Craps Strategy Investment Opportunity?
Everything from page 48 should be in another thread.
It is good stuff!
Will get run over when 98 returns.
Quote: guido11198steps thread has been hijacked again!
Everything from page 48 should be in another thread.
It is good stuff!
Will get run over when 98 returns.
I've yet to have it explained to me why so-called "thread hijacking" is more heinous than child molestation.
We haven't heard from 98steps for a while, so we are continuing a discussion on a related topic. When he wishes to post again, we will see that. If he has anything new to bring to the discussion, we will discuss that.
In the meantime, what's the problem?
Quote: goatcabin
Let's compare 100 $10 bets for each.
Why 100? Is it your idea of "long run"? :)
Try comparing 50 bets - you'll get about 38% of single-number bet sessions ahead vs. about 40% of even money bets.
Or how about 106 bets - 30.5% vs 32.8% ...
Simply put, the effect you are after is an illusion. Large standard deviation increases the fluctuations, but, like I repeatedly said before, it plays both ways. Sometimes, it helps you win, sometimes it makes you lose.
BTW, a more commonly accepted measure for the "amount of luck" you are (incorrectly) trying to approximate as ev/sd is known as probability ;)
Quote: goatcabinNo, the word "will" is inappropriate, as is "the more you bet" without distinguishing between betting more, once, and betting many times. If I bet $1000 on the pass line, my probability of winning is .4929. OTOH,
if I bet $10 100 times, my probability of coming out ahead is more like .44. etc. etc.
The formulas for the math are:
48.33067% chance of being even or ahead. (50 win/50 loss)
Excel formula:
=1-(BINOMDIST(49,100,244/495,TRUE))
40.45094% chance of being ahead $20 or more. (51 win/49 loss or more wins)
Excel formula:
=1-(BINOMDIST(50,100,244/495,TRUE))
More:
20.00554% chance of being ahead $80 or more. (54 win/46 loss or more wins)
Excel formula:
=1-(BINOMDIST(53,100,244/495,TRUE))
1.23093% chance of being ahead $220 or more. (61 win/39 loss or more wins)
Excel formula:
=1-(BINOMDIST(60,100,244/495,TRUE))
Yesterday I was lucky to have 61 wins and 39 losses just making $10 pass line bets.
Funny thing seeing Alan's example.
Took about 3 hours of play.
And I was up $220!
My buy-in was only $40. I had misplaced my wallet and only had $40 on me. Yes, I did later find my wallet.
I figured I was going to bust out real quick, but I was really surprised after I had won $200. I was so drunk and having fun I never even took odds.
The players at the table taking odds on the pass line did very, very well.
I spent the $220 wisely as you can imagine one can in Vegas. hehe
Quote: mkl654321I've yet to have it explained to me why so-called "thread hijacking" is more heinous than child molestation.
I don't think the problem is so much that this thread has been heinously highjacked. Sometimes that does happen in a counterproductive way, but I don't think that's the issue here.
In this case, it'd be good if your long-run related conversation went to a different thread for archival purposes. 4 months from now when others want to review it, it'll be a lot easier to find a separate thread on that topic than it will be to remember that your conversation on that topic occurred in the middle of this unrelated thread.
Quote: rdw4potusI don't think the problem is so much that this thread has been heinously highjacked. Sometimes that does happen in a counterproductive way, but I don't think that's the issue here.
No problem.
Since we are waiting for 98steps to appear from his voting in CA, as said in a past post...
How about some stats.
I did start a new thread.
| Member | Join Date | Days to Post | Posts | Day Avg. |
|---|---|---|---|---|
| Wizard | 14-Oct-09 | 386 | 2133 | 5.5 |
| DJTeddyBear | 2-Nov-09 | 367 | 1973 | 5.4 |
| pacomartin | 14-Jan-10 | 294 | 1714 | 5.8 |
| Nareed | 11-Nov-09 | 360 | 1445 | 4.0 |
| mkl654321 | 8-Aug-10 | 86 | 1423 | 16.5 |
| FleaStiff | 19-Oct-09 | 381 | 1059 | 2.8 |
Have had no further sessions to date. My trip home to California has unfortunately been extended. I am disapoined that my last trip relsuted in a loss, albeit a small loss. On the positive side, i did gather a lot of hopefully valuable information.
While the bus system in Vegas is Very Good, I have come to the conclusion that it will be necessary for me to have a vehicle when i return.
After discussions with my investors, we feel that the delay in actual start is acceptable. My 70 year old father has requested my help in preparing his home for winter. New timeline yet to be determined.
Quote: 98stepsStatus Update...........
Have had no further sessions to date. My trip home to California has unfortunately been extended. I am disapoined that my last trip relsuted in a loss, albeit a small loss. On the positive side, i did gather a lot of hopefully valuable information.
While the bus system in Vegas is Very Good, I have come to the conclusion that it will be necessary for me to have a vehicle when i return.
After discussions with my investors, we feel that the delay in actual start is acceptable. My 70 year old father has requested my help in preparing his home for winter. New timeline yet to be determined.
Seems living in Vegas would really help you out.
You are a good son for helping out Dad. I hope your Dad does not live in snow country.
Quote: weaselmanWhy 100? Is it your idea of "long run"? :)
Just a nice, round number, an example to counter your ridiculous example of one bet.
Quote: weaselmanTry comparing 50 bets - you'll get about 38% of single-number bet sessions ahead vs. about 40% of even money bets.
Or how about 106 bets - 30.5% vs 32.8% ...
