hit 2 points and then join the game, get past the come out roll and establish a point, I'd be at a 93.34% advantage of winning.
Do I have this correct?
MathExtremist is correct.
The Gamblers Fallacy is this:
"I'd be at a 93.34% advantage of winning."
OK
New shooter. 93.34% chance he will NOT hit 3 or more points.
He makes his first point.
Now he only need 2 points to hit 3 or more. The probability of NOT hitting 2 more points that is now 83.9%...NOT the 93.34%. probability is about what has NOT YET HAPPENED.
Now he hits his 2nd point. He only needs 1 more point to hit to have hit 3 or more.
So his probability of NOT hitting 3 or more in a row (already has hit 2 ) is $59.4%
because he only needs 1 more point to have hit 3 in a row.
Quote: JimmyMacIf this is the case, then why do casino's offer Fire Bets?
If you are just as likely to have 10 passes in a row as you are to have 1 pass, why would the house offer 999-1 odds against it happening?
The probability of hitting 10 in a row
or hitting 1 in a row ARE different.
From the very first roll.
As points are hit the probability of keeping the streak alive to win 10 points in a row get HIGHER as each point is hit because there are now less points needed to hit the 10 in a row.
Just because you "wait" until 2 points have hit in a row to make a don't pass, the question remains how far will the streak continue?
Can it go to 4 in a row? It now, after already hitting 2 in a row, hit 4 in a row with the same probability of only needing 2 in a row.
Quote: JimmyMacIf this is the case, then why do casino's offer Fire Bets?
If you are just as likely to have 10 passes in a row as you are to have 1 pass, why would the house offer 999-1 odds against it happening?
That's not it - the chances of 10 passes in a row are very different than just getting 1 pass. However, the chance of the *next* pass bet winning is 49.3% regardless of whether the previous nine pass bets were winners or losers or any combination. Each bet is independent.
Here is a nice webpage by Mr. Bluejay explaining the Gambler's Fallacy.
With examples.
The light should then go on for you.
I've played craps 100's of times and in the rarest of instances have I seen a shooter hit 5 or more points before passing the dice. Typically, the shooter either 7's out before even hitting his number or he hits one pass and then 7's out before hitting his second number.
The more passes he makes the more likely it seems to be that he will eventually 7 out. We all know that he will eventually 7 out. The house has the edge and they know he will eventually 7 out.
I like to play the Don't Pass side of the table. As I know the shooter will eventually 7 out and have to pass the dice, my goal is to simply get on the Don't Pass at the prime time before he does so.
If you were betting the don't side of the table and wanted to gain an advantage based on a series of passes, how many passes would you say you should wait until jumping in on the don't pass?
Thanks for all your help.
Quote: JimmyMac
If you were betting the don't side of the table and wanted to gain an advantage based on a series of passes, how many passes would you say you should wait until jumping in on the don't pass?
Thanks for all your help.
Now I see where you are coming from.
OK, say you decide to wait for 2 points to be hit from the same shooter, then bet the don't pass.
What would you do after an hour of play when NO shooter hit 2 points??
You would have made good money betting right from the get go.
So there is NO advantage to waiting, because you could have missed the win opportunity by then.
To have a 93% advantage, you bet right from the shooters first roll, do not wait, but bet for 3 points NOT to be hit. You will have to increase your bet after a loss somewhere to show a profit once you do win, so your bankroll needs to be even larger. But then, larger losses, smaller wins, and you end up going crazy and home still bankrupt!
For your question, here is the easiest way to explain it to you.
Quote: JimmyMac
If you were betting the don't side of the table and wanted to gain an advantage based on a series of passes, how many passes would you say you should wait until jumping in on the don't pass?
Thanks for all your help.
You will never gain an advantage on craps, it’s a negative expectation game, and that is all there is to it!
The shooter could hold the dice for 150 rolls, and you still would not gain an advantage. Sure the seven is coming, but you have no way of telling when the shooters will seven out!
You only have one shot at winning your bet in that 150 rolls that just happened, you just stood at the table for 3 to 4 hour to win your one bet, wasn’t that a lot of fun?
I am just trying to make a point, the don’t player has one shot of winning on every bet they make, the house always has the advantage on whatever bet you put on the table.
So don’t think that you are going to win, because you are betting with the house!
I have seen plenty of don’t players go broke trying to beat the tables when they didn’t have the brains to leave, when everybody was making one point.
