How do odds on the Don't Pass reduce the house advantage?

I don't see how laying odds on Don't Pass reduces the house advantage. I
understand how the entire DP bet ends up with a 1.40% house advantage: the
comeout has a big negative expectation with only 3 ways to win and 8 ways to
lose, and then money that gets past the comeout has a player advantage from
then on.
Here's my problem. The ONLY time I get to lay odds is when I have the
advantage. How does putting more money on the table with zero advantage
increase the player's advantage? I'm already playing with a big advantage
at this point, so just standing pat seems a better strategy than 'diluting'
my player advantage with more money with zero advantage.
Hope I explained that correctly.
I see how Pass/Come works because you're putting the odds out ONLY during
the part of the bet where you are at a disadvantage, but the opposite
scenario makes it seem wrong on Don't Pass.
--
GPC

The house advantage on the Don't Pass/Don't Come/Pass/Come portion of
the bet stays the same, at 1.4/1.41. All 'odds' bets have an
advantage of Zero. IF you add the amount wagered on odds to the
amount of the flat contract bet, and add the amount of money won on
both the flat and odds, then you can re-calculate the 1.4 disadvantage
to a lower level. I believe it was Alan Shank that pointed this out
many times back when this was a worthwhile place to visit. Do some
searches on this topic here, you should be able to find the
information.

Andrew

2012-09-23 14:40:02 UTC

On Fri, 21 Sep 2012 16:12:52 -0700 (PDT), TeddysDad

I don't see how laying odds on Don't Pass reduces the house advantage. I
understand how the entire DP bet ends up with a 1.40% house advantage: the
comeout has a big negative expectation with only 3 ways to win and 8 ways to
lose, and then money that gets past the comeout has a player advantage from
then on.
Here's my problem. The ONLY time I get to lay odds is when I have the
advantage. How does putting more money on the table with zero advantage
increase the player's advantage? I'm already playing with a big advantage
at this point, so just standing pat seems a better strategy than 'diluting'
my player advantage with more money with zero advantage.
Hope I explained that correctly.
I see how Pass/Come works because you're putting the odds out ONLY during
the part of the bet where you are at a disadvantage, but the opposite
scenario makes it seem wrong on Don't Pass.
--
GPC

The issue I always have with wrong-way betting is simply - Does it
really matter?

By which I mean, since most players will always fill the rails of
whatever table has the lowest table minimum, and since the avg.
session isn't likely to exceed 4 to 6 hours, then is the amount won or
lost really going to be all that different by betting wrong-way?

Technically, mathematically, I suppose yes. Realistically, NO.
If your goal is to shave every possible hundreths of a percentile of
House Advantage you'd be better off counting cards at Blackjack.
Does this mean any bet on the Craps table is fair game ? Clearly not.
But I wouldn't sweat 1.37 versus 1.41 - not worth the bother.

Just play your own flavor of a 3-Point Molly, Place a few when you're
ahead, throw some chips at the Hardways when way ahead, fall back
when losing, know your stop-loss, and have fun.

BTW, the best thing you can do for yourself is to take Teddy's Dad's
advice and read the old threads - they are a goldmine of info. And
don't forget to grab a copy of Wincraps to test what you learn.

Have Fun,
Andrew

Sancho Panza

2012-09-23 21:02:34 UTC

Post by WrongWayWade
I don't see how laying odds on Don't Pass reduces the house advantage. I
understand how the entire DP bet ends up with a 1.40% house advantage: the
comeout has a big negative expectation with only 3 ways to win and 8 ways to
lose, and then money that gets past the comeout has a player advantage from
then on.
Here's my problem. The ONLY time I get to lay odds is when I have the
advantage. How does putting more money on the table with zero advantage
increase the player's advantage? I'm already playing with a big advantage
at this point, so just standing pat seems a better strategy than 'diluting'
my player advantage with more money with zero advantage.
Hope I explained that correctly.
I see how Pass/Come works because you're putting the odds out ONLY during
the part of the bet where you are at a disadvantage, but the opposite
scenario makes it seem wrong on Don't Pass.
--
GPC

Among some of the major points that the estimable Mr. Shank made was
about increasing variance without increasing house edge, and he was not
a big fan of fixating on house edge. In essence, much of the message was
that the player had to decide between playing time and profit.
Variance tended to shorten playing time--either winning or losing. The
trick, as described in one thread in January 2008 or 2009 is to get out
when the getting is good.

WrongWayWade

2012-09-24 15:16:07 UTC

Post by WrongWayWade
I don't see how laying odds on Don't Pass reduces the house
advantage. I understand how the entire DP bet ends up with a 1.40%
house advantage: the comeout has a big negative expectation with
only 3 ways to win and 8 ways to lose, and then money that gets past
the comeout has a player advantage from then on.
Here's my problem. The ONLY time I get to lay odds is when I have the
advantage. How does putting more money on the table with zero
advantage increase the player's advantage? I'm already playing with
a big advantage at this point, so just standing pat seems a better
strategy than 'diluting' my player advantage with more money with
zero advantage.
Hope I explained that correctly.
I see how Pass/Come works because you're putting the odds out ONLY
during the part of the bet where you are at a disadvantage, but the
opposite scenario makes it seem wrong on Don't Pass.
--
GPC

But don't you see my argument? If you layed odds EVERY time you played
Don't Pass, I'd agree that you'd reduce the overall house edge on this
combination bet. But the ONLY time you get to lay odds is the instances of
the bet that you actually have a major advantage. So you are never laying
odds during the instances where the house edge worked against you, (losing
the comeout), but always are laying odds when you've got the positive player
advantage, thus 'diluting' your nice advantage by putting more money out
with zero advantage.

