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blount2000
blount2000
Joined: Oct 15, 2010
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April 20th, 2012 at 6:54:35 AM permalink
When playing craps I have traditionally just played a single $5 chip on the pass line and then waited to see how it turned out. I would like to shake things up a bit by increasing my bet to around 3 chips (I know, big spender!). I am considering trying out the following plays:


1. Playing a single chip on the pass line and then putting 2x odds behind it. (Pro = probably the best mathematical bet?)

2. Playing a single chip on the pass line and then making up to 2 come bets. (Pro = kind of a "hedge" if the 7 shows before I get all bets on the table?)

3. Playing a single chip on the pass line and then immediately placing the 6 and the 8. (Pro = seems like I could get more ongoing action?)


From a "making the bankroll last" standpoint, is any one of these plays superior over the others assuming a $250 buy-in?
You serious, Clark?
WongBo
WongBo
Joined: Feb 3, 2012
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April 20th, 2012 at 8:00:19 AM permalink
the methods listed are all valid and low edge.
your current method is 1.41% edge.
1. is .606% edge
2. is 1.41% edge
3. is 1.41% edge for the pass line chip and 1.52% edge for the other two.


any of these bets would be a fine way to get more involvement in the game.
i usually combine all of these bets, but if you are satisfied with less action and a lower risk per turn,
you are on the right track.
i would vote for the free odds bet.
In a bet, there is a fool and a thief. - Proverb.
MathExtremist
MathExtremist
Joined: Aug 31, 2010
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April 20th, 2012 at 8:30:12 AM permalink
You don't need $250 if that's how you're playing. But what I'd do in your case is a hybrid approach:
1) $5 pass
2) If a pass point is established, $5 come and $5 (or $4) in odds behind the passline.
3) If the come point is established, decide where you'd rather have your odds, behind the pass or come numbers and move it there. Now you have $15 in action.
4) One of two things will happen: you'll lose $15 on a 7-out, or you'll win one of your bets. If you win, take $5 from the winnings and rebet it as odds behind your other number. Make one more $5 line bet, either come or pass, depending on which one won. Keep the rest of the winnings if there are any (there won't be if
5) Repeat, adding $5 to the table after every win. Eventually you'll have 2 numbers working with full odds, and you'll also have some profit locked up.
6) Then expand to a 3rd number, and so forth. On a long roll, you'll make a lot more money (and have a much more exciting time) than if you just stand there with your one or two bets while everyone else is making money.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
RaleighCraps
RaleighCraps
Joined: Feb 20, 2010
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April 20th, 2012 at 8:37:43 AM permalink
Quote: WongBo

the methods listed are all valid and low edge.
your current method is 1.41% edge.
1. is .606% edge
2. is 1.41% edge
3. is 1.41% edge for the pass line chip and 1.52% edge for the other two.


any of these bets would be a fine way to get more involvement in the game.
i usually combine all of these bets, but if you are satisfied with less action and a lower risk per turn,
you are on the right track.
i would vote for the free odds bet.



The problem with doing the free odds bet is the amount of 'involvement' is the same, he is still just getting 2 numbers for a resolution. Either he waits until the point is made, or the 7. So he has more 'action', but the same amount of involvement.

As I read his post, I think he is looking for more 'action'. The come bet will provide more 'action' for at least 3 rolls. Then he will now have 4 numbers that will result in a possible resolution. But with a come bet, he could end up with the 4 & 10 as the points, which may not roll for quite some time, thus resulting in more money at risk for the 7, and still a lot of boredom in between.

