Dont pass with placing the 6 and 8-(Math question)

Hello, I was hoping I could get an answer to this based solely on the math. The bets are $5 on the DP and the point is 6. Player 1 makes a place bet of $6 on the 8(non point) and player 2 makes two place bets, $6 on the 6(point) and also $6 on the 8(non point). Both player have a $5 don't pass up.

So based on just the math I am trying to figure out if placing the point number will result in a bigger loss than just placing the non point number. This is assuming you are placing just those two numbers and if one of these numbers, the 6 or 8 is the point. Don't forget you also have a Don't pass of $5.

Also if the 6 or 8 is hit these numbers will come down after the one hit and not left up for the rest of the roll. For ex., the point is 6 and you place the 6 and 8 and the 6 is rolled. Player 2 would win $2(+7-5) and player 1 would lose $5 on the don't. BUT if a 7 is rolled before the 6 or 8 is rolled player 2 loses $7(-12+5) and player 1 loses just $1(-6+5).

So I guess my question is, is it better to place the point , say the 6 in this example with a place bet along with the 8 or just the other number only, which would only be the 8. Hope this makes sense to anyone out there. is there an answer based only on the math. Thanks.

Don't concern yourself with all this complicated betting. Each bet has an EV. The combinations do not matter unless you are hedging. In the case of hedging, they still do not effect EV or HE, but lower variance instead, which you don't want.

Yes, the more bets you place, the more you expect to lose. So placing only the 8 is better.

This is further reinforced by the maxim to not hedge, which is what placing the 6 does. Don't place the 6 to hedge your DP, that's just silly.

(I can do the math, but we don't need to in order to answer the question)

It is early
I see 3 players at the craps table

math for the expected value once a Dpass is on the 6

Player1 and
Player2 and
Sally

Sally:
$5 Dpass on 6 only. no other bets as a hedge
ev win = 6/11 * $5
ev loss = 5/11 * -$5
2.727272727
-2.272727273
add them up
ev: $0.454545455
or 5/11

Who can beat Sally with an average net win of $0.454545455

Is it Player1?
Player1 (call Player1 dog)
$5 Dpass on 6
$6 Place6
ev win = 5/11 * $2 = 0.909090909
ev loss = 6/11 * -$1 = -0.545454545
ev: $0.363636364
4/11

Player1 can not beat Sally on average comes up short by 1/11
Notice Sally and Player1 are only interested in the numbers 6 and 7

Player2 is interested in the numbers 6,7,8 (call Player2 dogwoman)
Player2
$5 Dpass on 6
$6 Place6
$6 Place8

ev 7 rolls = 6/16 * -$7 = -2.625
ev 6 rolls = 5/16 * $2 = 0.625
ev 8 rolls = 5/16 * $7 = 2.1875
add these up = $0.1875
(player1 beats player2 on average)


Sally wins on average

to answer your question
not every KISS begins with Kay

If you want get more action down after the come out my suggestion would be (in order of preference):

1) Make an odds bet
2) Make a come or don't come bet
3) Make a place/buy/lay bet on a point you do not currently have action on
4) Make a place/buy/lay bet on a point you do currently have action on as a hedge

Quote: endermike

If you want get more action down after the come out my suggestion would be (in order of preference):

1) Make an odds bet
2) Make a come or don't come bet
3) Make a place/buy/lay bet on a point you do not currently have action on
4) Make a place/buy/lay bet on a point you do currently have action on as a hedge

Care to explain why you like hedging at all?

odiousgambit, I don't.

It was mentioned in the OP and I thought my response would be taken more seriously by the OPer if I included to make it clear I had read the question.

Edit: Spelling, my arch-nemesis, we meet again. You may have beaten me this time...

What I am testing is when the 6 or 8 hits is to take the win(take the 6 and 8 down) and then lay single odds on the don't pass bet. In that way the laying of odds is mostly from profit made on the place bets. To make it simple a $5 DP bet and the point is a 5. Place the 6 and 8 for $6 each and if win take the $7 profit and lay single odds against the point. You would have to lay $9 to make $6 so you would have to put in $2 more of your money if the point is 5 or 9.

