Pass Line Odds .... does it change the house edge?

According to the wizardofodds website, the house edge for a Pass Line wager is 1.41%. The house edge for taking the odds behind that pass line wager, after a point is made is (obviously) 0%. If using 5x odds behind the pass line wager, the house edge is said to be 0.326%.....and that is where I get confused.

If there were a trillion simulations of craps calculated (assuming unlimited money, as this total amount would be negative), and person A bet $100 pass line only, while person B bet $100 pass line and backed with $500 odds, would these two people not have the exact same amount of money (in theory) after those trillion simulations? And if this is true, then how is the house edge changing with or without odds? I would appreciate any help or explanation.

and, if i may add a slight "part 2" to this question. If, in fact, there is no difference in that simulation example as mentioned above, would I be correct to say player A (betting just $100 pass line) would have less "swings" each time he was at the table (meaning, he would win less but lose less on average), than Person B who is betting $100 pass line and backing $500 odds?

Thank you to all who can help!
Joey

Yes, they would both end up with the same amount of money - but Person B bet much more than Person A in those trillion come-outs (which is what I am assuming you meant). Since 2/3 of the come-out rolls establishes a point, in those trillion come-outs, A bets a total of $100 trillion, while B bets a total of $433.333 trillion.

House edge = money expected to be lost / money expected to be bet.

Quote: JoeyE

According to the wizardofodds website, the house edge for a Pass Line wager is 1.41%. The house edge for taking the odds behind that pass line wager, after a point is made is (obviously) 0%. If using 5x odds behind the pass line wager, the house edge is said to be 0.326%.....and that is where I get confused.

If there were a trillion simulations of craps calculated (assuming unlimited money, as this total amount would be negative), and person A bet $100 pass line only, while person B bet $100 pass line and backed with $500 odds, would these two people not have the exact same amount of money (in theory) after those trillion simulations? And if this is true, then how is the house edge changing with or without odds? I would appreciate any help or explanation.

and, if i may add a slight "part 2" to this question. If, in fact, there is no difference in that simulation example as mentioned above, would I be correct to say player A (betting just $100 pass line) would have less "swings" each time he was at the table (meaning, he would win less but lose less on average), than Person B who is betting $100 pass line and backing $500 odds?

Thank you to all who can help!
Joey

There are a couple of really long threads about this here. If I understand them correctly, in summary:

It's a question of phraseology/terminology, primarily. The house edge does not change for the first hundred, in FACT. However, in EFFECT, the aggegate house edge is diluted proportionately to the even-money odds played.

Since you can't place odds without placing a bet subject to the HE first, most experienced players recommend you place the minimum line bet, backed by the maximum odds the table will allow, that equals the amount of money you want to risk on any one hand.

Part 2, I think you have completely correct. In the long run, they would pay the same amount in vig for the same rolls. However, the 2nd guy is exposing a much larger percentage (assuming they start with the same amount) of his funds at risk, so he's increasing his risk of ruin by playing odds. The first guy is much more likely to be around to recover from a bad run. But of course, he also wins a lot less money during a favorable run.

Welcome to the forum. I welcome any corrections to what I said as necessary.

Edit: I see TDG answered succinctly while I went on and on. Lol. But we seem to agree.

Quote: JoeyE

According to the wizardofodds website, the house edge for a Pass Line wager is 1.41%. The house edge for taking the odds behind that pass line wager, after a point is made is (obviously) 0%. If using 5x odds behind the pass line wager, the house edge is said to be 0.326%.....and that is where I get confused.

If there were a trillion simulations of craps calculated (assuming unlimited money, as this total amount would be negative), and person A bet $100 pass line only, while person B bet $100 pass line and backed with $500 odds, would these two people not have the exact same amount of money (in theory) after those trillion simulations? And if this is true, then how is the house edge changing with or without odds? I would appreciate any help or explanation.

and, if i may add a slight "part 2" to this question. If, in fact, there is no difference in that simulation example as mentioned above, would I be correct to say player A (betting just $100 pass line) would have less "swings" each time he was at the table (meaning, he would win less but lose less on average), than Person B who is betting $100 pass line and backing $500 odds?

Thank you to all who can help!
Joey

Welcome to the forum. The first guy and second guy on average will lose $1.41 per each come out roll. But the second guy will be betting more per come out roll
($100 when he rolls a 2,3,11,12, or 7 and $600 when he rolls a 4,5,6,8,9,or 10) So the second guy is betting $600 2/3 of the time and $100 1/3 of the time, or an average of $433. Both will lose the same $1.41. One way to look at it is the odds bettor loses 32.6 cents for every $100 he risks, while the non odds bettor loses 141 cents for every $100 he risks.

