Apologizing in advance...the do/don't with odds topic
November 3rd, 2013 at 8:14:41 AM
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Guys, I know this has been discussed here and on Michael's other site, I'm pretty sure I've read all that is to be reasonably found on the web discussing the issue. Still, I'm just not convinced it's a bad bet and before I put it into action, now that I've found what I believe is the ultimate site for asking this question - I'm willing to take a verbal beating if it means gaining a better understanding.
If I'm a traditional right side bettor, I'd bet one unit on the pass line, and one unit on the odds if there's a point. Maybe I get a come out 7 or 11, maybe not. It seems that the huge emphasis is on taking the odds anyway since there's no house edge. So I have a 2:1 shot of winning the first roll, and then it's on to a bet with a house edge and the odds.
If I bet a one unit do and a one unit don't, I have no chance of winning on the come out, and a 1:36 chance of losing on the 12. One way to look at that is I have less chance to lose than I would with a single do or don't bet. And, no chance of winning. However....
I think most people reading "how to play craps" material would agree that taking the odds, (or laying them) seems to be the main emphasis, once you get beyond the concept of making the best bets, because there's no house edge. Given that, my thinking is I'm willing to give up the chance to win a flat bet in order to reduce my risk of losing that flat bet and then place the same money I'd use for the flat bet AND the odds bet all on the odds. This makes far more sense to me.
I don't understand the logic that with two opposite bets I'm accepting twice the house edge... but more importantly, does it matter? Does it really matter what the edge is because I'm not looking to win those bets, I'm looking to offset them?
This is where the phrase,"in the long run" usually gets added...but we...well, at least I, don't play for the long run. I'm not looking to prove the math or the averages or the probabilities in the long run. As a right side bettor, I'm looking to take advantage of a series of numbers in the inevitable number line being rolled that is largely absent the 7. All of us know that "in the long run" we'd lose everything, controlled shooters notwithstanding. Once we've decided on the bet, the next goal is to walk away at the right time. Now there's a question for the ages, when should I leave...but I digress.
As I see it then, I'd bet table minimum on the do and don't. This is a wash result so really betting nothing except for the occasional 12 which I'll chalk up as the cost to do business. Now I have the opportunity to bet two units on the odds instead of what would have been a one unit pass line bet and a one unit odds bet. If a point is rolled and then a seven rolls, I'll lose the same amount I'd lose doing it the traditional way. If I win, I win more than the traditional way.
If I'm a traditional right side bettor, I'd bet one unit on the pass line, and one unit on the odds if there's a point. Maybe I get a come out 7 or 11, maybe not. It seems that the huge emphasis is on taking the odds anyway since there's no house edge. So I have a 2:1 shot of winning the first roll, and then it's on to a bet with a house edge and the odds.
If I bet a one unit do and a one unit don't, I have no chance of winning on the come out, and a 1:36 chance of losing on the 12. One way to look at that is I have less chance to lose than I would with a single do or don't bet. And, no chance of winning. However....
I think most people reading "how to play craps" material would agree that taking the odds, (or laying them) seems to be the main emphasis, once you get beyond the concept of making the best bets, because there's no house edge. Given that, my thinking is I'm willing to give up the chance to win a flat bet in order to reduce my risk of losing that flat bet and then place the same money I'd use for the flat bet AND the odds bet all on the odds. This makes far more sense to me.
I don't understand the logic that with two opposite bets I'm accepting twice the house edge... but more importantly, does it matter? Does it really matter what the edge is because I'm not looking to win those bets, I'm looking to offset them?
This is where the phrase,"in the long run" usually gets added...but we...well, at least I, don't play for the long run. I'm not looking to prove the math or the averages or the probabilities in the long run. As a right side bettor, I'm looking to take advantage of a series of numbers in the inevitable number line being rolled that is largely absent the 7. All of us know that "in the long run" we'd lose everything, controlled shooters notwithstanding. Once we've decided on the bet, the next goal is to walk away at the right time. Now there's a question for the ages, when should I leave...but I digress.
As I see it then, I'd bet table minimum on the do and don't. This is a wash result so really betting nothing except for the occasional 12 which I'll chalk up as the cost to do business. Now I have the opportunity to bet two units on the odds instead of what would have been a one unit pass line bet and a one unit odds bet. If a point is rolled and then a seven rolls, I'll lose the same amount I'd lose doing it the traditional way. If I win, I win more than the traditional way.
...in the long run, you'll lose money.
November 3rd, 2013 at 10:58:55 AM
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How many lifetime pass line bets are you planning to make?Quote: nowakezoneAs I see it then, I'd bet table minimum on the do and don't. This is a wash result so really betting nothing except for the occasional 12 which I'll chalk up as the cost to do business.
add up all the individual sessions you could make
3,600
7,200
36,000
more, less in only 360 maybe
does it even matter?
3,600 do/don't lifetime bets produces an average loss of 100 units ($500 at $5 bets)
on the come out roll with no average win. Zero chance of winning. I think you call this a "wash"
how do you plan on making that up? or it is just a wash.
easy, I think, just win more times with the odds bets.
is that really easier to do?
how about
7,200 total lifetime pass/don't bets
average loss of 200 units ($1000 at $5 bets)
it looks to me the more bets you make the loss amount increases.
that can not be a good thing.
unless, as many say, you have more fun playing that way.
not me! I hate losing $20
Sally
I Heart Vi Hart
November 3rd, 2013 at 11:53:28 AM
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You don't need to be doing something weird with your pass-line bets and Don'tPassLine bets.
