Theoretical Win and Pass/Don't Pass Combined Edge
I was reading through some of the Wizard's answers to questions and have read some of the other threads about how taking full odds against your pass/don't pass will not change the amount of money a player will lose, but it mentions the reduction of the house edge.
Taken from the website:
"Hi, if person A makes 1000 consecutive bets on the pass line without backing up his bet, and person B makes 1000 consecutive bets on the pass line and he takes 100X odds whenever possible, doesn’t each person lose the same amount of money?
BLAKE HAAS FROM THOUSAND OAKS
Yes. I can just imagine the follow up question to be why I recommend taking the odds if doing so doesn’t help to win more. What I suggest is betting less on the pass so that your need for action is mostly met by a full odds bet. For example if you are comfortable betting about $90 per bet, and the casino allows 5x odds, then I would drop the pass line bet to $15 and bet $75 on the odds. That will lower the overall house edge from 1.414% to 0.326%."
This is my confusion. As mentioned in other threads, the free odds has a 0% house edge and therefore the only money affected as far as house edge goes is the pass line (for example) at 1.41%. However, the Wizard stated that the house edge was lowered from 1.41% to .32%.
Does this mean that when we calculate the Theoretical Win (defined: Handle x House Edge (or Time Played x Avg bet x Decisions per Hour) x H/E) for the example from the quote, that the casino uses the adjusted .32% or the original 1.41%?
The obvious implication would be an expected loss of revenue on the craps table due to the lowering of the house edge. And subsequently, does offering greater odds actually decrease the amount of revenue the casino can make from the pass line bet?
Quote: AsheThanks in advance for those that take the time to help with this.
I was reading through some of the Wizard's answers to questions and have read some of the other threads about how taking full odds against your pass/don't pass will not change the amount of money a player will lose, but it mentions the reduction of the house edge.
Taken from the website:
"Hi, if person A makes 1000 consecutive bets on the pass line without backing up his bet, and person B makes 1000 consecutive bets on the pass line and he takes 100X odds whenever possible, doesn’t each person lose the same amount of money?
BLAKE HAAS FROM THOUSAND OAKS
Yes. I can just imagine the follow up question to be why I recommend taking the odds if doing so doesn’t help to win more. What I suggest is betting less on the pass so that your need for action is mostly met by a full odds bet. For example if you are comfortable betting about $90 per bet, and the casino allows 5x odds, then I would drop the pass line bet to $15 and bet $75 on the odds. That will lower the overall house edge from 1.414% to 0.326%."
This is my confusion. As mentioned in other threads, the free odds has a 0% house edge and therefore the only money affected as far as house edge goes is the pass line (for example) at 1.41%. However, the Wizard stated that the house edge was lowered from 1.41% to .32%.
Does this mean that when we calculate the Theoretical Win (defined: Handle x House Edge (or Time Played x Avg bet x Decisions per Hour) x H/E) for the example from the quote, that the casino uses the adjusted .32% or the original 1.41%?
The obvious implication would be an expected loss of revenue on the craps table due to the lowering of the house edge. And subsequently, does offering greater odds actually decrease the amount of revenue the casino can make from the pass line bet?
Essentially, it's like taking 500ml of eggnog that has 500 calories, adding 500ml of water to it, and saying now it has fewer calories because it used to be 1 calorie per 1 ml but now it's 0.5 calories per 1 ml.
The house edge on the pass bet is 1.41%. Doesn't matter if you have 2x odds, 10x odds, or 100x odds. That pass line bet still has a 1.41% HE. The odds bet, similarly, is always going to be 0%.*
If you bet an incorrect amount on odds, for example, an amount that isn't divisible by 5 when betting on the 6 or 8, or an odd amount when betting on 5 or 9, then there's going to be a slight house edge, because you're going to get shorted by however many cents when it wins because they round down.
IME, when rating players, a casino is going to use a "blended theo", which means they don't track which bets you make, just how much you're betting. They figure for every $1,000 on the table, some is going to be on pass line, some place bets, and other money on center action, and they may conclude the average or blended theoretical is (just pulling out a random number) 4%. So they'll just assign everything you bet to have a 4% theoretical loss (assuming 4% is the number they come up with).
The exception, of course, is the odds bet. That differs. Some casinos will rate your odds and some won't. Even at the same casino, some floormen will rate your odds and others won't. If at both casinos you always bet $10 pass with $50 odds, one casino might rate you as having $10 average bet while the other may rate you as having a $60 average bet.
Quote:
Essentially, it's like taking 500ml of eggnog that has 500 calories, adding 500ml of water to it, and saying now it has fewer calories because it used to be 1 calorie per 1 ml but now it's 0.5 calories per 1 ml.
The house edge on the pass bet is 1.41%. Doesn't matter if you have 2x odds, 10x odds, or 100x odds. That pass line bet still has a 1.41% HE. The odds bet, similarly, is always going to be 0%.*
This is where I was leaning as well. However, if the HE truly doesn't change, why does he even mention it or state that it gets lower? Doesn't that just create confusion?
Quote:IME, when rating players, a casino is going to use a "blended theo", which means they don't track which bets you make, just how much you're betting. They figure for every $1,000 on the table, some is going to be on pass line, some place bets, and other money on center action, and they may conclude the average or blended theoretical is (just pulling out a random number) 4%. So they'll just assign everything you bet to have a 4% theoretical loss (assuming 4% is the number they come up with).
The exception, of course, is the odds bet. That differs. Some casinos will rate your odds and some won't. Even at the same casino, some floormen will rate your odds and others won't. If at both casinos you always bet $10 pass with $50 odds, one casino might rate you as having $10 average bet while the other may rate you as having a $60 average bet.
This is a great example for how casinos lose money in promotions to players. If the odds bets are not generating revenue (due to having no HE) but are still being applied to the player's average bet against the "blended theo", then the player is being rated as more worth than they actually are. The correct course of action would be to not include odds bets in average bet calculations in player ratings.
Quote: AsheThis is where I was leaning as well. However, if the HE truly doesn't change, why does he even mention it or state that it gets lower? Doesn't that just create confusion?
Yes it causes confusion. But I guess a reason to mention it is, if say, someone wants to bet $100 on the passline....they'd be better off to bet $20 on the pass with $80 in odds, for instance. So you've gone from 1.41% of $100 which is $1.41 to 1.41% of $20 which is $0.282....but in both cases, you get the same amount of action (roughly speaking, since you don't have that $100 in action on every roll, only once a point is established), but the latter has a lower expected loss than the former.
Quote: AsheThis is a great example for how casinos lose money in promotions to players. If the odds bets are not generating revenue (due to having no HE) but are still being applied to the player's average bet against the "blended theo", then the player is being rated as more worth than they actually are. The correct course of action would be to not include odds bets in average bet calculations in player ratings.