Most players that just won that one bet when paying the right side of the table will lose on the next bet that they make! They have poor money management, look I just won, now I will put two more bets on the table, when I didn’t make that much in profit. Only now they will lose those bets that they just made!
The other thing is everybody thinks they have some type of system that can beat the game; this way of thinking has been going on for ever. The math of the game is in favor of the house anyway you bet the game!
Can you win at the game “sure” for most players, if they just get lucky, and know when to walk out the doors? You can win on both sides of the table if you have brains, and a bank roll to go along with your brains. Again for most players to get good at the game, it takes years of playing, or practicing at home.
You need more than the math of the game to win, like always I have to thank the guys that tell me I am crazy because I don’t bet like they think is the best way to bet!
These guys go through a lot of work coming up with the math of the game.
Try playing win craps at home or when you do play in the casinos, bet small so you do not lose too much money on your learning curve. When you read anything in the books try it at home first, before you take it into a casino!
Then keep asking the question that you think will help you to win!
Quote: JimmyMac
If you were betting the don't side of the table and wanted to gain an advantage based on a series of passes, how many passes would you say you should wait until jumping in on the don't pass?
Thanks for all your help.
I would Wait for none.
It is not having the highest advantage of winning that counts but how many wins you do get that makes you the most money.
If you only want to bet against a point you could do it by NOT betting the don't pass.
Try making a Lay bet instead.
https://wizardofodds.com/craps see;Lay Bets
has the HA for Lay bets with vig on bet and on win.
Looking between 1.6 to 2.2% HA with a vig paid on win only.
So the HA is a bit higher than the don't pass but as you know when shooters refuse to hit their points, the dont pass seems to still get hit from shooters hitting winner 7 and 11s.
So a Lay bet can be a better option for going just against the point.
Good Luck!
Quote: JimmyMac
I've played craps 100's of times and in the rarest of instances have I seen a shooter hit 5 or more points before passing the dice. Typically, the shooter either 7's out before even hitting his number or he hits one pass and then 7's out before hitting his second number.
83.5377% or 5 out of 6 shooters will do just that.
Hit 0 or 1 point then 7 out. New shooter
Quote: JimmyMacThe more passes he makes the more likely it seems to be that he will eventually 7 out. We all know that he will eventually 7 out. The house has the edge and they know he will eventually 7 out.
You are correct.
With every roll the shooter does get closer to a 7 out. We just do not know exactly when.
The "prime time" only has a degree of certainty to it.Quote: JimmyMacI like to play the Don't Pass side of the table. As I know the shooter will eventually 7 out and have to pass the dice, my goal is to simply get on the Don't Pass at the prime time before he does so.
93.336% or 14 out of 15 shooters will either get 0,1 or 2 points, then 7 out. But by waiting after the first point is made you just gave up 83.53% (5 out of 6) that made 0 or 1. That is a big percentage of wins to just throw away.
Quote: JimmyMacIf you were betting the don't side of the table and wanted to gain an advantage based on a series of passes, how many passes would you say you should wait until jumping in on the don't pass?
Thanks for all your help.
One can not gain an advantage based on a series of passes.
waiting to bet can miss many wins. and betting into a losing streak can and is even more dangerous. You wait for 1 point win, you start betting the don't and the shooter fires off 4 more point winners with 1 or 2 pass line winners in there.
Are you increasing your bets after a loss? Laying the odds? The don't side with lay odds needs a greater bankroll to start with compared to playing the pass.
How do you play the don't pass?
After 2 or 3 in a row losses on the come out roll do you increase your bet or do you just flat bet and start laying odds?
I'm trying to simulate exactly how i would play at a real craps table. I'm not rolling in the dough, so I can't break the bank with my bets. I'm simply trying to have fun and win a little bit of money.
To keep from getting bored, I place $10 on the pass line and then play my odds. I'm only taking single odds, simply to make my bankroll last longer. i know that the house has 0 advantage on my odds bets, but in order to keep my bankroll for a longer session, i choose to only take single odds. My theory behind this is that I have just under a 50% chance of winning. I should be able to play for a while hopefully breaking even just by doing this and keep myself entertained.
However, after I hit a point, I stop playing the Pass Line and then I switch sides and place $100 on the Don't Come.
If I win, I simply go back to the Pass line bet of $10 and wait for the next point to be hit so I can place another $100 on the Don't Pass.
If on the Don't Pass, a 7 or 11 is rolled, I lose my $100 and then place another $100 bet up hoping to establish a point and then hope the shooter does not hit his point. If the shooter hits another point, I simply put another $100 on the don't pass.