And I understand the argument about the expected $$ loss doesn't change by
laying odds. I'm talking about the percentage HA for the total amount bet.

Alan Shank

2012-09-24 22:27:00 UTC

On Mon, 24 Sep 2012 11:16:07 -0400, "WrongWayWade"

Post by WrongWayWade
I don't see how laying odds on Don't Pass reduces the house
advantage. I understand how the entire DP bet ends up with a 1.40%
house advantage: the comeout has a big negative expectation with
only 3 ways to win and 8 ways to lose, and then money that gets past
the comeout has a player advantage from then on.
Here's my problem. The ONLY time I get to lay odds is when I have the
advantage. How does putting more money on the table with zero
advantage increase the player's advantage? I'm already playing with
a big advantage at this point, so just standing pat seems a better
strategy than 'diluting' my player advantage with more money with
zero advantage.
Hope I explained that correctly.
I see how Pass/Come works because you're putting the odds out ONLY
during the part of the bet where you are at a disadvantage, but the
opposite scenario makes it seem wrong on Don't Pass.
--
GPC

But don't you see my argument? If you layed odds EVERY time you played
Don't Pass, I'd agree that you'd reduce the overall house edge on this
combination bet. But the ONLY time you get to lay odds is the instances of
the bet that you actually have a major advantage. So you are never laying
odds during the instances where the house edge worked against you, (losing
the comeout), but always are laying odds when you've got the positive player
advantage, thus 'diluting' your nice advantage by putting more money out
with zero advantage.
And I understand the argument about the expected $$ loss doesn't change by
laying odds. I'm talking about the percentage HA for the total amount bet.

How timely that I decided to check this newsgroup today! I am "the
estimable" Alan Shank. >:-)

Although it may seem that you are "diluting" an advantage you already
have by laying odds on a DP/DC point, this is a misconception.

Look at it this way:
Suppose you have survived the comeout with a $10 DP and have a point
of 4. You now have twice as much chance to win as to lose. The
expected value is +$3.33.

.666667 * 10 = +6.66667
.333333 * -10 = -3.33333
-------
+3.33

If you lay $20 odds, you get:

.66667 * 20 = +13.333
.33333 * -30 = -10.000
--------
+3.33

You haven't changed your expectation at all, which makes sense,
because the expectation on odds bets is always zero. You have,
however, increased your standard deviation, because there is a bigger
difference between the positive and negative outcomes. That is what
taking/laying odds is all about -- adding variance.

Now, suppose, instead of laying that $20 on the odds, you save it and
bet it on Don't Come. Whoops!! Now you have a negative expectation of
about 28 cents on that same money!

The point here is that the more of your money you bet on odds INSTEAD
of on the flat part, the lower your expected loss is. Always, every
time, period.

So, compare:

The (perhaps) late, lamented Mason always used to use the term
"commensurate risk". IOW, compare strategies using the same or very
similar amounts bet (i.e. at risk).

Let's compare two scenarios with very similar amounts risked over 60
comeouts, around two hours' play. One player bets $20 DP and lays no
odds. The other bets $10 DP and lays $12, $15 and $20 on points of
6/8, 5/9 and 4/10, respectively. Here are the average expectations:

$20 DP, no odds $10 DP, odds
ev: -$16.83 -$8.42
SD: $155 $148
p(even) .457 .477

It may surprise you that the standard deviation is lower for the odds
bettor; this is because of the reverse relationship of between the
amount bet and the payoff. In right betting (i.e. taking odds), the SD
comes out higher.

The essential point is that, for the same total amount bet, the
expected loss is only HALF for the odds bettor.

Nice to see some activity here. I will check back within a few days.
Cheers,
Alan Shank

WrongWayWade

2012-09-25 13:37:17 UTC

Post by Alan Shank
On Mon, 24 Sep 2012 11:16:07 -0400, "WrongWayWade"

Post by WrongWayWade
I don't see how laying odds on Don't Pass reduces the house
advantage. I understand how the entire DP bet ends up with a 1.40%
house advantage: the comeout has a big negative expectation with
only 3 ways to win and 8 ways to lose, and then money that gets
past the comeout has a player advantage from then on.
Here's my problem. The ONLY time I get to lay odds is when I have
the advantage. How does putting more money on the table with zero
advantage increase the player's advantage? I'm already playing with
a big advantage at this point, so just standing pat seems a better
strategy than 'diluting' my player advantage with more money with
zero advantage.
Hope I explained that correctly.
I see how Pass/Come works because you're putting the odds out ONLY
during the part of the bet where you are at a disadvantage, but the
opposite scenario makes it seem wrong on Don't Pass.
--
GPC

The house advantage on the Don't Pass/Don't Come/Pass/Come portion
of the bet stays the same, at 1.4/1.41. All 'odds' bets have an
advantage of Zero. IF you add the amount wagered on odds to the
amount of the flat contract bet, and add the amount of money won on
both the flat and odds, then you can re-calculate the 1.4
disadvantage to a lower level. I believe it was Alan Shank that
pointed this out many times back when this was a worthwhile place
to visit. Do some searches on this topic here, you should be able
to find the information.

