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PLEASE, no one reply to this message until JABoa has had a chance to read it and reply. Thanks!
JABoa,
Thank you for taking the time to drop by. We'd appreciate it if you'd consider the following proposition and tell us what you think of it:
You have a single die which you know is fair. You may roll the die as many times as you like  these rolls don't count. When you are ready, you make one and only one roll of the die (this one counts). If the die comes up 1, 2, 3, 4 or 5, you win $100 (one hundred dollars). If the die comes up 6, you lose $1000 (one thousand dollars).
Is this a "good bet?" Why or why not?
Thanks in advance, Rick and Jim

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I'll modify:
PLEASE, no one reply to this message until JABoa has had a chance to read it and reply. Thanks!
JABoa,
Thank you for taking the time to drop by. We'd appreciate it if you'd consider the following proposition and tell us what you think of it:
You have a single die which you know is fair. You may roll the die as many times as you like  these rolls don't count. When you are ready, you make one and only one roll of the die (this one counts). If the die comes up 1, 2, 3, 4 or 5, you win $100 (one hundred dollars). If the die comes up 6, you lose $1000 (one thousand dollars).
Rick would take the numbers 15 and make the bet with Jim, who would serve as the "house." Rick will take the bet only once, since he has a 5 in 6 chance of winning. Sure, if he hits the 6, he loses big, but he feels the one time "odds" are overwhelmingly in his favor. Rick would not consider taking the bet 100 times, which is enough time to let the normal distribution fill in and let statistics take over.
Is this a "good bet" for Rick? Why or why not?
Thanks in advance, Rick and Jim

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You are sampling from an infinitely deep population, namely all rolls of the fair die. While you, Rick, will win 5/6 of the time and lose only 1/6 of the time, your expected return is the same on a single roll as it is on a zillion rolls. That return is (5/6)*$100 + (1/6)*($1000) or $83.33.
Therefore, it seems to me, it is not a good idea for Rick to take the bet if his aim is to make money on that roll.
I'm sure you guys have discussed subjective aspects, and also things like how deep each player's pockets are. But for the limited question you ask, the key point is that the expected return per roll is the same whether it is one roll or many.

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Heathen Dogs! You worship at the feet of the false God, Newton. Recant and repent! Join the True Church, the Church of Einstein. The universe is expanding, space is curved, the shortest distance between two points is a curved line.
the key point is that the expected return per roll is the same whether it is one roll or many.  JABoa
No! The key point is that the player can pick his spot to attack! Yes, we know the mathematical expectation of each roll is always the same and in favor of the house. That is why the contest involves only one roll, to prevent the house from taking advantage of the expectation.
Get away from the math and get into an understanding of how gambling works. The house must constantly expose itself to loss. The player does not. The house designs each game to give itself an edge and relies on human frailty and the laws of probability to generate profit for the house because of large sample size.
The smart, tough, patient player uses the same laws of probability against the house. He enters and leaves the action as he sees fit. That is the key element here. That is how I defeated JK's 500 trial crap sequence. I picked my spot and hit the winner. I will continue to defeat his crap sequences because I hold the option of when to bet.
Raken

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JABoa said: But for the limited question you ask, the key point is that the expected return per roll is the same whether it is one roll or many.
This is the exact answer I expected, the same I've admitted to. But in practice, IT DOESN'T WORK. Roll the dice. Check yourself. IT DOESN'T WORK.
Besides, I'm a gambler, I'll take the bets I choose. Jim, I'll take this bet with you in person, if you throw in the plane ticket, probably about $250.
Rick

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One more thing, JABoa makes the claim: That return is (5/6)*$100 + (1/6)*($1000) or $83.33.
On one roll of the dice, this formula is incorrect. For it to hold, it must assume a full continuous distribution of rolls, and one roll is not that.
Rick