Yes, the player would be rated as having a higher rated theo than he really should. I don't really have an answer to that, or a better way to do it, other than having a floor person note how much each player has on each different type of bet.....which isn't feasible. As far as why they may rate odds, I really don't know. The only thing I can possibly think of is because people like being rated for a higher amount (and, thus, would be unhappy if they got rated on "only" their pass-line bet). Casinos don't always make the best decisions and sometimes it may be along the line of, "Yeah...it's not the right way to do it, but it's a whole hell of a lot easier, and it's all going to average out anyway. Some may get a benefit while others get a bit shorted, but whatever".
Quote: RS
Yes, the player would be rated as having a higher rated theo than he really should. I don't really have an answer to that, or a better way to do it, other than having a floor person note how much each player has on each different type of bet.....which isn't feasible. As far as why they may rate odds, I really don't know. The only thing I can possibly think of is because people like being rated for a higher amount (and, thus, would be unhappy if they got rated on "only" their pass-line bet). Casinos don't always make the best decisions and sometimes it may be along the line of, 's not the right way to do it, but it's a whole hell of a lot easier, and it's all going to average out anyway. Some may get a benefit while others get a bit shorted, but whatever".
We do it because it would cause a skewed representation of the player's actual worth to the casino. Most promotions utilize the theo in some way in their decision making process when giving back points because they are after all literally giving back a percentage portion of their revenue they just earned as a thank you. Most casinos may opt to reinvest 25% of the revenue earned from the player back into that player in various forms, whether it be in points, comps, mailers, etc.
Due to high variation of bets and different HE's on the craps table, the casino should have a rating system that encompasses slow, medium, and fast pace along with low, med, high action... a player playing the props would be more profitable than a player just playing the pass line and game pace affects decisions per hour. However, as you stated, most casinos find it too difficult for their boxpersons to rate players and watch the game and usually just assign a single blended theo. This means a low profit player gets more in promos than they deserve and the high profit players get less in promos than they deserve. It's a flaw that TG management should fix if they have it.
Quote: RS
Yes it causes confusion. But I guess a reason to mention it is, if say, someone wants to bet $100 on the ey'd be better off to bet $20 on the pass with $80 in odds, for instance. So you've gone from 1.41% of $100 which is $1.41 to 1.41% of $20 which is $0.282....but in both cases, you get the same amount of action (roughly speaking, since you don't have that $100 in action on every roll, only once a point is established), but the latter has a lower expected loss than the former.
What you've described is how much theo the casino is making off the pass line. Basically betting the entire amount or just $20 of it because literally whatever you place behind doesn't matter. But it has to right? Otherwise, Michael never would have specifically stated it lowered the house edge. So then, why is he stating the house edge is lowered from 1.41% to .32%? No one on the forums has yet to answer or explain well why he stated it is lowered to that. They just keep saying that the house edge stays 1.41%. Is anyone able to show the math that makes it fall to .32% like Michael stated?
for craps, you want to add "/basics" to the url that is connected to that link seen there now to get:
https://wizardofodds.com/games/craps/basics/
for how to figure the HE on line bets combined with odds,
Quote: lower linkCombined Pass and Buying Odds
The player edge on the combined pass and buying odds is the average player gain divided by the average player bet. The gain on the pass line is always -7/495 and the gain on the odds is always 0. The expected bet depends on what multiple of odds you are allowed. Lets assume full double odds, or that the pass line bet is $2, the odds bet on a 4, 5, 9, and 10 is $4, and the odds on a 6 or 8 is $5.
The average gain is -2×(7/495) = -14/495.
The average bet is 2 + (3/36)×4 + (4/36)×4 + (5/36)×5 + (5/36)×5 + (4/36)×4 + (3/36)×4] =
2 + 106/36 = 178/36
The player edge is (-14/495)/(178/36) = -0.572%.
The general formula if you can take x times odds on the 6 and 8, y times on the 5 and 9, and z times on the 4 and 10 is (-7 / 495) / [ 1 + ((5x + 4y + 3z) / 18) ]
see last para and see if you can figure it out for the one you are referring to
https://wizardofodds.com/games/craps/appendix/1/
the Wizard answer of 'yes' is so wrong.Quote: AsheI was reading through some of the Wizard's answers to questions and have read some of the other threads about how taking full odds against your pass/don't pass will not change the amount of money a player will lose, but it mentions the reduction of the house edge.
It is NOT how much actual money they lose, but the ratio of the net/total resolved wagers
will be very close to each other with more and more resolved wagers.
The Law of Large numbers at wikipedia
" the law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed."
the Wizard also stating the house edge is lowered means it is a 'combined house edge' over the average bet between the flat bet and the odds bet.
I simulated this years ago and I know someone did even a larger sim than mine
I had 2 players at the SAME craps table, where
one makes a $5 pass line bet 1000 times, no odds
the other makes a $5 pass line bet 1000 times, and 100 times odds on every point established
(it should be very clear their average resolved wager COMBINED is way different)
this was then done 10,000 times and the data collected.
How many times did they both end up at the same amount of money after the 1000 wagers were completed?
-155
145
-55
-145
-115
5
-215
-155
-145
25
-45
-45
-65
now we added all the sessions together for each player
who lost way more actual money than the other?
or did they lose the same amount of money?
(over 1000 wagers done 10,000 times, should be easy to guess - 10 million bets for each player)
odds player lost: $998,620
not EVEN the same
this shows clearly the Wizard answer of 'yes' in that Ask the Wizard is wrong.
my code is correct as I used the WinCraps risk of ruin code that is free at the website
(I also tested it to see if it was 100% accurate, and it is)
odds adds variance without adding to the -ev (that is good for the player)
we would need to show some math to explain this
but that is a total waste of time.
most all craps players believe what they want to believe
no more no less
Quote: 7crapsthe Wizard answer of 'yes' is so wrong.Quote: AsheI was reading through some of the Wizard's answers to questions and have read some of the other threads about how taking full odds against your pass/don't pass will not change the amount of money a player will lose, but it mentions the reduction of the house edge.
It is NOT how much actual money they lose, but the ratio of the net/total resolved wagers
will be very close to each other with more and more resolved wagers.
The Law of Large numbers at wikipedia
" the law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed."
the Wizard also stating the house edge is lowered means it is a 'combined house edge' over the average bet between the flat bet and the odds bet.
I simulated this years ago and I know someone did even a larger sim than mine
I had 2 players at the SAME craps table, where
one makes a $5 pass line bet 1000 times, no odds
the other makes a $5 pass line bet 1000 times, and 100 times odds on every point established
(it should be very clear their average resolved wager COMBINED is way different)
this was then done 10,000 times and the data collected.
