Hourly expected loss for craps

Okay kids, first I apologize as I know this question has been asked in its various forms over and over again, but I can't seem to dig out the forum post or "Ask the Wizard" page where it's answered in the way I'm asking.

Given that I am following the strategy of one pass line bet, along with maximum 3 come bets, what is my expected loss per hour, assuming a moderately full table, so 120 rolls or so per hour? Assume a $25 unit. Note that odds don't factor in to the expected loss as they are 0 EV either way.

For my upcoming trip, I'm trying to make sure I don't lose too much EV playing recreationally. I'm actually thinking of just keeping track of every individual bet I make, as I know each $25 bet is about 35 cents (or 34 cents if I'm playing the wrong way). That way when I hit my allotted -EV, I can just stop (assuming I can, lololololol). Not sure how the mechanics of this would work without cell phones allowed at the table, but I could probably track it with dollar chips or something.

The house edge per roll is 0.42%. If you make 120 rolls per hour with $25 on the pass line, that's an expected loss of about $12.60 per hour.

Quote: JB

The house edge per roll is 0.42%. If you make 120 rolls per hour with $25 on the pass line, that's an expected loss of about $12.60 per hour.

Aha! Thank you JB. The key to simplify the whole thing is to use the edge per roll, as you just did. I actually did just stumble on to this way of looking at it myself, after reading just the very top of the Wizard's page on craps.

For taking into account come bets, I'm just going to assume I always have 4 units on the table, even though this will be over-estimating my expected loss since you start with just one unit.

Just bear in mind that you may easily experience 20-30 times EV with the variance you get with typically allowed free odds, which I imagine you max. I have experienced up to 40 times EV.

Quote: odiousgambit

Just bear in mind that you may easily experience 20-30 times EV with the variance you get with typically allowed free odds, which I imagine you max. I have experienced up to 40 times EV.

Yep, totally understand. This is really just a long-term thing for me, I want to start tracking, in some way, my "expected losses." The problem is (okay, a real first-world problem coming) that I'm a lifetime winner at craps, and I don't want to fool myself into thinking I'm "lucky" or some poppycock. So I want to start playing less of it in the future, while increasing the amount of +EV gambling I do, and track the EV of both of them.

Anyway, long-winded answer to your question. Yep, I'm prepared to lose up to 40 times the expected loss. That would suck, but would not (on its own) bust out my bankroll for the weekend :).

Quote: AcesAndEights

Given that I am following the strategy of one pass line bet, along with maximum 3 come bets,
what is my expected loss per hour,
assuming a moderately full table, so 120 rolls or so per hour?
Assume a $25 unit. Note that odds don't factor in to the expected loss as they are 0 EV either way.

You can have a pencil and small notepad with you. I do it all the time.
Now, Not 120 rolls per hour unless everyone is making just one bet on a cold table.

One can calculate the Expected loss and standard deviation from Alan's two posts

https://wizardofvegas.com/forum/questions-and-answers/math/1636-what-is-my-average-bet/#post12528

https://wizardofvegas.com/forum/gambling/tables/1130-how-many-come-bets-is-too-many/#post8852

My sims show the number of resolved come bets per resolved pass line bet for max 3 come bets made is:
per 1000 rolls: 1.85
per 100 rolls: 1.83
Alan for 200 rolls is 1.91
We are close.
I did a few million sessions where he did only 2k.

The EV for max 3 come bets should be the easy calculation.
Do not stop there.

EV by itself can be very misleading and does not lead one to make very good decisions for playing time,
bet size, bankroll requirements to the bet sizes for example.

Hate to see you quit at the EV value and as you are walking away
you keep hearing the word "Winner", "7Winner", "Front line winner", "Coming out, same hot shooter"

The Standard Deviation will tell you how far from the mean (EV)
you can expect to be after a number of rolls/bets made and
your Risk of Ruin for a number of bets or rolls and a starting bankroll.

IMO, If you show your work first, I will think others may show theirs after.
Have fun!

The trouble is that them dealers never take away your expected loss... they take away the whole bet!

At roulette if you put five dollars down on Red, the house's expected win is twenty-six cents but if it comes up Black, the dealer takes five dollars. If it comes up Red, the house pays five dollars but keeps twenty-six cents of what it should have paid you.

At craps an odds bet may have a zero house edge but if you lose, the house still takes all of it and its totally gone from your bankroll.

