Don't Pass but Play Numbers Too

Hey guys, what do you think about betting the don't pass line with odds, but also playing the 5,6,8 and 9? I noticed while observing some craps tables over the weekend, that before the dice "crap out", shooters rolled about 3-4 numbers (other than the point). Do you think by playing some numbers alone with the don't pass, it would minimize the expected loss?

I had thought about this at one point, as a way to play minimal amounts of money. The idea would be a $5 Don't Pass, and should a 4,5,9,10 be the point, lay double odds on the number. That would be either $15 or $20. Then you can Place the 6 and 8 for $6 each. Total layout would be $32 or $37. And, the 3 most common numbers (6,7,8) would be working for you. Naturally, if there's a 7-out before you roll any of your Place numbers, you lose those, but since your payout on the Don't Pass would be $15 dollars, you essentially make $3.

The downside of course is the same whenever you play the Don't Pass, which is that the 7 or 11 on the comeout results in a loss. Plus the possible stigma associated with playing the Don't from other players at the table.

EDIT: If the point is 6 or 8, then you can still lay double odds, and then play either $6 or $12 on the other number. Or, I suppose, $6 on the other, and $5 on either the 5 or 9. Whatever.

As long as you're not also betting the point, it's a reasonable strategy.

If you're also placing the point, that's Hedging, and that violates one of the Wiz' 10 Commandments.

Quote: IAchance5

I noticed while observing some craps tables over the weekend, that before the dice "crap out", shooters rolled about 3-4 numbers (other than the point).

I'm not picking on you, just setting the record straight.
I think the proper term is to say "Seven Out". In my 30 years of dice dealing this is the most common craps term mistake made at a craps table.
To "Crap Out" means a shooter rolling a 2,3 or 12 on a come out roll causing ALL pass line wagers to lose. When a shooter rolls a 7 and loses the dice it is called a "7 out".

Quote: IAchance5

Do you think by playing some numbers alone with the don't pass, it would minimize the expected loss?

great math question! or a good wincraps simulation for member goatcabin to run.
You are now betting more, making more bets that have a house advantage. Common thinking would be that your expected loss would increase.

25% of the time when you place the inside numbers you win none of them before you "seven out" and lose all of them.
44% of all inside place bets go down after winning either none or just one.

time for a good simulation run
good luck

Quote: 7winner

44% of all inside place bets go down after winning either none or just one.

Well, I would also take down the place numbers that win. (i.e. If I put $6 on the 8 and it is rolled, then I would take my $7 win and also my $6 bet)

I don't mean to be a downer, but don't forget the worst case scenarios where you lose all of your bets. You could get a series of points 4,5,9,10, and the shooter keeps hitting those until you A) run out of money; B) get tired of losing and quit betting the don't; C) switch to the passline. And of course, the next result is a 7 out.

There's always a perfect betting strategy for every sequence of rolls. The trouble is you don't know it until it is over. If you're lucky you got it right and won some money, but you HAVE to be comfortable with the prospect that you got it exactly wrong -- I know this is an major aspect of gambling that is sometimes overlooked, especially be overly exuberant gamblers.

Often strategies like this force people to put more money on the table than they normally would with the thought that they are "protecting" or "minimizing" and then suddenly, there are no more chips in front of them and they haven't even got to shoot yet and waitress has not come back with that beer ... and yes - there is tomorrow's gambling budget, so why not use a hundred or two of that, after all the strategy has to work eventually -- it can't always turn out wrong, can it? (Yes, I have gone through that thought process a number of times :-)

By the way, if I KNEW the situation would be as originally described - hitting 3-4 numbers before the 7-out - I think I would go with placing the numbers and making repeating don't come bets so that I win when the number first hits (take down the place bet) and then win again on the 7!!

Quote: seattledice

I don't mean to be a downer, but don't forget the worst case scenarios where you lose all of your bets. You could get a series of points 4,5,9,10, and the shooter keeps hitting those until you A) run out of money; B) get tired of losing and quit betting the don't; C) switch to the passline. And of course, the next result is a 7 out.

There's always a perfect betting strategy for every sequence of rolls. The trouble is you don't know it until it is over. If you're lucky you got it right and won some money, but you HAVE to be comfortable with the prospect that you got it exactly wrong -- I know this is an major aspect of gambling that is sometimes overlooked, especially be overly exuberant gamblers.

Often strategies like this force people to put more money on the table than they normally would with the thought that they are "protecting" or "minimizing" and then suddenly, there are no more chips in front of them and they haven't even got to shoot yet and waitress has not come back with that beer ... and yes - there is tomorrow's gambling budget, so why not use a hundred or two of that, after all the strategy has to work eventually -- it can't always turn out wrong, can it? (Yes, I have gone through that thought process a number of times :-)

By the way, if I KNEW the situation would be as originally described - hitting 3-4 numbers before the 7-out - I think I would go with placing the numbers and making repeating don't come bets so that I win when the number first hits (take down the place bet) and then win again on the 7!!

If the shooter keeps hitting points, then all I would lose is my don't pass bets, I would not lose any place bets. Also, if the shooter 7's out, I would lose my place bets but win on my don't pass. Is it more probability that a shooter will hit a number or two before a 7 out? If so, then it seems like this would minimize the expected loss in the long run, right?

The series of bets doesn't minimize expected loss, your expected loss is just the sum of the expectations of each of the bets.