Make which ever line bet you wish ... then take or lay odds as you wish if a point is generated.
If the dollar level is not to your enjoyment then get thee to Hendertucky and those one and two dollar games.
Make which ever line bet you wish ... then take or lay odds as you wish if a point is generated.
If the dollar level is not to your enjoyment then get thee to Hendertucky and those one and two dollar games.
November 3rd, 2013 at 11:57:05 AM
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I'm thinking about your reply Sally. Thanks for the response.
Like you, I hate to lose $20. I don't have fun losing. Love the game, hate to lose. I'm not a math whiz, although not a dolt either...however your very excellent illustration has given me a solid place to start.
My first thought, without any nice example like yours, is yes, that 12 deficit would be made up by the 2X odds bet.
Btw, when we say the 12 "will show" 1 of 36 times, are we talking about the independent probability of each roll, and not the odds of three 12's showing in a row for example?
Like you, I hate to lose $20. I don't have fun losing. Love the game, hate to lose. I'm not a math whiz, although not a dolt either...however your very excellent illustration has given me a solid place to start.
My first thought, without any nice example like yours, is yes, that 12 deficit would be made up by the 2X odds bet.
Btw, when we say the 12 "will show" 1 of 36 times, are we talking about the independent probability of each roll, and not the odds of three 12's showing in a row for example?
...in the long run, you'll lose money.
November 3rd, 2013 at 12:17:17 PM
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I also play in the short run, but they start to add up after the very first session played. Now I play with OPM. more fun this way (I know it is my husband's money, but he likes it that way too)Quote: nowakezoneI'm thinking about your reply Sally. Thanks for the response.
"will" is not allowed.Quote: nowakezoneBtw, when we say the 12 "will show" 1 of 36 times,
the 12 shows, on average, 1 in 36 times (1/probability)
with a probability of 1 in 36 (1/36)
yesQuote: nowakezoneare we talking about the independent probability of each roll,
3 in a row is not an independent event.Quote: nowakezoneand not the odds of three 12's showing in a row for example?
any run (streak) greater than length 1
is a dependent event.
The probability of
3 in a row in 3 trials
is different from
3 in a row in 4 trials
is different from
3 in a row in N trials
The average number of rolls to see 3 12s in a row = 47,988.00
1/p^3 + 1/p^2 + 1/p^1
not
1/p^3 = 46,656 = the number of trials it takes
but each trial is not 1 roll in length
(they are rare events indeed)
added
to get 3 in a row we need 1 roll
and 1/36 of the time we get the 2nd 12 on the 2nd roll
and 1/36^2 we get the 3rd 12 in a row on the 3rd roll
1+(1/36)+(1/36)^2 = 1.028549382716049382716049382716
that # * 46,656 = 47,988 (try it!)
here is an easier example to see and calculate in one's head
many many many
say the average number of coin flips to see 3 heads in a row is x = (1/2)^3 and (1/x) = 8
a simulation shows this to not be correct. why?
well it is 8 trials long but for a run of 3
we need the average length of the trial
p = 0.50
1 + p + p^2 = 1.75
1.75 * 8 = 14
(same as 1/p + 1/p^2 + 1/p^3
and matches simulation results)
more here on that
https://wizardofvegas.com/forum/questions-and-answers/math/8141-on-average-how-many-trials-will-it-take-to-see-a-streak-of-8-qs-for-fun/
to get 3 in a row we need 1 roll
and 1/36 of the time we get the 2nd 12 on the 2nd roll
and 1/36^2 we get the 3rd 12 in a row on the 3rd roll
1+(1/36)+(1/36)^2 = 1.028549382716049382716049382716
that # * 46,656 = 47,988 (try it!)
here is an easier example to see and calculate in one's head
many many many
say the average number of coin flips to see 3 heads in a row is x = (1/2)^3 and (1/x) = 8
a simulation shows this to not be correct. why?
well it is 8 trials long but for a run of 3
we need the average length of the trial
p = 0.50
1 + p + p^2 = 1.75
1.75 * 8 = 14
(same as 1/p + 1/p^2 + 1/p^3
and matches simulation results)
more here on that
https://wizardofvegas.com/forum/questions-and-answers/math/8141-on-average-how-many-trials-will-it-take-to-see-a-streak-of-8-qs-for-fun/
of course, betting the do/don't makes a starting bankroll last longer, so
maybe more fun for the player.
But,
Your 2X odds have to win more times in N attempts
to make up for the horrible "loss only - never win" on the come out roll situation
that becomes more difficult to do the more bets are made.
right?
like trying to guess the next coin flip and being correct 53% of the time instead of the 50%.
Becomes harder the longer it is attempted.
Have fun
Sally
I Heart Vi Hart
November 3rd, 2013 at 12:21:52 PM
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Quote: mustangsallyHow many lifetime pass line bets are you planning to make?
add up all the individual sessions you could make
3,600
7,200
36,000
more, less in only 360 maybe
does it even matter?