As near as really matters, the PASS LINE and the DON'T PASS LINE are equally likely to prevail. (Do you REALLY want to split an already fine hair?). Now that makes it pretty much a coin toss but until its resolved you will also be making COME/DON'TCOME bets and you will also be making odds bets. Usually you will get THREE of those down and then that dreaded Seven rears its ugly head.

So it works out to two bets a minute.

Quote: 7craps

You can have a pencil and small notepad with you. I do it all the time.
Now, Not 120 rolls per hour unless everyone is making just one bet on a cold table.

One can calculate the Expected loss and standard deviation from Alan's two posts

https://wizardofvegas.com/forum/questions-and-answers/math/1636-what-is-my-average-bet/#post12528

https://wizardofvegas.com/forum/gambling/tables/1130-how-many-come-bets-is-too-many/#post8852

My sims show the number of resolved come bets per resolved pass line bet for max 3 come bets made is:
per 1000 rolls: 1.85
per 100 rolls: 1.83
Alan for 200 rolls is 1.91
We are close.
I did a few million sessions where he did only 2k.

The EV for max 3 come bets should be the easy calculation.
Do not stop there.

EV by itself can be very misleading and does not lead one to make very good decisions for playing time,
bet size, bankroll requirements to the bet sizes for example.

Hate to see you quit at the EV value and as you are walking away
you keep hearing the word "Winner", "7Winner", "Front line winner", "Coming out, same hot shooter"

The Standard Deviation will tell you how far from the mean (EV)
you can expect to be after a number of rolls/bets made and
your Risk of Ruin for a number of bets or rolls and a starting bankroll.

IMO, If you show your work first, I will think others may show theirs after.
Have fun!

Thanks, those threads were some of the ones I was looking for!

Good post.

I hit Vegas over the weekend. Hit $3k Royal Flush in first 5 minutes playing video poker. Great luck!
I played craps at $3 min and 10x odds. I entered with $500 and played only pass bet with max odd and 1-2 come bets with max odd. I left in 45 minutes down $450. Playing with 10x the house edge is 0.184%, however that didn't matter with the 7's appearing too often.

So I had my best luck ever in VP and worse luck ever in craps close to each other. Oh well still a great weekend and went home a winner :)

Quote: DenDexter

Good post.

I hit Vegas over the weekend. Hit $3k Royal Flush in first 5 minutes playing video poker. Great luck!
I played craps at $3 min and 10x odds. I entered with $500 and played only pass bet with max odd and 1-2 come bets with max odd. I left in 45 minutes down $450. Playing with 10x the house edge is 0.184%, however that didn't matter with the 7's appearing too often.

So I had my best luck ever in VP and worse luck ever in craps close to each other. Oh well still a great weekend and went home a winner :)

Welcome to the forum - sounds like a typical Vegas trip. You win some, you lose some!

Quote: 7craps

You can have a pencil and small notepad with you. I do it all the time.
Now, Not 120 rolls per hour unless everyone is making just one bet on a cold table.

One can calculate the Expected loss and standard deviation from Alan's two posts

https://wizardofvegas.com/forum/questions-and-answers/math/1636-what-is-my-average-bet/#post12528

https://wizardofvegas.com/forum/gambling/tables/1130-how-many-come-bets-is-too-many/#post8852

My sims show the number of resolved come bets per resolved pass line bet for max 3 come bets made is:
per 1000 rolls: 1.85
per 100 rolls: 1.83
Alan for 200 rolls is 1.91
We are close.
I did a few million sessions where he did only 2k.

The EV for max 3 come bets should be the easy calculation.
Do not stop there.

EV by itself can be very misleading and does not lead one to make very good decisions for playing time,
bet size, bankroll requirements to the bet sizes for example.

Hate to see you quit at the EV value and as you are walking away
you keep hearing the word "Winner", "7Winner", "Front line winner", "Coming out, same hot shooter"

The Standard Deviation will tell you how far from the mean (EV)
you can expect to be after a number of rolls/bets made and
your Risk of Ruin for a number of bets or rolls and a starting bankroll.

IMO, If you show your work first, I will think others may show theirs after.
Have fun!

Okay 7craps, taking this data provided by goatcabin in the linked post, if I'm going Pass + maximum 3 come bets, my average units on the table will be 2.91.
I'm going to use this number and go with
$25 * 2.91 * .0042 * 100 = $30.55/hour expected loss.