And yes, swings are far greater when you use the odds bet. It increases variance. Which can be good or bad.....

The terminology has caused a lot of arguments, so I'll make it an equation, V = H * A, where V stands for expected value, H stands for house edge, A stands for amount bet.

You can change H by adding free odds to your bet, but not V. So the argument is often made that free odds are a big waste of time.

But notice I said "adding free odds" to a Pass Line bet that is already made. The circumstance can arise where the free odds are valuable, and that is when you bet more on the Pass Line than the minimum bet. Using your example of 5x odds, at a table where the minimum bet is $10 you could instead decide to bet $60 on Pass. Or you could continue to bet $10 and wait until a point is established, then put $50 on the odds. This is taking advantage of the free odds. In theory, A, if it stood for total amount put in action of all your bets, including that put on the free odds, could stay the same.* In that way V, a negative amount for sure, would be smaller.

PS: I certainly see players put the larger full amount on the line and fail to put any on the free odds instead, especially darkside players.

*one of the ways the casino gets us is that we might lower H, but A , if defined as total action, we might let go up, which can defeat the purpose at a certain point.

Indeed the verbiage has caused many of confusion... so just write it out in black and white.

EV = Expected Value
HE = House Edge

TotalEV = SUM(individual EV's from individual bets)

Individual EV's... EV = AmountBet*HE... you can do this for the PL bet, and the odds bet... Examples of the SUMMARY:

EV(Player 1, $100 PL, no odds) = AmountBet*HE (FOR EACH INDIVIDUAL BET)... thus this player has ONE bet...
EV = 100*(-.0141) = -$1.41

EV(Player 2, $100 PL, $500 odds) = AmountBetPL*HEPassLine+ AmountBetOdds*HEOdds...
EV = 100*(-.0141) + 500*(0) = 100*(-.0141) + 0 = -$1.41

Betting odds doesn't change your expected value, at all. The "idea" is that it dilutes your total action, but I don't buy in to that give the House Edge on the odds bets are ZERO, so you're always multiplying your odds amounts (no matter how much) by 0...

The big thing it does change is your VARIANCE! This is what any player playing a losing game wants though, because given enough variance you can come out ahead over "the long run." When you bet more and more on your odds you're upping your variance and standard deviation windows. Examples:

*Disclaimer: I did this once before with actual math, this is to get the idea across and I did not take the time to do the actual math*

Player 1, no odds... over the "long run" they might have -$500,000 in EV +/- $250,000 in Standard Deviations... so the BEST they could ever do is -$250,000.

Player 2, big odds... over the "long run" they might have the same -$500,000 in EV, but they have +/- $750,000 in Standard Deviations due to their odds creating more variance. This means Player 2 could do WAY WORSE, but it also gives Player 2 a realistic chance at being AHEAD as the best they could do is +$250,000.

Variance is variance though... Player 1's "worst case" is -$500,000 over his lifetime. Player 2's "worst case" is -$1,250,000... big difference for a chance to be "up" over your lifetime!

1.41% + 0% = 1.41%

But like everyone saying, it does affect variance quite a bit.

I created a simulator of 1 million rolls for 100 players playing passline/odds at both 1x and 10x.

Players who play single odds are statistically guaranteed to lose over that many rolls.

10x odds changes things quite a bit though, around 20% of those players will come out ahead after 1 million rolls. However, the bankroll swings and potential loss are absolutely enormous.

Quote: gamerfreak

1.41% + 0% = 1.41%

But like everyone saying, it does affect variance quite a bit.

I created a simulator of 1 million rolls for 100 players playing passline/odds at both 1x and 10x.

Players who play single odds are statistically guaranteed to lose over that many rolls.

10x odds changes things quite a bit though, around 20% of those players will come out ahead after 1 million rolls. However, the bankroll swings and potential loss are absolutely enormous.

Hey can I get a copy of your simulator?
How does the number of players effect the simulation unless they're all playing different games concurrently?

Romes, nothing you said is incorrect but you ignore the chance to put part of your total action on the zero percent HE and the potential effect of it.

by 'total action' I mean what is sometimes called 'coin-in.'

I will admit that much of this opportunity is squandered by the temptation to do increased betting, but in theory a player who bets only the pass line could have the same coin-in as the player who bets the odds as well. In these cases the second player has a lower expected loss.