But don't you see my argument? If you layed odds EVERY time you
played Don't Pass, I'd agree that you'd reduce the overall house
edge on this combination bet. But the ONLY time you get to lay odds
is the instances of the bet that you actually have a major
advantage. So you are never laying odds during the instances where
the house edge worked against you, (losing the comeout), but always
are laying odds when you've got the positive player advantage, thus
'diluting' your nice advantage by putting more money out with zero
advantage.
And I understand the argument about the expected $$ loss doesn't
change by laying odds. I'm talking about the percentage HA for the
total amount bet.

But STOP RIGHT HERE. What's the HA on these two scenarios as a percentage
over the total amount bet? That's my problem. Without odds the HA is more
positive for the player than with odds. 3.33/10 or 3.33/30 (or however you
compute the HA).

Post by Alan Shank
You haven't changed your expectation at all, which makes sense,
because the expectation on odds bets is always zero. You have,
however, increased your standard deviation, because there is a bigger
difference between the positive and negative outcomes. That is what
taking/laying odds is all about -- adding variance.
Now, suppose, instead of laying that $20 on the odds, you save it and
bet it on Don't Come. Whoops!! Now you have a negative expectation of
about 28 cents on that same money!
The point here is that the more of your money you bet on odds INSTEAD
of on the flat part, the lower your expected loss is. Always, every
time, period.
The (perhaps) late, lamented Mason always used to use the term
"commensurate risk". IOW, compare strategies using the same or very
similar amounts bet (i.e. at risk).
Let's compare two scenarios with very similar amounts risked over 60
comeouts, around two hours' play. One player bets $20 DP and lays no
odds. The other bets $10 DP and lays $12, $15 and $20 on points of
$20 DP, no odds $10 DP, odds
ev: -$16.83 -$8.42
SD: $155 $148
p(even) .457 .477

Do this again with $10 DP, no odds vs. $10 DP plus odds. That's what I want
to see compared, because that's the decision I'm interested in. With the
same flat bet, add odds or don't add odds.

Post by Alan Shank
It may surprise you that the standard deviation is lower for the odds
bettor; this is because of the reverse relationship of between the
amount bet and the payoff. In right betting (i.e. taking odds), the SD
comes out higher.
The essential point is that, for the same total amount bet, the
expected loss is only HALF for the odds bettor.
Nice to see some activity here. I will check back within a few days.
Cheers,
Alan Shank

Alan Shank

2012-09-28 21:19:52 UTC

On Tue, 25 Sep 2012 09:37:17 -0400, "WrongWayWade"

Post by Alan Shank
How timely that I decided to check this newsgroup today! I am "the
estimable" Alan Shank. >:-)
Although it may seem that you are "diluting" an advantage you already
have by laying odds on a DP/DC point, this is a misconception.
Suppose you have survived the comeout with a $10 DP and have a point
of 4. You now have twice as much chance to win as to lose. The
expected value is +$3.33.
.666667 * 10 = +6.66667
.333333 * -10 = -3.33333
-------
+3.33
.66667 * 20 = +13.333
.33333 * -30 = -10.000
--------
+3.33

You are hung up on the same issue most of the books about craps are:
the HA on a combination of two different bets.

The expected value (expected loss) and the variance are the two most
important factors. The books tell you that you lower the HA by taking
or laying odds, and, strictly speaking, that's true, but it's
misleading.

When you take or lay odds, you are making a new bet, with an HA of 0%
and an expected value of zero. The expected value of the flat bet DOES
NOT CHANGE! You add variability, because the odds bets are paid off at
uneven money and you have more money out there.

Do this again with $10 DP, no odds vs. $10 DP plus odds. That's what I want
to see compared, because that's the decision I'm interested in. With the
same flat bet, add odds or don't add odds.

You have missed the essential point, I see. >:-)

OK. Have it your way.

$10 DP vs. $10 DP, laying $12, $15 or $20 (this is single odds, based
on how much you win)
$10 DP w/odds
ev: -$8.42 -$8.42
SD: $77.45 $148.20
-2 SD -$163 -$305
+2 SD +$147 +$288
-1 SD -$ 83 -$157
+1 SD +$ 66 +$140

You expected loss remains the same in either case, but the
consequences of being lucky or unlucky are considerably magnified by
laying odds. You win more and you lose more. The ratio of the expected
loss to the standard deviation tells you how lucky you have to be to
overcome the negative expectation and break even (or better). Since
the expected values are the same, the increased variance of laying
odds gives you a slightly better chance of doing that, a probability
of .477, compared to .457.

However, I recommend that you read over the part about SHIFTING money
from the line to the odds; this is the only way to reduce the expected
loss.

Of course, if you are already at the table minimum on your flat bet,
you cannot lower the flat bet and put the excess on the odds.

Most places you can lay considerably more on the odds, increasing your
volatility even more, for bigger wins and losses and a slightly better
chance to break even or better.
Cheers,
Alan Shank

WrongWayWade

2012-10-01 19:09:29 UTC

Post by Alan Shank
On Tue, 25 Sep 2012 09:37:17 -0400, "WrongWayWade"

Post by Alan Shank
How timely that I decided to check this newsgroup today! I am "the
estimable" Alan Shank. >:-)
Although it may seem that you are "diluting" an advantage you
already have by laying odds on a DP/DC point, this is a
misconception.
Suppose you have survived the comeout with a $10 DP and have a point
of 4. You now have twice as much chance to win as to lose. The
expected value is +$3.33.
.666667 * 10 = +6.66667
.333333 * -10 = -3.33333
-------
+3.33
.66667 * 20 = +13.333
.33333 * -30 = -10.000
--------
+3.33

But STOP RIGHT HERE. What's the HA on these two scenarios as a
percentage over the total amount bet? That's my problem. Without
odds the HA is more positive for the player than with odds. 3.33/10
or 3.33/30 (or however you compute the HA).

the HA on a combination of two different bets.
The expected value (expected loss) and the variance are the two most
important factors. The books tell you that you lower the HA by taking
or laying odds, and, strictly speaking, that's true, but it's
misleading.

