Buying variance

Discussion in 'Advanced Craps' started by goatcabin, Feb 10, 2010.

  1. goatcabin, Feb 10, 2010

    goatcabin

    goatcabin Member

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    Suppose you could get a couple of million people to visit their favorite casino and play craps for a couple of hours, betting nothing but the pass line. Each person would have enough bankroll to cover two hours' worth of betting, so no one would bust before two hours. Each person would report his/her net win or loss to me, and I would record and analyze the results. What would a graph of all the outcomes look like?

    Since the player would never bust, and would always play the full two hours, the graph would be a symmetrical bell-shaped curve, its peak height at about -1 unit, its slope such that almost 70% of the area under the curve would be between about -8.5 units and about +7 units and 95% between about -16 units and +14.5 units.

    The fact that the peak of the curve comes in the negative part of the graph is because of the house advantage of 1.414% built into the pass (and come) bet. It is actually the mathematical average of all the individual net outcomes, and is called the "expected value" or, in gambling, the "expected loss". The "house advantage" is usually expressed as a percentage, the "expected loss" as a number of units or a dollar value if we're talking about a specific bet amount.

    This "expected loss" is always there; think of it as a "handicap" that you have to overcome in order to break even or better. If your luck is exactly average, then you will lose about one unit in two hours' pass play, and something like one quarter of those couple of million players would be expected to end up between -2 and +2 units. But others would experience widely different results, perhaps winning or losing up to 30 units, or even a bit more. Unfortunately, for any given number of units we expect more people to lose that many than to win that many. That's because of the built-in "handicap" everybody faces.

    The fact that different people's results from a couple of hours' pass play will be so varied is due to what is called "variance". Without it, there would be no point to gambling. If you knew you were going to lose 1.414% of your total bet amount, you probably wouldn't play, right? But it's not only possible, it's expected that "your results will vary". When you make bets in craps, or any other game of chance, the house advantage is what you pay to buy variance. Variance is what gives you the chance to overcome that handicap and break even or win money from the casino.

    Most books on craps compare different types of bets on the basis of their house advantages. We all know the numbers: 1.414%, 1.515%, 4.0%, all the way up to 16.67% for the "Any Seven". The higher the house advantage, the bigger the "handicap" you have to overcome to break even or come out ahead. However, the probability of overcoming that handicap for any given number of bets depends heavily on the degree of variance built into the bet. Let's compare a couple of bets:

    pass line boxcars
    win one unit 30 units
    probability .4929 .0278
    lose one unit one unit
    probability .5071 .9722

    As a general rule, the larger the difference between the probabilities of winning and losing a bet, and the larger the difference between the payoff and the bet amount, the greater the variance. Remember that we are buying variance with the house advantage, and the more variance we buy, the more we have to pay. That is why the proposition bets and hardways carry a higher house advantage than the line and place bets. Higher variance means higher risk, for the player but also for the casino.

    To see how this affects one's chances of overcoming the HA handicap, let's examine 72 bets on the passline vs. the boxcars:

    expectation: pass 35W 37L = - 2 units
    12 2W 70L = -10 units

    one better pass 36W 36L = breakeven
    12 3W 69L = +20 units

    Because of the high variance, for the bet on the 12 all it takes is beating the expectation by one outcome to go from losing 10 units to winning 20 units! Of course, there's a flip side to that: variance works both ways.

    More later.
    Cheers,
    Alan Shank
     
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  2. goatcabin, Feb 10, 2010

    goatcabin

    goatcabin Member

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    To continue, the probability of overcoming the house advantage for any given number of bets is a function of both the expected loss and the variance, expressed as what we call the "standard deviation" (SD). I will not go into the calculation of the SD in this post, but if you get WinCraps from www.cloudcitysoftware.com, it gives you the SD for each bet, expressed as a percentage of the bet amount (I use a decimal fraction instead).

    pass bet boxcars
    expected loss -.0141 .1389
    SD .9999 5.0944

    In other words, the SD of a passline bet is essentially the amount of the bet, while the SD of the bet on the 12 (or any single combination paying 30:1) is over five times the amount of the bet. That's for one bet, but while the expected loss increases with the number of bets, the SD increases more slowly, with the square root of the number of bets.

    For any number of bets (n):
    pass bet boxcars
    expected loss -.0141 times n .1389 times n
    SD .9999 times sqrt n 5.0944 times sqrt n

    Remember that about 70% of the area under that curve was between -8.5 and +7 units? That is one standard deviation worse and better than the mean outcome. The 95% of the area between -16 and +14.5 units represents two standard deviations worse/better than the mean.