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On one roll of the dice, this formula is incorrect. For it to hold, it must assume a full continuous distribution of rolls, and one roll is not that.
I'm sure that this has been asked before, but...
If you can't use that type of analysis, what type of analysis do you use to determine the expected outcome of a single roll of the die?
Albaby

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Albaby said: If you can't use that type of analysis, what type of analysis do you use to determine the expected outcome of a single roll of the die?
I don't think there is one that is accurate outside stating the obvious, that there is a 5 in 6 chance that the numbers 15 will come up. Yeah, the loss is big if I lose, but as long as I can afford it, it's a good bet for me. I'll take it once.
Rick

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I'll say this again, that the predictions of expected values are only verified in large numbers of samples. I can calculate the expected value of the energy of a single molecule, for example, and sit at a viewing port and not see that energy in the first 100 molecules. If I make a million measurements, I'll see it emerge as the mean, but not in the first 1oo views. On the other hand, if I look for the maximum in the distribution, themost probably value, I still won't see it for sure in that first 100, but it is more likely than seeing the mean value.
Rick

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Rick said:
I don't think there is one that is accurate outside stating the obvious, that there is a 5 in 6 chance that the numbers 15 will come up. Yeah, the loss is big if I lose, but as long as I can afford it, it's a good bet for me. I'll take it once.
But that's what I'm confused about. How do you know that there is a 5 in 6 chance, except by reference to the overall distribution of outcomes? And if you can meaningfully say that there is a 5 in 6 chance that the numbers 15 will come up (and concomitantly a 1 in 6 chance that #6 will come up)  why doesn't JaBoa's analysis hold up?
Alan

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Albaby said: And if you can meaningfully say that there is a 5 in 6 chance that the numbers 15 will come up (and concomitantly a 1 in 6 chance that #6 will come up)  why doesn't JaBoa's analysis hold up?
There are 6 sides to the dice, that is what I see. I guess JABoa's analysis is right, but I guess I differ with the terminology that the expectation value of the bet is 83. I cannot possibly get a payout of $83, so it cannot be a real expected value. I guess this could be just semantics, but I think it also implies too much truth. It's the problem I run into with my students when I try and teach them some truths about the quantum mechanical world. We can calculate energies and all that for a collection of molecules (or samples, it matters not what we are discussing) but you won't get that measurement all day.
Does this make sense, Al?
Rick

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JABoa,
Thanks so much for your clearly explained answer!
"Well," as the frustrated egg said to the chicken lying next to it in bed, the chicken smoking a cigarette with a very satisfied expression on its face, "I guess we've finally settled that question."
Regards, Jim

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I think what Rick is trying to say, is that for one roll of the die, his "real expected value" is either +100 or 1000. Statistically, he has a 5/6 chance of getting the +100 result (83.3%), and a 1/6 chance of getting the 1000 result (16.7%).
If 6,000 "Ricks" come in and bet one roll each in a single day, 5,000 "Ricks" walk away $100 richer, while 1,000 "Ricks" crawl away $1,000 dollars poorer. The house ends up $500,000 ahead at the end of the day.
Jeff

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<If you can't use that type of analysis, what type of analysis do you use to determine the expected outcome of a single roll of the die?>
I don't think there is one that is accurate outside stating the obvious, that there is a 5 in 6 chance that the numbers 15 will come up. Yeah, the loss is big if I lose, but as long as I can afford it, it's a good bet for me. I'll take it once.
You use the binomial distribution instead of the normal distribution. Flipping a coin is binomial, tossing a die is, ahh, hexomial.
http://www.cmh.edu/stats/simple/prob400.htm
or, cutting ahead to the good stuff:
http://www.cmh.edu/stats/simple/prob423.htm
I don't really know how to expand that to get the probability of specific outcomes from multinomial trials (like rolling a die). Throw in some exponents basically. You can do logistic regression to tell if a pattern of outcomes deviates from what you would expect under particular probability distributions.