That last is way, way off. In 106 single-number bets, you need to win three to be ahead, and the probability of winning three or more is actually .53. However, there is something else going on here, due to the skew of the single-number bet. Every 36 bets, you need one win to break even. So, just before you get to 36n trials, your chances of getting that "extra" win are maximized, then right after that your chances of being ahead drop, because you need another win. For example:
37 bets - need two to be ahead, probability is about .25
50 bets - need two to be ahead, prob ~.38
71 bets - need two to be ahead, prob ~.56
73 bets - need three to be ahead, prob ~.30
So, these particular bets are not the best examples. Somewhere, either in this thread or another, I think I compared passline w/single odds to passline w/3,4, 5X odds, bets that are resolved in parallel. Here again, the expected losses are the same for both strategies, but the SD is much higher for 3, 4, 5X odds. I can't find that post, so I ran some numbers again. For the straight passline, no odds, the SD and ev are almost exactly equal after 5000 bets. So, I compared single vs. 3, 4, 5X odds for that many bets. The ev's, of course, are the same, -$353.54, and the SD's are $692 and $1738. So, the ev/SD figures are .511 and .203. This means that the gambler needs to come out about 1/2 of a standard deviation better than expectation to break even or better with single odds, but only .2 SD taking the larger odds multiples. The probabilities associated with those, taken from the usual table of "Z values", are about .31 and .42. This means that the player taking more odds has a higher probability of breaking even or better, simply because he/she takes more advantage of good luck. Look at it another way -- if a million players play each way for 5000 bets, some 11,000 more of the high-SD players would be ahead than in the single-odds group.
Quote: weaselmanSimply put, the effect you are after is an illusion. Large standard deviation increases the fluctuations, but, like I repeatedly said before, it plays both ways. Sometimes, it helps you win, sometimes it makes you lose.
To quote you from another post:"High standard deviation (or, equivalently, variance) does increase the likelihood of a win, it also increases a likelihood of a large loss. You are actually more likely go bankrupt betting progressively on a high-SD game."
To quote myself, from another post:"The higher odds multiple, the more the variance, without adding any expected loss. Just don't forget: variance works both ways!
In the above case, the ev for 60 $10 passline bets, taking full 6, 8, 10X odds, is just -$8.48, but the standard deviation is $696. The probability of coming out one standard deviation worse than the expectation is about 16%, so the player has about a one-in-six chance of busting a $700 bankroll within a couple of hours of this play. If the player starts with $1000, it would take about 1.4 SD below expectation, with a probability of about .08, or about one chance in 12 of busting. The player needs to decide what "risk of ruin" he/she is prepared to accept and set the bankroll/odds level accordingly."
I don't think you understand the "effect" I am "after". I have never equated the probability of breaking even or better, for a session or a lifetime, with having a positive expectation, "defeating" the ev". The percentage of winning sessions, or the percentage of individual players who are ahead after any number of bets, increases with higher variance. That does not imply that the mean session outcome, or the net of all the players combined, is positive. Of course, high variance increases the magnitude of losses when luck is bad. Of course, high variance increases the "risk of ruin". It's the other side of the coin. You don't get something for nothing. One of my favorite expression on rec.gambling.craps was "you squeeze the balloon on one side, it just bulges out on the other". You put in a loss limit, you get more, but lower, losses. You put a win goal, you get more, but lower wins. etc.etc.
Quote: weaselmanBTW, a more commonly accepted measure for the "amount of luck" you are (incorrectly) trying to approximate as ev/sd is known as probability ;)
You are not understanding this, either. The ev/SD tells you how lucky you have to be to break even or better, i.e. what multiple or fraction of a standard deviation better than expectation will equal the expected loss. The "amount of luck" is the multiple or percentage of a standard deviation by which one's results differs from the ev.
Here's a more practical illustration of what I'm talking about. My friend Paul and I recently spent three days in Reno, mostly playing craps. He is a more aggressive player than I. While we both bet the passline with some odds, he also makes hardway bets and yo bets, which I mostly avoid. When the dice are passing, we both win, but he is winning more than I am. We are experiencing the same dice rolls, the same W-L percentage on the line, the same point wins, but he is winning the passline plus the 15:1 on a comeout 11, while I am just winning even money. Some of the point wins are also on hardways that he has money on, while I am winning the line and odds only. We both had roughly the same "amount of luck", but his higher variance took better advantage of it. This doesn't mean he was a "better" craps player than I. We both know that, if we had had bad luck, he would have lost a lot more than I would have. We also both know that, if he plays that way long enough, the higher HA on those center bets is going to become more and more significant.
Cheers,
Alan Shank
Woodland, CA
Quote: goatcabin
That last is way, way off. In 106 single-number bets, you need to win three to be ahead,
Nope, need 4. 350*3 = 1050 < 106*10
Quote:I don't think you understand the "effect" I am "after".
I think, I do ;)
Quote:I have never equated the probability of breaking even or better, for a session or a lifetime, with having a positive expectation, "defeating" the ev".
And I have never accused you of anything like that either.
Quote:The percentage of winning sessions, or the percentage of individual players who are ahead after any number of bets, increases with higher variance.
Well ... sometimes it does, and sometimes it doesn't (as you have just seen).
Quote:
You are not understanding this, either. The ev/SD tells you how lucky you have to be to break even or better,
You are the one not understanding it. The probability of breaking even is what tells you how lucky you need to be to break even.
ev/SD maybe an OK (yet, still inaccurate) estimate for it for some distributions, but is way off for the others.
More importantly, there is simply no need to bring this new invention into the picture, because everything that needs to be quantified already is with existing statistics.
Quote: weaselmanNope, need 4. 350*3 = 1050 < 106*10
Sorry, if you win 3 of 106, then you lose 103, not 106, right? 103 * = 1030, so you're net +20.
My quote:"I don't think you understand the "effect" I am "after"."
You:I think, I do ;)
OK, how would you describe this effect I'm after?
Quote: weaselmanAnd I have never accused you of anything like that either.
I quote you:"If the distribution is symmetric, increasing SD raises your chances of winning as much (actually, slightly less) as your chances of loosing. Yes with a huge SD you can win a lot, but you can lose a lot as well, so on average, it's the same."
To me, that implies you think I don't understand that the ev is going to be the same.
Also, I am not talking about HOW MUCH you win in the winning sessions, just that the probability of having a winning session (or being a winning player) is increased with a higher SD. You can't increase the probability of winning AND the probability of losing.
Another quote:"It's misleading though for the reason I mentioned above. If SD is large, you may not need as much "luck" to win, but at the same time it doesn't take much "lack of luck" to go bankrupt either."