My loss limit per roller is $300. If I hit a $300 loss, I wait for the next shooter.
I also have a stop win limit of $300. My goal is to play for a little while, win $300 and then take a break. When in Vegas, it is hard to simply stop gambling. But, my goal is to win multiple sessions and remove my winnings before the house gets their money back.
It seems to be working. I've been playing Wizard's craps game for the last week for a few hours each night and have had very good success. But, i'm wondering if Wizard's game is acurate because I'm actually winning at this strategy. I know that no strategy is supposed to work so I keep playing trying to lose and overall have had still a positive sessions.
I'm trying to simulate exactly how i would play at a real craps table. I'm not rolling in the dough, so I can't break the bank with my bets. I'm simply trying to have fun and win a little bit of money.
To keep from getting bored, I place $10 on the pass line and then play my odds. I'm only taking single odds, simply to make my bankroll last longer. i know that the house has 0 advantage on my odds bets, but in order to keep my bankroll for a longer session, i choose to only take single odds. My theory behind this is that I have just under a 50% chance of winning. I should be able to play for a while hopefully breaking even just by doing this and keep myself entertained.
However, after I hit a point, I stop playing the Pass Line and then I switch sides and place $100 on the Don't Come.
If I win, I simply go back to the Pass line bet of $10 and wait for the next point to be hit so I can place another $100 on the Don't Pass.
If on the Don't Pass, a 7 or 11 is rolled, I lose my $100 and then place another $100 bet up hoping to establish a point and then hope the shooter does not hit his point. If the shooter hits another point, I simply put another $100 on the don't pass.
My loss limit per roller is $300. If I hit a $300 loss, I wait for the next shooter.
I also have a stop win limit of $300. My goal is to play for a little while, win $300 and then take a break. When in Vegas, it is hard to simply stop gambling. But, my goal is to win multiple sessions and remove my winnings before the house gets their money back.
It seems to be working. I've been playing Wizard's craps game for the last week for a few hours each night and have had very good success. But, i'm wondering if Wizard's game is acurate because I'm actually winning at this strategy. I know that no strategy is supposed to work so I keep playing trying to lose and overall have had still a positive sessions.
Quote: JimmyMac
However, after I hit a point, I stop playing the Pass Line and then I switch sides and place $100 on the Don't Come.
If I win, I simply go back to the Pass line bet of $10 and wait for the next point to be hit so I can place another $100 on the Don't Pass.
If on the Don't Pass, a 7 or 11 is rolled, I lose my $100 and then place another $100 bet up hoping to establish a point and then hope the shooter does not hit his point. If the shooter hits another point, I simply put another $100 on the don't pass.
My loss limit per roller is $300. If I hit a $300 loss, I wait for the next shooter.
Interesting style of play. Switching sides?
You should program WinCraps and see how often you "bust" and hit your win goal.
As a don't bettor you may like to know how often you can lose before that 7 out makes you a winner per each shooter.
The below table is from a 1 millon shooter WinCraps simulation.
It shows the "score" or result from a shooter.
P=pass
either a natural winner or a point winner
M=miss a pass line lost on the come out roll or a 7out...
every shooter has at least 1 miss, that being a 7out
The highest result or "score" is 0-1 (0 pass/1 miss) 39.85% of the time. The 0-1 "score" happened twice as often as the next highest result.
2nd highest result or "score" is 1-1 (1 pass/1 miss)and 55% of the time the 1 pass was a point winner
3rd highest result or "score" is 2-1 (2 pass/1 miss)and 49% of the time the 2 pass was 1 point winner and 1 natural winner, 30% the 2 pass was 2 point winners.
So if you really want to "predict" what the current or next shooter will do and when he will 7out...