How many times did they both end up at the same amount of money after the 1000 wagers were completed?only 13 times out of 10,000 sessions played-155
145
-55
-145
-115
5
-215
-155
-145
25
-45
-45
-65
now we added all the sessions together for each player
who lost way more actual money than the other?
or did they lose the same amount of money?
(over 1000 wagers done 10,000 times, should be easy to guess - 10 million bets for each player)no odds player lost : $695,770
odds player lost: $998,620
not EVEN the same
this shows clearly the Wizard answer of 'yes' in that Ask the Wizard is wrong.
my code is correct as I used the WinCraps risk of ruin code that is free at the website
(I also tested it to see if it was 100% accurate, and it is)
odds adds variance without adding to the -ev (that is good for the player)
we would need to show some math to explain this
but that is a total waste of time.
most all craps players believe what they want to believe
no more no less
It’s completely unsurprising that the player taking odds and the player not taking odds would hardly ever finish with the same win/loss while playing a session at the tables together. It’s surprising that the person taking odds lost so much more in the aggregate, but I suspect it’s not statistically significant. Am I missing something?
I’m also not sure that you and the Wiz disagree. If you play more than the table min at a pass line, you are better off jamming the extra above table min on the odds rather than on an increased pass line bet. You should only increase the PL from the table min if you are planning to max odds, otherwise you are losing more EV on an amount wagered basis.
Quote: 7crapsI had 2 players at the SAME craps table, where
one makes a $5 pass line bet 1000 times, no odds
the other makes a $5 pass line bet 1000 times, and 100 times odds on every point established
This is a wrong experiment. If one makes a $5 Pass bet 1,000 times without taking odds, the other should make a $1 Pass bet 1,000 times with 6 times odds (approximately). Can you try simulating it again with these parameters? The one who takes odds should lose less money.
Quote: unJonIt’s surprising that the person taking odds lost so much more in the aggregate, but I suspect it’s not statistically significant. Am I missing something?
It's unsurprising when the person who takes odds bet so much more.
Quote: BlackjackLoverThis is a wrong experiment. If one makes a $5 Pass bet 1,000 times without taking odds, the other should make a $1 Pass bet 1,000 times with 6 times odds (approximately). Can you try simulating it again with these parameters? The one who takes odds should lose less money.
It's unsurprising when the person who takes odds bet so much more.
It’s very surprising. The “more” that was bet was odds at zero EV. That’s the whole point of odds. If you think taking odds means losing more, then I’m not sure where to begin.
Quote: unJonIt’s very surprising. The “more” that was bet was odds at zero EV. That’s the whole point of odds. If you think taking odds means losing more, then I’m not sure where to begin.
I don't think that at all. It's about variances. If you think that betting $1,000,000 per round 1,000 times and betting 1$ per round 1,000 times on a game with a 100% expected return will have the same results, then you ignore the existence of variances.
Even if a game has no house edge, it doesn't mean that if you play it 1,000 times, you will win 500 times and lose 500 times. If you win 498 times and lose 502 times, you will lose $4,000,000 with a $1,000,000 bet per round and lose $4 with a $1 bet per round. If you bet more, you will lose or win more. It's simple. If the results will be exactly the same, what's the point of betting more?
Quote: BlackjackLoverI don't think that at all. It's about variances. If you think that betting $1,000,000 per round 1,000 times and betting 1$ per round 1,000 times on a game with a 100% expected return will have the same results, then you ignore the existence of variances.
Even if a game has no house edge, it doesn't mean that if you play it 1,000 times, you will win 500 times and lose 500 times. If you win 498 times and lose 502 times, you will lose $2,000,000 with $1,000,000 bet per round and lose $2 with $1 bet per round.
Now you are agreeing with me. My initial point was that the difference wouldn’t be statistically significant. I get variance. I’m not sure you get that the EV is the same for someone betting $1 per round and someone betting $1 per round with $100 odds. Over a large number of trials, we expect both rollers will lose the same amount of $ on average. The player with odds will have wildly higher and lower results (variance) but the average is the same.
Quote: unJonI’m not sure you get that the EV is the same for someone betting $1 per round and someone betting $1 per round with $100 odds. Over a large number of trials, we expect both rollers will lose the same amount of $ on average.
No, the amount of money loss will be different as I explained. If you bet more, you will lose more. The loss percentage will be less if you take odds, but the amount of money you are expected to lose will be much more if you bet much more.
answer from the WizardQuote: AsheThanks in advance for those that take the time to help with this.
I was reading through some of the Wizard's answers to questions and have read some of the other threads about how taking full odds against your pass/don't pass will not change the amount of money a player will lose, but it mentions the reduction of the house edge.
Taken from the website:
"Hi, if person A makes 1000 consecutive bets on the pass line without backing up his bet, and person B makes 1000 consecutive bets on the pass line and he takes 100X odds whenever possible,
doesn’t each person lose the same amount of money?
BLAKE HAAS FROM THOUSAND OAKS
Yes.
He goes on the say something else, but the YES answer is what so many players nowadays refer to.
doesn’t each person lose the same amount of money?
same amount of money?
Wizard answer is a simple YES
I disagree with that 100% and simulations for the exact question he answered and calculations prove both the Wizard and I can not be 100% correct.
I go with my gathered data
both players (call them Player 1 and Player 2)
WILL NOT lose the same amount of money over 1000 resolved pass line wagers.
I call this out every time some new person brings up this point.(well the Q&A)
again, ask 100 players standing at a Craps table the same question.
almost all say the same thing.
each person will lose the same amount of money
the experiment I did was the exact one asked in an"Ask The Wizard" column.Quote: BlackjackLoverThis is a wrong experiment.
the Wizard answered that question in the OP with a simple
YES
"I was reading through some of the Wizard's answers to questions and have read some of the other threads about how taking full odds against your pass/don't pass will not change the amount of money a player will lose, but it mentions the reduction of the house edge.
Taken from the website:
"Hi, if person A makes 1000 consecutive bets on the pass line without backing up his bet, and person B makes 1000 consecutive bets on the pass line and he takes 100X odds whenever possible,
doesn’t each person lose the same amount of money?
BLAKE HAAS FROM THOUSAND OAKS"
https://wizardofodds.com/ask-the-wizard/136/
added: I simulated the odds player at a different table than the no odds player
just because some might say this is not fair...
and looked at each session for an equal result
only 2 out of the 10,000 sessions were equal
doesn’t each person lose the same amount of money?
Wizard
YES
*****
the odds player over these 10 million resolved wagers lost
$680,060
LESS than the no odds player in the 1st simulation
$695,770
Quote: BlackjackLoverNo, the amount of money loss will be different as I explained. If you bet more, you will lose more. The loss percentage will be less if you take odds, but the amount of money you are expected to lose will be much more if you bet much more.
That’s just mathematically incorrect. Here try this one. My friend and I play craps together. He doesn’t take odds but just bets the PL. I just take his max odds 100x every chance I get. I make no other bets. What’s my EV?