Here again, you are lecturing me as if I didn't understand the whole picture of the graph, i.e. that high variance increases the AMOUNTs of losing sessions.
Quote: weaselmanYou are the one not understanding it. The probability of breaking even is what tells you how lucky you need to be to break even.
ev/SD maybe an OK (yet, still inaccurate) estimate for it for some distributions, but is way off for the others.
More importantly, there is simply no need to bring this new invention into the picture, because everything that needs to be quantified already is with existing statistics.
So, how would you calculate the probability of breaking even for 5000 passline bets with 3, 4, 5X odds?
I use existing probability calculations to get the ev and SD, divide and look the resulting number up in the Z table.
Alternately, I could fire up WinCraps and run 10,000 5000-bet sessions, and I would get very similar figures.
Do you have a method for directly calculating the probability of coming out ahead for 5000 passline bet with 3, 4, 5X odds?
Cheers,
Alan Shank
Woodland, CA
Quote: goatcabin
Another quote:"It's misleading though for the reason I mentioned above. If SD is large, you may not need as much "luck" to win, but at the same time it doesn't take much "lack of luck" to go bankrupt either."
Here again, you are lecturing me as if I didn't understand the whole picture of the graph, i.e. that high variance increases the AMOUNTs of losing sessions.
I am not lecturing you, just pointing out that it is misleading to focus on the positive deviation and ignore the fact that large SD spreads the distribution both ways. I am not saying that you don't understand it, just that the statements you make ignore the left side of the curve, and are misleading because of that.
Quote:So, how would you calculate the probability of breaking even for 5000 passline bets with 3, 4, 5X odds?
I use existing probability calculations to get the ev and SD, divide and look the resulting number up in the Z table.
Alternately, I could fire up WinCraps and run 10,000 5000-bet sessions, and I would get very similar figures.
Do you have a method for directly calculating the probability of coming out ahead for 5000 passline bet with 3, 4, 5X odds?
I don't know craps, so not sure exactly what you are asking, but I am guessing that it should not be much different fom how it works everywhere else - you figure the probability of a single win, how many total wins you need, and use binomial distribution to get the final result.
Or you could run a simulation if you find it easier - the fraction of sessions won in your simulation is an approximation of the probability.
Z table (if I understand correctly what you mean by that) is for normal distribution. It's ok to use it for others as long as you understand that what you are getting is a more or less rough estimate of the real thing, which is the probability, and do not attempt to put the picture upside down, and pretend that your ev/sd has meaning of its own.
Quote: weaselmanI am not lecturing you, just pointing out that it is misleading to focus on the positive deviation and ignore the fact that large SD spreads the distribution both ways. I am not saying that you don't understand it, just that the statements you make ignore the left side of the curve, and are misleading because of that.
I quote myself, again:
"The higher odds multiple, the more the variance, without adding any expected loss. Just don't forget: variance works both ways!
In the above case, the ev for 60 $10 passline bets, taking full 6, 8, 10X odds, is just -$8.48, but the standard deviation is $696. The probability of coming out one standard deviation worse than the expectation is about 16%, so the player has about a one-in-six chance of busting a $700 bankroll within a couple of hours of this play. If the player starts with $1000, it would take about 1.4 SD below expectation, with a probability of about .08, or about one chance in 12 of busting. The player needs to decide what "risk of ruin" he/she is prepared to accept and set the bankroll/odds level accordingly."
I have focused on the fact that high variance takes better advantage of good luck, which results in a higher probability of a player being even or ahead after any number of bets, compared to a lower-variance strategy. I used the example of pass plus single odds vs. pass plus 3, 4, 5X odds, because the ev's are the same, only the "variance varying". Here's an interesting picture:
-3 SD -2SD -1SD ev +1SD +2SD +3 SD
-2429 -1738 -1046 -354 +338 +1030 +1722 pass line, single odds
-5568 -3830 -2092 -354 +1384 +3122 +4860 pass line, 3, 4, 5X odds
Those outcomes are equally likely between the two players. Now the point I am trying to make is that the "zero points", the points at which each player breaks even, are in different places. If we blow up part of the above:
ev +1SD
-354 0 +338
-354 0 +1384
Quote: weaselmanI don't know craps, so not sure exactly what you are asking, but I am guessing that it should not be much different fom how it works everywhere else - you figure the probability of a single win, how many total wins you need, and use binomial distribution to get the final result.
Or you could run a simulation if you find it easier - the fraction of sessions won in your simulation is an approximation of the probability.
Now we're getting somewhere. Well, when you are taking odds, you cannot just do a binomial to get percentage of wins, because it depends on how you win, comeout, which points win and which lose at what rates. That is EXACTLY why I use the method I do. And it is supported when I do a simulation. When you have a strategy with flat betting, you can just calculate the ev and SD using n * ev and n^.5 * SD and get a good estimate. When you get more complicated, with different types of bets and/or progressive/regressive betting, you just have to simulate. I have been doing these analyses for many years, using both the math and WinCraps or programs I have written myself (I am a retired "C" programmer).
It's very true that, for skewed bets like the 12 in craps or single-number roulette, the binomial is the way to go, although, again, if the bet amounts change you just have to simulate.
Quote: weaselmanZ table (if I understand correctly what you mean by that) is for normal distribution. It's ok to use it for others as long as you understand that what you are getting is a more or less rough estimate of the real thing, which is the probability, and do not attempt to put the picture upside down, and pretend that your ev/sd has meaning of its own.
Well, if you cannot directly calculate the probability of being ahead, which you cannot do for the example I'm using, then ev/SD is a shortcut you can use. I don't know what you mean by it having a "meaning of its own".
Cheers,
Alan Shank
Woodland, Ca
Quote: goatcabinI quote myself, again:
"The higher odds multiple, the more the variance, without adding any expected loss. Just don't forget: variance works both ways!
The second sentence directly contradicts the last part of the first one.
Quote:Well, when you are taking odds, you cannot just do a binomial to get percentage of wins, because it depends on how you win, comeout, which points win and which lose at what rates.
Just take an average, it'll be good enough. Better than ev/sd anyway.