3,600 do/don't lifetime bets produces an average loss of 100 units ($500 at $5 bets)
on the come out roll with no average win. Zero chance of winning. I think you call this a "wash"
how do you plan on making that up? or it is just a wash.
easy, I think, just win more times with the odds bets.
is that really easier to do?
how about
7,200 total lifetime pass/don't bets
average loss of 200 units ($1000 at $5 bets)
it looks to me the more bets you make the loss amount increases.
that can not be a good thing.
unless, as many say, you have more fun playing that way.
not me! I hate losing $20
Sally
Not an argument here Sally, just another way of looking at the game. Kudo's by the way for explaining "mean" and average.
People play for their own reason's, I say you pay's your dues you takes your chances.
So, if a player has 7200 12's on the comeout roll and everything else washed and they liked "free" drinks and the glitz in a casino. That 1000 dollars over a lifetime is cheap entertainment. Try going to movies for twenty years for a thousand.
Imagine a player thinking they can get the free odds by way of a do/don't pass strategy. If a player wins and loses an equal number of hands in a lifetime and takes max. odds, I think what they will have won will be the difference of the odds pay, versus flat? If a craps player was to go strictly by the math, they wouldn't play at all.
I think when the stats were written about no house edge on the free odds, the thinking included the pass/don't pass wager?? What about intuition [superstition I know] or apparent trends/streaks? Then there's are old friend variance.
I don't know how the player is being rated, but maybe they can get more than 1000 dollars value in comps in lifetime play to make up for the 12's.
November 3rd, 2013 at 12:23:10 PM
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I don't think bypassing the come out roll in order to get the entire bet working at true odds is weird.
Our local casino sure seems a lot slower than this but enough writers go with 100 rolls per hour to allow me to say, okay let's presume 100 rolls per hour. In fact for the sake of easier math, let's say 108 rolls per hour and the 12 shows 3 times per hour. Maybe it would more, maybe less. And let's be generous and say those three showings ALL happen on a come out roll. I'll point out that while I was searching for an average number of rolls reference I saw that the Wizard says just under 30% of rolls per hour are come out rolls. Now if we do some simple fudging we could say 30% is one third and one third of 108 is 36 which would mean we could expect the 12 to show on ONE come out roll per hour.
Empirically, the entire bet working at true odds, or even the traditional bet, should be able to make up the one unit loss per hour created by the come out 12.
Our local casino sure seems a lot slower than this but enough writers go with 100 rolls per hour to allow me to say, okay let's presume 100 rolls per hour. In fact for the sake of easier math, let's say 108 rolls per hour and the 12 shows 3 times per hour. Maybe it would more, maybe less. And let's be generous and say those three showings ALL happen on a come out roll. I'll point out that while I was searching for an average number of rolls reference I saw that the Wizard says just under 30% of rolls per hour are come out rolls. Now if we do some simple fudging we could say 30% is one third and one third of 108 is 36 which would mean we could expect the 12 to show on ONE come out roll per hour.
Empirically, the entire bet working at true odds, or even the traditional bet, should be able to make up the one unit loss per hour created by the come out 12.
...in the long run, you'll lose money.
November 3rd, 2013 at 12:36:01 PM
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This I can wrap my head around, as they say.Quote: petroglyph
Imagine a player thinking they can get the free odds by way of a do/don't pass strategy. If a player wins and loses an equal number of hands in a lifetime and takes max. odds, I think what they will have won will be the difference of the odds pay, versus flat?
Consider a win on the 6. If one unit is $10, then the traditional one flat unit and one odds unit would return $22 v a potential loss of $20. With the two unit odds bet and no flat bet the 6 winner returns $24 v a potential $20 loss. The 5 returns $25 traditionally, or $30 the "weird" way. ;) The 4 returns $30 traditionally, or $40 via the bypass.
A $2 difference, eh....tip the dealer. However, those 5's and 10's....for a $10 bettor like myself, not so bad.
...in the long run, you'll lose money.
November 3rd, 2013 at 12:37:57 PM
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expect = average = long termQuote: nowakezoneNow if we do some simple fudging we could say 30% is one third and one third of 108 is 36 which would mean we could expect the 12 to show on ONE come out roll per hour.
1 in 36 means:
36.2710% chance of NOT seeing a 12 per 36 come out rolls
37.3073% chance of seeing exactly 1 (one) 12 per 36 come out rolls
26.42168% (1 in 4) chance of seeing at least 2 (2 or MORE) 12s per 36 come out rolls
7.76802% (1 in 13) chance of seeing at least 3 (3 or MORE) 12s per 36 come out rolls
are you ready for those 2 and 3 and more come out 12s??
I do not think any one expects it when they think about a
long term average of 1 in 36
but that is me
Sally
I Heart Vi Hart
November 3rd, 2013 at 12:46:26 PM
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yes, I agree too.Quote: petroglyphPeople play for their own reason's, I say you pay's your dues you takes your chances.
and some have more fun than others playing craps playing different ways.
I see this all the time.
but Maybe? I think that IS a BIG but.Quote: petroglyphI don't know how the player is being rated, but maybe they can get more than 1000 dollars value in comps in lifetime play to make up for the 12's.
For my $5 example, The expected loss over 7200 pass line bets only bets is just over $500.(I had $100 by error)
Will any casino even double that for playing both sides at the same time?
I doubt it.
If yes, I want to play at that casino
Sally
I Heart Vi Hart
November 3rd, 2013 at 12:57:08 PM
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Quote: mustangsallyyes, I agree too.
and some have more fun than others playing craps playing different ways.