My assertion is that by LAYING odds I am INCREASING the HA, not lowering it.

Post by Alan Shank
When you take or lay odds, you are making a new bet, with an HA of 0%
and an expected value of zero. The expected value of the flat bet DOES
NOT CHANGE! You add variability, because the odds bets are paid off at
uneven money and you have more money out there.

You've explained numerous times that the expected dollar value of the bet
doesn't change with odds and I understood that the first time, and agree
with that.

The question I still want answered is that the book says when I lay odds the
house advantage (comparing expected value won/lost to the total amount
wagered) goes down. I disagree. From my argument the player advantage
(because I've gotten past the come-out) should be REDUCED when I lay odds.

I may be 'hung up' on this issue, but it still seems to me the books have it
wrong. I'm making my expectation worse (as a percentage of the total wager)
by laying odds, not making it better. Where am I wrong? (Can you please try
to answer this question and not re-explain how the expected $$ lost stays
the same?) I'm not interested in arguments about standard deviation or
increased risk of ruin at this point, either.

Post by Alan Shank
Suppose you have survived the comeout with a $10 DP and have a point
of 4. You now have twice as much chance to win as to lose. The
expected value is +$3.33.
.666667 * 10 = +6.66667
.333333 * -10 = -3.33333
-------
+3.33
.66667 * 20 = +13.333
.33333 * -30 = -10.000
--------
+3.33

First scenario: Player advantage is 3.33 / 10 = +33.3%
Second scenario: Player advantage is 3.33 / 30 = +11.1%

+33.3% > 11.1%. Haven't I just REDUCED my expected return by a 22.2%? I
could have just bet the full $30, ignored the odds and got the +33.3% over
the full $30.

Everything else you've posted is perfectly accurate, but still hasn't
answered my question. I assert that I've made my position worse (in terms
of HA%) by laying odds, not improved it. That's my question. But the
'book' numbers always say the HA% is reduced by laying odds.

I'm sure the 'book' answer just takes then known HA of the DP bet (-1.40%)
and adds the 0% odds and based on the amount of odds comes up with the new
combined figure. Whether we SHOULD care about this number or not is not
what I'm trying to debate. The fact that we ONLY take odds during the
player advantage part of the bet means this methodology of coming up with
the HA is invalid.

<><><><>
Some more info from the Wizard of Odds: (he almost answers the question
then sort of fudges at the end): Note this part: "So, yes, the player edge
as a percentage drops by making the odds bet."

QUESTION: I do not understand why you should lay the odds on the don't pass
or don't come bets. It seems that you have already dodged the 7 and ll
bullet, so the bet is now in your favor. Why would you dilute a bet that is
already heavily in your favor with a large (relative speaking)bet at true
odds? It seems that you are working in the houses favor by reducing the
house edge on the entire bet.
I understand that taking the odds on the pass side reduces the overall house
edge, however I don't understand how laying the odds can reduce the house
edge on the don't side. I'm very curious? By the way, I discussed this with
several casino bosses and dealers yesterday and they all had opinions, but
not reasons for these opinions. Thanks for your time.
- Mike

ANSWER: Let's say you have a $10 don't pass bet and the point is a 4. You
have a 2/3 chance of winning the bet, so the expected value is (2/3)*$10 +
(1/3)*-$10 =$ 10/3 = $3.33. Now consider adding a $40 odds bet on top of it.
Now you have a 2/3 chance of winning $30 and a 1/3 chance of losing $50. The
expected value of both bets combined is (2/3)*$30 + (1/3)*-$50 = $10/3 =
$3.33. So either way your expected gain is 3 dollars and 33 cents. With the
don't pass alone the player edge is $3.33/$10 = 33.33%. With the don't pass
and odds the player edge is $3.33/$50 = 6.67%. So, yes, the player edge as a
percentage drops by making the odds bet. However that player edge is
effective over more money. The way I think gamblers should view the house
edge is as the price to pay for entertainment. If you want to pay as little
as possible then taking or laying the odds is getting entertainment for
free.

The Midnight Skulker

2012-10-04 05:36:26 UTC

Post by WrongWayWade
The question I still want answered is that the book says when I lay odds the
house advantage (comparing expected value won/lost to the total amount
wagered) goes down. I disagree. From my argument the player advantage
(because I've gotten past the come-out) should be REDUCED when I lay odds.
I may be 'hung up' on this issue, but it still seems to me the books have it
wrong. I'm making my expectation worse (as a percentage of the total wager)
by laying odds, not making it better. Where am I wrong?

You're not wrong, but your calculation is meaningless because it ignores the comeout phase of the flat Don't bet, during which the house has an advantage, and jumps directly to the point phase, during which the player has an advantage. There is no bet on the table that offers the player an advantage from bet placement to bet resolution, so once a Don't bet establishes a point, ->any<- additional bet, even another Don't bet, will "dilute" the player's advantage ->on the combination of those bets<-. To give any meaning to the HA you calculate you would have to compare it to the HA of the combination of the established flat Don't bet and the other bet you would make with the money you are not using for odds. So what bet other than odds do you propose to make that will not dilute the player's advantage as much as if that player had laid odds?