Try this:
68.89% (2 out of 3 shooters) will have a "score" of:
0-1
1-1
or
2-1
| Pass/Miss | relative freq | or less |
|---|---|---|
| 0-1 | 0.398592216 | |
| 1-1 | 0.194300663 | 0.592892879 |
| 2-1 | 0.096040798 | 0.688933677 |
| 3-1 | 0.047781188 | 0.736714865 |
| 1-2 | 0.043327786 | 0.780042651 |
| 0-2 | 0.042900259 | 0.82294291 |
| 2-2 | 0.032634531 | 0.855577441 |
| 4-1 | 0.023142422 | 0.878719863 |
| 3-2 | 0.021020058 | 0.899739921 |
| 4-2 | 0.013044651 | 0.912784572 |
| 5-1 | 0.011558487 | 0.92434306 |
| 5-2 | 0.00781254 | 0.932155599 |
| 1-3 | 0.007329028 | 0.939484627 |
| 2-3 | 0.007084727 | 0.946569353 |
| 3-3 | 0.005618921 | 0.952188275 |
| 6-1 | 0.005568025 | 0.9577563 |
| 0-3 | 0.004651897 | 0.962408196 |
| 6-2 | 0.004259997 | 0.966668194 |
| 4-3 | 0.004168384 | 0.970836578 |
| 5-3 | 0.002977417 | 0.973813995 |
Note:
0-2
0-3 "score" for example does not mean that the don't pass won 3 times. It could have. A 12 craps on the come out roll causes a miss but not a don't pass win. That would be a job for another simulation.
But it means the pass line LOST 3 times for that 1 shooter.
Too bad for the score of 0-5 for pass line bettors.
Quote: guido111
As a don't bettor you may like to know how often you can lose before that 7 out makes you a winner per each shooter.
The below table is from a 1 millon shooter WinCraps simulation.
It shows the "score" or result from a shooter.
P=pass
either a natural winner or a point winner
M=miss a pass line lost on the come out roll or a 7out...
every shooter has at least 1 miss, that being a 7out
The highest result or "score" is 0-1 (0 pass/1 miss) 39.85% of the time. The 0-1 "score" happened twice as often as the next highest result.
2nd highest result or "score" is 1-1 (1 pass/1 miss)and 55% of the time the 1 pass was a point winner
3rd highest result or "score" is 2-1 (2 pass/1 miss)and 49% of the time the 2 pass was 1 point winner and 1 natural winner, 30% the 2 pass was 2 point winners.
So if you really want to "predict" what the current or next shooter will do and when he will 7out...
Try this:
68.89% (2 out of 3 shooters) will have a "score" of:
0-1
1-1
or
2-1
I have never seen statistics like those before.
So I played a few games in WinCraps from actual dice rolls that I have tracked over the years.
Funny how those above 3 results show up for the average shooter.
I had a streak of 13 and another at 17 in a row where the shooters never do anything but those 3 results.
And the final results were within +/-2 after just a few hours of play.
Just to make sure I am reading this correctly, I am interpreting your data correctly?
68.9% of shooters will do 1 of the 3 below:
0-1 39.9%
Shooter will not his pass (either natural winner or hit their point)
1-1 19.4%
Shooter will hit 1 pass (either natural winner or hit their point)
2-1 9.6%
Shooter will hit 2 pass (either natural winner or hit their point)
out of this one 49% of the time the pass was a natural winner and hit thier point.
ouf of this one 30% of the time they hit 2 points.
I'm pretty sure it looks like I just copied our data again, but i want to make sure I am understanding it correctly.
Does that mean there is a 9.8% better chance that the shooter will not go 2-1 after they have gone 1-1? (19.4 - 9.6)
Thanks again!
Quote: JimmyMac
Does that mean there is a 9.8% better chance that the shooter will not go 2-1 after they have gone 1-1? (19.4 - 9.6)
Thanks again!
You have the math right.
But when a shooter goes 1-1:
meaning he made 1 pass and then a 7out. He is done. New Shooter.
anytime you see a 1 for the miss, that is a 7out.
example:1-1
2-1
3-1
4-1
all the above examples mean the shooter 7out after a certain number of pass line wins.
example: 2-2
means the shooter made 2 passes, and 2 misses, 1 being the ending 7out.
These are just probabilities of what can happen. Start writing down the "scores" and rolls of shooters, you will find that the math gets fairly close to actual outcomes the longer you observe them.
Quote: guido111
These are just probabilities of what can happen. Start writing down the "scores" and rolls of shooters, you will find that the math gets fairly close to actual outcomes the longer you observe them.
I have been doing just that while watching the baseball playoffs.
Seems when the first 3, 0-1,1-1 and 2-1 have not occurred within the last 5 shooters - watch out- here they come. You can get aggressive with your don't bets. Of course, bigger bets means larger wins and larger losses.
How about the table for the don't pass scores?
I would like to see that also if you have it available.
Since the one above is for the pass line.
Thank you
Quote: 7winner
How about the table for the don't pass scores?
I would like to see that also if you have it available.
Since the one above is for the pass line.