I see this all the time.
but Maybe? I think that IS a BIG but.
For my $5 example, The expected loss over 7200 pass line bets only bets is just over $100.
Will any casino even double that for playing both sides at the same time?
I doubt it.
If yes, I want to play at that casino
Sally
I don't know how they figure the comps, but if I play enough they send me enough junk mail to heat my home if I burn it in the fireplace.
And, and they do send free room offers [which I don't like at all] but also bonus this or that, meals, shows, etc.
The offers are directly proportional to bet size. I don't work the ap on comps. If I break even or better and they buy all the coffee and feed me, that's a win. However small it's still a victory.
November 3rd, 2013 at 2:40:06 PM
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Really? Sally not that I'll ever get to where I can calculate such a thing but really? 7200 pass line bets of $5 and one should expect to lose that bet only 20 times? 7200 bets, 20 losers? Doesn't that mean that one would expect 7180 winners?Quote: mustangsally
For my $5 example, The expected loss over 7200 pass line bets only bets is just over $100.
Aren't the chances for the 4 and 10 better?Quote: mustangsallyexpect = average = long term
1 in 36 means:
36.2710% chance of NOT seeing a 12 per 36 come out rolls
37.3073% chance of seeing exactly 1 (one) 12 per 36 come out rolls
26.42168% (1 in 4) chance of seeing at least 2 (2 or MORE) 12s per 36 come out rolls
7.76802% (1 in 13) chance of seeing at least 3 (3 or MORE) 12s per 36 come out rolls
But, if you never answered that and only answered the first question... whoa!
...in the long run, you'll lose money.
November 3rd, 2013 at 3:15:50 PM
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Sure you can.Quote: nowakezoneReally? Sally not that I'll ever get to where I can calculate such a thing but really? 7200 pass line bets of $5 and one should expect to lose that bet only 20 times? 7200 bets, 20 losers? Doesn't that mean that one would expect 7180 winners?
You would have seen my error.
To calculate the expected value (expected loss in this case)
of many bets (especially if they are all the same amount)
house edge * avg $bet * # of bets made
-7/495 * $5 * 7,200 = -$509.09 (a long term average, the theoretical value)
it would be rare to lose exactly that amount because that is an average.
The range around that is $424.22 (this is called the standard deviation - related to variance)
So not impossible by any means to play that many pass line bets
and come away a winner.
(ev / sd = -1.2
above and beyond -3.0 would suck - and this can come from more bets and/or a combination of higher amounts wagered)
The winning probability of the pass line bet is p = 244/495 or about 49.3%
expected (average) number of wins = 3,549 (rounded) 7200*p
3,651 that lose
even with that 49.3% chance of a win on one pass line bet
over 7,200 such bets I would not be at all surprised if that winning percentage came out to be between
51.0605% to 47.5253%
even higher good and bad luck could push out even further to
51.6497% to 46.9361%
Sally
I Heart Vi Hart
November 3rd, 2013 at 3:43:30 PM
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Sally you're very good at explaining the math and how it applies practically. I really appreciate you taking the time to do so. Also, you've not only given me a fish, but shown me how to fish as well.
I'm so much further along now. Math tools to use that I understand. I'm glad it's still early in the evening so I have plenty of time to think about it and work up some ideas. This is just what I needed.
Thank you Sally.
I'm so much further along now. Math tools to use that I understand. I'm glad it's still early in the evening so I have plenty of time to think about it and work up some ideas. This is just what I needed.
Thank you Sally.
...in the long run, you'll lose money.
November 3rd, 2013 at 5:46:27 PM
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OKQuote: nowakezoneThis I can wrap my head around, as they say.
Consider a win on the 6. If one unit is $10, then the traditional one flat unit and one odds unit would return $22 v a potential loss of $20.
parameters:
$10 flat and $10 odds for the pass line
Now let us consider the probability of winning and losing the 6 or 8.
5 ways to win and 6 ways to lose.
IF that holds true in the long run
5 winners = 5*$22 = $110
6 lost = 6*$20 = -$120
for a net LOSS of $10
You can NEVER win and only lose on the come out roll.
against the wind even more
but one may still have fun playing this way.
in order to show a profit from the 11 wagers, you NEED 6 wins and 5 that lost, minimum.
the odds of that, for you, have gotten rotten, or not better.
Now let us consider the probability of winning and losing the 5 or 9.
4 ways to win and 6 ways to lose.
IF that holds true in the long run
4 winners = 4*$25 = $100
6 lost = 6*$20 = -$120
for a net LOSS of $20
we be going the wrong way!
it now gets even harder over those 10 bets to pull out a win profit (the best kind)
should we even consider the probability of winning and losing the 4 or 10?
3 ways to win and 6 ways to lose.
IF that holds true in the long run
3 winners = 3*$30 = $90
6 lost = 6*$20 = -$120
for a net LOSS of $30
ouch!
we be going the wrong way still!
but as long as you understand the process and the consequences
it may still be the way you want to play.
but you may also find out other ways to play
that increase your wins and decrease your losses.
Have fun!
just food for thought
I am hungry now
Sally
I Heart Vi Hart
November 3rd, 2013 at 6:26:46 PM
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@mustangsally
Thanks for explaining that so well in layman's terms.