Quote:
Well, if you cannot directly calculate the probability of being ahead, which you cannot do for the example I'm using, then ev/SD is a shortcut you can use. I don't know what you mean by it having a "meaning of its own".
I still don't see why you can't calculate the probability (if you are varying bets, the sd doesn't stay constant either, so, that's irrelevant).
But if you can't, and don't want to use a simulator, then yeah, you can use ev/sd for an estimate ... in the same sense as in which you can use EV as an estimate of how much you will lose :) These are both "long term" statistics becoming increasingly accurate as number of trials approaches infinity.
Quote: weaselmanThe second sentence directly contradicts the last part of the first one.
No, it doesn't. I quote:""The higher odds multiple, the more the variance, without adding any expected loss. Just don't forget: variance works both ways!"
You apparently do not understand craps. The odds bets behind the line bets pay true odds, so no expected loss is involved. The ev of pass betting is always just -.01414 time the amount of the bet(s). That is why, in the example I gave, the ev's were the same for single odds and 3, 4, 5X odds; they would have been the same for no odds, 10X odds or 100X odds. Variance works both ways, i.e. on either side of the ev, which is the same regardless of odds.
Quote: weaselmanJust take an average, it'll be good enough. Better than ev/sd anyway.
Then you're really ignoring variance.
Quote: weaselmanI still don't see why you can't calculate the probability (if you are varying bets, the sd doesn't stay constant either, so, that's irrelevant).
But if you can't, and don't want to use a simulator, then yeah, you can use ev/sd for an estimate ... in the same sense as in which you can use EV as an estimate of how much you will lose :) These are both "long term" statistics becoming increasingly accurate as number of trials approaches infinity.
I am not interested in infinity. We all know what happens when the number of trials gets very, very large, but no player ever plays that much. When I first started using WinCraps or a program of my own for this stuff, I just fired it up and ran it for more trials than anyone could ever possibly experience, but the results weren't that interesting, because that's not the way players experience gambling. They don't have one big, lifetime bankroll and just play, play, play until it's gone. Mostly, they take a trip to Vegas and play craps, as mkl suggests on that other thread, for 2-3 sessions/day for a couple of hours, then go back home and work some more, earning enough to have some extra for recreation, and start all over. The experience is all about sessions, and I doubt that most people keep track of where their overall net is at any point in time. Players don't get to the "long term", so it seems to me they are more interested in knowing what a distribution of their typical sessions will look like. That's what WinCraps can do now. Steen introduced the "Games Log", which keeps track of lots of different parameters (# of rolls, # of bets, bet handle, net) for each session and presents these either as data (including mean, median, mode, SD, skew, kurtosis) or in histogram form. Of course, it also keeps track of the totals, so you get both pictures. You can also look at pieces of the distribution. For example, you can find out how many of the sessions resulted in wins of a certain amount or more, how many resulted in losses of a certain amount or more, how many busted, etc. etc. This gives you a reasonable idea of what is likely to happen in a given session, in the same way that you know you have a .4929 probability of winning a pass bet, or that there's a probability of 1/6 that the next roll of the dice will be a seven.
Another approach is to estimate how many hours you might play in your "craps career", then simulate that many hours' play 10,000 times or so and see what the distribution looks like.
Cheers,
Alan Shank
Woodland, CA
Quote: goatcabinNo, it doesn't. I quote:""The higher odds multiple, the more the variance, without adding any expected loss. Just don't forget: variance works both ways!"
Quoting the same thing over and over, doesn't make it less contradictory.
I guess, I'll just have to quote myself too then ;)
"The second sentence directly contradicts to the last part of the first one"
Quote:Then you're really ignoring variance.
No, I am not.
Quote:I am not interested in infinity.
I know. You keep saying that you don't like the EV as an estimate of your potential loss for that reason. Yet, you insist on using the EV/SD to estimate the probability of a win.
It is the other contradiction in your position, because both estimates are based on the Central Limit Theorem, and work well at large number of trial, with accuracy rapidly decreasing in "short run".
You are trying to "have the cake and eat it too" here. Either you accept the "long run" stuff and asymptotic statistics or you don't.
It's not my argument, but I am trying to understand what this is all about.Quote: weaselmanQuoting the same thing over and over, doesn't make it less contradictory.
I guess, I'll just have to quote myself too then ;)
"The second sentence directly contradicts to the last part of the first one"
weaselman, it appears that you are saying that "variance works both ways" contradicts "without adding expected loss". Pulling out the phrases as quotes like that makes the sentence a little awkward. I think you are saying that "variance works both ways" contradicts a claim that "wagering greater odds behind the pass line does not add anything to (nor reduce) the expected loss." Did I interpret this correctly? If not, please help me understand your position a little better.
If I did interpret your position correctly, let me express what I think is goatcabin's view: The odds bet is paid/collected without any house advantage or disadvantage. Whether you bet $5 or $1000 behind the line, the expected value (long term average outcome or however you want to describe it) of wins and losses on the odds bet itself is exactly $0. The expected loss of the total line-plus-odds bet is strictly the expected loss on the line bet alone.
Individual outcomes can be and will tend to be different from the expected value and we only see the long-term average value after you average them all together. Increasing an odds bet increases the variance associated with the odds bet, not the expected value ($0). The expression that "variance works both ways" is not a comment about expected value or long term average outcome. With increased variance, those individual outcomes will be more widely distributed -- the wins will be bigger wins and the losses will be bigger losses, but still with the same long-term average. It "works both ways" means you can't count on the idea that increasing odds bets (and variance) will necessarily result in either better results or worse results. You pay your money and take your chances.
I think this disagreement may be related to the facts that expected value is (like median, mean, and mode) a measure of "central tendency" while variance is a measure of "dispersion".
Did I represent both weaselman's and goatcabin's positions accurately? Where is the remaining disagreement? Before I struggle through reading more of these posts, I would like to understand what the argument is, if any.
Quote: DocThe odds bet is paid/collected without any house advantage or disadvantage.
Yes, with this correction there is no contradiction. If expected loss is zero, it is indeed not affected by the amount of the bet.