Anyone want to take a stab at that 2.91 number if you're playing DP/DC? I guess I have a Windows machine at work now, I could go and grab WinCraps...

Quote: DenDexter

Good post.

I hit Vegas over the weekend. Hit $3k Royal Flush in first 5 minutes playing video poker. Great luck!
I played craps at $3 min and 10x odds. I entered with $500 and played only pass bet with max odd and 1-2 come bets with max odd. I left in 45 minutes down $450. Playing with 10x the house edge is 0.184%, however that didn't matter with the 7's appearing too often.

So I had my best luck ever in VP and worse luck ever in craps close to each other. Oh well still a great weekend and went home a winner :)

Oh, my, that is nothing. Many times I would be happy to be down only $450 playing that way. The variance in the odds bet should not be underestimated -- it is brutal, but is also what allows you to win big sometimes!

Congrats on the royal hit. I assume it was three lines of quarter play?

Quote: DenDexter

Good post.

I hit Vegas over the weekend. Hit $3k Royal Flush in first 5 minutes playing video poker. Great luck!

Dealt royal in quarter triple play? I'm trying to figure out how you can have a royal for $3k... And I easily take a -$450 craps session for that Royal. :D

Quote: AcesAndEights

Okay 7craps, taking this data provided by goatcabin in the linked post, if I'm going Pass + maximum 3 come bets, my average units on the table will be 2.91.
I'm going to use this number and go with
$25 * 2.91 * .0042 * 100 = $30.55/hour expected loss.

Nice work!

What is missing are the expected odds you plan on making.

I can show results for NO odds and 345X odds and 2X odds for you.
This is a fast lesson. Good MLB today. Go Tigers!

I will show how to do this per decision.

This will give one more data for the standard deviation calculations.
Since you can easily know how may bets, on average, you will make.

1) 100 rolls / (557/165) = 29.62 pass line decisions (This goes towards the EV)
557/165 = average number of rolls per pass line decision

2) 19.75 pass line odds bet (29.62*(2/3))
2/3 of the time an odds bet can bet made.

3) 29.62 * 1.91 = 56.6 average come bet decisions (This goes towards the EV)
The 1.91 is the average number of come bet decisions per pass line decision
that came from a simulation and is a little higher than what a nasty calculation shows. Close enough for what we do here.

4) 37.7 come odds bet decisions (56.6*(2/3))
86.22 total decisions for the EV calculation

Now step 5)
(Add #1 and #3 results = 86.22)
86.22 * $25 * (-7/495) = 30.48
Simple. (#bets * avg bet * House Edge = EV)

Now the Standard Deviation.
link here for formulas...
https://wizardofvegas.com/forum/gambling/tables/1213-variance-in-craps/
SD tells us how far from that -$30 we can except to end up at, as long as we have enough starting bankroll to survive the 100 rolls.
SD can also tell us how much a starting bankroll to have for our bet sizes to have a certain session Risk of Ruin or a certain net gain to our bankroll.

================================================
We NEED the TOTAL number of decisions.
29.62 + 19.75 + 56.6 + 37.7 = 143.67

1) Pass/Max3Come with NO odds: 1unit SD = 0.999900005 (or 1 rounded)
$Bet ($25) * 1 (1 unitSD) * square root of 86 -# of decisions- (9.27) = $232
A 1SD (68%) range would be -$30 +/- $232 (-$262 to +$202)
That -$30 gets buried in the range for one session.

"The probability that the session outcome will be within one standard deviation is 68.26%.
The probability that the session outcome will be within two standard deviations is 95.46%.
The probability that the session outcome will be within three standard deviations is 99.74%."

A) Add your EV and 2*SDs to get a starting bankroll that will give you about a 5% (1 in 20 bust) session Risk of Ruin.
B)Add your EV and 1*SD to get a starting bankroll that will give you about a 32% (1 in 3.1 bust) session Risk of Ruin.

For longer sessions, one MUST re-calculate EV and SD.
Easy math for Excel to do.

Quote: AcesAndEights

Anyone want to take a stab at that 2.91 number if you're playing DP/DC? I guess I have a Windows machine at work now, I could go and grab WinCraps...

I can say to subtract about .07 (by memory) from the pass/come decisions to get you close.