In the past when you were on hiatus from the wov I went back and read some of your posts as you were so well respected from other members.
I'm all ears [eyes] now, what else you got?
I'd like to read what you have to say about other ways to increase wins and decrease losses.
As an aside, I usually won't bet the pass line unless I'm shooting. I do sometimes after the fact place the point. Not as some cheesy defense against an indefensible position, but in case that might be part of your next help.
Thanks for explaining that so well in layman's terms.
In the past when you were on hiatus from the wov I went back and read some of your posts as you were so well respected from other members.
I'm all ears [eyes] now, what else you got?
I'd like to read what you have to say about other ways to increase wins and decrease losses.
As an aside, I usually won't bet the pass line unless I'm shooting. I do sometimes after the fact place the point. Not as some cheesy defense against an indefensible position, but in case that might be part of your next help.
November 3rd, 2013 at 8:16:25 PM
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Sally I just signed on to mention that your previous post led to further examination of some of the terms like where did the -7/495 come ffrom.This has been an excellent evening and now I see you've posted another buffet of good things for me and the rest of us. Fantastic! I can't wait to load up on it.
I'd like to mention that you have a good way of saying how it really is, while allowing that a person might have fun playing a certain way even though it isn't an optimal way, for lack of a better term. I, along with everyone else I'm sure, appreciate your graciousness very much.
And now, I'm hungry too.
I'd like to mention that you have a good way of saying how it really is, while allowing that a person might have fun playing a certain way even though it isn't an optimal way, for lack of a better term. I, along with everyone else I'm sure, appreciate your graciousness very much.
And now, I'm hungry too.
...in the long run, you'll lose money.
November 3rd, 2013 at 8:20:54 PM
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November 3rd, 2013 at 8:51:11 PM
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Okay all this stuff has my brain working finally. Sally your recent post helped me begin to understand the process and will allow me to attempt to consider the consequences.
In your recent post you outline using the standard pass line and odds bets, which result in net long run losers regardless of point number. And you mention this scenario includes NEVER winning on the come out etc.
My scenario puts all $20 on the odds instead of splitting them. However, because you extended the pplan by posting expected wins and losses, it made me realize three important facts. The long run expectation of true odds on win v losses for each point number with my plan is breakeven. Example - 5 ways to make the 6 for +$120, 6 ways to lose the 6 for -120. Long run = Box number Breakeven. Also, your post made me see that given a long run expected breakeven on box numbers, there would be a net deficit created by the 12, as you mentioned. And finally (so far) your post affirmed that as far as I have seen so far....the long run is a loser (no mystery there)...and a profit playing craps can result only from short run periods where reality and probability deviate in the player's favor and the player walks away while net positive.
Still learning...still thinking.... awesome stuff Sally. As the previous poster did, I'll be reading thru your posts. I'm certain there will be plenty of nuggets.
In your recent post you outline using the standard pass line and odds bets, which result in net long run losers regardless of point number. And you mention this scenario includes NEVER winning on the come out etc.
My scenario puts all $20 on the odds instead of splitting them. However, because you extended the pplan by posting expected wins and losses, it made me realize three important facts. The long run expectation of true odds on win v losses for each point number with my plan is breakeven. Example - 5 ways to make the 6 for +$120, 6 ways to lose the 6 for -120. Long run = Box number Breakeven. Also, your post made me see that given a long run expected breakeven on box numbers, there would be a net deficit created by the 12, as you mentioned. And finally (so far) your post affirmed that as far as I have seen so far....the long run is a loser (no mystery there)...and a profit playing craps can result only from short run periods where reality and probability deviate in the player's favor and the player walks away while net positive.
Still learning...still thinking.... awesome stuff Sally. As the previous poster did, I'll be reading thru your posts. I'm certain there will be plenty of nuggets.
...in the long run, you'll lose money.
November 4th, 2013 at 10:28:19 AM
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To compare the pass/odds with $10/$10 against the do/don't system with $10/$10/$10 pass odds onlyQuote: nowakezoneOkay all this stuff has my brain working finally. Sally your recent post helped me begin to understand the process and will allow me to attempt to consider the consequences.
In your recent post you outline using the standard pass line and odds bets, which result in net long run losers regardless of point number. And you mention this scenario includes NEVER winning on the come out etc.
The key is to account for every possible outcome. Now there will be no secrets.
this can be done with pencil and paper or the use of a spreadsheet or a simple program.
add up the winning and losing probabilities to see what has changed
between the two different (not by much at first glance) betting methods.
The first is just $10/$10 pass/odds
net prob 1 in
-10 0.111111111 9.00
-20 0.395959596 2.53
10 0.222222222 4.50
22 0.12626263 7.92
25 0.08888889 11.25
30 0.05555556 18.00
sumproduct = -0.14141
net prob 1 in
0 0.305555556 3.27
-10 0.027777778 36.00
-20 0.395959596 2.53
22 0.126262626 7.92
25 0.088888889 11.25
30 0.055555556 18.00
sumproduct = -1.53030
The winning probabilities are 49.29% to 27.07%
The losing probabilities are 50.7% to 42.37%
chance of showing a net loss after x rounds of play
pass/odds
do/dont/pass odds
36 rounds (your 1 hour example)
51.8% <<< about a coin flip
69.5%
108 rounds (your 3 hour example gives or takes)
53.1%
81.0% <<< this one is increasing too fast for me
360 rounds (your 10 hour example gives or takes more)
55.6%
94.5%
how about losing at least $100 over a good weekend of play
44.7%
90.6% <<< yeeeoowwwww!