But since it was not mentioned in the original quote, it was self-contradictoryas stated.
Good. I think the misunderstanding was probably because goatcabin assumed that you already knew that a craps odds bet has no advantage or disadvantage (in terms of expected value) for either the house or player and didn't bother to mention it. This is why it is often called a "free odds" bet.Quote: weaselmanYes, with this correction there is no contradiction. If expected loss is zero, it is indeed not affected by the amount of the bet.
But since it was not mentioned in the original quote, it was self-contradictoryas stated.
Issue resolved, I hope.
Quote: weaselmanQuoting the same thing over and over, doesn't make it less contradictory.
Now, this is really dishonest. You completely left out my explanation of WHY it isn't contradictory. I will add it here:
"You apparently do not understand craps. The odds bets behind the line bets pay true odds, so no expected loss is involved. The ev of pass betting is always just -.01414 time the amount of the bet(s). That is why, in the example I gave, the ev's were the same for single odds and 3, 4, 5X odds; they would have been the same for no odds, 10X odds or 100X odds. Variance works both ways, i.e. on either side of the ev, which is the same regardless of odds."
I even saw in a later exchange between you and "Doc", that you agreed. Why did you omit my explanation, then agree with his?
Quote: weaselmanNo, I am not.
This needs some context.
goatcabin: Well, when you are taking odds, you cannot just do a binomial to get percentage of wins, because it depends on how you win, comeout, which points win and which lose at what rates.
weaselman: Just take an average, it'll be good enough. Better than ev/sd anyway.
goatcabin: Then you're really ignoring variance.
Here's why you are ignoring variance. Where do you think the additional variance comes from, when you take odds or higher odds? The ev of all odds bet is zero, because the payoff matches the odds against winning. The additional variance comes in because, even if you have the same W-L record as someone standing beside you who's only betting the flat part, you may win the outside points more often, you may win more comeouts and lose more points, etc. etc. If you take an average, you are ignoring that variance, and the probability of coming out ahead will be the same, regardless of whether you take odds at all, or in what multiples. That is clearly not the case.
Quote: weaselmanYou keep saying that you don't like the EV as an estimate of your potential loss for that reason. Yet, you insist on using the EV/SD to estimate the probability of a win.
The ev is like a "handicap" that you start with, whether it's for one bet or thousands. If there were no variance, then you would always lose. If you bet pass and don't pass at the same time, you eliminate almost all variance, the only remaining factor being how many 12s you get, where you lose the pass and the DP is a push.
Cheers,
Alan Shank
Woodland, CA
Geeez! Did I bump into a hornet's nest? I was just trying to help resolve what seemed to be an excessively long-running disagreement. How about this as an answer to your question: the postings had gotten so long and convoluted that I couldn't read through them -- maybe weaselman got caught in the same trap and overlooked your explanation in a rush to post his comment. In any case, I think (hope) that the two of you agree on things now.Quote: goatcabin... I even saw in a later exchange between you and "Doc", that you agreed. Why did you omit my explanation, then agree with his? ...
Quote: DocGeeez! Did I bump into a hornet's nest? I was just trying to help resolve what seemed to be an excessively long-running disagreement.
I have no problem with your post at all, Doc. I objected to weaselman's implying that I simply repeated my statement, without any reasoning behind it, then doing the same thing. I think that is dishonest.
Quote: DocHow about this as an answer to your question: the postings had gotten so long and convoluted that I couldn't read through them -- maybe weaselman got caught in the same trap and overlooked your explanation in a rush to post his comment.
Perhaps. I hope he actually reads and thinks about my explanation of why averaging the comeout/point outcomes simply eliminates the variance before he just posts again, "No, I'm not." >:-)
Cheers,
Alan Shank
Woodland, CA
Quote: Docthe postings had gotten so long and convoluted that I couldn't read through them -- maybe weaselman got caught in the same trap and overlooked your explanation in a rush to post his comment. In any case, I think (hope) that the two of you agree on things now.
A good point.
This thread is still about 98steps Craps Investment Opportunity and this becomes a large hassle in keeping up with the "related topic" posts that keep going and going.
goat and weasel, no one would complain if you moved over to a new thread.
I'm interested in what you have to say and I know others would be also.
yes, off to Europe!
Quote: goatcabinI I objected to weaselman's implying that I simply repeated my statement, without any reasoning behind it, then doing the same thing. I think that is dishonest.
Look, we had a discussion about -EV games. mkl said that if you are playing a -EV game, you will lose. You objected to that statement.
Three pages later it turns out that you are talking about 0 EV. If you can convince yourself somehow, that this is me being "dishonest", fine, you can have it.
Quote: weaselmanLook, we had a discussion about -EV games. mkl said that if you are playing a -EV game, you will lose. You objected to that statement.
Three pages later it turns out that you are talking about 0 EV. If you can convince yourself somehow, that this is me being "dishonest", fine, you can have it.
goatcabin, the way I understand his statements, was talking about 2 different wagers that can be made at a Craps table..
1) pass line bet that has a HA of 7/495 (1.4141%)
AND
2)pass line ODDS bet that has a 0% HA.
#2 can not be made unless #1 is made first.
the EV is based off of 1) since 2) does nothing to change it. A -EV bet.
goatcabin also objected to the statement "you will lose" made by mlk and then showed why he objected. No need to put on any boxing gloves, save those for the system and dice control sellers!
Keep up the great examples.
Im sure 98steps is getting ready for his return trip to Vegas. Full bankroll, proper system bets and win goals.
Good Luck 98!
Quote: weaselmanLook, we had a discussion about -EV games. mkl said that if you are playing a -EV game, you will lose. You objected to that statement.
Three pages later it turns out that you are talking about 0 EV. If you can convince yourself somehow, that this is me being "dishonest", fine, you can have it.
The problem here is that you do not know much, if anything, about craps. The zero-ev bet, the "free odds", as they are called, can only be made when you have a line bet, which is a -ev bet. Most books about craps express the HA as ev/total bet handle, including the "flat" part and the odds, so that line betting with different levels of odds betting have the same ev, but different HAs.