Quote: unJonThat’s just mathematically incorrect. Here try this one. My friend and I play craps together. He doesn’t take odds but just bets the PL. I just take his max odds 100x every chance I get. I make no other bets. What’s my EV?
$0. Over millions of rolls I should expect to break even while my friend has negative EV so should lose money. Odds keep your EV flat. Someone betting no odds vs full odds with the same PL has the same $EV. So over millions of rolls we should expect them to lose the same amount of money. Even though the odds players bankroll will fluctuate much more wildly.
It's not about EVs. You ignore variances again. No, you and your friend won't lose the same amount of money. You can try simulating it. If you flip a coin one million times, you won't get 500,000 heads and 500,000 tails. If you get heads 499,500 times and tails 500,500 times, the person who bet more amount of money on heads will lose more amount of money.
correct. the Ask the Wizard question says NOTHING about ev.Quote: BlackjackLoverIt's not about EVs.
https://wizardofodds.com/ask-the-wizard/136/
"Hi, if person A makes 1000 consecutive bets on the pass line without backing up his bet, and person B makes 1000 consecutive bets on the pass line and he takes 100X odds whenever possible, doesn’t each person lose the same amount of money?
BLAKE HAAS FROM THOUSAND OAKS"
Answer
Yes.
doesn’t each person lose the same amount of money?
simulations shows YES is not the correct answer
BLAKE HAAS FROM THOUSAND OAKS
should have been told this and only the expectation would be the same (on paper)
actual results (of any one session) is a function of expectation and variance
and both players
same -ev
way different variance
no way
each person lose the same amount of money?
added again
the expectation for the no odds player to break even or be ahead over 1k resolved pass line wagers is about 1 in 3 (32.7%)
for the 100x odds player, about 1 in 2 (49.8%)<<< I would have to check this one
(using expected value and standard deviation)
sure this can be calculated very easily using a Markov chain
break even or be ahead over 1k resolved pass line wagers
.339 for no odds player (1 in 3)
.497 for 100x odds player (1 in 2)
this only is true when every bet can be made (risk of ruin = 0)
I have a related question. If I flip a coin 1 million times, what is the probability that I will get heads exactly 500,000 times?
I’m not ignoring variance. I said “Even though the odds players bankroll will fluctuate much more wildly.”Quote: BlackjackLoverQuote: unJonThat’s just mathematically incorrect. Here try this one. My friend and I play craps together. He doesn’t take odds but just bets the PL. I just take his max odds 100x every chance I get. I make no other bets. What’s my EV?
$0. Over millions of rolls I should expect to break even while my friend has negative EV so should lose money. Odds keep your EV flat. Someone betting no odds vs full odds with the same PL has the same $EV. So over millions of rolls we should expect them to lose the same amount of money. Even though the odds players bankroll will fluctuate much more wildly.
It's not about EVs. You ignore variances again. No, you and your friend won't lose the same amount of money. You can try simulating it. If you flip a coin one million times, you won't get 500,000 heads and 500,000 tails. If you get heads 499,500 times and tails 500,500 times, the person who bet more amount of money on heads will lose more amount of money.
And it’s unarguable that both players have same EV so over the long term expect to do the same. In other words, what’s the chance that after 1,000,000 rolls that the player betting odds is losing less than the person betting no odds?
And the bottom line that you are ignoring is that odds are better than simply variance if taking odds allows you to lower your PL bet while keeping you at the same total at risk that scratches your gambling risk. The real comparison is the person betting $25 PL and no odds vs $15 PL and $10/12 odds.
Quote: unJonAnd it’s unarguable that both players have same EV so over the long term expect to do the same. In other words, what’s the chance that after 1,000,000 rolls that the player betting odds is losing less than the person betting no odds?
If both players have the same negative EV, the player who bets more amount of money will lose more amount of money. If you argue that taking odds has no house edge so it doesn't count, we'll go back to square one: variances. You should try simulating coin flipping. The results may surprise you.
Quote: unJonAnd the bottom line that you are ignoring is that odds are better than simply variance if taking odds allows you to lower your PL bet while keeping you at the same total at risk that scratches your gambling risk. The real comparison is the person betting $25 PL and no odds vs $15 PL and $10/12 odds.
I don't ignore this, but this is not what Wizard said.
the actual number of Heads is really meaninglessQuote: BlackjackLoverI have a related question. If I flip a coin 1 million times, what is the probability that I will get heads exactly 500,000 times?
what we are after is what is the ratio of Heads to all flips?
is it approaches 50%, but not exactly 50%
using pari/gp calculator
(14:07) gp > x=binomial(1000000.,500000)*(1/2)^500000*(1/2)^500000;
(14:08) gp > print(x)
0.00079788436133175008908724559691299459776
(14:08) gp > print(1/x)
1253.3144506440737461007789574188421238
answer your question
about 1 in 1,254
a one person experiment, doubt it will happen
1 million persons trying this
we have lots of winners! (but as a percentage, not that high)
that might be true but the OP contained a question from the OP (original post) askerQuote: unJonThe real comparison is the person betting $25 PL and no odds vs $15 PL and $10/12 odds.
showing the Ask the Wizard question
the Wizard answer before any other explanation was
YES
it should have been NO
with an explanation why it is NO
just wanted to point out why so many craps players these days say ODDs bet do not matter at all
they come back to that question and the simple answer
YES
No is the correct answer
https://wizardofodds.com/ask-the-wizard/136/
"Hi, if person A makes 1000 consecutive bets on the pass line without backing up his bet, and person B makes 1000 consecutive bets on the pass line and he takes 100X odds whenever possible, doesn’t each person lose the same amount of money?
BLAKE HAAS FROM THOUSAND OAKS"
Quote: BlackjackLoverIf both players have the same negative EV, the player who bets more amount of money will lose more amount of money. If you argue that taking odds has no house edge so it doesn't count, we'll go back to square one: variances. You should try simulating coin flipping. The results may surprise you.
Quote: unJonAnd the bottom line that you are ignoring is that odds are better than simply variance if taking odds allows you to lower your PL bet while keeping you at the same total at risk that scratches your gambling risk. The real comparison is the person betting $25 PL and no odds vs $15 PL and $10/12 odds.
I don't ignore this, but this is not what Wizard said.
Bolded for the part that crystallizes our disagreement. When I say same negative EV it means EV denominated in $ not in % of bet. So if two people have the same negative EV then they will be expected to lose the same amount irrespective of how much they are betting. Because the size of the bet was already factored in to get to the EV.
Do you agree or disagree with this statement: someone betting $10 PL with full 100x odds will be expected to lose the same amount as someone betting $10 PL and no odds over the same number of resolved bets.
are you trying to say in so many wordsQuote: unJonDo you agree or disagree with this statement:
someone betting $10 PL with full 100x odds
will be expected to lose the same amount
as someone betting $10 PL and no odds over the same number of resolved bets.
that the mathematical expected value would be the same for each player?
