For this question, let's stay in a purely theoretical world. So the variables of villians ability, hero's focus, mental state, confidence, emotional state, comfort level, and play style and ability all remain the same. Allso, assume bankroll is not an issue. If I missed something, it remains stable also. The only thing that changes is the amount of the blinds.

Question:

If a player has an infinant sample size win rate of 2.00 bb/100 and over a smaller recent sample has seen a win rate of 5.00 bb/100, would it be logical to step down a level and play at a lower blind assuming regression to the mean should lead to a downswing of -1.00 bb/100 over the an equal sample size?

At the same time, would it be best for said hero to step up following a -2.00 bb/100 downswing over a small sample expecting regression to the mean to equate to a 6.00 bb/100 upswing in the near future?


This is a debate I have had in my head for a long time and would like to see what people think. Obviously this would not work in practicality because you skill is constantly changing, your confidence level changes, bankroll restictions, the importance of you $/hr and other issues. But in theory, would it be smart?

I work in investments, and this is similar to the contrarian style of investing.

Thanks

This has nothing at all to do with regression to the mean, what you are suggesting is just the gambler's fallacy.

If your long-term expectancy is 2.00 bb/100, then you will tend to run at 2.00 bb/100 over any future sample, irrespective of whether you have run badly or well recently.

http://en.wikipedia.org/wiki/Gambler%27s_fallacy

Yes Pyro is correct. Regression to the mean is much more often seen as a between person effect. I believe it explain phenomenon like "the sophomore slump". Take Major League Baseball for example. Say there are 50 rookies this year (this is a made up number, I have no idea what the actual number is). One of them will win "Rookie of the Year", likely the player with the best statistical performance.

However, there is a certain amount of random noise (or luck) involved in a single season's baseball statistics. Thus, the numbers used to determine the "rookie of the year" award include some unknown quantity of good luck for the winner. Because the luck is random, that same player may not be as lucky the next season, and in fact, if he was extremely lucky (say in the 99th percentile of all luck), he is likely to be less lucky next year (less lucky b/c his expectation each season is the 50th percentile, not b/c he was lucky the year before).

In any case, if his luck is at expectation the following season, you will see a drop in his numbers. Typically analysts attribute this to teams "adjusting" to the player and "figuring him out". But realistically, the player who wins the rookie of the year also likely "ran well" in terms of luck that season, and if his luck is closer to expectation next season he will regress to the mean.

However, you don't see teams sending the rookie of the year down to the minor leagues before the next season starts because they expect him not to be as lucky next year. Because the luck is random, it makes no sense to do that. The player may end up in the 99th percentile of luck again next year, or he is at least just as likely to as he was next year.

Sherman

PS - I'm not really sure if this example makes it more clear to anyone else, but it does for me. :-\


edit: The basic idea is that if what made a player do well over a stretch of hands, at-bats, tests, etc. was partly luck/random you cannot predict the portion of the performance that was due to randomness on the next test (by definition).

I don't think so. It shouldn't matter whether the hands are in the future or in the past. The expectation for any block of hands is 2BB/100 - past, present, or future. The greater the number of hands in any one block, the more likely the win rate for that block is 2BB/100. That's what regressing to the mean is. If you know the long term expectation is 2BB/100, then you know that as the number of hands grows, the win rate for all the hands will approach 2BB/100. This is regardless of what's happened up until now. It's past, present, and future. The more hands you look at the more likely the WR will be 2BB/100.

The problem is you can't know the WR. WR evolves as you play. And you are only guaranteed, assuming the WR is 2BB/100, that the observed WR will be 2BB/100 after an infinite number of hands.

So you never actually know your WR and you never can play an infinite number of hands. So the question is a foul - no solution. But as far as thought goes, then yes. But reality will never mirror the thought process.

I'm open to being wrong ....

I think regression to the mean applies, but you better be careful how you use it. The result only has to be inline with the expectation after an infinite number of hands. The shorter the block of hands, the weaker your assertion about what the total WR will be over that block.

if probability worked like this i'd be rich

That is nothing at all to do with regression to the mean. It's not even true, either. If you consider a continuous variable with expectation 2bb/100, for instance, the mean of a trial of any size no matter how big will never be 2bb/100.