I am done with this.
Cheers,
Alan Shank
Woodland, CA
Sorry, I don't have enough time (or energy) to muster through the rest of the thread. can someone give me the Cliff Notes?
-TheArchitect
Quote: TheArchitectDid he ever take the challenge? Did he just go straight to the casinos and lose?
No and yes.
Quote: TheArchitectWhat ever happened? I read this tread till page 23-24 and then I dunno, maybe a case of ADD kicked in, but I didn't have it in me to read the next 30+ pages. As far as I know, 98Steps submitted his system to JB, had it coded into WinCraps by himself and two others, and had one member proofread/correct poor English and instructions. fast forward to page (What is this one 55ish), and looking back a few, He was playing this method at some casinos and there were some tables of earnings posted. Lot's of $150 wins, and always one $2000+ loss. Did he ever take the challenge? Did he just go straight to the casinos and lose?
Sorry, I don't have enough time (or energy) to muster through the rest of the thread. can someone give me the Cliff Notes?
-TheArchitect
1. 98steps posted that he had just invented a new craps system based on progressive betting as losses occured. He was interested in taking the "system challenge" posted by MichaelBlueJay.
2. Much posting ensued. Seventeen years later, MichaelBlueJay and 98steps reached tentative agreement on the precise nature of the challenge.
3. For whatever reason, 98steps decided not to take the challenge after all, but instead, gathered up some "investors"' money, and trooped off to Vegas to try out his system under battle conditions.
4. A rogue virus escaped from the secret government lab, and now zombies walk the land.
5. 98steps reported results that were very consistent with a Martingale: many small wins, more than offset by a horrendous loss.
6. 98steps then left Vegas to attend to family matters. When we last heard from our hero, he was in California.
7. We await further news of 98craps' struggles with bated breath.
Quote: goatcabin
The problem here is that you do not know much, if anything, about craps.
That's no news. I told you before I did not know much about craps. But I do know this much about free odds.
The statement you disputed applied to -EV bets. The more you bet -EV, the more you will lose.
If the expectation is not negative, then, this statement obviously does not apply, but nobody ever said that it did, so now I just don't know what you've been arguing with.
Quote: mkl654321I'll answer the specific question that was asked: yes, there is almost certainly someone out there who would consider that.
However, your chances of finding that someone on this site are essentially zero. That's because the persons frequenting this site are either a) persons who realize that every betting system ever devised with the goal of defeating a negative expectation game is a steaming pile of horsecrap, or b) persons who believe in such systems, and as a consequence, don't have any money to invest in YOUR system.
Well said.
Or like a dealer was saying last Sat at the craps table while I played the DON'T, " Nobody can win at this game, nobody, the house always wins no matter what you do." The dealer was having a bad day. I have known him for years and he was complaining about life.
What prompted me to write now is that I just overhauled the rules for my $30,000 Betting System Challenge, and now it requires only 200,000 hands per session rather than a billion. You'd said that you weren't certain your system could last for a billion rounds but you were confident it could work for shorter runs, so maybe the revised challenge will be of interest to you. Of course, I have to advise you again though to do a proper computer test of your system before risking $1000 or more with me. Either way, good luck.
Quote: weaselmanThe more you bet -EV, the more you will lose.
The problem with that statement is that it does not include the all-important phrase "in the long run." If you do not think that a -EV game in some short runs will have the player ahead, you are sadly mistaken. If you agree that the player might be winning at some points, then it all boils down to the question of at what point do you stop playing.
Quote: SanchoPanzaThe problem with that statement is that it does not include the all-important phrase "in the long run." If you do not think that a -EV game in some short runs will have the player ahead, you are sadly mistaken. If you agree that the player might be winning at some points, then it all boils down to the question of at what point do you stop playing.
Of course, the degree of "long" in "long run" for the casinos and the players, taken as aggregates, does guarantee that the casinos win and the players lose, and the amount is going to be close to edge * action. However, it's extremely unlikely that any individual player has enough action to reduce the probability of being ahead anywhere near zero.
What is true is that, the more you bet -EV, the lower the probability that you will be ahead.
BTW, I will be surprised if 98Steps posts here again.
Cheers,
Alan Shank
Woodland, CA
If you look up my posts, I have a system that simulates the pass and come bets and comes up with 1.41% as expected over tens of millions of dice rolls.
The sim takes about 15 seconds to run on my pc and it uses a decent rng and the whole thing took me 30 minutes to implement after I realized probably nobody else would do it.
You might even be able to do it with some other math software, but no language is difficult for this level of programming. You can learn enough programming to modify my code in a matter of a few minutes.
But I would also agree that short of an infinite bankroll, there is no betting system that I know of that works.
You need to know something of the outcome of the roll to make a system that earns money.
Quote: goatcabinHowever, it's extremely unlikely that any individual player has enough action to reduce the probability of being ahead anywhere near zero.
Depends on what you mean by "unlikely", and "near" :)
The probability of being ahead after 5000 tosses of an unfair coin with a 2% bias is about 0.21% (using normal distribution approximation).
This is pretty "near zero" if you ask me.
Is 5000 trials too long? It's about 17 5-hour sessions of blackjack. About half a year if you play twice a month. How unlikely is it that "any player" gets this much action? Well, again, in my book, this is pretty likely.
And after 10000 trials, the probability is less than 0.0029%
Quote: weaselmanDepends on what you mean by "unlikely", and "near" :)
The probability of being ahead after 5000 tosses of an unfair coin with a 2% bias is about 0.21% (using normal distribution approximation).
This is pretty "near zero" if you ask me.
Is 5000 trials too long? It's about 17 5-hour sessions of blackjack. About half a year if you play twice a month. How unlikely is it that "any player" gets this much action? Well, again, in my book, this is pretty likely.