Instead, you are 'minimizing' variance, which is actually a bad thing. You gamble with a negative expectation because you are trading it for higher variance (which is actually good) and the chance to win some money. If you gamble with a negative expectation and no variance you just lose money.

Of course, if your goal is to get some comps, get some free beer, watch the pretty people and make some temporary friends in exchange for knowing you will lose a small amount of money, then you DO have a good strategy.

What if you played the Don't Pass with the Iron Cross? Theoretically, after the comeout roll, every number would be working for you. If you played the Iron Cross with $6 on the 6 and 8, $5 on the 5 and Field, I'm sure you could figure out how to work the Don't Pass with Odds to cover this bet should the 7-out come immediately.

I just got back from the casino betting the $5 don't pass, taking minimum odds, and betting 6 and 8 as well.....I doubled my money....I'm not saying that it will work everytime, but so far, betting the 6, 7, and 8 are paying off....

Quote: konceptum

What if you played the Don't Pass with the Iron Cross? Theoretically, after the comeout roll, every number would be working for you. If you played the Iron Cross with $6 on the 6 and 8, $5 on the 5 and Field, I'm sure you could figure out how to work the Don't Pass with Odds to cover this bet should the 7-out come immediately.

$5 field, place the 6 for $6, place the 8 for $6, $31 NO 5

now you win $2 on 16 combinations, win $5 on 14 combinations, win $10 on 1 combination, win $15 on 1 combination, and lose $36 on 4 combinations.

2(16/36) + 5(14/36) + 10(1/36) + 15(1/36) - 36(4/36)

for an expected loss of 47.22cents per roll

A regular iron cross bet ($5 field, $5 place 5, $6 place 6, $6 place 8), only has an expected loss of 25 cents per roll.

What about the iron cross with a no-4, or no-10? How would that pan out, number-wise. (I can't figure out the no bets payouts, they always confuse me.)

Quote: konceptum

A regular iron cross bet ($5 field, $5 place 5, $6 place 6, $6 place 8), only has an expected loss of 25 cents per roll.

What about the iron cross with a no-4, or no-10? How would that pan out, number-wise. (I can't figure out the no bets payouts, they always confuse me.)

$22 iron cross with a $41 no 4 or no 10.

2(14/36)+5(11/36)+10(1/36)+15(1/36)-3(6/36)-36(3/36)

which works out nicely to -18/36 (90/36-108/36) for an expected loss of 50cents per roll.

So your no-5 was still better.

I played the other day, and wanting to play as cheaply as possible, I did this:

$5 Don't Pass, with double odds.

Once the point was made, place the 6 and 8 for $6 each. If the point was a 6 or 8, place the other for $12. At this point, a 7-out nets you a win of $3. Rolling the point is bad, of course.

When the 6 or 8 hits, go to $12 on that number. If the place is already at $12, then take the $14 and put $5 on the 5, and $5 on the field, to get your iron cross.

Personally, I like the iron cross with $12 on each of the 6 and 8, $5 on each of the field and 5. So, basically, I'm shooting for this configuration. If I won enough on the iron cross to cover my potential loss on the Don't Pass bet, then I'd start pressing the 5,6,8, one unit at a time.

Quote: rudeboyoi

$22 iron cross with a $41 no 4 or no 10.

2(14/36)+5(11/36)+10(1/36)+15(1/36)-3(6/36)-36(3/36)

which works out nicely to -18/36 (90/36-108/36) for an expected loss of 50cents per roll.

What do you mean no 4 or no 10?

Quote: IAchance5

What do you mean no 4 or no 10?

its like the reverse of a buy bet.

youre betting a seven will come up before the number shows.

they charge you a 5% commission up front on the total amount you can win so...

a NO 4 or NO 10 would be $41 to win $20.

a NO 5 or NO 9 would be $31 to win $20.

a NO 6 or NO 8 would be $25 to win $20.

Well, I'm 2 for 2 now! Just went back to the casino over the weekend.....bet the don't pass with single odds, and placed the 6 & 8.
I bought in for $80 and cashed out with $115 after 2 hours of play.....I think what has been helping me is that every time I win on my 6 or 8 place bet, I will take it down so that I am no longer risking it until the next point is rolled....

I use a similar strategy to this...

I start with a $200 bankroll. I place a $5 bet on the Dont Pass Bar. When the point is established, I lay $30 behind it (3x-4x-5x) no matter what the point is. I then place the 5 for $5, the 6 and 8 for $6 each and put $5 in the field.

I like this alot better than a standard iron cross strategy because your rooting aginst the point instead of the more likely 7.

This strategy pays well if you get a point set, see a few numbers and then hit your 7.

I know the house advantage is higher than playing strictly dont pass with odds, but I play craps for fun, comradrie, and beer as much as I o for money, so I 4 hour session where I may have lost a few bucks isn't going to make me real mad. Ya know?

Quote: mwalz9

on the Dont Pass Bar.

That does still make me laugh a little.

Mwalz, it's Don't Pass, Bar 12. Meaning 12 is a push, but your mistake is very common...

I like the $15 Dont Pass with $6 on 6 and 8... it seems a bit "better", but really its $15 at -1.36% and $12 on -1.515% an algebraic -1.43%. Breaking it up as $5 DP + $10 odds (presuming 4,5,9,10 only) is better, and I think it was the OP way. THATS a really good idea.