3600 rounds (not all at one time)
I can't look.
3600 rounds
pass/odds only about 67.3% of showing a net loss
the other shows
100% but it must be less than that. maybe the computer can not get that close to 100% without calling it 100%
how about losing at least $2000 over 3600 rounds played
9.5%
99.94%
these percentages are rounded
pass/odds only about 67.3% of showing a net loss
the other shows
100% but it must be less than that. maybe the computer can not get that close to 100% without calling it 100%
how about losing at least $2000 over 3600 rounds played
9.5%
99.94%
these percentages are rounded
I now have my math and programs all working again
see you soon
Sally
I Heart Vi Hart
November 4th, 2013 at 4:11:16 PM
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Well, the 10/10/10 doesn't work very well. Mine is a 10/10/20. I wonder how that would work. The purpose of of the 10 do / 10 don't is to allow me "free access" to the odds bet AND take 20 odds. Free access of course isn't free because of the 12. The 20 comes from being able to use what would have been the old pass line bet for odds since I don't have to use it to get to the odds plus the old odds bet....so 10/10/20.
However, even without a program run, I have seen from our posts yesterday that given the probabilities and edge, "in the long run" the method is a loser, as they all are in the long run, in this case because of the 12. The 10/10/20 on its own is a break even venture. With the deficit created by the 12, it's a loser.
As we've both mentioned, neither of us play for the long run. Overall, I'm not sold that considering the long run is the way to go. At the same time we can't hope to be in position to catch the series where snake eyes roll 3X consecutively and our strategy is to parlay it because the math doesn't support ever betting on the aces!
Essentially, the math doesn't support playing the game at all. The numbers from the Appendix show that the total probability of winning on the Don't side is 2847/5940 or 47.9%, and the Do side is 244/495 or 49.3%. That was surprising to learn since the entire craps-playing world seems to say that the Don't side is better. Maybe that comes from the fact that when playing the Do side one will lose 251/495. There's that darn 12 again.
However, even without a program run, I have seen from our posts yesterday that given the probabilities and edge, "in the long run" the method is a loser, as they all are in the long run, in this case because of the 12. The 10/10/20 on its own is a break even venture. With the deficit created by the 12, it's a loser.
As we've both mentioned, neither of us play for the long run. Overall, I'm not sold that considering the long run is the way to go. At the same time we can't hope to be in position to catch the series where snake eyes roll 3X consecutively and our strategy is to parlay it because the math doesn't support ever betting on the aces!
Essentially, the math doesn't support playing the game at all. The numbers from the Appendix show that the total probability of winning on the Don't side is 2847/5940 or 47.9%, and the Do side is 244/495 or 49.3%. That was surprising to learn since the entire craps-playing world seems to say that the Don't side is better. Maybe that comes from the fact that when playing the Do side one will lose 251/495. There's that darn 12 again.
...in the long run, you'll lose money.
November 4th, 2013 at 5:49:40 PM
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The example I gave is a simple one I think.Quote: nowakezoneWell, the 10/10/10 doesn't work very well. Mine is a 10/10/20. I wonder how that would work. The purpose of of the 10 do / 10 don't is to allow me "free access" to the odds bet AND take 20 odds. Free access of course isn't free because of the 12. The 20 comes from being able to use what would have been the old pass line bet for odds since I don't have to use it to get to the odds plus the old odds bet....so 10/10/20.
However, even without a program run, I have seen from our posts yesterday that given the probabilities and edge, "in the long run" the method is a loser, as they all are in the long run, in this case because of the 12. The 10/10/20 on its own is a break even venture. With the deficit created by the 12, it's a loser.
The do/dont example also was removing the don't pass after the point being established and using that to make the pass odds bet.
I gather this is what is done. One could leave the don't pass and take odds but this seems to be even a larger loser over time.
Your 10/10/20
after the point is made, are you removing the don't pass and then placing $20 on the pass line odds?
That would leave you with $10 pass and $20 odds.
you should be able to duplicate the outcome table I made earlier
it should look like this if I have your method correct
net prob 1 in
0 0.305555556 3.27
-10 0.027777778 36.00
-30 0.395959596 2.53
34 0.126262626 7.92
40 0.088888889 11.25
50 0.055555556 18.00
sumproduct = -1.53030303
expected value, variance and standard deviation easily expands the picture of the bet method over time.
I think it is more fun to compare to another method(s).
I made that just pass/odds
I could compare with another like don't pass and odds
or maybe one large don't pass with odds against one small don't pass/one small don't come with odds.
so true that many play craps to win and length of time is very important too.
the math makes clear that playing against a house edge at Craps
lowers the chance to win the longer one plays (more bets made)
I suppose if the pass/don't were one roll bets, (I doubt they ever were)
many would not make them, even if they were the better bets.
Sally
I Heart Vi Hart
November 4th, 2013 at 6:01:52 PM
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The pass line and don't pass both stay until the end. That way, except for the 12, it's essentially a null bet. Either the pass line wins or the don't pass wins. Both being flat bets, it's a wash. If you take the Don't bet after establishing a point and the Pass bet loses, you lose the whole bet. Leaving the Don't covers the Do on a loser after a point is established and you only lose the Odds.