EV by itself, IF you know the Standard Deviation, can get you to a bankroll,
otherwise figure the SD and be happy.

Good Luck!

Quote: 7craps

...
Yep. The 0% house odds bet sure adds no EV but LOTS of variance/standard deviation.

A) Add your EV and 2*SDs to get a starting bankroll that will give you about a 5% (1 in 20 bust) session Risk of Ruin.
B)Add your EV and 1*SD to get a starting bankroll that will give you about a 32% (1 in 3.1 bust) session Risk of Ruin.

For longer sessions, one MUST re-calculate EV and SD.
Easy math for Excel to do.
...
EV by itself, IF you know the Standard Deviation, can get you to a bankroll,
otherwise figure the SD and be happy.

Good Luck!

Thanks man, although let the record show I didn't ask for all that work :).

I know it sounds shortsighted, but I really don't care about SD for this trip since craps is going to be a very small portion of my gambling action. I sized my trip bankroll mainly for blackjack, and am comfortable with my risk of ruin playing 10ish hours of BJ over the weekend. For craps, I really just wanted to know what my EV would be, so I can keep my expected losses in check over the LONG TERM (e.g. my lifetime of gambling).

I hadn't thought about it, but I should set a stop-loss limit for craps, so I don't lose my whole BJ bankroll before the weekend is out! I will use your EV+2*SD calculation above...only 5% chance I will hit that stop-loss. It wouldn't wipe out my bankroll, but it would hurt. Aye, there's the point of gambling, no?

Quote: AcesAndEights

I hadn't thought about it, but I should set a stop-loss limit for craps, so I don't lose my whole BJ bankroll before the weekend is out!
I will use your EV+2*SD calculation above...only 5% chance I will hit that stop-loss.

I just did all the simple calculations (many steps)
for SD so one post can show how it is done for line bets with X number of come or Dcome bets.

Many others will follow and read.

only 5% chance I will hit that stop-loss...
ONLY 5% chance
means 1 in 20 session bust outs on average

Expect, on average, to bust a EV+2*SD Bankroll one time in 20 sessions.
That 1 in 20 average comes from a distribution of about 36% probability of busting 0 times in 20 sessions
64% probability of busting at least 1 time in 20 sessions
26% probability of busting at least 2 time in 20 sessions (This is 1 in 4)

Good Luck!
Hope to read about your Craps winnings in a news article!

one more item about SD.
One has the same chance of ending up 1SD as being down 1SD
or up XSD as being down XSD.
In Craps, this does not apply to any DI roll sequences.

Quote: AcesAndEights

Note that odds don't factor in to the expected loss as they are 0 EV either way.

I have a question and I hope you will give this some thought before answering:

Do you really exclude your odds money from your "expected loss" when you play?

In all honesty, when I bet full odds I don't look at that money on the table and consider it to be a "push" or a "wash." It's either going to be money won or money lost and in all cases it is money at risk. I would like to think that if I play long enough that the odds bets will all just even-out in the end. But when is the end?

Look guys, I've been playing craps for almost 20 years and I am no where close to be even on odds bets. I probably could have bought a new car for what is considered to be a "no house advantage" bet.

That's what happens when you hitch a star to a wagon instead of the other way around

The odds bet doesn't exist without the negative expectation it is tied to
It only helps to bring the theoretical average loss down
But the amounts are so much higher your losses may seem to have increased.
Let's say you throw in 5000 a night at 1.4% just on pass line is roughly -$70
Or play 3x4x5x at .374% for 25000 pass with odds is roughly -$93.50

Quote: AlanMendelson

Look guys, I've been playing craps for almost 20 years
and I am no where close to be even on odds bets.
I probably could have bought a new car for what is considered to be a "no house advantage" bet.

Look Alan, I have been playing Craps for over 30 years and have a friend that has been playing for over 40 years.
We are both way down on the Place bet and Hardways,
my friend even admits more since he used to chase his losses.

The last 10 years or so, we play now to have most of our money on the odds bets, any odds bets.
We both bet pass and don't.
I still make a few $5 hardways and parlay them.
Most times with my winnings. It is entertainment.

I know for a fact my don't odds are well up after 10 years and my pass odds slightly down.
Not 0% or 0EV.
Because of variance.