So, he can Place the 6/8. It offers no safety on the 7 come out rolls (like the Come bet does), and it costs more from a HE perspective. But it has the advantage of picking 2 numbers that should be rolled more frequently than any other number but the 7. This would potentially result in more 'involved decisions' per shooter than the other 2 solutions.
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!
Juyemura
Juyemura
Joined: Apr 8, 2012
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April 20th, 2012 at 9:06:33 AM permalink
If you are craving action, I would place the 6 & 8. It gives you a couple of extra numbers quickly, with relatively HA. If your point is 6 or 8, how about placing the other number and giving yourself 1X odds behind your point.
Lottery:  A tax on people who are bad at math.
pacomartin
pacomartin
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April 20th, 2012 at 10:26:51 AM permalink
Quote: blount2000

When playing craps I have traditionally just played a single $5 chip on the pass line and then waited to see how it turned out. I would like to shake things up a bit by increasing my bet to around 3 chips (I know, big spender!).



If your goal is to stay at the table as long as possible, then your current strategy of waiting for a $5 pass line to resolve is the smartest strategy.

But I would suggest that you go with a 3X,4X,5X strategy to make it more exciting. Now the house edge is about 0.4%. If a 4,10 point is hit, add $15 odds, a 5,9 add $20 odds, and 6,8 add $25 odds. Each time the point is made then you will win $35.

That level of excitement will make it a lot more interesting without increasing house edge.
Ahigh
Ahigh
Joined: May 19, 2010
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April 20th, 2012 at 10:46:50 AM permalink
I will make one more observation for those seeking low edge strategies that do not depend on what the dice do on previous rolls:

If you are shooting bet the $5 pass line or minimum just as the cost to play. Ignore it from then on. If you're not rolling, don't bet the line.

If you want to risk $25 or less, man up and get it to $25, and buy the 4 or 10. You can do this in almost any casino. The vig is a buck and the edge is lower than the passline or the dont pass line and the pay is double minus vig. (Edge is 1.33% for a $25 buy).

Man up again, and buy the 4 AND the 10, and you've not got an even 50/50 chance to win $49 from a $50 bet -- at which point you can take them both down.

Half the problem that I observe with players isn't just the edge, but the problem with keeping track of things like how often they should be winning.

When you are supposed to win 5 out of 11 events on the 6/8, I think some people are happy enough to win 3 or 4 out of 11, and wonder why they are still losing money.

If you can man up to a buy bet, and understand that you'll lose most of the time, but get paid nearly double when you don't .. you'll win bigger and quicker with less vig payments.

You generally have to pay $.50 -- even for a $5 5/9. Why not pay a buck to get $50 instead of $.50 to get $7?

Get a few hits .. maybe press it all the way a time or two and you can win quick and be on your way with only a couple bucks for the house.

Stiff them the tip and save a couple more. That the real way to win. The edge matters a little bit .. but if you're at 1.41 or below, shaving off those last fractions matter more for the chronic gambler like me. Not for normal people!

Get in, win, get out. That's what I say. And do it in a way that you get more than 7/6th's what you bet and without depending on what the dice just did.
TIMSPEED
TIMSPEED
Joined: Aug 11, 2010
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April 20th, 2012 at 10:51:55 AM permalink
Keep betting $5 on the Pass..
If the point is 5 or 9, place bet the 6 & 8 for $6 each ($17 total outlay)
If the point is 4 or 10, put $10 odds behind the Pass Line ($15 total outlay)
This should give you a little more excitement for not too much more investment.
Gambling calls to me...like this ~> http://www.youtube.com/watch?v=4Nap37mNSmQ
7craps
7craps
Joined: Jan 23, 2010
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April 20th, 2012 at 11:00:55 AM permalink
Quote: blount2000

When playing craps I have traditionally just played a single $5 chip on the pass line and then waited to see how it turned out. I would like to shake things up a bit by increasing my bet to around 3 chips (I know, big spender!). I am considering trying out the following plays:


1. Playing a single chip on the pass line and then putting 2x odds behind it. (Pro = probably the best mathematical bet?)

2. Playing a single chip on the pass line and then making up to 2 come bets. (Pro = kind of a "hedge" if the 7 shows before I get all bets on the table?)

3. Playing a single chip on the pass line and then immediately placing the 6 and the 8. (Pro = seems like I could get more ongoing action?)


From a "making the bankroll last" standpoint, is any one of these plays superior over the others assuming a $250 buy-in?