Thank you
OK, the 2 tables are pretty much the same.
I included the Bar12 for the don't pass.
Enjoy.
I dont know what you can do with it.
The below table is for the don't pass "scores" per shooter.
Pretty much the same as the pass line table.
example: 1-1-0
1 don't pass win for the shooter (7out)
1 Push for the Don't Pass (Bar 12 on the come out roll)
0 for 0 pass line wins
| . | Score Per Shooter | Don't Pass | . |
|---|---|---|---|
| . | Win/Push/Lose | relative freq | or less |
| 1 | 1-0-0 | 0.395001909 | . |
| 2 | 1-0-1 | 0.195694109 | 0.590696017 |
| 3 | 1-0-2 | 0.097498409 | 0.688194427 |
| 4 | 1-0-3 | 0.046708233 | 0.734902659 |
| 5 | 2-0-0 | 0.033052551 | 0.767955211 |
| 6 | 2-0-1 | 0.032543581 | 0.800498791 |
| 7 | 1-0-4 | 0.023789286 | 0.824288077 |
| 8 | 2-0-2 | 0.023692582 | 0.847980659 |
| 9 | 2-0-3 | 0.015538873 | 0.863519532 |
| 10 | 1-0-5 | 0.011507825 | 0.875027357 |
| 11 | 1-1-0 | 0.010779997 | 0.885807355 |
| 12 | 1-1-1 | 0.010505153 | 0.896312508 |
| 13 | 2-0-4 | 0.01004199 | 0.906354498 |
| 14 | 1-1-2 | 0.007889044 | 0.914243542 |
| 15 | 2-0-5 | 0.005588497 | 0.91983204 |
| 16 | 1-1-3 | 0.0055376 | 0.92536964 |
| 17 | 1-0-6 | 0.005522331 | 0.930891971 |
| 18 | 3-0-1 | 0.004290622 | 0.935182593 |
| 19 | 3-0-2 | 0.003695127 | 0.93887772 |
| 20 | 2-0-6 | 0.003343937 | 0.942221657 |
| 21 | 1-1-4 | 0.00329813 | 0.945519786 |
| 22 | 3-0-3 | 0.003201425 | 0.948721211 |
| 23 | 2-1-2 | 0.002896043 | 0.951617254 |
| 24 | 1-0-7 | 0.002824787 | 0.954442041 |
| 25 | 3-0-0 | 0.002794249 | 0.95723629 |
| 26 | 2-1-1 | 0.002641557 | 0.959877847 |
| 27 | 3-0-4 | 0.00259066 | 0.962468507 |
| 28 | 2-1-3 | 0.00224965 | 0.964718158 |
| 29 | 1-1-5 | 0.001873012 | 0.966591169 |
| 30 | 2-1-0 | 0.001766128 | 0.968357297 |
Pass Line Decisions per shooter.
The expected number of Pass Line decisions per seven out is 495/196 or 2.5255
The average from the sample below is: 2.5221
1 decision per shooter should be obvious. Just a 7out.
| decisions | relative | cumulative |
|---|---|---|
| per shooter | frequency | frequency |
| 1 | 0.396221374 | (or less) |
| 2 | 0.238030534 | 0.634251908 |
| 3 | 0.145664122 | 0.779916031 |
| 4 | 0.087515267 | 0.867431298 |
| 5 | 0.051889313 | 0.919320611 |
| 6 | 0.032167939 | 0.95148855 |
| 7 | 0.019068702 | 0.970557252 |
| 8 | 0.011767176 | 0.982324427 |
| 9 | 0.006977099 | 0.989301527 |
| 10 | 0.004167939 | 0.993469466 |
| 11 | 0.002580153 | 0.996049618 |
| 12 | 0.001564885 | 0.997614504 |
| 13 | 0.000889313 | 0.998503817 |
| 14 | 0.000629771 | 0.999133588 |
| 15 | 0.000335878 | 0.999469466 |
| 16 | 0.000229008 | 0.999698473 |
| 17 | 0.000122137 | 0.999820611 |
Quote: JimmyMacIf this is the case, then why do casino's offer Fire Bets?
If you are just as likely to have 10 passes in a row as you are to have 1 pass, why would the house offer 999-1 odds against it happening?
You seriously, seriously need to read my blog on this subject:
https://wizardofvegas.com/member/goatcabin/blog/#post16
If you confuse the probability of a series of events at its beginning with the shifting probabilities as you go through the series, you fall into an abyss of illogic.