Quote: JoeyE

According to the wizardofodds website, the house edge for a Pass Line wager is 1.41%.
The house edge for taking the odds behind that pass line wager, after a point is made is (obviously) 0%.
If using 5x odds behind the pass line wager, the house edge is said to be 0.326%.....and that is where I get confused.

that value is a 'combined house edge' over a total average bet that most can not calculate.

Quote: JoeyE

If there were a trillion simulations of craps calculated (assuming unlimited money, as this total amount would be negative),
and person A bet $100 pass line only,
while person B bet $100 pass line and backed with $500 odds,
would these two people not have the exact same amount of money (in theory) after those trillion simulations?

they would NOT have the SAME money left.
those that say the would do not understand the Law of Large Numbers.

The Law of Large Numbers says it is the ratio one will be concerned about, not to a number of actual $$$ won or lost.
From Understanding Probability
by Henk Tilms
"It is the case, however, that as the number of tosses increases,
the fractions of heads and tails should be about equal, but that is guaranteed only in the long run.
In the theory of probability, this fact is known as the law of large numbers.
Just
as the name implies, this law only says something about the game after a large
number of tosses.
This law does not imply that the absolute difference between
the numbers of heads and tails should oscillate close to zero. On the contrary."

*****
so total $$$ net / total resolved $$$ wagers
converges to the HE of 1.41% and not to the EV.

The EV is an AVERAGE of many sessions.
say, 1 trillion times 1 trillion, the ev will be close to that value. Actual $$$ won or lost is meaningless as to the house edge value. (now ev per roll or decision will be close to the ev, that is an average)

that value, whatever it is, is a result of variance and standard deviation, not EV (expected value)

yep
shocking!!
Sally

almost forgot
1 million actual dice rolls
2 players at same table (in Wincraps betting $1 with 0 or 5x odds)
note actual loss is far from ev for both players

Quote: discflicker

Hey can I get a copy of your simulator?
How does the number of players effect the simulation unless they're all playing different games concurrently?

I think you can do the same type of simulation with WinCraps

Quote: ThatDonGuy

Yes, they would both end up with the same amount of money

you should rethink or rephrase your reply, imo.

this is the same as saying, imo,
over 1 million fair coin flips
we will be very close to 500,000 heads and tails, their difference getting close to 0.
and making 1 trillion coin flips we should be even closer to 500 million heads and tails, no variance.

that is not even close to the Law of Large Numbers as you should know.

here are 2 players again at the same craps table
pass 0 x odds and pass 5x odds
over 1 million resolved bets they did NOT lose the same amount of $$$
(1 million bets would take about 8 hours a day * 30 bets per hour and over 11 years)

but the averages per roll or decision is about right on.
(have to do the math yourself to see that)

pass 0x odds


pass 5x odds


Sally

Quote: Romes

*Disclaimer: I did this once before with actual math, this is to get the idea across and I did not take the time to do the actual math*

Player 1, no odds... over the "long run" they might have -$500,000 in EV +/- $250,000 in Standard Deviations...
so the BEST they could ever do is -$250,000.

you can not really believe that.

Quote: Romes

Player 2, big odds... over the "long run" they might have the same -$500,000 in EV, but they have +/- $750,000 in Standard Deviations due to their odds creating more variance.
This means Player 2 could do WAY WORSE, but it also gives Player 2 a realistic chance at being AHEAD as the best they could do is +$250,000.

the best they can do??? just 3 standard deviations??
never 4 sd?? or 5 sd??

come on
you now sound just like me when I was in math class years ago
thinking the normal distribution was only 3sd in width.

here is simple example to show 3 sd can not be the best..:

7 billion people flip a fair coin 20 times.
How many heads?

math time:
mean = 10 (n*p or 20*0.50)
variance = n*p*q (20*0.5*0.5) = 5
standard deviation = square root of variance = 2.236
so 3 sd from the mean = 10 + (3*2.236) = 16.708 Heads
"so the BEST they could ever do is" 17 Heads

but we know to expect about 6,700 of the 7 billion to
flip ALL 20 Heads (4.47 sd)
and all of them end up
on Oprah!

next
Sally

Okay, now we look at from a different point of view:

Odds bet: sort of like women's fashions. Used to both entice and conceal.
Lots of 'mystery' No place on the layout for it (the house doesn't encourage it so its bad for them, good for the player.

Cruise ships (that can tell you to walk to the next casino) rarely give good odds. 2x at best.
Known Sweat The Money joints such as SouthPoint only allow 2x.