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Sorry for shouting.
YOU CANNOT PREDICT WHICH FACE WILL END UP ON TOP WHEN YOU ROLL THE DICE ONE TIME!
Ok, now I feel better.
Rick

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C'mon, man, do you really believe that either of those formulas would yield the exact right answer in one roll? If it's right, it gets the correct answer every time. If it doesn't it's not right. Right?
Well, there is no exact right answer for a single roll, with 6 possible outcomes, (divided unequally across 2 possible gaming results).

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Albaby said: If you can't use that type of analysis, what type of analysis do you use to determine the expected outcome of a single roll of the die?
Rick said: I don't think there is one that is accurate outside stating the obvious, that there is a 5 in 6 chance that the numbers 15 will come up. Yeah, the loss is big if I lose, but as long as I can afford it, it's a good bet for me. I'll take it once.
I still don't understand this. Having more scenarios in which you win than in which you lose can't itself be sufficient to make you bet, even if you can afford to lose the $1,000. Are you saying that this is a better bet than one in which you would receive $400 for rolling a one or two and nothing for three through five (still a losing proposition of course, but one with an expected value of "only" $33.33)? Or that you would refuse a bet that cost you $100 for rolling a one through five, but paid you $1,000 for rolling a six, if you could only play once? Alternatively, what if, in the original scenario, the payoff was $50? Or $.50? Would you still take the bet, just because of its fiveinsix chance of winning?
What I'm getting at, obviously, is there must be some function  explicit or not  which relates the minimum winning payoff which would induce you to play to the chance of winning and to that possible $1,000 loss. How do you decide that something is a "good bet"?

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magpie7 said: Or that you would refuse a bet that cost you $100 for rolling a one through five, but paid you $1,000 for rolling a six, if you could only play once?
That's Jim's bet, and although he'd take it in a second he says, I won't. Bad one time bet, great bet for the casino.
Magpie again: How do you decide that something is a "good bet"?
That's the question, isn't it! It's a hard one to answer explicitly for one time rolls, it comes down to payoff, I think. I certainly wouldn't do it for $10, I might for $50. The main point I am making here, is not what is a good bet, but that a calculation of an expected value is always wrong for low sample numbers. Whether a proposition is a good bet or not is another question.
Rick

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Warning, long post. I sure hope AOL doesn't kick me off.
Raken, #401: The house must constantly expose itself to loss. The player does not.
Sure, but this is immaterial when the house has infinitely deep pockets and you are playing a pure game of chance. (I.e not against other players and where the probabilities are fixed.)
Trick, #402: ...if you throw in the plane ticket, probably about $250.
You talkin to me? You talkin to me? Fuggedabout it, you have tenure and I don't.
Trick, #403: On one roll of the dice, this formula is incorrect...
No it isn't. But this has to do with definitions.
Trick, #405:...as long as I can afford it, it's a good bet for me.
Aha, now we see that your attitude depends on how deep your pockets are. This, it seems to me, is why people play the lottery. Their expected win per dollar ticket is 43 cents or something (and is never a whole number of dollars, Rick), but they've got the dollar for the ticket and they might hit the jackpot.
Suppose the penalty for losing were that you lost your job, your house was torched, and girls wouldn't talk to you any more. Would you take the bet then, even though you had a 5 in 6 chance of winning $100?
Trick, #406: ..the maximum... is more likely than seeing the mean value.
Sure, this happens all the time if you have a skew distribution like Maxwell or lognormal. An extreme would be a distribution like a 2 humped Bactrian camel. Then you'd never see the mean. [Rick knows this; I think he just didn't express himself so well here.]
Trick, #407: I guess this could just be semantics...
Yes I think so. It is perfectly all right to say, according to the definitions, that the average American has 1 testicle. (Well, almost perfectly since we know about John Kruk and Lance Armstrong.)
JKorenthal, #409: Thanks so much for your clearly explained answer!
You're welcome. But in your little story, am I supposed to be the chicken or the egg?
Trick, #413: YOU CANNOT PREDICT...
Exactly, and that is the point of having probability theory.