Cheers,
Alan Shank
Woodland, CA
Your blog was excellent and now I fully understand what you have been trying to explain to me. Thanks again for all the help. I've realized the only way to guarantee a win at craps is simply not to play. But, if you are going to play and want to minimize your losses, stick to the bets with the lowest HA and try your best to quit a winner when you have the opportunity.
Quote: JimmyMacAlan,
Your blog was excellent and now I fully understand what you have been trying to explain to me. Thanks again for all the help. I've realized the only way to guarantee a win at craps is simply not to play. But, if you are going to play and want to minimize your losses, stick to the bets with the lowest HA and try your best to quit a winner when you have the opportunity.
Well, not playing does not guarantee a win, just that you won't lose. There is NO way to guarantee a win, but you can give yourself a reasonable chance to win by minimizing your expected loss and accepting some variance, always keeping in mind that variance works both ways. If you are lucky, variance is your friend, giving you a better chance to overcome the expected loss that is always, always there; if you are not lucky, then the more variance, the more you lose. You have to decide what your tolerance is and adjust your strategy accordingly. The way to get variance without increasing your expected loss is to take or lay odds behind your line bet.
Cheers and good luck,
Alan Shank
Woodland, CA
Quote: WizardYou're right, I answered the wrong question.
Let x be the answer. The probability of eventually making the next point is (6/24)*(1/3)+(8/24)*(2/3)+(10/24)*(5/11) = 201/495 = 0.460606.
The expected number of points made is 0.460606/(1-0.460606) = 201/294 = 0.683673469.
201/495 = 0.460606 should be corrected to 201/495 = 0.406060606
Why the chart? Why the simulations of 1,000,000 shooters? Why all the deciphering the numbers when all the probability begins anew with the very next roll???
Quote: JimmyMacSo, if I'm reading this right.. if I was to walk up to a full crap's table, I could use that typically the shooter will not hit his point? If he hits one of his points he is doing average. For him to even hit two points before crapping out and passing the dice would be above average.
Would it be fair to say then that if I go to a table and wait for a roller to hit 1 number and then play the don't pass that I would have a little better shot of winning? (not counting the come out roll where I am at a disadvantage because of the 7)
I know if you run scenario's with a million rolls, the numbers probably tell you that will not have any advantage because the dice have no memory.
But, in real life when I go to a table, I may play long enough for the dice to go around the table about 3 times.
I'm just looking for an edge for the 2 hours I'm at the table.
This site is a great forum. I really appreciate all your advice.
What you have to keep in mind is that the player could throw a couple of 7's and 11's on his next come out rolls and you'll lose your Don't Pass bet. Sure the odds that he'll make 2 passes are low but that doesn't change the fact that he still has decent odds of having a pass line winner or two.
So there's nothing wrong with your strategy but of course it's not guarantee you'll win either.
Quote: WizardThe probability of eventually making the next point is (6/24)*(1/3)+(8/24)*(2/3)+(10/24)*(5/11) = 201/495 = 0.460606.
The expected number of points made is 0.460606/(1-0.460606) = 201/294 = 0.683673469.
If I understand what you're driving at in this post, may I suggest just a couple of corrections to what are apparently some typo's in your above calculation?
The probability, p, expression, I believe, should actually read (6/24)*(1/3) + (8/24)*(2/5) + (10/24)*(5/11) = 201/495 = 0.4060606...., with the implied change, (1-0.4060606....). Everything else should stand as it is.
Quote: goatcabinThere is NO way to guarantee a win,
Unless you are the House ...
This morning I was trying to figure out some odds progression to use on the Do-side but they all require hitting 2 or 3 points and when 5/6 of shooters can only make 1 point or less, those progressions will be instant losers unless there's one shooter with the luck hitting 6 points twice in a row. I recently had an odds progression where I won big after hitting 5 points, and I only needed 3 points for it to be a significant win. I just got beginner's luck and I shouldn't try that again because it will be such a drain unless used on Hot Shooters.
I've always hesitated to play the DP when others are at the machine or table because it's just anti-social, but sometimes I don't care and need a winning streak. I could take it to the DC, but if I'm shooting I have to make a line bet.
On the Do-side, I count on my come-out rolls winning and I make a progression out of those successfully, but the odds bets, idk; I think I'd better just make a PB 6 and/or 8 bet instead of an odds bet and make a progression out of those too. I only need to hit one PB number twice to make one or both of them to come out above even.