1 2
| The Midnight Skulker
9 * 3 aka Van Lewis
aka ***@gmail.com
6

WrongWayWade

2012-10-04 19:26:42 UTC

Post by WrongWayWade
The question I still want answered is that the book says when I lay
odds the house advantage (comparing expected value won/lost to the
total amount wagered) goes down. I disagree. From my argument the
player advantage (because I've gotten past the come-out) should be
REDUCED when I lay odds.
I may be 'hung up' on this issue, but it still seems to me the books
have it wrong. I'm making my expectation worse (as a percentage of
the total wager) by laying odds, not making it better. Where am I
wrong?

You're not wrong, but your calculation is meaningless because it
ignores the comeout phase of the flat Don't bet, during which the
house has an advantage, and jumps directly to the point phase, during
which the player has an advantage. There is no bet on the table that
offers the player an advantage from bet placement to bet resolution,
so once a Don't bet establishes a point, ->any<- additional bet, even
another Don't bet, will "dilute" the player's advantage ->on the
combination of those bets<-. To give any meaning to the HA you
calculate you would have to compare it to the HA of the combination
of the established flat Don't bet and the other bet you would make
with the money you are not using for odds. So what bet other than
odds do you propose to make that will not dilute the player's
advantage as much as if that player had laid odds?

(To your last sentence, ANY further bet other than odds would dilute the
advantage even more, of course. But that's a side issue.)

You agree that when I'm lucky enough to get past the comeout, laying odds
DOES dilute (reduce) the player advantage (in other words it INCREASES the
House Advantage). That is my main point.

If that's a fact, why does the 'book' say that the House Advantage DECREASES
when the player lays the odds on the don't pass? That's the paradox and
what I'd like explained. I believe the 'book' has it all wrong on this
particular detail.

From the help in WinCraps:
Bet True Odds Payoff Odds House Adv%
Don't Pass / Don't Come * 1.0285:1 1:1 1.4
w/ Single odds 0.69
w/ Double odds 0.46
w/ Triple odds 0.34
w/ 5x odds 0.23
w/ 10x odds 0.124
w/ 20x odds 0.065
w/ 100x odds 0.014

I suggest these percentages listed are inaccurate.

Can someone do a breakdown using the perfect 1980 on $10 don't pass with
single odds vs. just $10 don't pass? Can you figure the net HA using that?

The Midnight Skulker

2012-10-04 20:08:59 UTC

... So what bet other than odds do you propose to make that will not
dilute the player's advantage as much as if that player had laid odds?

(To your last sentence, ANY further bet other than odds would dilute the
advantage even more, of course. But that's a side issue.)

Well, not really. If you ever plan to make another bet after your established Don't bet is resolved then it is a real issue because, as you have just admitted, you can't do better HA-wise than to lay odds.

Post by WrongWayWade
You agree that when I'm lucky enough to get past the comeout, laying odds
DOES dilute (reduce) the player advantage (in other words it INCREASES the
House Advantage). That is my main point.

OK, I will stipulate that when you have a bet in action that has an advantage, ->any<- bet you make, including odds, "dilutes" that advantage as a percentage of the combination of the two bets. My main point is, so what?

Post by WrongWayWade
If that's a fact, why does the 'book' say that the House Advantage DECREASES
when the player lays the odds on the don't pass? That's the paradox and
what I'd like explained. I believe the 'book' has it all wrong on this
particular detail.

OK, I see your problem. The percentages in the literature apply to a ->strategy<- of always laying odds after a point is established. Those percentages account for comeout wins and losses when no odds can be laid, and therefore also that odds can be laid only 2/3 of the time. Hence, the line that says the house advantage on DP/DC w/2x odds is .46% means that you are expected to lose .46% of the total amount you bet -- called your "bet handle" BTW -- playing DP/DC and laying 2x odds once a point is established. Your approach jumps into the middle of this situation, assuming the flat bet has survived the comeout, where it was at a severe disadvantage. Hence, the tables in the literature do not apply to the situation ->you<- are analyzing.

WrongWayWade

2012-10-05 12:50:31 UTC

... So what bet other than odds do you propose to make that will
not dilute the player's advantage as much as if that player had
laid odds?

(To your last sentence, ANY further bet other than odds would dilute
the advantage even more, of course. But that's a side issue.)

Well, not really. If you ever plan to make another bet after your
established Don't bet is resolved then it is a real issue because, as
you have just admitted, you can't do better HA-wise than to lay odds.

Post by WrongWayWade
You agree that when I'm lucky enough to get past the comeout, laying
odds DOES dilute (reduce) the player advantage (in other words it
INCREASES the House Advantage). That is my main point.

OK, I will stipulate that when you have a bet in action that has an
advantage, ->any<- bet you make, including odds, "dilutes" that
advantage as a percentage of the combination of the two bets. My
main point is, so what?

Post by WrongWayWade
If that's a fact, why does the 'book' say that the House Advantage
DECREASES when the player lays the odds on the don't pass? That's
the paradox and what I'd like explained. I believe the 'book' has
it all wrong on this particular detail.