    For any bet, the probabilities of coming out one SD (or 2, or any multiple) better (or worse) than expectation are THE SAME. This means that the same amount of good luck is worth five times as much on a 30:1 bet as it is on a line bet (pass, come, don't pass, don't come).

    Let's look at those 72-bet comparisons again:

    pass boxcars
    expected loss -1.018 -10.0
    SD 8.484 43.2

    The expected losses are 72 times those for a single bet, but the SDs are only 8.485 (the square root of 72) times those for a single bet.

    Let's try to draw a picture of what this means:

    ------3--------2---------1--------E0------1-------2-------3-----

    The '0' indicates breaking even, the 'E' the expected value (loss) and the '1', '2' and '3' indicate outcomes 1, 2 and 3 standard deviations better, and worse, than the expected loss.

    For any given number of bets, the probability of overcoming the expected loss is a function of the magnitudes of the expected loss and the standard deviation. For 72 pass bets, the expected loss is about 12% of the standard deviation, so one doesn't have to be extremely lucky to break even or better.

    However, since the expected loss increases faster than the SD, the more bets you make, the larger the expected loss relative to the SD, and the luckier you have to be to overcome the handicap.

    For a pass bet, the expected loss is the greater than the SD after 5001 bets, see below:

    -.01414 * 5001 = -70.714
    square root of 5001 = 70.717 * .9999 = 70.710

    After 5001 pass bets, a player has to come out a whole standard deviation better than expectation just to break even, and mathematics tells us that the probability of doing that, or better, is under .16. Still, that means that if our two million friends all made 5001 pass bets, we would expect well over 300,000 of them to be even or better.

    You often hear people say, or see written, "in the long run, you can't beat the house edge", but that's NOT TRUE. What is true is that, the longer you play, the less likely you are to be even or ahead. After our 200-roll experiment, we would expect over 950,000 of our two millions friends to be even or better.

    Remember that more variance usually costs more, in terms of house advantage. However, there is a way to get more variance for free: the odds bet behind the line bets. More on that later.
    Good luck,
    Alan Shank
     
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  3. dustedone, Feb 11, 2010

    dustedone

    dustedone Member

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    The reason I play craps,and why I tell everyone, is that in the game of craps the odds on the roll of the dice are always the same; and I am trying to take advantage of the roll dice not the game.

    When and how to bet at the crap table has an advantage w/ the more knowledgeable player vs. the novice. In any simple play to the most of sophisicated of "systems" the interaction of player and dealer(s) puts probability into the game. Because the speed of the game of craps can be quite frantic, a single missed bet or miscommunication w/ the dealer could be costly. And as you stated, "beating the expectation by one outcome".

    But than I would have to say to your original question of the couple of million people just playing the pass line w/ free odds, based on the "numbers", would find losses. I believe that the novice player(s) would rackup a slightly higher loss than the more knowledgeable player(s).

    I also glanced at the embedded web page cloudycitysoftware.com and found it very interesting. Have you worked with the software to support some of your craps theory of buying variance?

    I think more of a probability question would be more like, "What is the chance of rolling the dice for an hour and a half w/out sevening out?"
     
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  4. goatcabin, Feb 11, 2010

    goatcabin

    goatcabin Member

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    Not sure what you mean here. Are you saying they would all lose?

    1. In my example, the players played only the flat bet, never odds.
    2. In that case, how would a novice be expected to lose more than a "more knowledgeable player"?


    Yes, I have used WinCraps very extensively, running tens if not hundreds of simulations of various kinds. I have about 200 "auto-bet" files, which tell WinCraps:

    1. what bets to make, in what amounts, under what conditions
    2. when to stop betting, record the results and start another session
    3. when to stop the sessions
    4. how much bankroll to start with
    etc., etc.

    That's actually quite easy. If you assume 100 rolls/hour, you're talking 150 rolls. A pass bet takes, on average, 3.375 rolls to resolve, so you would expect about 45 decisions to occur. Of course, this can vary substantially depending on the number of players at the table, the bets they're making and the skill of the dealers. But let's assume 45.

    Per the "perfect 1980" distribution, there are 784 seven-outs, so the probability of a seven-out on any given decision is .39596. Therefore, the probability of NOT sevening out is the complement of that (1 - .39596 = .60404). Therefore, the probability of not sevening out for 45 decisions is .60404 raised to the 45th power, which is 1.406 times 10 to the minus 10th power, or odds of over seven billion to one against this happening.