-7/495 * $bet * number of bets
https://wizardofodds.com/ask-the-wizard/136/
example OP
1000 resolved pass line wagers
or are you trying to say that after 1000 resolved pass line wagers
BOTH players will lose the same amount of real money (in this case US $s)
simulations show both players rarely lose the SAME AMOUNT OF MONEY
here are those 1 million session sims I knew was already completed
The Wizard still answered the OP question as
YES
data shows something different actually occurs.
Player 1 no odds
Simulation of Craps Pass Line Wagers
Odds Multiplier . . . . = 0
Session Bankroll . . . = 1000.00
Win goal to quit session= 1000.00
Max. Decisions to quit = 1000
No. Sessions simulated = 1000000
Starting Random seed . = 54321
------------------------------------
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1000.00
Avg. No. games won . . = 492.94
Avg. No. games lost . . = 507.06
Avg. No. dice rolls . . = 3375.71
Avg. Total amount bet . = 1000.00
Bankroll was busted . . = 0.000% of the time ( 0)
Win goal was met . . . = 0.000% of the time ( 0)
Bankroll decreased . . = 66.067% of the time
Bankroll increased . . = 31.637% of the time
Avg (mean) end bankroll = 985.89 (change of -14.11)
Std-dev ending bankroll = 31.61
Average expectation EVI = -1.411%
Player 2 100x odds every chance a point was established
Simulation of Craps Pass Line Wagers
Odds Multiplier . . . . = 100
Session Bankroll . . . = 101000.00
Max. Decisions to quit = 1000
No. Sessions simulated = 1000000
Starting Random seed . = 12345
------------------------------------
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1000.00
Avg. No. games won . . = 492.95
Avg. No. games lost . . = 507.05
Avg. No. dice rolls . . = 3375.81
Avg. Total amount bet . = 1000.00
Avg. amount bet on Odds = 66668.73
Bankroll was busted . . = 0.000% of the time ( 0)
Win goal was met . . . = 0.000% of the time ( 0)
Bankroll decreased . . = 50.186% of the time
Bankroll increased . . = 49.791% of the time
Avg (mean) end bankroll = 100990.23 (change of -9.77)
Std-dev ending bankroll = 3185.31
Avg Odds change in b/r = 4.32
Average expectation EVI = -0.977%
Average expectation EVR = -0.014%
this is now extremely boring
time for a new flavor of the month
yes, very boring.Quote: Dalex64Extremely boring?
just "they will lose the same amount"
"No, they will not lose the same amount"
ugly 2 sided coin
I am wrong most times coding and doing math. I have to check my work a lot and correct my errors mostly I am always in a hurry.Quote: Dalex64Does that mean you see where you have gone wrong?
The data I presented is from simulations that I and others have done
proof is in the data
both players over 1000 bets will not lose the same amount of money
the last 1 million sim covers 1 BILLION resolved wagers for each player
did they both lose the same?
looks like they did not
share and share alike
up to the reader to find out why
both players did not lose the same amount of money after 1 million sessions of 1,000 wagers
when the expected values are exactly the same.
added: a clue
how about standard deviation (wtf is that?)
for 1 billion bets at $1 flat bet (call this a unit bet)
for 0 odds player
ev = -14,141,414.14
sd = 31,619.61
6 sigma
-14,046,555.30
to
-14,236,272.98
nice range
100x odds player
ev = -14,141,414.14 <<< same as 0 odds player
sd = 3,187,378.80 <<< ok
6 sigma
-4,579,277.73
to
-23,703,550.55
The fact of the matter is, you don't know if they're going to win or lose the same amount of money. I can go play 1,000 hands of FPDW (0.76% edge) and no one can say whether I will or will not lose money.
But if asked if I will make money playing that FPDW game, no one can say 'yes' or 'no' if I'll win in a 1,000 hand session. However, my net average for every 1,000 hands of FPDW will be a positive (long run expectation).
So I say the question shouldn't be interpreted in a 'prediction' fashion, as if anyone can say "Yes, the players will lose the same amount" nor "No, they will lose different amounts", but it should be read as what is the EV or long term expectation of each players, in which case no one can deny the answer is "Yes, both players will lose the same amount."
I'd also say it's kind of stupid to interpret a question as if the asker is asking Wizard whether or not both players are going to lose the same amount. I mean, that's kind of asinine, isn't it? AFAIK, Wizard can't read the future. It'd almost be as if I flipped a coin, asked you what the chance is it landed on heads, you (generic you) saying it's 50%, then me saying nahhh it's 100% because it landed on heads.
Edit: As an added bonus -- if the two players do not lose the same amount, which player will lose more?
Quote: RSI'd also say it's kind of stupid to interpret a question as if the asker is asking Wizard whether or not both players are going to lose the same amount. I mean, that's kind of asinine, isn't it? AFAIK, Wizard can't read the future. It'd almost be as if I flipped a coin, asked you what the chance is it landed on heads, you (generic you) saying it's 50%, then me saying nahhh it's 100% because it landed on heads.
No, it's not stupid if that is what is asked. However, I don't think that the person who asked the question meant to ask if the results would be exactly the same. I believe that he meant to ask if the results would be approximately the same. Also, no, the person who asked this question didn't ask for a long term (one million or one billion bets) expectation. His question was very specific. He asked for the results of 1,000 bets. It's unlikely that the results will be roughly the same as I mentioned that according to the standard normal table, the values less than 0.1 standard deviation away from the mean account for about 7.97% of the set.
Quote: RSIf the question is "will both players lose the same amount of money?" and you argue the answer should be "no", then you're saying both players will NOT lose the same amount of money. Of course, it's possible both players lose the same amount of money. Likewise, it's possible both players do not lose the same amount of money, meaning a "yes" answer is also incorrect.
In my opinion, this is too pedantic. When this kind of question is asked, it's assumed that it's qualified by "every time" or "most of the times." It's like asking if you can win a negative expectation game in the long run with a betting system. Do you think what the correct answer is? If you're pedantic, the correct answer is yes. It's possible that you will win a negative expectation game in the long run if you're very lucky. When people answer no, they don't mean that it's absolute impossible, but it's very unlikely.
Anyway, technically, you're right. Wizard's "yes" answer is incorrect, and a "no" answer is also incorrect. The correct answer is "Yes and no, but it's much more likely that they won't."
from here we goQuote: BlackjackLoverAnyway, technically, you're right. Wizard's "yes" answer is incorrect, and a "no" answer is also incorrect. The correct answer is "Yes and no, but it's much more likely that they won't."
https://wizardofvegas.com/forum/gambling/craps/8329-craps-pass-no-odds-or-pass-100x-odds-same-loss-revisited/#post123111
the Ask the Wizard asker replied in another thread and says this, that is actually not right, not correct. a total misunderstanding what variance is all about
"if $5 was bet on the pass line every time,
you will lose the same amount (not percentage) of money on average whether you back the bet up or not.