The problem here is all gambling experts/writers say the house edge and the expected loss (EV) are set in stone,
the only values any gambler needs to consider, screw the variance and standard deviation,
and after any kind of length of play one can not help to be too far away from that HE% or $EV.

Most have bought into that BS because they do not understand how HE and Standard Deviation work together over N bets.
The math is too hard.
And because it is too hard and too much math,
most gamblers are lemmings and follow the leader and others over the edge of the cliff.
They do not know what really to expect!

Newsflash folks!!
There IS standard deviation for any House Edge value for any number of bets.
Pass line at 4000 bets
HE = 1.41% Everyone agrees here...
Standard Deviation of the HE = 1.58% (1unitSD/#bets^0.5)(0.9999/63.2455532)
1.58% * 3 is now the range around that 1.41%

Learn about EV and SD.
Life's outcomes will become less of a surprise.

Quote: WongBo

That's what happens when you hitch a star to a wagon instead of the other way around
The odds bet doesn't exist without the negative expectation it is tied to
It only helps to bring the theoretical average loss down
But the amounts are so much higher your losses may seem to have increased.

Yes, but only due to variance and not house edge when one compares equal action.

Quote: WongBo

Let's say you throw in 5000 a night at 1.4% just on pass line is roughly -$70
Or play 3x4x5x at .374% for 25000 pass with odds is roughly -$93.50

Nice example.

How about this example.

Let's say you throw in $5,000 a night at 1.4% just on pass line is roughly -$70 EV
Or play 3x4x5x at .374% for $18,717 pass with odds is roughly -$70 EV

Same EV... Really???

Same Entertainment??
NOT!!

Way more money bet, risked.

If the same bankrolls were risked,
the 345X odds player has a way higher probability of ending a session up than the player with no odds.
Most better understand this fact of Craps play.

Too easy just to get stuck on the EV and EV only.

"Hey, I was only supposed to lose freekin' $70 playing the best way.
Line plus Odds.
S**t!
Better to go back to just Place all the numbers, hardways and parlay horn bets.
No more odds bets for me. I never lose just freekin' $70.
I can win more and have way more fun playing this way than losing just freekin' $70"

See you all at the Craps tables
Have a Nice Day

Delete

I was not at all arguing against odds.
I frequently cover my friends odds as well.
Odds bet is the best bet in the house.

$5 flat and $100 odds makes me a very happy camper (ala: MainStreetStation)

Quote: tringlomane

Dealt royal in quarter triple play? I'm trying to figure out how you can have a royal for $3k... And I easily take a -$450 craps session for that Royal. :D

I played 3 line Ultimate X. Hit something that gave me the 3X on a line then hit the RF on that 3X line with the next hand. I also had a 1X line and a 2X line so happy it was the 3X that hit it.

Also won $20 on the Broncos and $120 with BJ. So all good except for craps.

I read that for craps to take 7-10x your bet to weather any storms. So going in with $500 and placing 1 pass line with max odds and 1 come bet with max odds was a total of $66. Should I have gone in with $660 and try to work my way back?

Quote: AlanMendelson

I have a question and I hope you will give this some thought before answering:

Do you really exclude your odds money from your "expected loss" when you play?

Yes.

Quote:

In all honesty, when I bet full odds I don't look at that money on the table and consider it to be a "push" or a "wash." It's either going to be money won or money lost and in all cases it is money at risk. I would like to think that if I play long enough that the odds bets will all just even-out in the end. But when is the end?

The end is when I die, or when I stop gambling, whichever comes first.

Quote:

Look guys, I've been playing craps for almost 20 years and I am no where close to be even on odds bets. I probably could have bought a new car for what is considered to be a "no house advantage" bet.

That sucks, it sounds like variance has deposited you on the negative side of expectation.

I have only been playing craps for roughly 3 years, so clearly I am just a whippersnapper and haven't had my will to gamble crushed yet :). I am a net winner on craps lifetime, and am hoping to ratchet down my play over time such that it stays that way.

But I am taking a purely mathematical view of it. If I play blackjack at an expected hourly win of X for Y hours, and play craps with an expected hourly loss of W for Z hours, then my expected win/loss for the trip will be N. It's an expected value and my real results will be wildly different, but over time, a long time, they will begin to converge. Cheers!

Quote: AcesAndEights

It's an expected value and my real results will be wildly different,

Real actual $$$ results being the Absolute Frequency will be wildly different but NOT the Relative Frequency expressed as a percentage.
This is the difference between Relative versus Absolute Frequency


Most get the Law of Large Numbers wrong on this exact point.