    Keep in mind that this is based on a number of decisions, and the number of decisions in 150 rolls can vary considerably, as can the number of rolls per hour. If, for example, only 35 decisions took place, the odds against would be "just" 46 million to one.

    BTW, since the average number of rolls for a pass bet to resolve is 3.375, you can derive the average number of rolls in a hand (i.e. for a shooter) by dividing 3.375 by .39596; you get 8.5235. I ran a WinCraps simulation a few years ago, recording the number of rolls in each hand, and the results supported that number, ranging from two (the minimum) to I don't remember how many.

    I once rolled the dice 56 times without rolling a seven, and you can use the same technique to figure the probability of that happening:

    p(7) = .1667
    p(not 7) = 1 - .1667 = .8333
    .8333^56 =.00003679, or odds against of over 27,000 to one.
    Cheers,
    Alan Shank
     
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  5. dustedone, Feb 12, 2010

    dustedone

    dustedone Member

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    Well, if all players were placeing only "PASS LINE BETS" w/out free odds than neither novice or knowledged player who have an advantage. And by odds not probablity the house advantage would make them all losers.

    It seem as though it would be a simple system to input into the WinCraps software; but unfortunately in reguards to random accesss generators in computer systems, they are not as random as one my think.

    The easiest way to explain this is a CD player. If one would were to hit the random button on their CD player; they would find that the list of song playing would eventually start to repeat itself. I have found this out myself in practicing craps on my video and computer simulators. And have found many patterns that repeat quite regularly. The same thing with video slot and poker games; And that a whole nother subject.

    This is one reason I chose not to invest my denaro on on-line craps. Also, I really love the interaction w/ people and the casino dealers; And believe me I have dealer that enjoy my tipping.

    I guess, how I look at odds and probability maybe differnt in my head. I'll look up the definitions tonight. I must say that I hope that your math skills don't keep you up at night, but keep poundin' out those number; may the dice scorch in your hand and and burn up the felt.
     
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  6. goatcabin, Feb 12, 2010

    goatcabin

    goatcabin Member

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    No way. We are talking about approximately 60 passline decisions here, only, for each player. If you take all their results as an aggregate, then, yes, the casinos would be ahead, but many of the individual players would have won. Surely you've had good sessions.

    I assure you that WinCraps has very sophisticated random number generators. Yes, there are flawed ones. Nobody is going to bother about making the "random button" in a CD player truly random, because it doesn't really matter. A decent random-number generator will not repeat for a long, long, long time. Also, in WinCraps you have the option of re-seeding the random-number generator between games. In long-term simulations in WinCraps, the overall numbers come out like you would expect, but it also exhibits all the vagaries of truly random results - long streaks of this and that, etc.

    Cheers,
    Alan Shank
     
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  7. The Midnight Skulker, Feb 12, 2010

    The Midnight Skulker

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    I've been running a dark side system through WinCraps manually (because I'm too lazy to learn WinCraps' programming language). I have attached the stats page (and have no idea what it's going to look like here). Note the average rolls/hand of 8.35, close enough to the theoretical to validate it given the statistically small number of hands I have run.
     

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  8. goatcabin, Feb 12, 2010

    goatcabin

    goatcabin Member

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    If you send me details of your system via e-mail, I will create an auto-bet file to implement it, Van. Then you can put WinCraps into hyperdrive and get a significant number of trials very quickly. Do you have version 5.1?
    Cheers,
    Alan Shank
     
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  9. The Midnight Skulker, Feb 13, 2010

    The Midnight Skulker

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  10. goatcabin, Feb 13, 2010

    goatcabin

    goatcabin Member

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    OK, Van, I ran that down and saved it in a text file. I will attempt to program this into WinCraps, but do not hold your breath! >:) It is quite complex.

    Do want this to run continuously, or in sessions. If continuous, I will just tell WinCraps to allow negative bankroll. If in sessions, tell me a starting bankroll and one or more stopping conditions.
    Cheers,
    Alan Shank
     
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  11. The Midnight Skulker, Feb 14, 2010

    The Midnight Skulker

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    That is why I got lazy and haven't tried to do it myself. BTW I trust you caught that I was being facetious about quickly forgetting.
    Let's go with 1000-roll sessions (a little over 8 hours, which is about the maximum I would play in a day) and $1000 bankroll, which is what I try to take on a trip. If it makes it easier I would accept a hard stop at 1000 rolls since established don't bets can be taken down. However, I'd prefer to let a session go past that until the count went to or below zero. Of course a bust is a bust, so no trips to the ATM (i.e. game over when the bankroll is gone).
     