Backing the bet up has no house edge and no player edge,
so it won't affect the amount of money won or lost. That's all the wizard was saying and that's all I was asking about."
because the Wizard answered YES to that Ask the Wizard question
so many get it wrong (not correct) and will never believe any simulation or calculation results... NEVER
14 different simulations can NOT be all wrong from 4 different programs
you will lose the same amount (not percentage) of money on average whether you back the bet up or not.
LOL
goes to show
people believe what they want to
when they want and do not care what others believe
because they know they are right and others are not
quite the story
Quote: 7crapsfrom here we go
https://wizardofvegas.com/forum/gambling/craps/8329-craps-pass-no-odds-or-pass-100x-odds-same-loss-revisited/#post123111
the Ask the Wizard asker replied in another thread and says this, that is actually not right, not correct. a total misunderstanding what variance is all about
"if $5 was bet on the pass line every time,
you will lose the same amount (not percentage) of money on average whether you back the bet up or not.
Backing the bet up has no house edge and no player edge,
so it won't affect the amount of money won or lost. That's all the wizard was saying and that's all I was asking about."
because the Wizard answered YES to that Ask the Wizard question
so many get it wrong (not correct) and will never believe any simulation or calculation results... NEVER
14 different simulations can NOT be all wrong from 4 different programs
you will lose the same amount (not percentage) of money on average whether you back the bet up or not.
LOL
goes to show
people believe what they want to
when they want and do not care what others believe
because they know they are right and others are not
quite the story
I’m not following you.
My friend (call him Tom) bets pass line bets with no odds. I stand next to him and take full odds on his PL bet. Neither one of us make any other bets.
My other friend (call him Pete) plays at the same table and also only does PL with no odds. No one takes his odds (I’m tapped out of bankroll comfort maxing our Tom’s odds).
At the end of a simulated large number of rolls:
1) Do you think my results are positive or negative (on average)? Remember I’m only making a bet at fair odds.
2) Does Tom or Pete do better (on average)? Or is it necessarily the same?
3) Do the combined bets of me/Tom do better/worse/same than Pete (on average)?
I was gonna say something similar.Quote: Dalex64Try using the same Starting Random seed for both cases.
When running the simulations, both players must be betting on the same dice results.
1st set of sims was exactly thisQuote: DJTeddyBearWhen running the simulations, both players must be betting on the same dice results.
"I had 2 players at the SAME craps table, where
one makes a $5 pass line bet 1000 times, no odds
the other makes a $5 pass line bet 1000 times, and 100 times odds on every point established"
https://wizardofvegas.com/forum/gambling/craps/31955-theoretical-win-and-pass-dont-pass-combined-edge/#post685725
same dice rolls, same craps table
13 out of the 10,000 simulations
they ended up the same net win/loss
over 10 million resolved wagers, they did not lose the same or even have the same average loss
why?
think about it
0 odds vs. 100x odds
variance
The 1 million sim of 1k wagers I did again and used the same dice rolls
same exact results (have not posted those yet)
***** ***** (10 stars)
does everyone think that 1000 pass line wagers is a very very very large number?
only 1000 times that of 1
some stats:
(1 unit bet = $1)
pass line wager 0 odds
1000 resolved wagers
ev: -14.14141414
standard deviation: 31.6196
(one can do the math on this)
3 standard deviations: 94.8588
6 sigma
-109.00021414
80.71738586
pass line wager with 100x odds
ev: -14.14141414
standard deviation: 3,187.38
3 standard deviations: 9562.14
6 sigma
-9576.28141414
9547.99858586
Quote: BlackjackLoverI don't understand why it seems so difficult to understand to some people that when you flip a coin 1,000 times, it's unlikely that you will get heads 500 times. In theory, the odds that you will get heads are 50%, but in practice most of the times the results will not be exactly 50% even if you flip it one billion times.
Nobody in this thread is saying that’s likely that it will be exactly the same. But people in this thread have argued that the odds player should be expected to do worse. That’s false.
I don’t understand why it’s difficult for some people in this thread to understand that the distribution of outcomes for the full odds and no odds players are centered on the same value.
Quote: unJonBut people in this thread have argued that the odds player should be expected to do worse. That’s false.
Who said that?
The player who takes odds may lose more or less money or win more or less money. It doesn't matter. The point is that in practice it's unlikely that the losses and wins from taking odds will perfectly balance out, which means that it's unlikely that a player who doesn't take odds and a player who takes odds will lose or win the same amount of money.
Quote: unJonI don’t understand why it’s difficult for some people in this thread to understand that the distribution of outcomes for the full odds and no odds players are centered on the same value.
Not sure what you're talking about. Anyway, people understand that if you play a game with no house edge, the expected return is 100%. In actuality, however, if you make 1,000 bets, it's unlikely that you'll win exactly 500 times.
Quote: BlackjackLover
It's unsurprising when the person who takes odds bet so much more.
That’s not correct. On average he loses the same. Sometimes much more and sometimes much less.
Quote: unJonI agree with your post above. But you are the one saying that the person taking odds should lose more because he bet more:
That’s not correct. On average he loses the same. Sometimes much more and sometimes much less.
You misinterpreted my post. Generally, people who bet much more will win or lose much more than people who bet much less. It's possible that they will win or lose the same amount of money, but it's unlikely. So it's unsurprising when people who bet much more win or lose much more. I hope this is clear.
Ah. Got it. And agree with it. I would now like back the hour I put into posting on this thread.Quote: BlackjackLoverYou misinterpreted my post. Generally, people who bet much more will win or lose much more than people who bet much less. It's possible that they will win or lose the same amount of money, but it's unlikely. So it's unsurprising when people who bet much more win or lose much more. I hope this is clear.
it gets LESS LIKELYQuote: BlackjackLoverI don't understand why it seems so difficult to understand to some people that when you flip a coin 1,000 times, it's unlikely that you will get heads 500 times.