The ratio or average or percentage converges but not the actual (absolute) numbers.
The absolute difference converges to 0
Those, actual (absolute) numbers, actually can increase with the square root of the number of trials.
The absolute differences do not and never can converge to 0.

Show and tell
http://demonstrations.wolfram.com/LawOfLargeNumbersComparingRelativeVersusAbsoluteFrequencyOfC/
http://www.tinafad.com/coin.php
========================
Coin flip example
10 flips and 6 heads
EV = 5
Absolute Difference = 1
% heads = 60%

at 100 flips we have 55 heads.
EV = 50
Absolute Difference = 5 (Increasing, for this example just 1SD from EV)
% heads = 55% % is decreasing, exactly what is is expected to do by the law of Large Numbers

at 1000 flips we have 515 heads.
EV = 500
Absolute Difference = 15 (Increasing but just 1SD from EV) because the standard deviation is increasing and over ime is way more likly to increase with the square root of the number of trials.
% heads = 51.5% is decreasing, exactly what is is expected to do by the law of Large Numbers

Getting closer to 50% HE value,
with more trials as a Relative Frequency but not as the Absolute Frequency.

Quote: AcesAndEights

but over time, a long time, they will begin to converge. Cheers!

Not to actual $$$s. Absolute Frequency
Only as a percentage. Relative Frequency

if one thinks you will lose only actual X$$$s over N+ trials,
many and many will be in for a very big shock!

Your actual $$ losses will be close to the house edge as a %
but variance still controls the actual EV (real $$s) in absolute values.

Don't believe me.
Most do not grasp, and never will, this fact... too confusing for the majority.
Others have pointed this out also.
http://forumserver.twoplustwo.com/25/probability/running-100-buy-ins-below-ev-almost-normal-after-10-000-flips-1220722/
BruceZ
"Your total winnings do not converge to your EV as the sample becomes large.
That's a common misconception.
The difference between your actual winnings and your EV becomes infinitely large, and it actually makes larger and larger fluctuations above and below your EV.
But if you divide that difference by your actual EV, that goes to zero.
If you divide your total winnings by the number of hands to get winnings per hand, that will converge to your EV per hand."

Concept: Relative Vs Absolute Frequency

Good Luck living and dying by EV and EV alone

Quote: DenDexter

I read that for craps to take 7-10x your bet to weather any storms.
So going in with $500 and placing 1 pass line with max odds and 1 come bet with max odds was a total of $66.
Should I have gone in with $660 and try to work my way back?

Your question would make for a good thread on it's own. (I would show my results in your thread and not this one)
Because of the high odds taken.
10X odds variance and higher does not fit easy math formulas.
Simulations are easier to run for accurate answers,
unless you do not care how accurate an answer is.

Quick answer, 7-10X is just an old gambler's tale.
It is a "one fits all" answer that only fits one out of maybe 118.4 players and sessions.

Get good with WinCraps, there are FREE auto-bet files to download from the site
the riskofruin.bet comes to mind.
Prove to yourself how often 7-10X is actually correct for any Risk of Ruin calculation (bankroll busting) and bankroll requirements based from bet sizes.

Have a Nice Day

Quote: 7craps

Your actual $$ losses will be close to the house edge as a %
but variance still controls the actual EV (real $$s) in absolute values.

I'm not going to pretend that I understand everything you just said, but it seems like that point is the most important. Over time, as you play more and more, the absolute difference of your results compared to the expected result won't converge to zero, they will continue to swing wildly above/below expectation. But as you play more and more, if you divide that difference by your number of rolls/samples/whatever, THAT number will approach zero. So the more you play, the closer you will be to expectation taking into account how long you have been playing. Does that sound about right?

Quote: 7craps

Most get the Law of Large Numbers wrong on this exact point.

I was one who was clueless on this, but I think I've got it now. At least the nature of the fallacy, which takes the form of "OK the HE is zero so in the long run I will break even". Wrong.

In the long run, your variance from zero will lose its magnitude by percentage, but the dice or whatever will feel no urge to get back your money, if you have been losing. If you are down thousands of dollars with 10x odds and $10 table, chances are this will change little towards zero in a longer run.

Yes, you could be up, too, looking at it from the other side; just somewhat more unlikely in a negative expectation game.