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JABoa,
You're welcome. But in your little story, am I supposed to be the chicken or the egg?
You, sir, are in the enviable position of being the farmer. :)
 Jim

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I should add that mere expected value is not necessarily why we play or do not play. Everybody knows this but it gives me the opportunity to tell a pretty good story.
Our company was one of 3 reentry houses at the time, and the military tended to give 1/3 of the contracts to each of us to keep us alive and competitive. But we were small and the other 2 parts of huge companies that could absorb losses. So our Bid & Proposal expenditures became more important. Management wanted to curtail them in order to plow the money back to HQ as profit, while engineering wanted to go ahead, knowing that if you have no new work, you are "out of business."
Something came in where our probability of putting together a winning proposal was quite small. A strategy meeting was held, including my supervisor and a young female hotshot who had just been hired up from Washington as a strategic thinker. There were a lot of worries about how she would react to the corps of male engineer fossils.
I will call my supervisor P and the hotshot H.
Anyhow, P put his analogy forth, given the low win probability. "Look, suppose you are sitting at a bar next to a goodlooking tootsie [he said that a lot] and you want to put the moves on her. Then in comes someone and sits on her other side, and he's taller than you, and better looking, and better dressed, and he orders an expensive drink and flashes a roll of bills and he drove up in a Caddylac. Will you then not make your pitch?"
He then smiled at H who gave the only gracious response: "F*** you, P." They got along fine thereafter.
As I recall the proposal was made and lost.

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I should add that mere expected value is not necessarily why we play or do not play. Everybody knows this but it gives me the opportunity to tell a pretty good story.
It's true, it was a good story, and I tell a lot more of them about the unexpected values. But, I take a lot of single rolls.
Rick

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Magpie7 wrote:
How do you decide that something is a "good bet"?
That question directly refers to Expected utility. Utility changes from person to person and is a very subjective thing to measure, but it is a metric to track the "worthiness" of risk.
A riskaverse person has decreasing utility with increased risk. They are satisfied with JABoa's answer that any negative expected return is a "bad" bet. Mathematically that is correct.
But expected return does not account for risklovers, whose utilty does not decrease as quickly with increased risk. For those people, they view an "expected utility" that is greater than an expected return. From their view the extra risk is "worth" the possible returns.
Repeat the bet with increasing payouts of $1,100, $1,200 etc. Sooner or later every person hits a dollar amount where they won't take the bet. That point is where their expected utility turned negative.
The bet offered has greater risk than reward. That is mathematically proven. But it may not have greater risk than utility for every individual.

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Raken, #401: The house must constantly expose itself to loss. The player does not.
Sure, but this is immaterial...  JABoa
No! It is the heart of the matter.
This is the bone of contention between the green eyeshade boys like JK and the combat vets like Raken. I have gone toe to toe with the casinos and beaten them. I have stood hour after hour, night after night at the Vegas crap tables and walked away with more money than I started with.
I could not have done this without the two great advantages of the play/no play option and money mgmt. Be careful you do not lock yourself into the rigid mind set of 'you can't beat the numbers.' Else, you may end up like Victor Hugo's Inspector Javert.
Javert's value system was a rigid adherence to the black letter of the law, 'do the crime, do the time.' Having hunted the parole violator Jean Valjean with the determination of the True Believer, he was unable to cope when he finally realized that the black letter of the law must be subservient to the human spirit of a good man. He let Valjean go and with his life long value system shown to be false, he threw himself into the River Seine.
The first rule of life is selfpreservation. But, not of the body, rather, of the self concept. That is why a mother will rush into a burning house to save a child. That is a mother's self concept, someone who protects her young. That self concept must be protected even at risk of her life.
If you acolytes who worship at the Church of The Holy Numbers ever experience an epiphany. If you realize that numbers do not determine reality only describe it. If you finally see that what matters is the results and not the odds, I hope you don't throw yourselves into the nearest river. Meanwhile, take Northwestern +11 at TCU tomorrow.
Raken