OK, I see your problem. The percentages in the literature apply to a
->strategy<- of always laying odds after a point is established.
Those percentages account for comeout wins and losses when no odds
can be laid, and therefore also that odds can be laid only 2/3 of the
time. Hence, the line that says the house advantage on DP/DC w/2x
odds is .46% means that you are expected to lose .46% of the total
amount you bet -- called your "bet handle" BTW -- playing DP/DC and
laying 2x odds once a point is established. Your approach jumps into
the middle of this situation, assuming the flat bet has survived the
comeout, where it was at a severe disadvantage. Hence, the tables in
the literature do not apply to the situation ->you<- are analyzing.

It seems the 'book' is describing exactly the situation I am analyzing, but
it still feels like a paradox. When you just look at it that the 2X odds
will be layed 2/3 of the time and the overall bet has a 1.4% HA, then it's
pretty easy to come up with the 0.46% number. But when you realize you are
reducing your current player advantage EVERY time you lay odds, it seems
strange that the HA would still go down. I guess it must be because the bet
handle goes up even though the raw player advantage percentage is reduced.

Andrew

2012-10-05 14:42:27 UTC

On Fri, 5 Oct 2012 08:50:31 -0400, "WrongWayWade"

... So what bet other than odds do you propose to make that will
not dilute the player's advantage as much as if that player had
laid odds?

(To your last sentence, ANY further bet other than odds would dilute
the advantage even more, of course. But that's a side issue.)

Well, not really. If you ever plan to make another bet after your
established Don't bet is resolved then it is a real issue because, as
you have just admitted, you can't do better HA-wise than to lay odds.

Post by WrongWayWade
You agree that when I'm lucky enough to get past the comeout, laying
odds DOES dilute (reduce) the player advantage (in other words it
INCREASES the House Advantage). That is my main point.

OK, I will stipulate that when you have a bet in action that has an
advantage, ->any<- bet you make, including odds, "dilutes" that
advantage as a percentage of the combination of the two bets. My
main point is, so what?

Post by WrongWayWade
If that's a fact, why does the 'book' say that the House Advantage
DECREASES when the player lays the odds on the don't pass? That's
the paradox and what I'd like explained. I believe the 'book' has
it all wrong on this particular detail.

OK, I see your problem. The percentages in the literature apply to a
->strategy<- of always laying odds after a point is established.
Those percentages account for comeout wins and losses when no odds
can be laid, and therefore also that odds can be laid only 2/3 of the
time. Hence, the line that says the house advantage on DP/DC w/2x
odds is .46% means that you are expected to lose .46% of the total
amount you bet -- called your "bet handle" BTW -- playing DP/DC and
laying 2x odds once a point is established. Your approach jumps into
the middle of this situation, assuming the flat bet has survived the
comeout, where it was at a severe disadvantage. Hence, the tables in
the literature do not apply to the situation ->you<- are analyzing.

I'm not going to give you an argument why you're right or wrong.
Instead, I'm going to suggest you print out the entire thread and
re-read it. It really seems to me Midnight has answered your basic
issue not once but several times. In particular, his last para that
begins with, "OK, I see your problem.", really seems to sum it up.

Like I said, I'm not going to insert myself into this to argue with
you. I'm just trying to offer helpful advice that **any** forum thread
of any kind in any group will tend to become more and more diffuse and
argumentative as times goes on. But the point is the content. That's
all.

Hope this helps,

Andrew

WrongWayWade

2012-10-08 15:40:16 UTC

Post by Andrew
On Fri, 5 Oct 2012 08:50:31 -0400, "WrongWayWade"

... So what bet other than odds do you propose to make that will
not dilute the player's advantage as much as if that player had
laid odds?

(To your last sentence, ANY further bet other than odds would
dilute the advantage even more, of course. But that's a side
issue.)

Well, not really. If you ever plan to make another bet after your
established Don't bet is resolved then it is a real issue because,
as you have just admitted, you can't do better HA-wise than to lay
odds.

Post by WrongWayWade
You agree that when I'm lucky enough to get past the comeout,
laying odds DOES dilute (reduce) the player advantage (in other
words it INCREASES the House Advantage). That is my main point.

OK, I will stipulate that when you have a bet in action that has an
advantage, ->any<- bet you make, including odds, "dilutes" that
advantage as a percentage of the combination of the two bets. My
main point is, so what?

Post by WrongWayWade
If that's a fact, why does the 'book' say that the House Advantage
DECREASES when the player lays the odds on the don't pass? That's
the paradox and what I'd like explained. I believe the 'book' has
it all wrong on this particular detail.

OK, I see your problem. The percentages in the literature apply to
a ->strategy<- of always laying odds after a point is established.
Those percentages account for comeout wins and losses when no odds
can be laid, and therefore also that odds can be laid only 2/3 of
the time. Hence, the line that says the house advantage on DP/DC
w/2x odds is .46% means that you are expected to lose .46% of the
total amount you bet -- called your "bet handle" BTW -- playing
DP/DC and laying 2x odds once a point is established. Your
approach jumps into the middle of this situation, assuming the flat
bet has survived the comeout, where it was at a severe
disadvantage. Hence, the tables in the literature do not apply to
the situation ->you<- are analyzing.

It seems the 'book' is describing exactly the situation I am
analyzing, but it still feels like a paradox. When you just look at
it that the 2X odds will be layed 2/3 of the time and the overall
bet has a 1.4% HA, then it's pretty easy to come up with the 0.46%
number. But when you realize you are reducing your current player
advantage EVERY time you lay odds, it seems strange that the HA
would still go down. I guess it must be because the bet handle goes
up even though the raw player advantage percentage is reduced.

