JoeyE
Joined: Feb 20, 2018
• Posts: 1
February 20th, 2018 at 9:48:44 AM permalink
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
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4473
February 20th, 2018 at 10:01:49 AM permalink
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.
beachbumbabs
Joined: May 21, 2013
• Posts: 14230
February 20th, 2018 at 10:11:21 AM permalink
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

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.
If the House lost every hand, they wouldn't deal the game.
SOOPOO
Joined: Aug 8, 2010
• Posts: 7219
February 20th, 2018 at 10:16:19 AM permalink
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.....
odiousgambit
Joined: Nov 9, 2009
• Posts: 8409
February 20th, 2018 at 10:35:11 AM permalink
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.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
Romes
Joined: Jul 22, 2014
• Posts: 5490
February 20th, 2018 at 12:44:16 PM permalink
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!
Playing it correctly means you've already won.
gamerfreak
Joined: Dec 28, 2014
• Posts: 3157
February 20th, 2018 at 12:54:51 PM permalink
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.
discflicker
Joined: Jan 1, 2011
• Posts: 453
February 20th, 2018 at 4:28:20 PM permalink
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?
The difference between zero and the smallest possible number? It doesn't matter; once you cross that edge, it might as well be the difference between zero and 1. The difference between infinity and reality? They are mutually exclusive.
odiousgambit
Joined: Nov 9, 2009
• Posts: 8409
February 20th, 2018 at 4:34:21 PM permalink
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.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
February 20th, 2018 at 6:52:07 PM permalink
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
pass 0 odds 1 million rolls

pass 5x odds same 1 million rolls as pass with 0 odds

the same actual \$\$\$ lost was not the same
and simulations of 1 billion rolls shows the same
so would 1 trillion rolls

the Law of Large Numbers is right there
Last edited by: mustangsally on Feb 20, 2018
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