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FractalWalk,
Fascinating... thanks for the clearest explanation of the "other side" of this argument so far. As I've stated, my goal in pursuing this was to gain insight into a different way of thinking than my own, and you've helped a great deal.
I intend to explore what you've said in detail (I agree with most of it), but if I did so now it would come out gobblegook. I'll do so sometime this weekend if I can get my head above water for a while. (And Raken, same goes for our experiments  gonna be working both weekend days, hard to steal time for the fun stuff at the moment.) (Hey, you think it's easy making the best online bookstore in town even better?) (LOL, talk about provocative statements!)
 Jim

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JK:
I'm not sure that the "other side" would agree that I have explained their position. It seems from what I have read that they are trying to bend some of the math to support their views. However, in reality they are follwing the exact same math but substituting utility for return.
The end result is that their long term distribution of results will be shifted to the left of the person who minimizes the odds against them and thus have a lower expected return. However, their subjective utility distribution will exactly match that of a riskaverse person.
I have skimmed over many of Trick's posts and the focus of his objections seems to be that expectation does not equal outcome. That is 100% correct, but it doesn't have anything to do with his argument that walking away increase the chances of winning. The math does not attempt to predict the outcome rather it assigns a probability to that outcome. If you stay or walk, those probabilities do not change. Assuming a nonchanging strategy, the only thing you can control is the number of times that you play. But the distribution of the games outcomes doesn't change at all if you play 10, 1,000,000 or 0 times.
Trick has mentioned that he has consistently won at craps over a 15 year period. Well, that isn't really that impressive. Since we don't have the specifics of his game (units, style, sessions etc.) it is impossible to calculate the true probability, but the math predicts that a little less than half of all players will win in the long run. This is far from conclusive proof that walking away will make you a longterm winner. My take is that merely showing up gives you almost a coin flip chance of being a logterm winner.
I myself am a craps player and have been a long term winner going on 6 year. For about 2 years I played 34 times a week although I play a lot less now than I used to (1 a month). But the key is that I haven't realized a return that is statistically significantly different then the games expected return. Or in other words, there is no evidence that my return is anything other than luck (which makes sense). I'm not quite sure how Trick feels confident making the assertions that he does after admitting that he doesn't have a good understanding of statistics. Without that knowledge there is no way to attach any meaningful evidence to his claims and if he simply rejects the mathematical arguments out of hand, then the only proof to him is his personal experience.
Therefore, extrapolating his claim to its logical conclusion, let's assume a world where everyone played with his strategy and use that distribution as proof of his claim. He is implying that the distribution of actual realized returns would shift away from the game's current actual. I would agree with that as many players have different strategies so a unified strategy would change the shape of the outcome curve and move the mean. In addition, he claims the mean would shift to the right increasing actual return as many players currently make stupid bets or don't walk away. This point assumes that his bets have less of a house edge than the current population's bets. Even though I don't know his betting style, I will accept this as well because he is right that there are a lot of players who make "dumb" bets. Personally, it's not hard for me to beat the average player and I'll assume he can do it as well. Now if we assume his conclusion that the walk away strategy will make him a long term winner, then it should make everyone a longterm winner. As such the mean of our "Trick" distribution should be positive or in other words, actual return should be positive. Since we can't use stats and we can't use math, what are we left with? Only his personal experience, which represents a single data point which can not help us define the distribution. A single data point is insufficient proof in any branch of science, so by definition we can not prove the distribution. The only thing we can do is collect as many data points as possible and view the distribution. That can easily be done with a simulation. All we need to know is Trick's strategy. What bets to place, what amounts and when do we walk away. I think that it is obvious to most that the resulting distribution from that simulation will have a negative actual mean.
BTW, don't think I can't see the ensuing comments, "how can you have an actual return from a simulation?", "playing in a casino with real money is a lot different than playing on a computer" etc.