Method A is 10/10/20 and Method B is 10/10... say a person is using the 5 Count to begin play... or anything to begin play. And say you and I are betting, I use method A, you use method B. Say the dice pass for what...what would be a good number of rolls to use...what have you experienced? That's what I'm talking about. We're both standing there grinding it out, waiting and waiting.... then boom we're both there for a 5 point run. Who makes more money? The differences will show up as a result of how many come out 7's and 11's are thrown and how many come out craps. We know in advance that betting the same dollars per bet the 2X odds bettor beats the 1X odds with a flat bet bettor.
Without seeing the numbers, I've just answered my own question about who would win more, or in other words...which method is better during a hot streak. The answer is dependent on the number of come out 7's and 11's and the number of come out 2's and 3's. If there were none of either then method A makes more money. To whatever combination of come out 2's, 3's, and 7's and 11's there are will decide the winner of that round.
Bankroll drawdown (variance) between Methods A and B will also be decided by the come out 2's, 3's, 7's and 11's. In between those hot multi-point rolls the more come out 7's and 11's there are and the fewer 2's and 3's there are Method B will rule. Any "likely" combination of 2's, 3's,7's and 11's could be looked at to provide an idea of how many points it takes for A to catch B in the case of more come out 7's and 11's than 2's and 3's showing. In the case of more come out 2's and 3's than 7's...Method A automatically wins. Hmmm...the conclusion would be that Method B outperforms Method A only when more come out 7's are rolled than come out 2's and 3's before point winners. To the extent that occurs before point winners will decide how long Method B maintains a lead once point winners start occurring.
Method A is 10/10/20 and Method B is 10/10... say a person is using the 5 Count to begin play... or anything to begin play. And say you and I are betting, I use method A, you use method B. Say the dice pass for what...what would be a good number of rolls to use...what have you experienced? That's what I'm talking about. We're both standing there grinding it out, waiting and waiting.... then boom we're both there for a 5 point run. Who makes more money? The differences will show up as a result of how many come out 7's and 11's are thrown and how many come out craps. We know in advance that betting the same dollars per bet the 2X odds bettor beats the 1X odds with a flat bet bettor.
Without seeing the numbers, I've just answered my own question about who would win more, or in other words...which method is better during a hot streak. The answer is dependent on the number of come out 7's and 11's and the number of come out 2's and 3's. If there were none of either then method A makes more money. To whatever combination of come out 2's, 3's, and 7's and 11's there are will decide the winner of that round.
Bankroll drawdown (variance) between Methods A and B will also be decided by the come out 2's, 3's, 7's and 11's. In between those hot multi-point rolls the more come out 7's and 11's there are and the fewer 2's and 3's there are Method B will rule. Any "likely" combination of 2's, 3's,7's and 11's could be looked at to provide an idea of how many points it takes for A to catch B in the case of more come out 7's and 11's than 2's and 3's showing. In the case of more come out 2's and 3's than 7's...Method A automatically wins. Hmmm...the conclusion would be that Method B outperforms Method A only when more come out 7's are rolled than come out 2's and 3's before point winners. To the extent that occurs before point winners will decide how long Method B maintains a lead once point winners start occurring.
...in the long run, you'll lose money.
November 4th, 2013 at 8:00:03 PM
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OK now
I was way off
what was I thinking!
sure, do/ dont and odds
almost doubles the expected value
here goes
let us see if you grasp my math results
these are the probabilities of a net LOSS after X rounds played
and I threw in a few $ amounts too
as to the variance and bankroll draw-downs
that will be later
I read you last post that got extended too
Sally
I was way off
what was I thinking!
sure, do/ dont and odds
almost doubles the expected value
here goes
let us see if you grasp my math results
pass 1x odds
net prob 1 in
-10 0.111111111 9.00
-20 0.395959596 2.53
10 0.222222222 4.50
22 0.12626263 7.92
25 0.08888889 11.25
30 0.05555556 18.00
sumproduct = -0.1414
pass 2x odds
net prob 1 in
-10 0.111111111 9.00
-30 0.395959596 2.53
10 0.222222222 4.50
34 0.12626263 7.92
40 0.08888889 11.25
50 0.05555556 18.00
sumproduct = -0.1414
do/dont 1x pass odds
net prob 1 in
0 0.305555556 3.27
-10 0.027777778 36.00
-10 0.395959596 2.53
12 0.126262626 7.92
15 0.088888889 11.25
20 0.055555556 18.00
sumproduct = -0.2778
do/dont 2x pass odds
net prob 1 in
0 0.305555556 3.27
-10 0.027777778 36.00
-20 0.395959596 2.53
24 0.126262626 7.92
30 0.088888889 11.25
40 0.055555556 18.00
sumproduct = -0.2778
these are the probabilities of a net LOSS after X rounds played
and I threw in a few $ amounts too
| method | 36 rounds | 108 rounds | 360 rounds | down $100+ | 3600 rounds | down $500+ | down $1000+ |
|---|---|---|---|---|---|---|---|
| pass/1xodds | 51.80% | 53.10% | 55.60% | 44.60% | 67.30% | 50.36% | 33.30% |
| pass/2xodds | 51.30% | 52.10% | 53.80% | 46.60% | 61.70% | 50.27% | 38.80% |
| do/dont/1xpass odds | 57.50% | 61.70% | 70.00% | 50.30% | 94.60% | 80.00% | 50.00% |
| do/dont/2xpass odds | 54.30% | 56.70% | 60.40% | 50.30% | 80.00% | 66.20% | 50.00% |
as to the variance and bankroll draw-downs
that will be later
I read you last post that got extended too
Sally
I Heart Vi Hart
November 4th, 2013 at 10:19:18 PM
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Sally I'm glad you think this is fun, I do too. Not because you're doing the work, rather it's because I love asking "what if" questions that can be solved with math scripts. Thank you for taking the time to not only do the work but also making such a nice presentation. I'm beginning work on an excel that will allow me to consider different ideas, thanks to your suggestions.