Quote: AlanMendelson

I have a question and I hope you will give this some thought before answering:

Do you really exclude your odds money from your "expected loss" when you play?

In all honesty, when I bet full odds I don't look at that money on the table and consider it to be a "push" or a "wash." It's either going to be money won or money lost and in all cases it is money at risk. I would like to think that if I play long enough that the odds bets will all just even-out in the end. But when is the end?

Look guys, I've been playing craps for almost 20 years and I am no where close to be even on odds bets. I probably could have bought a new car for what is considered to be a "no house advantage" bet.

You got hit. Someone had to. Think of it this way - if you and a friend are betting on a coin flip, what's the expected value? Zero. But one of you will leave ahead, and one of you will leave behind. Over a long string of coin flips, it may be that one of you is well ahead of the other, and that's essentially what's happened to you. You're not going to even out, either; over sufficient bets, you're about as likely to go up as down (although you'll always be slightly more likely to go down because of the positive skew).

The entire point of odds bets is to put you further from the expected value, since the expected value is negative. Unfortunately, for every action...

Quote: 7Craps

Your actual $$ losses will be close to the house edge as a %
but variance still controls the actual EV (real $$s) in absolute values.

Quote: AcesAndEights

I'm not going to pretend that I understand everything you just said,
but it seems like that point is the most important.

#1) Over time, as you play more and more, the absolute difference of your results compared to the expected result won't converge to zero, they will continue to swing wildly above/below expectation.

Yep.
Everyone needs to understand this point or the Law, and the reason why it happens.
(EV = total $ wagered * (net $ gain / total $ wagered)
or
(EV = total $ wagered * (House edge)
"The difference between your actual winnings and your EV
becomes infinitely large, and it actually makes larger and larger fluctuations above and below your EV."

Most ALL gambling writers and public opinion/belief is the FALSE belief that
HE AND EV both will converge to expectation.
That is a % and actual $s lost.
A total misunderstanding of the Law of Large Numbers.

Now, the EV is an absolute value expressed in $s. -$1500 for example.
Not a percentage.
The Law of Large Numbers does not deal at all with absolute values, only Relative Frequencies, in other words, ratios or percentages.
Again for all that will read this far into the future
House edge is simply a ratio (expressed as a % of $s)
HE = net $ gain / total $ wagered
EV = net $ wagered * ($ gain / total $ wagered)
EV = net $ wagered * HE
EV = actual $s
EV is not a ratio, it is a product, an absolute value.

Quote: AcesAndEights

#2) But as you play more and more, if you divide that difference by your number of rolls/samples/whatever, THAT number will approach zero.

yes, because you are now converting actual $s, EV, into a ratio of actual $s of rolls/samples/whatever.

Now, The percentage converges, not actual $s, the EV, for whatever unit you so desire.

Quote: AcesAndEights

#3)So the more you play, the closer you will be to expectation
taking into account how long you have been playing. Does that sound about right?

Only by comparing percentages to percentages.
the closer you will be to the HE expectation and NOT the EV expectation.
An EV is NOT a percentage.
It is in $units
In the real world, NO ONE EVER converts EV to percentages for common casino games.
It is always stated as EV = X$s

One gets closer to the HE the more one plays and not the EV, when EV is expressed in $s.
Because the HE is the ratio between $net/$action
House edge is an apple
EV is an orange
"The difference between your actual winnings and your EV
becomes infinitely large, and it actually makes larger and larger fluctuations above and below your EV."

This is why many get stuck about actual gambling losses.
They THINK and come to a conclusion, incorrectly, because of the EV value and that value only.
Standard deviation shows how far one can be after many bets or a long time of play,
and that distance has a way higher probability of being FURTHER from expected EV than closer to IT.

But the math for variance and standard deviation is way harder to do for most than
EV = HE * $Bet * # of Bets

But the effort to calculate var/sd, IMO, is worth it at least 1 billion times more than EV,
since the results from both EV and SD now paints a very useful and more complete
picture, in most cases, of what one can expect after any length of play.

================================
It is like players crying at a Craps table
"The shooter just Crapped out. He rolled a 7"

truth is, he roiled a 7 while a point was established.
But the incorrect public and many dealers,
say that shooter "Crapped out" instead of the correct "Sevened Out"
This example does not cause confusion in what actually happened.
The shooter lost his point and a new shooter starts.