And after 10000 trials, the probability is less than 0.0029%
Well, as usual, you and I are posting at cross-purposes. Although craps is a negative-ev game, it also affords the opportunity to make zero-ev bets alongside low-ev ones (1.4%), and in amounts up to 100 times the flat bet. This vastly increases the probability of an individual player being ahead of the game after a large number of bets. Flipping a biased coin is (presumably) an even-money bet, so it provides very little variance. Same thing, to a lesser extent, with blackjack. Also, in craps 5000 passline decisions translate to almost 17,000 dice rolls, roughly 170 hours' play, about twice what you figure for blackjack. Even at 10,000 decisions, craps passline with 3, 4, 5X odds has a standard deviation 3 1/2 times the expected loss. In a large group of players betting this way, we would expect over 35% of them to be even or better. And, yes, we all know that the variance is symmetrical about the ev, not the zero point.
But even a probability of .0021 I don't consider "near zero". Here again, keep in mind that there are a hell of a lot of gamblers out there. So even in that low-variance scenario of the biased coin, somewhere in the region of 2100 out of a million players would be expected to be ahead, per your figures, which I believe are based on an HA of 4%, i.e. 48 wins, 52 losses expected out of 100. If you use a 2% HA, 49 vs. 51, the probability is close to .08, and .023 for 10,000 decisions. For craps, even with no odds at all, for 10,000 decisions the probability of being even or better is almost .08.
Happy New Year,
Alan Shank
Woodland, CA
Quote: goatcabinFlipping a biased coin is (presumably) an even-money bet, so it provides very little variance.
The variance is proportional to the bet amount, you can make it as high as you want.
Quote:But even a probability of .0021 I don't consider "near zero". Here again, keep in mind that there are a hell of a lot of gamblers out there.
It doesn't matter how many gamblers are out there. If the probability of being up after half a year of playing is 0.0021, it means that you will be up about once in 238 years.
Quote: goatcabinFlipping a biased coin is (presumably) an even-money bet, so it provides very little variance.
Quote: weaselmanThe variance is proportional to the bet amount, you can make it as high as you want.
Yes, and for flipping a coin, the standard deviation is just about the same as the bet amount, whereas in my examples it's considerably higher.
Do you agree that your example is a 4% HA, BTW?
Cheers,
Alan Shank
Woodland, CACheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
With 1% HA, the probability of breaking even after 60000 trials is 0.7%. 60000 trials is about 100 sessions - still not something a typical player is unlikely to reach in his lifetime - I'd say, it is between 2 and 4 years of reasonably intensive playing.
(Note, this is not 2-4 years before you lose, it's 2-4 years before it becomes extremely unlikely that you are up).
Of course, if you keep playing, you can still expect to come up ahead once in every 143 4-year intervals :)
Quote: weaselmanThe variance is proportional to the bet amount, you can make it as high as you want.
Of course, as you raise the bet amount and the variance, you also raise the expected loss, so the relationship between the ev and the SD does not change at all, nor does the probability of breaking even or better, which is dependent on that ratio.
Cheers,
Alan Shank
Woodland, CA
Quote: weaselmanYes, I agree it is 4% HA.
With 1% HA, the probability of breaking even after 60000 trials is 0.7%. 60000 trials is about 100 sessions - still not something a typical player is unlikely to reach in his lifetime - I'd say, it is between 2 and 4 years of reasonably intensive playing.
(Note, this is not 2-4 years before you lose, it's 2-4 years before it becomes extremely unlikely that you are up).
Of course, if you keep playing, you can still expect to come up ahead once in every 143 4-year intervals :)
And, of course, you are entirely ignoring my point about craps as opposed to coin-flipping or blackjack.
Cheers,
Alan Shank
Woodland, CA
Quote: goatcabinAnd, of course, you are entirely ignoring my point about craps as opposed to coin-flipping or blackjack.
No, I am not. But I don't know what I can do with the statement "SD is higher". It seems a bit too vague to be meaningful.
Quote: weaselmanNo, I am not. But I don't know what I can do with the statement "SD is higher". It seems a bit too vague to be meaningful.
In your examples, the SD is very close to the amount of the bet, since it's an even money bet. In my craps example of pass line w/3, 4, 5X odds, the standard deviation is almost five times the amount of the flat bet, the only part with a negative ev.
In your examples, you took the difference between the number of wins to break even and the expected number of wins, and divided by the standard deviation. This gives you a Z score -- the number of standard deviations or fraction thereof required to reach that number of wins (or more). The area under the standard normal curve to the "right" of that figure gives you the probability of breaking even or better. I get the same figure you do, so we're both on the same page. BTW, this is the same result you get if you just divide the expected value by the standard deviation.
For craps pass line w/3, 4, 5X odds, the expected value is -.01414 of the FLAT BET, and the standard deviation is 4.915 times the amount of the flat bet, or 347.6 times the ev. Since the ev increases in magnitude proportionally with the number of bets, while the SD increases proportionally with the square root of the number of bets, the ratio of ev/SD gets higher, eventually reaching unity, where the probability of breaking even or better (p0+) is about .16, two, where p0+ = .023, and three, where p0+ = .0013, etc. With craps pass line w/3, 4, 5X odds, it takes 120,837 decisions for the ev/SD to reach unity, 483,351 to reach two and 1,087,540 to reach three. And some casinos allow 5X, 10X or even 100X odds.
It's this ability to pile up variance without increasing expected loss that makes craps so attractive to many gamblers. For 10,000 bets (around 340 hours' play) it doesn't even take 1/3 of a standard deviation of "good luck" to break even or better. Without the odds bets, it takes +1.4 SD to break even.
Cheers,
Alan Shank
Woodland, CA
Quote: goatcabinIn your examples, the SD is very close to the amount of the bet, since it's an even money bet.
I don't know what you mean by this.
The SD of 60,000 bets at 1% HA is 122.47. The SD of 1 bet is almost exactly 0.5.
Neither of those numbers seem "very close tp the amount of the bet" (60000 or 1 respectively) .
Quote:
In my craps example of pass line w/3, 4, 5X odds, the standard deviation is almost five times the amount of the flat bet, the only part with a negative ev.
Can you show me the calculations you use to arrive at this number?
Why are you only considering the initial bet? Even though it is the only one with -ev, you are putting the whole amount at risk, and the whole amount is factored into the calculation of the final low EV.