A trillion rolls? Okay, but that is a bit like them angels dancing on the head of a pin or an honest man or ...well, you get the idea.

In the real world where most craps games are played, its not a Trillion Rolls that count, its anywhere from 5 to 20 rolls that count.

If you win those first five rolls you are "playing with the casino's money" (yeah, yeah, I know .... that is why I put it in quotes). Odds bets on those first five rolls are great for keeping you at the table. Odds on the first five losing bets can dent your bankroll. So the odds bet is just a way to "cap" your bet lawfully. You still gotta win that underlying line bet.

The best upthread comment was the use of the slot term "coin in".

The odds bet will clean out a short stacked player faster and make for more room at the table for the next sheep to be shorn.
The player who is lucky with his first few bets will spend more time at table.

All them math types say its best for our fearless leader to walk in, put his unicycle on the Don't Pass line and then accept the consequences from that first and only roll and walk away from the table (or ride away, depending). Since most players do not trek to Vegas, Baby for just one roll, no one does what is mathematically correct. And no one gets the results of a trillion rolls.

Its the same way with that sweet young thing that edges in next to you. If you take her up to your room, your goose is cooked courtesy of a Roffie and a Strait Edge razor and it only takes a few minutes. If instead you marry her, your goose is cooked over the next umpteen years.

Its Time at Table that counts for the pass line and its the next few rolls that counts for the odds bet, not the next trillion rolls.

Students of Teddys know also of another way casinos put themselves "in a spot"* by offering 10x or more odds...

I won't say "at risk" because it isn't +EV*. On the other hand I don't care to say what it is.

*but that depends on how you are comped too

Quote: odiousgambit

Romes, nothing you said is incorrect but you ignore the chance to put part of your total action on the zero percent HE and the potential effect of it.

The action was accounted for... In the example of $500 odds... I did 500*0... which gets you the expected value of the odds. IF you can find a place that rates your odds in to your average bet, then I agree there's potential for greater comps/return through comp hustling... but odds don't change your expected values, only the variance and standard deviation windows of the game.

Quote: mustangsally

...the best they can do??? just 3 standard deviations??
never 4 sd?? or 5 sd??

Hmmm, when you run your sims, how often (over a very large sampling size) does one come out 4 or 5 SD, or more? I'm willing to wager less than half a percent of the time =)... Probably because 3SD is 99.7%... and again, that's 3SD, not 4SD, 5SD, 6SD, etc. I don't think it's wise for anyone (gambling or AP) to ever plan or hope to be in the 4th, 5th, etc SD's...

Quote: mustangsally

come on
you now sound just like me when I was in math class years ago
thinking the normal distribution was only 3sd in width.

I think it's a very good window to run math/sims off of... I fully understand it's not 100% of the entire result set. Adding more standard deviations to my previous example would only widen the windows. It wouldn't change the outcome one bit. Each player would have the same expected loss, and the player who bets big odds would have much larger variance and SD windows.

So instead of -$500,000 +/- $750,000... you're right, 6SD could be something like -$500,000 +/- $2,000,000... again it doesn't change the reality that both carry the same negative expectation, the player with odds has a "chance" to be positive due to the larger standard deviation windows, and it's at a huge risk because the player with the larger standard deviation windows can absolutely lose all that much more to boot.

Quote: Romes

T when you run your sims, how often (over a very large sampling size) does one come out 4 or 5 SD, or more?

"over a very large sampling size"
is they key there.

"How often" depends on the size of p (p=probability of success, most times)

one can run a sample size
very large that is larger than large
producing a result point at least 4sd higher ,for example.

this easily happens when the value of p is very small

we use terms like very small and very large
in a relative way
other terms too
like house edge and house edge
instead of house edge and combined house edge being two totally different things (yuk)

there lies the problem.

the OP did not know how the house edge went lower,and may still not know,
and it did not actually go lower as you pointed out before.

-7/495 and 0% never changes
Sally

added:
1 billion rolls for the OP
2 players at same table.
still not losing exactly the same $$$$
but looks closer on the HE

Quote: Romes

The action was accounted for... In the example of $500 odds... I did 500*0... which gets you the expected value of the odds.

You are implying that the 'total coin-in position' of mine is incorrect.

I will pose this question to the Wizard and ask him to come down on one side or the other, but only if you agree to a bet. I will bet the Wizard says I am correct, do you want the other side of that bet?