| flips | 50% Heads | prob 50/50 |
|---|---|---|
| 50 | 25 | 0.112275173 |
| 100 | 50 | 0.079589237 |
| 150 | 75 | 0.065038514 |
| 200 | 100 | 0.056348479 |
| 250 | 125 | 0.050412213 |
| 300 | 150 | 0.046027514 |
| 350 | 175 | 0.042618271 |
| 400 | 200 | 0.039869302 |
| 450 | 225 | 0.037591749 |
| 500 | 250 | 0.035664646 |
| 550 | 275 | 0.034006451 |
| 600 | 300 | 0.032559931 |
| 650 | 325 | 0.031283573 |
| 700 | 350 | 0.030146433 |
| 750 | 375 | 0.029124915 |
| 800 | 400 | 0.028200665 |
| 850 | 425 | 0.027359167 |
| 900 | 450 | 0.026588765 |
| 950 | 475 | 0.025879982 |
| 1000 | 500 | 0.025225018 |
here is a table of data flipping a fair coin
while Heads always having an excess of them (more than 25)
at 1st, the pct of Heads is at 100% but as each session shows an excess number of heads
the actual % of heads is going down and moving to 50% as expected and predicted by the Law of Large Numbers.
| session flips | Heads | pct Heads | total flips | total Heads | pct Total Heads | excessive Heads |
|---|---|---|---|---|---|---|
| 50 | 50 | 100.00% | 50 | 50 | 100.00% | 25 |
| 50 | 35 | 70.00% | 100 | 85 | 85.00% | 35 |
| 50 | 34 | 68.00% | 150 | 119 | 79.33% | 44 |
| 50 | 33 | 66.00% | 200 | 152 | 76.00% | 52 |
| 50 | 32 | 64.00% | 250 | 184 | 73.60% | 59 |
| 50 | 31 | 62.00% | 300 | 215 | 71.67% | 65 |
| 50 | 30 | 60.00% | 350 | 245 | 70.00% | 70 |
| 50 | 29 | 58.00% | 400 | 274 | 68.50% | 74 |
| 50 | 28 | 56.00% | 450 | 302 | 67.11% | 77 |
| 50 | 27 | 54.00% | 500 | 329 | 65.80% | 79 |
| 50 | 26 | 52.00% | 550 | 355 | 64.55% | 80 |
| 50 | 26 | 52.00% | 600 | 381 | 63.50% | 81 |
| 50 | 26 | 52.00% | 650 | 407 | 62.62% | 82 |
| 50 | 27 | 54.00% | 700 | 434 | 62.00% | 84 |
| 50 | 28 | 56.00% | 750 | 462 | 61.60% | 87 |
| 50 | 29 | 58.00% | 800 | 491 | 61.38% | 91 |
| 50 | 30 | 60.00% | 850 | 521 | 61.29% | 96 |
| 50 | 29 | 58.00% | 900 | 550 | 61.11% | 100 |
| 50 | 28 | 56.00% | 950 | 578 | 60.84% | 103 |
| 50 | 27 | 54.00% | 1000 | 605 | 60.50% | 105 |
| 50 | 26 | 52.00% | 1050 | 631 | 60.10% | 106 |
| 50 | 26 | 52.00% | 1100 | 657 | 59.73% | 107 |
| 50 | 26 | 52.00% | 1150 | 683 | 59.39% | 108 |
| 50 | 27 | 54.00% | 1200 | 710 | 59.17% | 110 |
| 50 | 28 | 56.00% | 1250 | 738 | 59.04% | 113 |
| 50 | 29 | 58.00% | 1300 | 767 | 59.00% | 117 |
| 50 | 30 | 60.00% | 1350 | 797 | 59.04% | 122 |
| 50 | 29 | 58.00% | 1400 | 826 | 59.00% | 126 |
| 50 | 28 | 56.00% | 1450 | 854 | 58.90% | 129 |
| 50 | 27 | 54.00% | 1500 | 881 | 58.73% | 131 |
| 50 | 26 | 52.00% | 1550 | 907 | 58.52% | 132 |
| 50 | 26 | 52.00% | 1600 | 933 | 58.31% | 133 |
| 50 | 26 | 52.00% | 1650 | 959 | 58.12% | 134 |
| 50 | 27 | 54.00% | 1700 | 986 | 58.00% | 136 |
| 50 | 28 | 56.00% | 1750 | 1014 | 57.94% | 139 |
| 50 | 29 | 58.00% | 1800 | 1043 | 57.94% | 143 |
| 50 | 30 | 60.00% | 1850 | 1073 | 58.00% | 148 |
| 50 | 29 | 58.00% | 1900 | 1102 | 58.00% | 152 |
| 50 | 28 | 56.00% | 1950 | 1130 | 57.95% | 155 |
| 50 | 27 | 54.00% | 2000 | 1157 | 57.85% | 157 |
the is no Law of Small Numbers as I know of
this relates to the Ask the Wizard question of 1000 wagers
doesn’t each person lose the same amount of money?
NO, and we do not expect them to either
it may happen, but that probability is very small
some say, thanks for sharing YOUR OPINION HERE.Quote: unJonI agree with your post above. But you are the one saying that the person taking odds should lose more because he bet more:
That’s not correct. On average he loses the same.
"On average he loses the same"
I say no way
NOT over the 1000 wagers asked in the question that I brought up.
that is TOO SMALL OF A SAMPLE SIZE with the high odds
look at my all my sim data (I know too boring and useless)
not one time did they lose the same (even on average)
over 1000s of sessions
not once
why?
variance
standard deviation is a bit more than 100 times that with 100x odds than 0 odds.
DO YOU KNOW WHAT THAT DOES TO RESULTS OF 1000 WAGERS?
we are NOT talking about comparing of X number of wagers
of RED and 1 number at 00 Roulette where the 1 number is close to 6 times the standard deviation of RED.
this is craps and 100x odds vs 0 odds
The Wizard knows most will never understand the concept of variance and the roll it plays over X number of lifetime bets that is smaller than infinite.
my 2 sims I posted that had 1000 wagers 1 million times (1 BILLION total wagers) still did not have an equal average loss.
TOTALLY 100% OPPOSITE OF WHAT YOU SAID WITHOUT ANY DATA BACKING YOUR OPINION.
that is what most opinions are
your opinion is still noted and accepted here.
In my opinion, Wizard gave the simple answer of yes and then tried to offer a suggestion that had nothing to do with the Ask the Wizard question to keep from getting into a long winded not understood explanation on something most will NEVER understand.
that being variance
(I first understood variance and how to use it back in 1999, after finding The Wizard of Odds website as I taught casino dealing in Reno at the time and wanted some easy answers to more difficult gambling questions. Back then, the Wizard was really good at taking the time with nice explanations of difficult concepts. I think that lacks today, even tho the attempt is there.)
better means ?Quote: DeMangoOkay, next question: Laying odds on the DP. Is more better?