From a "making the bankroll last" standpoint or Longest Survival method of play.
(you need to decide which one is *superior* by what you want *superior* to mean)
From short term risk of ruin formula and simulation results:

1. $5 pass 2X odds
11655 average number of rolls until ruin
7691 median number of rolls until ruin
3453 average number of decisions until ruin
2372 median number of decisions until ruin
>>>>>>>>>>>>>>

2. $5 pass and max 2 - $5 come bets (NO Odds)
4864 average number of rolls until ruin
2789 median number of rolls until ruin
3474 average number of decisions until ruin
1990 median number of decisions until ruin
>>>>>>>>>>>>>>

3. $5 pass and $6 place 6&8 ($5 odds with a point of 6 or 8)
4480 average number of rolls until ruin
1628 median number of rolls until ruin
3252 average number of decisions until ruin
1195 median number of decisions until ruin
>>>>>>>>>>>>>>

No surprise to me in the above order for From a "making the bankroll last" standpoint

FYI. The $5 pass with $10 odds had 2.27% of the players going well past 100,000 total dice rolls (1 in 44) before eventual ruin.
That is a nice surprise.
winsome johnny (not Win some johnny)
Ahigh
Ahigh
Joined: May 19, 2010
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April 20th, 2012 at 11:17:10 AM permalink
If you want to make ANY bankroll last the longest on a crap table, that's super easy. You bet minimum bet on the don't pass line and nothing else.

In general, it's nearly impossible to lose all of your money betting the minimum line bet and nothing else. You will have to go home first or go to sleep.

If you go to Joker's Wild, you can be $1 on the passline, and you could probably spend every waking spare moment you have at the table and still afford to do that.

But seriously, you guys are contemplating how to play the longest. That's like the easiest thing for anybody to tell you. If you want last longer, go with a friend and pony up their odds money. Want to last even longer? Don't bet and just keep your chips on the rail .. even when the cocktail waitress brings you a drink.

Is this really how you guys spend time? This isn't hard to figure out, really. I mean how about "don't bet."

LOL.

Quote: 7craps

Quote: blount2000

When playing craps I have traditionally just played a single $5 chip on the pass line and then waited to see how it turned out. I would like to shake things up a bit by increasing my bet to around 3 chips (I know, big spender!). I am considering trying out the following plays:


1. Playing a single chip on the pass line and then putting 2x odds behind it. (Pro = probably the best mathematical bet?)

2. Playing a single chip on the pass line and then making up to 2 come bets. (Pro = kind of a "hedge" if the 7 shows before I get all bets on the table?)

3. Playing a single chip on the pass line and then immediately placing the 6 and the 8. (Pro = seems like I could get more ongoing action?)


From a "making the bankroll last" standpoint, is any one of these plays superior over the others assuming a $250 buy-in?

From a "making the bankroll last" standpoint or Longest Survival method of play.
(you need to decide which one is *superior* by what you want *superior* to mean)
From short term risk of ruin formula and simulation results:

1. $5 pass 2X odds
11655 average number of rolls until ruin
7691 median number of rolls until ruin
3453 average number of decisions until ruin
2372 median number of decisions until ruin
>>>>>>>>>>>>>>

2. $5 pass and max 2 - $5 come bets (NO Odds)
4864 average number of rolls until ruin
2789 median number of rolls until ruin
3474 average number of decisions until ruin
1990 median number of decisions until ruin
>>>>>>>>>>>>>>

3. $5 pass and $6 place 6&8 ($5 odds with a point of 6 or 8)
4480 average number of rolls until ruin
1628 median number of rolls until ruin
3252 average number of decisions until ruin
1195 median number of decisions until ruin
>>>>>>>>>>>>>>

No surprise to me in the above order for From a "making the bankroll last" standpoint

FYI. The $5 pass with $10 odds had 2.27% of the players going well past 100,000 total dice rolls (1 in 44) before eventual ruin.
That is a nice surprise.


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