Quote: for wizardly arbitration

My position is that the free odds in Craps offer a chance to lose less money given two players with the same total action or "coin-in". Player #1 will put some of his total action in Pass line betting and some on the free odds. Player #2 will bet only on the pass line. Again, total betting put in action, total coin-in, is to be the same. Player #1 will have an absolute value of EV that is less. Am I correct?

Quote: odiousgambit

Romes, nothing you said is incorrect but you ignore the chance to put part of your total action on the zero percent HE and the potential effect of it.

by 'total action' I mean what is sometimes called 'coin-in.'

others call it 'handle'. Casinos use it, Wincraps uses it to name a few.

right there lies the confusion the OP talked about.
everyone and their grandmas calls the same thing by a different name.
yes, so confused.
From Wincraps:
"Action
A bet which reaches a winning or losing decision is said to have had action.
The amount of money that is utilized when the bet wins or loses is the amount of action.
This amount can also be called the handle or turnover."

Quote: odiousgambit

but in theory a player who bets only the pass line could have the same coin-in as the player who bets the odds as well. In these cases the second player has a lower expected loss.

this is about making a fair comparison. equal or close to equal 'handle'

a $100 pass line bettor making no odds has an average bet of $100
a $100 pass line bettor making 5x odds has an average bet of $433 1/3 (can one prove that?)

to attempt to have the same 'handle' (action) for both players
no odds player makes a $433 flat bet (funny!)

yes, different EV now for about the same 'handle'

Sally

"coin-in" at craps sounds funny too!

Quote: odiousgambit

You are implying that the 'total coin-in position' of mine is incorrect.

I will pose this question to the Wizard and ask him to come down on one side or the other, but only if you agree to a bet. I will bet the Wizard says I am correct, do you want the other side of that bet?

You're mixing and changing the betting to suit your outcome.

Quote: wiz arbitration

My position is that the free odds in Craps offer a chance to lose less money given two players with the same total action or "coin-in". Player #1 will put some of his total action in Pass line betting and some on the free odds. Player #2 will bet only on the pass line. Again, total betting put in action, total coin-in, is to be the same. Player #1 will have an absolute value of EV that is less. Am I correct?

If Player 1 is betting $100 on the PL and Player 2 is betting $10 on the PL and $90 in odds, of course player 2 is going to lose less, because he's exposing less to the house edge. Player 2 would have the EXACT SAME EV if he just bet $10 on the PL and NO odds... so the ODDS aren't what's helping him lose less. The odds are a 0 EV bet, since they carry 0 house or player edge.

EV(Player1) = $100*(-.0141) = -$1.41
EV(Player2) = $10*(-.0141) + $90*(0) = $10*(-.0141) = -$0.14

Are you willing to make a wager that says: Betting THE SAME AMOUNT on the pass line, a player betting odds vs a player with no odds has a reduced house edge?

Quote: mustangsally

...a $100 pass line bettor making no odds has an average bet of $100
a $100 pass line bettor making 5x odds has an average bet of $433 1/3 (can one prove that?)...

I think the reason you said "can one prove that" is because most casinos "correctly" don't rate your odds bets (because they're a 0% HE bet).

Quote: Romes

If Player 1 is betting $100 on the PL and Player 2 is betting $10 on the PL and $90 in odds, of course player 2 is going to lose less,

this is a very poor example, imo.
because over X number of resolved wagers,
the players do not have the same or very close to the same action or handle.

$100 player with 0 odds has an average bet of exactly $100

a $10 with $90 odds has an average bet of
(1/3)*$10 + (2/3)*100 = 10/3 + 200/3 = $210/3 = $70

why (1/3)*10? because the probability of any pass line wager resolving on the comeout roll is 1/3

over 100 bets
the action is different between the players
0odds has $10,000 in action
9x odds has $7,000 in action

equal action means
we compare apples to apples
not apples to rocks.

btw,
$20 flat with 6X odds should produce an average bet of exactly $100

now we can easily see who loses more (or loses less)
and why that is so for the same amount of action!

house edge is house edge not combined house edge
as the OP was questioning
Sally

way upthread I confessed that most of us at least diminish the benefit of free odds used in this way, if not wiping it out altogether. Although it is also true that many who could benefit do not at all, including newbies, but also including players who've played for years who are darksiders.

The principal can apply without the total put in action being exactly the same.

I will say for myself that I have vowed to keep my total action the same when at a 10x vs, say, a 3x4x5x table, and have done so, but it is not instinctive. 'Grinding away' is instinctively what we want as players.