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  12. goatcabin, Feb 14, 2010

    goatcabin

    goatcabin Member

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    Yes, no problem. In fact, I've been forgetting lots of stuff the last couple of years; too many "senior moments".

    That's fine. I'm pretty sure you are not going to take down established DC bets, so the program shouldn't, either. Do you remember that whole thing with ACDOC about Steen's mod to WinCraps to finish any outstanding bet after reaching a roll limit? This was on the pass line, and ignoring the bet was understating the HA, while abandoning it would be stupid, since it might be won. I sent you some questions via e-mail, too.
    Cheers,
    Alan Shank
     
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  13. goatcabin, Feb 16, 2010

    goatcabin

    goatcabin Member

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    Here's another illustration of the interplay of house advantage (the "handicap") with variance. If our two million friends, instead of playing for two hours, had each played 495 pass decisions, the graph of the distribution of their wins would have peaked at 244, as 244 wins vs. 251 losses is the result of a "perfect 495", i.e. the expected result. Out of those two million, however, "only" about 72,000 would be expected to actually lose exactly seven units. The standard deviation of the net units is about 22.2 units, so we would expect 68% of the people to win between 233 and 255 bets, and 95% of them to win between 222 and 266. The variation is symmetrical, but centered on 244, not exactly half of the decisions.

    It's equally likely for someone to have won 248 bets (4 more than expectation) as 240 (4 fewer), 252 bets as 236. So for every person who was just lucky enough to win one unit (248-247), there should be one who lost 15 units. The probability of winning exactly 248 bets is .0336, and the probability of winning 248 bets OR MORE is .3765. The same (almost exactly) goes for winning exactly 240 bets and for 240 OR FEWER. Since the expected loss is seven units, it takes four more wins than expected (for each additional win, there's one loss fewer, of course) to come out ahead. At the same time, with four fewer wins you lose 15 units.

    I like to refer to "degrees of luck", good or bad: one degree of good luck is coming out one standard deviation better than expectation, etc. etc. For 495 passline bets, we have:

    two degrees of good luck: 266 - 229, a net win of 37 units
    one degree of good luck: 255 - 240, a net win of 15 units
    average luck : 244 - 251, a net loss of 7 units
    one degree of bad luck: 233 - 262, a net loss of 29 units
    two degrees of bad luck: 222 - 273, a net loss of 51 units

    So, variation of a given magnitude in either direction is equally likely, but in either direction from the expected loss, not from zero.

    However, all this assumes that the session lasts the full two hours (assumed 200 rolls), or 495 decisions, or whatever session we're talking about. It assumes that none of those two million people run out of money before two hours, or get tired of playing, or get hungry, or succumb to bladder pressure or even win enough money to cash out and walk. If any of those things happen, the shape of the distribution is going to change.

    More later.
    Cheers,
    Alan Shank
     
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  14. ChumpChange, Feb 17, 2010

    ChumpChange

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    So I'd make a progression/regression where I buy-in for $100 and if I win 7 even-money bets ahead right off the bat, I've doubled my money; if I lose 13 bets right off the bat I lose $100 - I would create a 6 win advantage by doubling my money and resetting my progression by setting my winnings aside. The problem with this is rarely do I complete my win/loss goals in short order so I lose coin bouncing up & down the progression/regression on an extended basis. At some point I'll lose so much coin that to reach my +7 win goal, I may have to be +15 ahead to get there.

    I could win +7 three times in a row in quick order and have an advantage of +39 wins in set aside win$. Then I could increase my betting session level to $150 and knock the +39 down to +26 for two rounds to play instead of three. I'd still have a +5 advantage over the +21 it took to get there, if those were quick winning sessions. So I could increase my session stakes on a 3:2 or a 4:3 ratio as winning sessions stack up. If I win $150 4X, I could split that into 3x $200 sessions. If I lose those three $200 sessions, then I get bumped back to what's left of the two $150 session levels I had from my three wins at the $100 session level. I'd have to create certain gambling balance charts to decide which session I'm in and also what to bet in each of those session levels via progression/regression.

    I'd say more but I don't want to give my game away too much.
     
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