Quote:
IBTW, this is the same result you get if you just divide the expected value by the standard deviation.
Again, I am not sure what you are talking about here. Dividing ev of a single bet (-0.01) by its SD (0.5), I get -0.02
Dividing the EV of 60K bets (-600) by its respective SD (122.47), I get -4.9.
Are you saying these numbers signify some mathematical quantity to you? What is "the same result" you are referring to?
In general, I agree with you that the probability of breaking even in craps is higher because you are putting more money at risk to increase your SD, but respectively higher is the probability of losing a big fortune.
Quote: weaselmanI am afraid, I don't understand very much of what you are saying (beyond the qualitative point you already made that SD is higher). Perhaps, if you showed some calculations that you used to arrive to these conclusions, it might help me see the light, but so far it's not making very much sense. Here are some of my points of contention:
Quote: goatcabinIn your examples, the SD is very close to the amount of the bet, since it's an even money bet
Quote: weaselmanI don't know what you mean by this.
The SD of 60,000 bets at 1% HA is 122.47. The SD of 1 bet is almost exactly 0.5.
Neither of those numbers seem "very close to the amount of the bet" (60000 or 1 respectively) .
OK, I see our problem (or one of them). You are talking about the standard deviation of the number of wins, while I am talking about the standard deviation of the BET OUTCOME.
In an even-money bet, the standard deviation of the outcome, per bet, is very close to the amount of the bet, as figured below:
std = (x + 1)* sqrt(p*91-p))
where x is the payoff, in this case, 1.
.495 * .505 = .249975^.5=.4999749*2 = .9999498
So, if the bet amount is $5, the SD is $4.999749.
If you do it longhand, it's the same thing. The ev is -.05, so
5 - -.05 = 5.05^2 = 25.5025 * .495 = 12.6237
-5 - -.05 = 4.95^2 = 24.5025 * .505 = 12.3738
-----------
24.9975 = variance, stddev = 4.99975
ev for 60,000 bets is -$3000
SD for 60,000 bets is 4.99975 * sqrt(60,000) = $1224.68
ev/SD is 2.449, which yields a probability of .0071.
In these cases (coin flips with various biases or other even-money bets) the same result is achieved whether you treat the number of wins or the outcomes. If the bet is paid off at other than even money, you can still use the number of wins, then adjust for the payoff. However, with the craps line bets backed by odds, the payoffs depend on whether the bet is resolved on the comeout or after a point, and if after a point, by which point, since the payoffs are different for 6/8, 5/9 and 4/10. So you have to deal with the bet outcomes.
Quote: goatcabinIn my craps example of pass line w/3, 4, 5X odds, the standard deviation is almost five times the amount of the flat bet, the only part with a negative ev.
Quote: weaselmanCan you show me the calculations you use to arrive at this number?
Why are you only considering the initial bet? Even though it is the only one with -ev, you are putting the whole amount at risk, and the whole amount is factored into the calculation of the final low EV.
Here is how I calculated the standard deviation of the bet outcomes for pass line, 3, 4, 5X odds:
outcome - mean diff squared ways gross variance
5 - -.0707 = 5.0707 25.712 440 11313.28
-5 - -.0707 = 4.9293 24.298 220 5345.56
-25 - -.0707 = 24.9293 621.4700 264 164068.07
-20 - -.0707 = 19.9293 397.1770 220 87378.94
-30 - -.0707 = 29.9293 895.7630 300 268728.89
35 - -.0707 = 35.0707 1229.954 536 659255.34
----------
1196090 / 1980 = 604.086
Variance is 607.09, standard deviation is square root = 24.578.
The weights applied are based on the "perfect 1980", the smallest number of decisions where all the distinct outcomes can be expressed as integers. With 3, 4, 5X odds, any winning point pays $5 flat and $30 odds; this makes it easier for the dealers.
Once you have the SD for a single bet, you just multiply times the square root of the number of bets to get the SD for any given number of bets. You multiply the single-bet ev by the number of bets.
Quote: goatcabinBTW, this is the same result you get if you just divide the expected value by the standard deviation.
Quote: weaselmanAgain, I am not sure what you are talking about here. Dividing ev of a single bet (-0.01) by its SD (0.5), I get -0.02
Dividing the EV of 60K bets (-600) by its respective SD (122.47), I get -4.9.
Are you saying these numbers signify some mathematical quantity to you? What is "the same result" you are referring to?
Here again, you are using numbers of wins, while I am using outcomes. Actually, just above you have mixed an outcome (EV) with a number of wins (122.47), so that figure is meaningless. In your example before, you divided the DIFFERENCE between the number of wins to break even (30,000) and the expectation (29700) by the SD of the number of wins, getting 2.45, then looking that up in the "Z table", right?
I am doing essentially the same thing. In order to break even, one's result must be better than the ev by the amount of the ev. That difference is then divided by the SD to get the "Z value". It's the same principle.
So, here is how I derived the figures for the number of bets required to get Z values of 1, 2, and 3:
one-bet ev: -$.0707
one-bet SD: $24.578
For any number of bets, n, the ev is -.0707 * n, while the SD is 24.578 * sqrt(n).
If we want to find the number of bets where the ev is the same as the SD:
-.0707 * n = 24.578 * n^.5 divide both sides by n^.5
-.0707 * n^.5 = 24.578 divide both sides by -.0707
n^.5 = -347.64 square both sides
n = 120,852 (this is somewhat different, due to rounding, from my example)
Of course, for two or three SD, you just multiply 24.578 by 2 or 3.
BTW, I have verified this method by many, many simulations using WinCraps.
Quote: weaselmanIn general, I agree with you that the probability of breaking even in craps is higher because you are putting more money at risk to increase your SD, but respectively higher is the probability of losing a big fortune.
Yes, it cuts both ways, that's for sure. For every set of players between breaking even and winning $1750 over 10,000 decisions, we expect there's another set losing between $1400 and $3165 (between +.29 and +1 SD and -.29 and -1 SD). Sort of like insurance, no?
Cheers,
Alan Shank
Woodland, CA