sure I have sims for that
dpass with 0 lay odds
and
dpass with 100x lay odds every point
both players at same table
0 odds player over 1 BILLION bets lost $13.66 million
100 odds player over 1 BILLION bets lost $16.84 million
about 1 in 5000 sessions did both players end up with the same net session result
the average per session is NOT the same
Simulation of Craps Don't Pass Wagers
Odds Multiplier . . . . = 100
Session Bankroll . . . = 1000000.00
Win goal to quit session= 1000000.00
Max. Decisions to quit = 1000
No. Sessions simulated = 1000000
Starting Random seed . = 54321
------------------------------------
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1000.00
Avg. No. games won . . = 479.28
Avg. No. games lost . . = 492.94
Avg. No. games tied . . = 27.78
Avg. No. dice rolls . . = 3375.71
Avg. Total amount bet . = 972.22
Avg. amount bet on Odds = 99997.29
Bankroll was busted . . = 0.000% of the time ( 0)
Win goal was met . . . = 0.000% of the time ( 0)
Bankroll decreased . . = 50.103% of the time
Bankroll increased . . = 49.885% of the time
Avg (mean) end bankroll = 999983.16 (change of -16.84)
Std-dev ending bankroll = 3185.40
Average expectation EVI = -1.732%
Average expectation EVR = -0.017%
Simulation of Craps Don't Pass Wagers
Odds Multiplier . . . . = 0
Session Bankroll . . . = 1000000.00
Win goal to quit session= 1000000.00
Max. Decisions to quit = 1000
No. Sessions simulated = 1000000
Starting Random seed . = 54321
------------------------------------
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1000.00
Avg. No. games won . . = 479.28
Avg. No. games lost . . = 492.94
Avg. No. games tied . . = 27.78
Avg. No. dice rolls . . = 3375.71
Avg. Total amount bet . = 972.22
Bankroll was busted . . = 0.000% of the time ( 0)
Win goal was met . . . = 0.000% of the time ( 0)
Bankroll decreased . . = 66.388% of the time
Bankroll increased . . = 32.459% of the time
Avg (mean) end bankroll = 999986.34 (change of -13.66)
Std-dev ending bankroll = 31.19
Average expectation EVI = -1.405%
wonder how many more sessions than 1 MILLION do we need to see before both players average loss is the same as many claim it should be?
Quote: 7crapssome say, thanks for sharing YOUR OPINION HERE.
"On average he loses the same"
I say no way
NOT over the 1000 wagers asked in the question that I brought up.
that is TOO SMALL OF A SAMPLE SIZE with the high odds
look at my all my sim data (I know too boring and useless)
not one time did they lose the same (even on average)
over 1000s of sessions
not once
why?
variance
standard deviation is a bit more than 100 times that with 100x odds than 0 odds.
DO YOU KNOW WHAT THAT DOES TO RESULTS OF 1000 WAGERS?
we are NOT talking about comparing of X number of wagers
of RED and 1 number at 00 Roulette where the 1 number is close to 6 times the standard deviation of RED.
this is craps and 100x odds vs 0 odds
The Wizard knows most will never understand the concept of variance and the roll it plays over X number of lifetime bets that is smaller than infinite.
my 2 sims I posted that had 1000 wagers 1 million times (1 BILLION total wagers) still did not have an equal average loss.
TOTALLY 100% OPPOSITE OF WHAT YOU SAID WITHOUT ANY DATA BACKING YOUR OPINION.
that is what most opinions are
your opinion is still noted and accepted here.
In my opinion, Wizard gave the simple answer of yes and then tried to offer a suggestion that had nothing to do with the Ask the Wizard question to keep from getting into a long winded not understood explanation on something most will NEVER understand.
that being variance
(I first understood variance and how to use it back in 1999, after finding The Wizard of Odds website as I taught casino dealing in Reno at the time and wanted some easy answers to more difficult gambling questions. Back then, the Wizard was really good at taking the time with nice explanations of difficult concepts. I think that lacks today, even tho the attempt is there.)
We must be using the word “average” differently. Here’s what I mean: Both players have the same EV. Nothing more.
If I offered to wager you that the odds player would win more than the no odds player after a thousand rolls (if they win the same bet pushes), do you agree that 1:1 is the fair odds for that bet? If so, it’s because on average they win the same.
Quote: 7crapsbetter means ?
sure I have sims for that
dpass with 0 lay odds
and
dpass with 100x lay odds every point
both players at same table
0 odds player over 1 BILLION bets lost $13.66 million
100 odds player over 1 BILLION bets lost $16.84 million
about 1 in 5000 sessions did both players end up with the same net session result
the average per session is NOT the sameSimulation of Craps Don't Pass Wagers
Odds Multiplier . . . . = 100
Session Bankroll . . . = 1000000.00
Win goal to quit session= 1000000.00
Max. Decisions to quit = 1000
No. Sessions simulated = 1000000
Starting Random seed . = 54321
------------------------------------
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1000.00
Avg. No. games won . . = 479.28
Avg. No. games lost . . = 492.94
Avg. No. games tied . . = 27.78
Avg. No. dice rolls . . = 3375.71
Avg. Total amount bet . = 972.22
Avg. amount bet on Odds = 99997.29
Bankroll was busted . . = 0.000% of the time ( 0)
Win goal was met . . . = 0.000% of the time ( 0)
Bankroll decreased . . = 50.103% of the time
Bankroll increased . . = 49.885% of the time
Avg (mean) end bankroll = 999983.16 (change of -16.84)
Std-dev ending bankroll = 3185.40
Average expectation EVI = -1.732%
Average expectation EVR = -0.017%
Simulation of Craps Don't Pass Wagers
Odds Multiplier . . . . = 0
Session Bankroll . . . = 1000000.00
Win goal to quit session= 1000000.00
Max. Decisions to quit = 1000
No. Sessions simulated = 1000000
Starting Random seed . = 54321
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All bets are a single unit
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Simulation Results per Session
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Avg. No. games played . = 1000.00
Avg. No. games won . . = 479.28
Avg. No. games lost . . = 492.94
Avg. No. games tied . . = 27.78
Avg. No. dice rolls . . = 3375.71
Avg. Total amount bet . = 972.22
Bankroll was busted . . = 0.000% of the time ( 0)
Win goal was met . . . = 0.000% of the time ( 0)
Bankroll decreased . . = 66.388% of the time
Bankroll increased . . = 32.459% of the time
Avg (mean) end bankroll = 999986.34 (change of -13.66)
Std-dev ending bankroll = 31.19
Average expectation EVI = -1.405%
wonder how many more sessions than 1 MILLION do we need to see before both players average loss is the same as many claim it should be?
Yeah, it is all right there.
With odds:
Bankroll decreased . . = 50.103% of the time
Bankroll increased . . = 49.885% of the time
Avg (mean) end bankroll = 999983.16 (change of -16.84)
Std-dev ending bankroll = 3185.40
Without odds:
Bankroll decreased . . = 66.388% of the time
Bankroll increased . . = 32.459% of the time
Avg (mean) end bankroll = 999986.34 (change of -13.66)
Std-dev ending bankroll = 31.19
Is your whole argument that in a limited sample size of 1,000,000, that the average loss isn't EXACTLY the same as the theoretical loss?