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FractalWalk,
Trick has mentioned that he has consistently won at craps over a 15 year period. Well, that isn't really that impressive. ... My take is that merely showing up gives you almost a coin flip chance of being a logterm winner.
I think it is extremely unusual for Rick to be ahead after playing craps for 15 years. Even if any individual roll gives him "almost a coin flip chance" of winning, after 10,000 such rolls he has much less than a coin flip chance of being ahead.

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Even if any individual roll gives him "almost a coin flip chance" of winning, after 10,000 such rolls he has much less than a coin flip chance of being ahead.
Not true. Using the coin flip example, if you flip 1,000,000 times. You have a 50% chance of flipping more than 500,000 heads or more. The distribution is evenly split between heads and tails and it would not be unusual at all for a flipper to amass more than the average number of heads. Truly impressive performance only comes when your realized mean moves far away from the true population mean.
Now craps is not a 50/50 proposition but with 2X, 10X, 100X odds it can be very close to that(using optimal strategy). Since the mean is very close to 0, true outperformacne is more than just being "up" in the long run because half of all players (assuming optimal strategy) will. The trick (Trick?) is to substantially beat the mean which can not be done with strategy. It can only be achieved by luck or cheating.
Perhaps you are referring to the fact that it is easier to achieve extremely high returns in the short run than in the long run. In other words it is easier to flip 100% heads if you flip once (1 out of 2) than if you flip 10 times (1 out of 1,024 chance). That is true, but there is a huge difference between beating the mean (winning in the long run) and beating the mean by a statistically significant margin. Since we don't know Trick's winnings as compared to the amount placed at risk, we can't determine where he falls in the distribution and therefore we can't assign a probability to him realizing his return. But merely being in the black is not a very strong statement at all. It is akin to saying I flipped heads.

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FractalWalk,
Now craps is not a 50/50 proposition but with 2X, 10X, 100X odds it can be very close to that(using optimal strategy).
I suppose the question is: "how close?" Since I've never played craps, I don't know. Previously on this board it was suggested that a craps player was a 49%51% underdog on a oneroll proposition. If that is literally correct, then after a million rolls it is extremely unlikely that the craps player will be ahead of the house.
On the other hand, if I took that too literally, and the craps player is only, say, a 49.99999%51.00001% underdog, then even after a million rolls it would not be too surprising if the player were ahead of the house.

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Hey Frac,
You said: Trick has mentioned that he has consistently won at craps over a 15 year period. Well, that isn't really that impressive. Since we don't have the specifics of his game (units, style, sessions etc.) it is impossible to calculate the true probability...
I think we can make a SWAG at Trick's claims. Here's what he said: It's hard for me to tell you my stakes and profits, they've changed over time. But today, given some time in Nevada, I can come out ahead, usually about 3040% of my stake. I hardly *ever* lose more than $50100 in a single encounter with a table, usually more like $2030. If I'm not winning, I don't play long. When I win, I usually walk away winning $75400 in a single encounter with a table. I've had a couple encounters where I made much more than that. I usually have about 48 encounters with a table a day, sometimes spanning 1 hour, sometimes 18 hours. I've never, ever lost more than $400 in a single day playing craps, and that was while drinking with a friend. I've lost more than that betting other games.
Okay, so he has 48 encounters a day. Let's average that to 6. When he wins, he wins $75400, so we'll average that to $237. When he loses, he usually loses $2030 and hardly ever more than $50100. Hmm... shall we say he averages $30 lost?
He doesn't says how often he wins  let's guess 2/3 of the time. So in his 6 sessions his expectation is to consistently win 4 x 237  2 x 30 = $888 per day.
Now according to him, that's 3040% of his stake. Let's say 35%. So his stake is $888 / .35 = $2537.
I'd wager most of us here could pony up twentyfive hundred bucks. All we need to do is take it to Vegas, play craps every day for a year, and we'll make, umm, lessee, $888 * 365 = $324,120 a year. Not too shabby!
If I were Trick I'd be seriously tempted to do this for a living, wouldn't you?
 Jim
P.S. What I really want to do is discuss your concept of "Utility" as it relates to risk aversion  I think this is fascinating and relates well to the stock market. Gotta get back to the grind now but maybe I'll have time for a serious post later.