When EV is thought of incorrectly,

"the EV was ONLY -$50" and the result was
"But I lost a fu**ing $1000 playing this game with the best bests and strategy"
now that player just says "FU** everyone, I do as a please from now on",
This player believed in the MYTH of EV only,
Myth = the longer one plays the closer the actual EV will get to the expected EV.

and did not understand or did not want to understand
the possible range of wins and losses from variance.

Looks like the Wizard also might believe that HE and EV are the only guiding lights...
both converges to expectation
from the last question on this Ask the Wizard.
https://wizardofodds.com/ask-the-wizard/228/
Of course, he has a few weasel clauses in there too.
I would have answered that Craps Q way differently... and will
Yeah, I do that Craps Q next thread next week while on vacation...

Quote: 7craps

Looks like the Wizard also might believe that HE and EV are the only guiding lights...
both converges to expectation
from the last question on this Ask the Wizard.
https://wizardofodds.com/ask-the-wizard/228/
Of course, he has a few weasel clauses in there too.
I would have answered that Craps Q way differently... and will
Yeah, I do that Craps Q next thread next week while on vacation...

The Wizard worships at the altar of H.E. and E.V. True, that is the most important thing, but I feel variance and standard deviation are somewhat neglected. (He does have a very good page on SD/V in Video Poker and also the SD/V of most table games.) It could be because the math is somewhat more advanced for those elements. A beginner would do well to evaluate a game/bet by E.V./"Element of Risk" alone at first.

The fact of the matter is that none of us are going to play enough craps to see the 98.59% return materialize on all of our pass line bets. (For $1m wagered, that's a loss of $14,100!) Better just to play the free odds and hope you get enough positive variance to stay above water. Its hard for craps players to admit we are playing a losing game.

Quote: 7craps

Looks like the Wizard also might believe that HE and EV are the only guiding lights...
both converges to expectation
from the last question on this Ask the Wizard.
https://wizardofodds.com/ask-the-wizard/228/
Of course, he has a few weasel clauses in there too.
I would have answered that Craps Q way differently... and will
Yeah, I do that Craps Q next thread next week while on vacation...

Hoo-hoo! Calling out the Wizard! He don't like that!

Let's take a look at that ask-the-W. The part about a system being worthless to try to increase, reliably, losses instead of gains would be correct, except I would say often you do lose more *if* the system makes you risk more money, which it would seem to me a normal Martingale does. However, the parameter stated is wagering overall $1 Million exactly, so I would say the Wiz is on solid ground so far.

The last sentence by the questioner:

Quote: Peter K. from Bellevue NE

Is he doomed to only give the house roughly 0.18% or $1800?

To instantly say Yes! as the Wizard did, would seem to validate the amount as well as the percentage, so I would guess Michael did not intend that validation, but intended to sort of ignore that part of the question, not realizing what it would look like. Huge variance was afoot, what with the stipulation that the 10x odds would be maxed. One complication, though, is that the $1 million would include the odds portion, so the player would be making only about 1200 or so line bets I'm guessing.

Running a Wincraps test on that now.

Quote: teddys

The Wizard worships at the altar of H.E. and E.V.

Ouch! I can feel the Wizardly wrath building way over here 2000 miles away!

Quote: teddys

It could be because the math is somewhat more advanced for those elements.

Teddy!

Well, maybe the Wizard will not see this if you get lucky.

Quote: teddys

For $1m wagered, that's a loss of $14,100!

If the player made no other bets than pass line without odds. I'll see what happens with Wincraps on that too.

PS: I am finished with Wincraps, see my blog post

Huh... naïvely, it seems like this guy's best chance at meeting his goal would be to bet $50k total at a time, pressing as much as possible when up, and with any "luck" he should be down roughly the right amount at some point. From there on out, he should just bet the rest of his million in the lowest increments he has the patience for, and he probably won't finish too far up or down. If he can't find a table where he can bet $50k at a time at 10x odds, though, he's screwed.

Thank you 7craps for your responses on this thread - I found them very enlightening. I have a picture in my head of a graph that explains this concept much more concisely than a bunch of text, but I'm too tired right now to actually draw it and post it. Anyway, I think I understand exactly what you mean now, but I still think you are overplaying the distinction - just my opinion, no need to get upset and repeat everything over again :).