Basically what I wanted to find out if the point is a 6 or an 8 would I want to place just one of the numbers(point is 6 place the 8 only) or place both the 6 AND 8. I would be hedging on the 6 if rolled but in the long run would it be worth it. Mustang in his answer says NO based on the math. I understand that because I would be placing another bet with a casino edge.

Mustang, just noticed on you math Player 1 would be placing only the 8 , the non point number. So she would have a $5 don't on the 6 and a $6 place bet on only the 8. That would result in a -5 if the 6 rolls and +7 if the 8 rolls. Player 2 's numbers are correct. How would the math change for player 1, thanks.

Quote: dogman

Mustang, just noticed on you math Player 1 would be placing only the 8 , the non point number. So she would have a $5 don't on the 6 and a $6 place bet on only the 8. That would result in a -5 if the 6 rolls and +7 if the 8 rolls. Player 2 's numbers are correct. How would the math change for player 1, thanks.

dog, Sally is telling you what SHE would bet as opposed to your system. Her math on player 1 would not change.

Beach, in MS post the point is 6 and the place bet should be on the 8 and not the 6, so the math would change, no?

dogman,

Sally can speak for herself, but she doesn't use 3 words when 1 will do. I read what she's telling you is, she did not accept your premise on betting your system. She's saying she would do what she wrote about herself, in contrast to what you were asking, which she positioned as players 2 and 3, to demonstrate the difference. I could be wrong.

Quote: dogman

So based on just the math I am trying to figure out if placing the point number will result in a bigger loss than just placing the non point number.

So I guess my question is, is it better to place the point

as others have pointed out without any math, the more bets you make with a house edge on them the greater the expected loss will be the more times this is made unless you know when the bet will win more times than not and only bet it then.

My player1 is different from
your player1 having a $5 Dpass on 6 and a $6 place8. so no hedge there

your player1 is concerned with the rolls of a 6,7 or 8 just like player2 is.
5 ways to roll an 8 and win $7 for a $35 ev
5 ways to roll a 6 and lose $5 for a -$25 ev
6 ways to roll a 7 and lose $1 for a -$6 ev
add them up and get 4 /16 or $.25 (see where the 16 came from?)
that is higher than 3/16 for player2

you do not seem to be concerned, from your writings, what happens to your Dpass when the come out rolls are 7,11 and then a 6 point.
only looking to fund the lay odds with a place bet winning.
If trying to win more money that way (or losing less) someone also mentioned to have more money bet on the odds bet before making another house edge bet.

I guess most do not have an understanding or even a clue to what the house edge does to your bankroll the more you play.
on a win you are shorted on the payoff, every time, on a loss, they take too much, every time according to the probabilities of winning and losing.
so again, I see that as leading us back to "winning more or losing less" is to simply increase your winning percentages for each and every bet you make.

but how will your system of play compare when making the same average bet to just bet on the Dpass with odds and forget the rest of the place bets?
now you need to consider every possible outcome in the math and that will be time consuming.
this is where a simulation or a transition matrix (if possible) comes in handy.

but I also suppose if simulations do show just a Dpass with odds out performs your system,
you might shrug it off, as most do, that it is just a computer simulation and not real dice craps rolls, so a very useless test.
because anything can really happen at a craps table in just a few hundred dice rolls.
30 7s in a row
16 Yos in row
17 9s and 5s in a row
my personal best is 12 Fields in a row

but we still need to consider the element of fun in our betting systems.
if most feel they are having fun (or more fun) playing a certain way, the math is meaningless
that is +ev in the long run
Sally

Sally, excellent post. I played craps for 30 years and then switched to poker for the last 10. I realized as everyone does, you can't turn a positive into a negative. When I was playing craps I just about tried every system in the book. Some would never get ahead and others would only to lose it back plus more. I know that the best bet is Pass line with Max odds or DP with max odds gives one the best way to play. If you get ahead then quit but also if get down early the best thing to do is quit also and take the loss. Less decisions the better.

The only reason I brought this way of betting was to fund the DP odds with winnings from the 6 or 8 being rolled. But of course instead of placing the 6 and 8 which has a casino edge the best thing to do after a point is rolled is to lay the odds which has no casino edge. Thank you for doing the math.