I'm not going to give you an argument why you're right or wrong.
Instead, I'm going to suggest you print out the entire thread and
re-read it. It really seems to me Midnight has answered your basic
issue not once but several times. In particular, his last para that
begins with, "OK, I see your problem.", really seems to sum it up.
Like I said, I'm not going to insert myself into this to argue with
you. I'm just trying to offer helpful advice that **any** forum thread
of any kind in any group will tend to become more and more diffuse and
argumentative as times goes on. But the point is the content. That's
all.
Hope this helps,

I think my last statement summed it up pretty well. When you lay odds, you
DO reduce the player's advantage on that section of the bet, but that
percentage fails to state how the total bet handle has increased on the
overall bet. So... the bet handle goes up as the raw player advantage
percentage is reduced, but the net result BECAUSE of the higher bet handle
is a lower HA (or a less negative player advantage).

If you come at it from the other direction it's quite simple: The expected
dollar loss doesn't change, the bet handle increases, thus the HA must be
reduced.

It still feels a but funny to me how it works :)))
Here's some numbers I cooked up. They demonstrate where I 'see' the
apparent paradox (see the larger player advantage numbers with no odds), but
when I calculated the net edge it does come out the way the book says it
should.

No odds Total Bet Player advantage
Point 4/10 10 33.333%
Point 5/9 10 20.000%
Point 6/8 10 9.091%
Comeout w/l 10 -45.455%
Net -1.403%

Double odds Total Bet Player advantage
Point 4/10 50 6.667%
Point 5/9 40 5.000%
Point 6/8 34 2.674%
Comeout w/l 10 -45.455%
Net -0.459%

Alan Shank

2012-10-10 02:08:35 UTC

On Mon, 8 Oct 2012 11:40:16 -0400, "WrongWayWade"

Post by Andrew
On Fri, 5 Oct 2012 08:50:31 -0400, "WrongWayWade"

It still feels a but funny to me how it works :)))
Here's some numbers I cooked up. They demonstrate where I 'see' the
apparent paradox (see the larger player advantage numbers with no odds), but
when I calculated the net edge it does come out the way the book says it
should.
No odds Total Bet Player advantage
Point 4/10 10 33.333%
Point 5/9 10 20.000%
Point 6/8 10 9.091%
Comeout w/l 10 -45.455%
Net -1.403%
Double odds Total Bet Player advantage
Point 4/10 50 6.667%
Point 5/9 40 5.000%
Point 6/8 34 2.674%
Comeout w/l 10 -45.455%
Net -0.459%

As I wrote before, the book method is correct, but misleading, because
it analyzes two bets (flat and odds) as though they were a single bet.

It is a common problem among craps players to try to analyze the
second stage of a two-stage, contract bet. Over and over and over,
John Patrick and others compare place bets to a come bet by ignoring
the big advantage the come bet has on the comeout roll.

You are doing the same thing with the DP/DC.

You have a point on the DP. Do you lay odds, or put that same money on
a DC bet? That is really the issue. You reduce your expected loss
(which is a lot better than reducing the theoretical HA!) by betting
money on odds, RATHER than on another DC or DP bet.

Odds are to increase variance; that is their purpose.
Cheers,
Alan Shank

Andrew

2012-10-10 21:04:54 UTC

Alan,

Not trying hijack this thread but I would like
to ask a question or two based on your latest
reply to Wrong-Way.

Comparing a 3-Point Molly to a single PL bet of
same amount, are you saying the Expected loss is
lower for the lone PL bet?
Even when combined with a progression on Line Bets ?

If true,could you explain why that is so ?

For example:
1 PL = $10 each with $20 in X2 Odds = 30 total.
Come Bet 1 = $20 with $40 in X2 Odds = 60 total.
Come Bet 2 = $30 with $60 in X2 Odds = 90 total.
Obviously, Odds bets would follow rolling points
so the progession is supposed to hedge the fast
7-out.

versus,...

PL Bet $60 with $120 in X2 Odds = $180.00 total.

[Now that I re-read your words carefully I see you
could be simply saying that the more money you can
apply to Odds bets versus Line Bets the better off
you'll be.]

Please accept my thanks in advance,

Andrew

Andrew

2012-10-11 13:04:27 UTC

On Wed, 10 Oct 2012 14:04:54 -0700 (PDT), Andrew

Post by Andrew
Alan,
Not trying hijack this thread but I would like
to ask a question or two based on your latest
reply to Wrong-Way.
Comparing a 3-Point Molly to a single PL bet of
same amount, are you saying the Expected loss is
lower for the lone PL bet?
Even when combined with a progression on Line Bets ?
If true,could you explain why that is so ?
1 PL = $10 each with $20 in X2 Odds = 30 total.
Come Bet 1 = $20 with $40 in X2 Odds = 60 total.
Come Bet 2 = $30 with $60 in X2 Odds = 90 total.
Obviously, Odds bets would follow rolling points
so the progession is supposed to hedge the fast
7-out.
versus,...
PL Bet $60 with $120 in X2 Odds = $180.00 total.
[Now that I re-read your words carefully I see you
could be simply saying that the more money you can
apply to Odds bets versus Line Bets the better off
you'll be.]
Please accept my thanks in advance,
Andrew

Belated P.S. -
I typed the above in haste from work and thru Goggle Groups which
is now in the process of upgrading. If I were actually playing this
3-point Molly I would be doing a $10/$10/$20 PL& Come bet
sequence, not $10/$20/$30 as I originally posted.