If you intended to write 'the more likely the win rate for that block will be close to 2bb/100, for some arbitrarily chosen definition of close', then you are correct, but it's still not regression to the mean, it's the law of large numbers.

http://en.wikipedia.org/wiki/Law_of_large_numbers

FWIW, we're not talking about a continuous variable. The variable will never take on an irrational value so it's not continuous. You're talking about a technical detail that has nothing to do with what I described. You want me to use "approach" instead of "is". Fine ... done.

You don't think the observed WR approaches the true WR as the number of hands goes to infinity?

Is WR not the mean? Is the mean not the center of the distribution?

wiki links don't make what you're saying any more or less valid.

http://en.wikipedia.org/wiki/Regression_to_the_mean

You want to tell me how the WR doesn't regress to the mean as more observations are made?

What am I missing?

You are missing that the OP said nothing that even resembled this, then the bit of the post I quoted where you started with 'I don't think so' had nothing at all to do with what I wrote in my post, and that your post was a simple restatement of the law of large numbers.

Yes, I do think this, and I also think it is *exactly* the statement of the law of large number.

"According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed."

If you don't believe Wikipedia, feel free to look it up on any other Stats/Maths website in the entire world.

Regression to the mean has nothing to do with arbitrary increasing sample sizes. 'Regression to the mean' states that extreme results in an initial sample (even of one trial) will tend to be closer to the mean in a second sample (even of one trial).

Going back to your initial post, the OP of the thread wrote this:

I said "what you are suggesting is just the gambler's fallacy." Do you really not think this is the Gambler's Fallacy or is this whole argument just a misunderstanding?

Ok. I understand how the gamblers fallacy applies here. That being said, it refers more to a logic statement. I like the coin toss because it is much simpler than poker.

Say in an extraordinarily large sample size (n) of coin flips it runs 60/40 tails. Even though the mathematical expectation of the next n flips is 50/50, why would regression to the mean, or the law of large numbers not apply here.

If over the entire history of coin flips, lol, that's a phrase you probably never thought you would hear, we can expect that the distribution would be 50/50. After n flips, would we not have to see at least 1 series of n flips where heads improves so that over after every flip is completed the distribution is 50/50?

I understand that past events do not influence the results of future, but it does seem that even the most high variance player with a very large sample size has a fairly linear graph and it does work in predicting returns for the stock mark when you look at annual return charts or graph/mountain charts.

Cool conversation. I know I am smart enough to understand it all, but I wish I had the proper schooling to add to it.

It is too bad that it doesn't apply to a gambling situation, because if it did, it would be easy to win lots of money. You could play roulette (or any other game for that matter) and play small even money bets for a long time. Any time that you ran bad for 1000 spins, you could triple your bets for the next 1000 spins and voila, you have lots of money. Unfortunately, the next 1000 spins have just as much chance of being bad as they do of being good.

In poker, running well or badly is not always a function of just bad or good luck. When you are running well, it may be because you are actually playing well (there is this skill thing) and when you are doing poorly you may be playing like crap. So if you are running badly, and you think it is because of how you are playing, you might want to change stakes. On here, a lot of people suggest dropping down - I personally sometimes move up (still within the range of where I normally play) because I feel I focus more and play better if the stakes are higher.

Okay - suppose we flip a coin 1000 times and we observe 60/40 tails. 600 tails and 400 heads.

We now flip the coin 1000 more times - the average result will be that those flips are 500 heads and 500 tails. In total, on average, we will have 1100 tails and 900 heads on average, a 55/45 ratio.

We keep flipping. 8000 more flips. The average result will be 4000 heads, 4000 tails. Added to the 2000 flips we already observed, we have 5100 tails, 4900 heads on average, a 51/49 ratio.

After a million flips we would have on average 500100 tails and 499900 heads, a 50.01 / 49.99 ratio, etc.

You can see that even if heads never 'improves', the ratio still tends toward 50/50. The streak we observed becomes an insignificant proportion of the total flips.

Prsumably because it does not make sense to describe a stock price purely as a series of indendent random trials. There are underlying reasons for why the price operates in a certain range based on the valuation of the company and its turnover etc.

If the stock price goes toward the bottom of that range, it becomes a more attractive investment to people looking to put money into that sector, and this tends to drive the price up again. As it becomes 'overvalued', people tend to sell their stock and invest in a different, similar company.

If a companies stock price was decided by flipping a coin and subtracting or adding 0.25 points it would indeed be totally pointless to review charts, try to find resistance and support points, draw bollinger bands, and all that other crap that chartists do

Exactly what I was looking for. I get it.