I feel very confident that I understand the results and they are very telling, perhaps even more so after I ask some questions. It's likely that I'll use incorrect terms but I'm also confident you'll get what I mean.
When you say "Rounds", do you mean individual rolls of the dice or resolved decisions?
Presuming you mean rolls, does your program presume a normal expected occurrence of all twelve numbers, ie., the 7 shows an average of 6 times per 36 rolls?
What I'm trying to ask in general is how were the numbers distributed? The reason I'm asking is it makes a huge difference when the numbers 2, 3, 7, 11, and 12 appear. Over the course of 108 rolls, the 7 is expected to appear 18 times and the 11 should appear 6 times. They will only benefit the pass line if they appear on come out rolls. We know that every 7 can't, shouldn't be "what if'd" to appear on the come out roll. If 30% of rolls are come out rolls, does your program assign about 16% of those come out rolls a 7, and so on?
I'm thinking that if any of the results are based on probability equations then the results are what I would term optimized. In other words, in a real casino over the course of 120 rolls where 36 rolls would be come out rolls...while we could expect 6 of those rolls to be 7's, maybe more would be or maybe less would be. Maybe during my period of play only 3 come out rolls are 7's and the rest are outs. Maybe the 2 and 3 are showing way more than their probabilities dictate.
I can see how a program can be written that reflects what the optimized outcomes would be using the probability equations and a few other "IF-THEN-ELSE" statements and all one does is change bet size and number of rolls. But that still forces the results to reflect predicted probability versus actual occurrence.
The better program, in my opinion, doesn't use predicted probability as a component. It randomly generates two numbers from 1 to 6, adds them together, and runs the sum through a set of simple statements and subroutines that mimic game play with user variables being bet size and roll number. Basically it's what the online practice craps tables do but without the graphics. Then check 100 1 hr sessions (108 rolls), and 100 2 hr sessions, and 100 3 hr sessions and so on.
I feel very confident that I understand the results and they are very telling, perhaps even more so after I ask some questions. It's likely that I'll use incorrect terms but I'm also confident you'll get what I mean.
When you say "Rounds", do you mean individual rolls of the dice or resolved decisions?
Presuming you mean rolls, does your program presume a normal expected occurrence of all twelve numbers, ie., the 7 shows an average of 6 times per 36 rolls?
What I'm trying to ask in general is how were the numbers distributed? The reason I'm asking is it makes a huge difference when the numbers 2, 3, 7, 11, and 12 appear. Over the course of 108 rolls, the 7 is expected to appear 18 times and the 11 should appear 6 times. They will only benefit the pass line if they appear on come out rolls. We know that every 7 can't, shouldn't be "what if'd" to appear on the come out roll. If 30% of rolls are come out rolls, does your program assign about 16% of those come out rolls a 7, and so on?
I'm thinking that if any of the results are based on probability equations then the results are what I would term optimized. In other words, in a real casino over the course of 120 rolls where 36 rolls would be come out rolls...while we could expect 6 of those rolls to be 7's, maybe more would be or maybe less would be. Maybe during my period of play only 3 come out rolls are 7's and the rest are outs. Maybe the 2 and 3 are showing way more than their probabilities dictate.
I can see how a program can be written that reflects what the optimized outcomes would be using the probability equations and a few other "IF-THEN-ELSE" statements and all one does is change bet size and number of rolls. But that still forces the results to reflect predicted probability versus actual occurrence.
The better program, in my opinion, doesn't use predicted probability as a component. It randomly generates two numbers from 1 to 6, adds them together, and runs the sum through a set of simple statements and subroutines that mimic game play with user variables being bet size and roll number. Basically it's what the online practice craps tables do but without the graphics. Then check 100 1 hr sessions (108 rolls), and 100 2 hr sessions, and 100 3 hr sessions and so on.
...in the long run, you'll lose money.
November 9th, 2013 at 10:20:49 AM
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The question seems to be how to minimize losses on the come out roll for the dark side players.
Are you going to lose more on the come out roll in the long run if you play pass/don't or just the don't. How much more likely is it that the 7/11 will show as opposed to the 2/3?
There is also a 1 unit loss when the point is established and a 7 is rolled that comes from your pass line bet that you otherwise wouldn't have if you just played don't.
Anyone able to weigh in on this?
Are you going to lose more on the come out roll in the long run if you play pass/don't or just the don't. How much more likely is it that the 7/11 will show as opposed to the 2/3?
There is also a 1 unit loss when the point is established and a 7 is rolled that comes from your pass line bet that you otherwise wouldn't have if you just played don't.
Anyone able to weigh in on this?