My Bad,

Andrew

The Midnight Skulker

2012-10-11 17:17:56 UTC

Post by Andrew
Comparing a 3-Point Molly to a single PL bet of
same amount, are you saying the Expected loss is
lower for the lone PL bet?

Something of a trick question, this. The house advantage on the flat portions of both Pass and Come bets is the same: 1.41%. Hence, regardless of how you divide your bet handle between those two bets your expected loss is the same: 1.41% of that bet handle. The two strategies do not produce identical expected results, however.

Difference #1: One big Pass bet has a higher variance than three smaller Come bets for the same total amount. Hence, a 3-Point Molly will have a lower risk of ruin and less profit potential than a single Pass strategy. IOW the theoretical distribution is more tightly packed around the mean with the Molly.

Difference #2: Your expected loss is the same for the two strategies only if your bet handle is the same. Does anybody use, or even keep track of, bet handle as a way to limit session length? Table time, loss limits, and win goals: yes; total amount wagered: I think not. Therefore, for sessions of equal length you will have a lower bet handle, and hence an lower expected loss, with a 3-Point Molly than with a single Pass strategy.

1 2
| The Midnight Skulker
9 * 3 aka Van Lewis
aka ***@gmail.com
6

Andrew

2012-10-12 12:11:21 UTC

On Thu, 11 Oct 2012 10:17:56 -0700 (PDT), The Midnight Skulker

Post by Andrew
Comparing a 3-Point Molly to a single PL bet of
same amount, are you saying the Expected loss is
lower for the lone PL bet?

Something of a trick question, this. The house advantage on the flat portions of both Pass and Come bets is the same: 1.41%. Hence, regardless of how you divide your bet handle between those two bets your expected loss is the same: 1.41% of that bet handle. The two strategies do not produce identical expected results, however.
Difference #1: One big Pass bet has a higher variance than three smaller Come bets for the same total amount. Hence, a 3-Point Molly will have a lower risk of ruin and less profit potential than a single Pass strategy. IOW the theoretical distribution is more tightly packed around the mean with the Molly.
Difference #2: Your expected loss is the same for the two strategies only if your bet handle is the same. Does anybody use, or even keep track of, bet handle as a way to limit session length? Table time, loss limits, and win goals: yes; total amount wagered: I think not. Therefore, for sessions of equal length you will have a lower bet handle, and hence an lower expected loss, with a 3-Point Molly than with a single Pass strategy.
1 2
| The Midnight Skulker
9 * 3 aka Van Lewis
6

Thank you for a very thoughtful answer.
I wish there was a math book specific to Craps that taught all this.

-Andrew

Alan Shank

2012-10-15 21:07:02 UTC

Post by Andrew
On Thu, 11 Oct 2012 10:17:56 -0700 (PDT), The Midnight Skulker

Post by Andrew
Comparing a 3-Point Molly to a single PL bet of
same amount, are you saying the Expected loss is
lower for the lone PL bet?

Something of a trick question, this. The house advantage on the flat portions of both Pass and Come bets is the same: 1.41%. Hence, regardless of how you divide your bet handle between those two bets your expected loss is the same: 1.41% of that bet handle. The two strategies do not produce identical expected results, however.
Difference #1: One big Pass bet has a higher variance than three smaller Come bets for the same total amount. Hence, a 3-Point Molly will have a lower risk of ruin and less profit potential than a single Pass strategy. IOW the theoretical distribution is more tightly packed around the mean with the Molly.
Difference #2: Your expected loss is the same for the two strategies only if your bet handle is the same. Does anybody use, or even keep track of, bet handle as a way to limit session length? Table time, loss limits, and win goals: yes; total amount wagered: I think not. Therefore, for sessions of equal length you will have a lower bet handle, and hence an lower expected loss, with a 3-Point Molly than with a single Pass strategy.
1 2
| The Midnight Skulker
9 * 3 aka Van Lewis
6

Thank you for a very thoughtful answer.
I wish there was a math book specific to Craps that taught all this.

I've been working on one for about 15 years!! >:-)

I agree 100% with what the Skulker says.

Instead of a 3-point Molly, if you put money on the odds behind the
pass bet, you do lower your expected loss.

One principle to keep in mind is that you only make a come bet when
there's a point established, then only another when there's a point on
the come AND the pass point is not resolved; that's why your bet
handle (and expected loss) will be substantially lower for the same
number of comeouts when comparing $30 pass to $10, $10, $10 3-point
Molly. However, if you do $10 pass, $20 odds, you reduce your expected
loss even more, because you are exposing much less $$ to the HA.
Cheers,
Alan Shank

Sancho Panza

2012-09-26 14:13:09 UTC

Post by Alan Shank
Let's compare two scenarios with very similar amounts risked over 60
comeouts, around two hours' play. One player bets $20 DP and lays no
odds. The other bets $10 DP and lays $12, $15 and $20 on points of
$20 DP, no odds $10 DP, odds
ev: -$16.83 -$8.42
SD: $155 $148
p(even) .457 .477
It may surprise you that the standard deviation is lower for the odds
bettor; this is because of the reverse relationship of between the
amount bet and the payoff. In right betting (i.e. taking odds), the SD
comes out higher.
The essential point is that, for the same total amount bet, the
expected loss is only HALF for the odds bettor.

I hadn't seen that calculation before. Would, say, tripling the odds
laid to $60, $45 and $36 be just as valid?

Alan Shank

2012-09-28 21:44:09 UTC