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I'd wager most of us here could pony up twentyfive hundred bucks. All we need to do is take it to Vegas, play craps every day for a year, and we'll make, umm, lessee, $888 * 365 = $324,120 a year. Not too shabby!
Suspension of disbelief ain't your strong suit, is it Jim? As you should know by now, the trick to beating a negative expectation game is to walk away. The best way to walk away is to never go back. Now, it's true that Rick does go back, but not 365 days year.
So, the real question is, assuming you can beat a negative expectation game by walking away, how many days a year can you do it?

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Fractal,
Okay, I'm gonna make a quick stab at commenting on your earlier post  I wish I could take more than a few minutes, but maybe I can get at least an approximation of my thoughts down.
The concept of "Expected utility" makes a whole lot of sense to me. I agree that it's very subjective, but it sure does make it easier to understand...
*** WAIT WAIT WAIT HOLD THE PRESSES ***
As I'm writing I'm thinking to myself, "Hey maybe 'expected utility' isn't just something invented by Fractal... Maybe I should pop on over to Google and check it out..."
Jim gets one big whup upside the head. <BONK!>
Never mind anything I was going to say about Expected utility and investing, here's a great link  I'm gonna read this and other related stuff before opening my mouth to change socks...
http://www.tiaacref.org/libra/RDS/RD56/expected.html
...and maybe consider taking a modern Economics course... <big goofy grin>
 Jim

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1crates wrote:
Previously on this board it was suggested that a craps player was a 49%51% underdog on a oneroll proposition. If that is literally correct, then after a million rolls it is extremely unlikely that the craps player will be ahead of the house.
On the other hand, if I took that too literally, and the craps player is only, say, a 49.99999%51.00001% underdog, then even after a million rolls it would not be too surprising if the player were ahead of the house.
Very true and well stated. Just for fun(?) I did a back of the envelope. In craps the best bet that you can make is a Don't Pass bet combined with as much odds as the house will allow. Typical Vegas casinos offer 2x odds. A 2x odds Don't Pass bet has a 0.455% house edge which translates into a binomial probability of 49.545%. I will assume that that is his strategy. If I recall correctly he said he plays about 34 times a year with sessions anywhere from 118 hours. So total playing time is about 15 years x 4 trips a year x 9 hours a session (avg) = 540 hours. The resolution of a bet may take a single roll or multiple rolls, so it could take 10 seconds or it could take 20 minutes (I've seen even longer!). I have assumed a 3 minute resolution time per bet average. This means he has laid about 10,800 bets.
With such a large number of trials, I will use the normal to approximate the binomial. So I get a mean of 5,350.86 (10,800 * .49545) and a standard deviation of 51.959 (5,350.86*(1.49545))^.5. To be up, he must have at least 5,400 wins. So the zscore is .9457. Which translates to a 17.22% chance of being ahead of the game after 15 years.
Where I live, you can get 10x odds which dilutes the house edge to .1240%. Using the same process, I get 39.83% probability of long term wins. 100x odds jacks the probability up to a 48.9% chance.
Of course without knowing his strategy and more accurate data, my analysis has a 100% chance of being worthless ;)

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JK:
"Hey maybe 'expected utility' isn't just something invented by Fractal . . .
Very resourceful of you to find the link. Very foolish of you to think I would come up with something original ;)
It is the subjectivity of utility that makes it pseudomath and each person must define their own curve. But once that is done you have developed a good model to mimic that person's particular risktaste.