Not true. You are confined by a lot of different things. First bankroll. I have a book that talks about this exact same example in roulette that has a flawless theroy of roulette in it (i believe it assumed and even payout instead of a waited payout giving the house an advantage). It said what you do is bet black, and then each time you lose place a bet = to double what you previously lost. So for example, you bet $1 black and lose your next bet is $2. You lose, and then bet $4, 8, 16, and so on.

If you get a 2:1 payout, and there is no green, then it still doesn't work. The problem is that at the end, when you do win you only win $1 and a prolonged losing streak can quickly criple a bankroll. And this leaves out the green spots, and the houses take.

If you read the op, I was speaking specifically of a theoretically perfect situation. In real life, moving up after a downswing would be suicide because you have already seen you roll go down, you are not going to be playing your best no matter how tilt averse you are and the level of competition changes. If you downswing, then move down, regain your confidence, and roll, then move back up.

If your win rate is independent of the stakes that you play at, then, if you are a winning player you should play at the highest stakes you can find.

Not if you have to factor in the risk of ruin. See http://en.wikipedia.org/wiki/Kelly_criterion

True. I should amend that to say you should play at the highest level that you're bankrolled for and you should move up whenever your bankroll is large enough to support it.

Either way the optimal play is not to move down when you think that a big downswing is coming, or to move up when you feel that your luck is about to turn around.

Would it be fair to say that if X is a random variable which represents your chance of winning for any given session, X is a memoryless random variable?

Assuming all things such as skill, focus, etc. are constant as per OP.

redundant

Not redundant, but I think he meant something like 'each trial is an identical independent random variable X', rather than 'X is memoryless', which is a very specific condition on X (X must be a geometric or exponential distribution to be memoryless).

I was aiming for the geometric argument, I just didn't make it clear. If you know with certainty the win rate is 2bb/100 and the standard deviation is some positive number s and every session is exactly 1,000 hands, at some point you will run at 4bb/100. I guess theoretically, since you have an infinite number of sessions, you will run at pretty much any winrate at some point.

If you look at it this way, it is exactly a geometric random variable, which as you point out, is memoryless. I just think this is a good way to look at gambler's fallacy (which I agree is exactly what is being disputed here). I

Let's say you use the geometric distribution and figure that you expect to "succeed" on your fourth trial at winning 4bb/100. If you play three sessions and have negative winrates, you don't expect to succeed on your next trial. I use expectation in the mathematical sense, which for a geometric variable is 1/p.

Reading all that, I'm still not sure it's clear so if someone could clarify it for me perhaps it would be helpful.

Spadebidder: http://en.wikipedia.org/wiki/Memorylessness
Memoryless and random are not the same

Yes if you define a random variable as "The number of sessions until you have a winning session", or, "The number of sessions until you have a 4bb/100 winning session" then this should be a geometric distribution, and thus memoryless. What you originally wrote though, was not memoryless:

The point of "regression to the mean" is not that you have a downswing coming. If you are up, say, $5 above your EV, your EV in a billion trials will still be $5 above where it was when you started the match.

What "regression to the mean" is pointing out, is simply that $5 above EV after a few hands represents a dramatic departure from your true win rate (5bb / 100 vs 2bb / 100 in your example). After a billion trials though, this exact same amount of excess win will represent a very tiny departure (perhaps, 2.0001bb / 100 vs 2bb / 100). The limit as n --> infinity = 2 bb / 100, even though the EV will always have you $5 ahead.

good luck,
eric

I don't see how it wasn't? All I said was X represents the chance of achieving some "success" on the next trial. In the case in point, it references a winning session. As the past n sessions have nothing to do with expected sessions until a winning session or the probability the next session is a winning session, is this not a case of a geometric variable?

Perhaps this is just a semantics argument. Regardless, it seems we agree on the answer to OP's question.

Edit: I now see that I wrote "any given session" which may be introducing our disagreement.

No? It's nothing at all like a geometric variable. It isn't even a random variable at all if your definition is "X represents the chance of achieving some "success" on the next trial", it's just a probability.

Good point, poorly worded by me. What I meant to say is that you would use a geometric distribution to determine the probability of success in some number of trials, be it the next trial or 2 from now or 3 from now. If what OP was saying is true, the probability of success in 2 trials would somehow be higher if we failed on the last 5 trials than if we succeeded in our most recent session. But we know that is not the case.